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845777 | Finitely generated left (right) unitary modules over left(right)-Artinian ring with identity are Artinian
Finitely generated left (right) unitary modules over left(right)-Artinian ring [imath]R[/imath] with identity are Artinian. How to prove it? | 272607 | Finitely generated modules over artinian rings have finite length
Suppose [imath]M[/imath] is an [imath]R[/imath]-module. Prove that [imath]M[/imath] has finite length if [imath]R[/imath] is artinian and [imath]M[/imath] is finitely generated. |
846494 | Any element [imath]x \in D_{2n}[/imath], represented as [imath]rx=xr^{-1}[/imath].
I am working through problems in Dummit and Foote and am having a hard time interpreting this question. Use the generators and relations to shat if if [imath]x[/imath] is any element in [imath]D_{2n}[/imath], which is not of power [imath]r[/imath], then [imath]rx=xr^{-1}[/imath] I thought this relation was already given in presentation of [imath]D_{2n}[/imath]. How do I interpret this problem? [imath]D_{2n}= < r,s | r^n=s^2=1, \hspace{2mm} \underline{rs=sr^{-1} } >[/imath] | 39718 | Results for elements in Dihedral Groups: [imath]x \notin \langle r\rangle \Rightarrow rx = xr^{-1}[/imath]
As a follow-up to my previous question, the next two exercises state: Use the given generators and relations to show that if [imath]x[/imath] is any element of [imath]D_{2n}[/imath] which is not a power of [imath]r[/imath], then: a) [imath]rx = xr^{-1}[/imath], and b) [imath]x[/imath] has order 2, and hence [imath]D_{2n}[/imath] is generated by [imath]s[/imath] and [imath]sr[/imath], each with order 2. Here, [imath]D_{2n}[/imath] has the "usual" presentation [imath]D_{2n} = \left\langle r, s \mid r^n = s^2 = 1, rs = sr^{-1} \right\rangle.[/imath] For part a), since [imath]x[/imath] is not a power of [imath]r[/imath], I have written [imath] x = sr^k[/imath] for some [imath]0 \leq k \leq n - 1[/imath]. So then [imath]rx = r(sr^k) = (rs)r^k = (sr^{-1})r^k = sr^{k - 1}[/imath], but this doesn't seem to be moving me in the right direction. Should I be using induction on the power of [imath]r[/imath] in the element [imath]x[/imath] ? Then for part b), I simply wrote [imath](sr^k)(sr^k) = s(r^ks)r^k = s(sr^{-k})r^k = s^2 = 1[/imath], but I doubt that I have sufficiently justified my work. From the comments to the previous question, I get the feeling that this is a very simple verification, but I'm having trouble. Thanks in advance for your help! |
235253 | Hartshorne's proof of Proposition 2.5, Chapter II of his book Algebraic Geometry
Let [imath]S = \sum_{n\ge 0} S_n[/imath] be a graded commutative ring. Let [imath]f[/imath] be a homogeneous element of [imath]S[/imath] of degree [imath]> 0[/imath]. Let [imath]D_+(f) = \{\mathfrak{p} \in\operatorname{Proj} S\mid f \notin \mathfrak{p}\}[/imath]. Let [imath]S_{(f)}[/imath] be the degree [imath]0[/imath] part of the graded ring [imath]S_f[/imath], where [imath]S_f[/imath] is the localization with respect to the multiplicative set [imath]\{1, f, f^2,\dots\}[/imath]. The proposition states that [imath]D_+(f)[/imath] is isomorphic to Spec [imath]S_{(f)}[/imath] as locally ringed spaces. Part of his proof is as follows. For [imath]\mathfrak{p} \in D_+(f)[/imath], let [imath]\psi(\mathfrak{p})=\mathfrak{p}S_f\cap S_{(f)}[/imath]. Then [imath]\psi(\mathfrak{p}) \in[/imath] Spec [imath]S_{(f)}[/imath]. He wrote that the properties of localization show that [imath]\psi\colon D_+(f) \rightarrow[/imath] Spec [imath]S_{(f)}[/imath] is bijective. I wonder why [imath]\psi[/imath] is surjective. | 54456 | The bijection between homogeneous prime ideals of [imath]S_f[/imath] and prime ideals of [imath](S_f)_0[/imath]
It is well-known that if [imath]S[/imath] is a graded ring, and [imath]f[/imath] is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization [imath]S_f[/imath] and the prime ideals of [imath]S_{(f)}[/imath], the subring of [imath]S_f[/imath] comprising the homogeneous elements of degree [imath]0[/imath]. This is proposition II.2.5b in Hartshorne, exercise 5.5B in Ravi Vakil's notes [imath][1][/imath] (p. [imath]130[/imath] of the February [imath]24[/imath], [imath]2012[/imath] version), and proposition 8.1.21 of Akhil Mathew's notes [imath][2][/imath] (p. [imath]136[/imath]). Unfortunately I cannot follow any of those proofs to my own satisfaction, perhaps because I'm not well-versed in commutative algebra. The crux of the proof appears to be to show that, given a homogeneous prime ideal [imath]\mathfrak{p}[/imath] of [imath]S[/imath] not containing [imath]f[/imath], the construction used to obtain a prime ideal [imath]\Psi (\mathfrak{q})[/imath] of [imath]S[/imath] from a prime ideal [imath]\mathfrak{q}[/imath] of [imath]S_{(f)}[/imath] will recover [imath]\mathfrak{p}[/imath] when [imath]\mathfrak{q} = S_f \mathfrak{p} \cap S_{(f)}[/imath]. To be precise, let [imath]\Psi (\mathfrak{q})[/imath] be the homogeneous ideal of [imath]S[/imath] generated by [imath]\bigcup_{d \in \mathbb{N}} \{ s \in S_d : s / f^d \in \mathfrak{q} \}[/imath] and let [imath]\Phi (\mathfrak{p}) = S_f \mathfrak{p} \cap S_{(f)}[/imath]. It's easy to see that [imath]\Phi \circ \Psi[/imath] acts as the identity on [imath]\operatorname{Spec} A[/imath] (or, for that matter, the set of all ideals of [imath]A[/imath]), but I cannot see any obvious reason why [imath]\Psi \circ \Phi[/imath] should act as the identity on the set of prime ideals of [imath]S[/imath] not containing [imath]f[/imath]. A detailed proof of this point would be much appreciated. References [imath][1][/imath] Foundations of Algebraic Geometry. [imath][2][/imath] Algebraic geometry notes (covers material at the level of the first and second volume of EGA): html page, and pdf file. |
745047 | What is the unmixedness theorem?
I want to show that [imath]R[/imath] is Noetherian local ring and [imath]S[/imath] is Gorestein local ring s.t. [imath]R=S/J[/imath] and [imath]f[/imath] is an [imath]R[/imath]-regular element, if for every [imath]p\in\operatorname{Ass}_{S}(R),\operatorname{ht}_{S}(p)=\operatorname{ht}_{S}(J)[/imath], then for every [imath]p\in\operatorname{Ass}_{S}(R/f),\operatorname{ht}_{S}(p)=\operatorname{ht}_{S}(J)+1[/imath]. But I am confused. Let [imath]R[/imath] be a Cohen-Macaulay ring. Then for every ideal [imath]I \subset R[/imath], [imath]p\in\operatorname{Ass}_{R}(R/I)[/imath], [imath]\operatorname{ht}(p)=\operatorname{ht}(I)[/imath]. Is this the unmixedness theorem? | 708101 | Height unmixed ideal and a non-zero divisor
Let [imath]R[/imath] be a commutative Noetherian ring with unit and [imath]I[/imath] an unmixed ideal of [imath]R[/imath]. Let [imath]x\in R[/imath] be an [imath]R/I[/imath]-regular element. Can we conclude that [imath]x+I[/imath] is an unmixed ideal? Background: A proper ideal [imath]I[/imath] in a Noetherian ring [imath]R[/imath] is said to be unmixed if the heights of its prime divisors are all equal, i.e., [imath]\operatorname{height} I=\operatorname{height}\mathfrak{p}[/imath] for all [imath]\mathfrak{p}\in \operatorname{Ass} I[/imath]. |
846861 | Continuous functions such that [imath]f(x)+f(x^2)=x[/imath]
Find all continuous [imath]f :[0,1]\to \mathbb R[/imath] such that [imath]\forall x\in [0,1], f(x)+f(x^2)=x[/imath] I suspect there are none. I made little progress so far, but it's worth noticing that [imath]f(0)=0[/imath] and [imath]\displaystyle f(x)+\lim_{n\to\infty}\left((-1)^n\sum_0^{n-1}(-1)^{k+1}x^{2^k}\right)=0[/imath] | 651784 | How prove there is no continuous functions [imath]f:[0,1]\to \mathbb R[/imath], such that [imath]f(x)+f(x^2)=x[/imath].
Prove that there is no continuous functions [imath]f:[0,1]\to R[/imath], such that [imath] f(x)+f(x^2)=x. [/imath] My try. Assume that there is a continuous function with this property. Thus, for any [imath]n\ge 1[/imath] and all [imath]x\in [0,1][/imath], \begin{align*} f(x)&=x-f(x^2)=x-\big(x^2-f(x^4)\big)=x-x^2+\big(x^4-f(x^8)\big)=\cdots\\ &=x-x^2+x^4-\cdots+(-1)^n\left(x^{2^n}-f\big(x^{2^{n+1}}\big)\right) \end{align*} since [imath]f(0)=0[/imath] and [imath]\displaystyle\lim_{n\to \infty}x^{2^{n+1}}=0[/imath] for any [imath]x\in(0,1)[/imath],it follows by the continuity of [imath]f[/imath] that [imath]\displaystyle\lim_{n\to \infty}f\big(x^{2^{n+1}}\big)=0[/imath], hence [imath]f(x)=x-x^2+x^4-x^8+\cdots+(-1)^nx^{2^n}+\cdots[/imath] for any [imath]x\in (0,1)[/imath] Why do I have prove that there is exists such a continuous functions [imath]f[/imath]? Maybe my example is wrong? Why? if my method is wrong,then How prove this problem ? Thank you. |
846654 | Linear algebra, matrix in [imath]\mathbb{C}^n[/imath]
, Can you help me with the following exercise? Show that doesn't exists matrix [imath]A,B\in M_n(\mathbb{C})[/imath] such that [imath]AB-BA=I[/imath] Has something to do with the annihilator? Thanks ! | 800234 | Proving that [imath]AB-BA=cI[/imath] for nontrivial [imath]c \in \mathbb{C}[/imath]
I have a homework question I can`t solve: Let [imath]X[/imath] be a normed linear space, [imath]A,B \in B(X)[/imath]. Show that there exists no nontrivial [imath]c \in \mathbb{C} [/imath] such that [imath]AB-BA=cI[/imath]. Thanks alot already guys! I seem to have a professor who gives impossible homework problems. The hint he gave: [imath]A^{n}B-BA^{n}=ncA^{n-1}[/imath], [imath]n \in \mathbb{N}[/imath] |
846462 | Characterizing quadratic number fields that are subfields of cyclic quartic number fields
Given a quadratic number field [imath]F = \mathbb{Q}(\sqrt{d})[/imath], is there a way to determine whether or not [imath]F \subset K[/imath] for some quartic numberfield [imath]K[/imath] with [imath]\operatorname{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}[/imath]? Moreover, how can we find that field [imath]K[/imath] if it exists? For instance, [imath]\mathbb{Q}(i)[/imath] is not contained in any cyclic quartic number field and neither is [imath]\mathbb{Q}(\sqrt{-3})[/imath]. I thought I heard somewhere that the intermediate subfield of a cyclic quartic number field must be totally real, but I am not sure. | 124878 | A characterisation of quadratic extensions contained in cyclic extensions of degree 4
Let [imath]D[/imath] be a squarefree integer. I am trying to prove that [imath]\mathbb Q[\sqrt D][/imath] is contained in a Galois extension of [imath]\mathbb Q[/imath] with Galois group [imath]\mathbb Z/4[/imath] if and only if [imath]D[/imath] is the sum of two squares, [imath]D = a^2 + b^2[/imath] with [imath]a,b\in \mathbb Q[/imath]. The hint in the exercise in Dummit and Foote suggests considering the extension [imath]\mathbb Q[\sqrt{s + s\sqrt D}][/imath] for the forward direction. However, I am unsure how to show that this extension is Galois, much less that its Galois group is cyclic of order 4. For the reverse direction, I expressed the given Galois extension as [imath]\mathbb Q[\sqrt{a+b\sqrt D}][/imath] for some [imath]a,b\in \mathbb Q[/imath]. I showed that this extension has [imath]\mathbb Q[\sqrt{c^2 - d^2 D}][/imath] as a subfield for some [imath]c,d \in \mathbb Q[/imath]. From this, I deduced that the [imath]\sqrt D[/imath] and [imath]\sqrt{c^2 - d^2 D}[/imath] generate the same extension, so [imath]\sqrt D = x \sqrt{c^2 - d^2 D}[/imath] for some [imath]x[/imath], and hence that [imath]D = \frac{x^2 c^2}{1 + x^2 d^2} [/imath]. It seems that there should be a way to decompose this as a sum of two squares, but I did not see how. I did not tag this as homework as I am studying it independently. |
847052 | finite dimention subspaces of infinite dimention hilbert space are closed?
Let H is a hilbert space of infinite dimention, and [imath] V \subset H [/imath] of finite dimention, can we show that V is closed? I know if H is of finite dimention, then V is closed, but what if H is infinite dimention? | 836511 | Topological Vector Space: [imath]\dim Z\text{ finite}\implies Z\text{ closed}[/imath]
Let [imath]V[/imath] be a Hausdorff topological vector space and [imath]Z[/imath] a linear subspace: [imath]Z\leq X[/imath] Is there a neat way to prove that: [imath]\dim Z\text{ finite}\implies Z\text{ closed}[/imath] |
846101 | Expected value equals sum of probabilities
Let [imath]X[/imath] be a random variable that takes non-negative integer values. Show that, [imath]E[X] = \sum^{\infty}_{k=1}P(X \geq k)[/imath] I'm having trouble following the solution. Could someone help clarify some steps? Thanks. By definition, [imath]P(X \geq k) = \sum^{\infty}_{i=k}p_{X}(i)[/imath] Therefore, we substitute to get [imath]\sum^{\infty}_{k=1}P(X \geq k) = \sum_{k=1}^{\infty}\sum_{i=k}^{\infty}p_{X}(i)[/imath] Now here is where I'm confused. [imath]\sum_{k=1}^{\infty}\sum_{i=k}^{\infty}p_{X}(i) = \sum_{i=1}^{\infty}\sum_{k=1}^{i}p_{X}(i) = \sum^{\infty}_{i=1}ip_{X}(i)[/imath] I don't understand how we are manipulating the summations in the first equality and how we derive [imath]ip_{X}(i)[/imath] in the second equality. | 843845 | Find the Mean for Non-Negative Integer-Valued Random Variable
Let [imath]X[/imath] be a non-negative integer-valued random variable with finite mean. Show that [imath]E(X)=\sum^\infty_{n=0}P(X>n)[/imath] This is the hint from my lecturer. "Start with the definition [imath]E(X)=\sum^\infty_{x=1}xP(X=x)[/imath]. Rewrite the series as double sum." For my opinion. I think the double sum have the form of [imath]\sum\sum f(x)[/imath], but how to get this form? And how to continue? |
840967 | Prove no existing a smooth function satisfying ... related to Morse Theory
i) Show that there does not exist a smooth function [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath], s.t. [imath]f(x) \geq 0[/imath], [imath]\forall x \in \mathbb{R}[/imath], [imath]f[/imath] has exactly two critical points, [imath]x_1,x_2\in\mathbb{R}[/imath] and [imath]f(x_1)=f(x_2) = 0[/imath]. (This part is easy). ii) Show that there does not exist a smooth function [imath]f:\mathbb{R}^2 \rightarrow \mathbb{R}[/imath], s.t. [imath]f(x,y) \geq 0[/imath], [imath]\forall (x,y) \in \mathbb{R}^2[/imath], [imath]f[/imath] has exactly two critical points, [imath](x_1,y_1),(x_2,y_2)\in\mathbb{R}^2[/imath] and [imath]f(x_1,y_1)=f(x_2,y_2) = 0[/imath]. I have tried several methods, however, it does not work, could anybody help me out? | 850207 | a question related with morse theory
Show that there exists no smooth function [imath]f:\mathbb{R}^2β\mathbb{R}[/imath],such that [imath]f(x,y)\geq 0[/imath] for any [imath](x,y)\in\mathbb{R}^2[/imath], with exactly two critical points[imath](x_1,y_1)\in\mathbb{R}^2[/imath], [imath](x_2,y_2)\in\mathbb{R}^2[/imath] and [imath]f(x_1,y_1 )=f(x_2,y_2 )=0[/imath]. |
847463 | Let [imath]f β L_1((0,1)),[/imath] and define [imath]g : (0,1) β \mathbb{R}[/imath] by [imath]g(x)= \int^1_x \frac{f(t)}{t}dt[/imath]
Let [imath]f β L_1((0,1)),[/imath] and define [imath]g : (0,1) β \mathbb{R}[/imath] by [imath]g(x)= \int^1_x \frac{f(t)}{t}dt[/imath] Prove that [imath]g β L_1((0, 1)).[/imath] Some help would be awesome. I tried doing this directly from definition and then using fubini's theorem for non negative measurable functions but had no luck. Thanks. | 839957 | Prove g is Lebesgue intergrable
Let [imath]f[/imath] be Lebesgue integrable on [imath](0, 1)[/imath]. For [imath]0 < x < 1[/imath] deο¬ne g(x) = [imath]\int_x^1t^{-1}f(t)dt[/imath] Prove that [imath]g[/imath] is Lebesgue integrable on [imath](0, 1)[/imath]. [imath]\int^1_0g(x)dx=\int^1_0f(x)dx.[/imath] I am not really getting idea from where should i start. I tried supposing f(x) as characteristics function then simple function and approximating f(x) by the simple function. I don't think this is the right idea. Could anyone give me some hint how to start up. |
847821 | Existence of [imath]p \times p [/imath] matrices [imath]A[/imath] and [imath]B[/imath] over the field [imath]\mathbb F_p[/imath], [imath]p[/imath] prime, such that [imath]AB-BA=I[/imath].
Let [imath]p[/imath] be a prime number. Prove or disprove that there exists [imath]p\times p[/imath] matrices [imath]A[/imath] and [imath]B[/imath] over a field [imath]\mathbb F_p[/imath] with [imath]AB-BA = I[/imath]. With the aid of MAPLE i was able to find out that this statement is valid for [imath]2[/imath] and [imath]3[/imath]. But hadn't been able to prove it theoretically. This is a homework question and we would receive extra marks if solved. Any hint(s) will be appreciated. Thank you. | 99175 | Solutions to the matrix equation [imath]\mathbf{AB-BA=I}[/imath] over general fields
Some days ago, I was thinking on a problem, which states that [imath]AB-BA=I[/imath] does not have a solution in [imath]M_{n\times n}(\mathbb R)[/imath] and [imath]M_{n\times n}(\mathbb C)[/imath]. (Here [imath]M_{n\times n}(\mathbb F)[/imath] denotes the set of all [imath]n\times n[/imath] matrices with entries from the field [imath]\mathbb F[/imath] and [imath]I[/imath] is the identity matrix.) Although I couldn't solve the problem, I came up with this problem, that, does there exist a field [imath]\mathbb F[/imath] for which that equation [imath]AB-BA=I[/imath] has a solution in [imath]M_{n\times n}(\mathbb F)[/imath]? I'd really appreciate your help. |
848379 | A function is discontinuous at all rational points and continuous at all irrational points
Define [imath]f(x)[/imath] for [imath]x\in[0,1][/imath] by [imath]f(\frac pq)=\frac1q[/imath] if [imath]p[/imath] and [imath]q[/imath] are relatively prime, and [imath]f(x)=0[/imath] if [imath]x[/imath] is irrational. How can we see that [imath]f[/imath] is discontinuous at all rational points and continuous at all irrational points? Furthermore, if [imath]g(\frac pq)=q[/imath] for [imath](p,q)=1[/imath] and zero otherwise, how can we see that [imath]g[/imath] is not bounded on any subinterval of [imath][0,1][/imath]? | 207118 | Prove continuity on a function at every irrational point and discontinuity at every rational point.
Consider the function: [imath]f(x)= \begin{cases} 1/n \quad &\text{if [/imath]x= m/n[imath] in simplest form} \\ 0 \quad &\text{if [/imath]x \in \mathbb{R}\setminus\mathbb{Q}[imath]} \end{cases} [/imath] Prove that the function is continuous at every irrational point and also that the function is not continuous at every rational point. Also, we can say that the function is continuous at some point [imath]k[/imath] if [imath]\displaystyle\lim_{x \to k} f(x)=f(k)[/imath]. I was thinking of doing an epsilon delta proof backwards using the fact that [imath]\mathbb{Q}[/imath] is dense in [imath]\mathbb{R}[/imath] for rational points and irrational points. Any ways on how to expand on this are welcome. |
848392 | When writing a proof, why do we want to assume a different but equivalent condition given in the proposition?
In the proof for the inductive step, we start by assuming [imath]k \ge 10[/imath]. But along the way, the author mentions [imath]k \ge 1[/imath] and [imath]k \ge 7[/imath] to justify the inequality. Why do we bother to do this instead of just sticking with [imath]k \ge 10[/imath]? | 848384 | In proof by induction, what does it mean when condition for inductive step is lesser than the propsition itself?
My question is regarding the question posed at the end of the proof. My answer is that the result does not hold for all [imath]m \ge 7[/imath] because when [imath]m=7[/imath], the result is [imath]343 \le 128[/imath], which is false. Is there another way to answer this sort of question besides directly trying different values until you get a false result? EDIT: Another question I have regarding this solution is: In the proof for the inductive step, we start by assuming [imath]k \ge 10[/imath]. But along the way, the author mentions [imath]k \ge 1[/imath] and [imath]k \ge 7[/imath] to justify the inequality. Why do we bother to do this instead of just sticking with [imath]k \ge 10[/imath]? |
848565 | Could someone explain me this "ownership" of the arctangent
someone could explain to me this: [imath]\int { \arctan { \left( \frac { 1 }{ { u }^{ 2 } } \right) } } \,du=\int { \frac { \pi }{ 2 } } -\arctan { \left( { u }^{ 2 } \right) } \, du[/imath] | 720517 | Verify the identity: [imath]\tan^{-1} x +\tan^{-1} (1/x) = \pi /2[/imath]
Verify the identity: [imath]\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x > 0[/imath] [imath]\alpha= \tan^{-1} x[/imath] [imath]\beta = \tan^{-1} (1/x)[/imath] [imath]\tan \alpha = x[/imath] [imath]\tan \beta = 1/x[/imath] [imath]\tan^{-1}[\tan(\alpha + \beta)][/imath] [imath]\tan^{-1}\left [{\tan\alpha + \tan\beta\over 1 - \tan\alpha \tan\beta} \right][/imath] [imath]\tan^{-1}\left[ {x + 1/x\over 1- x/x }\right][/imath] [imath]\tan^{-1}\left[{x + (1/x)\over 0} \right][/imath] I can't find out what I'm doing wrong.. |
849191 | Interesting Mathematical Fallacies
I recently volunteered to help with a summer math program at a local high school for which I thought would be a breeze. As it turns out, it isn't a program for those catching up (summer school) like I thought, it's a small group of kids that have strong math skills and seem to have an endless desire for knowledge. I have been able to keep them reasonably in check intellectually, but I'm nearing the end of my bag of tricks. For better or worse, they love mathematical fallacies. Of course they were familiar with showing [imath]0=1[/imath] (all numbers equal all other numbers fallacy), and after I showed them [imath]1=-1[/imath] through complex numbers I was out of ideas. I knew of the, all triangles are isosceles triangles fallacy, but couldn't remember it at the time. Does anyone have any interesting fallacies that I can show them? Their math knowledge goes through most of calculus but fallacies outside of that are fine too. If this is a ridiculous question, I apologize. I figured there were more fallacies out there than what I could find on wikipedia and a few of the other top search results. Thanks in advanced. Note: I have another 3 weeks with them so no major rush. | 348198 | Best Fake Proofs? (A M.SE April Fools Day collection)
In honor of April Fools Day [imath]2013[/imath], I'd like this question to collect the best, most convincing fake proofs of impossibilities you have seen. I've posted one as an answer below. I'm also thinking of a geometric one where the "trick" is that it's very easy to draw the diagram wrong and have two lines intersect in the wrong place (or intersect when they shouldn't). If someone could find and link this, I would appreciate it very much. |
835886 | Finding the integral of [imath]\frac{x}{e^x + 1}[/imath]
I've having some difficulty with finding this integral: [imath] \int_0 ^{\infty} \frac{x}{e^x + 1}[/imath] Now usually I would use the monotone convergence theorem to write (using geometric series): [imath]f_n (x) = \sum_0 ^n (-1)^k x e^{-(k+1)x},[/imath] but this isn't a sequence of positive terms, so how do we justify moving the integral inside? Thanks. | 798895 | How to find [imath] \int_0^\infty \dfrac x{1+e^x}\ dx[/imath]
How to find [imath] \int_0^\infty \dfrac x{1+e^x}\ dx=\ ...? [/imath] I don't know where should I start with. The correct answer from my textbook is [imath]\frac{\pi^2}{12}[/imath]. This is my homework with 10 questions but I can only answer 9 questions, this one I'm stuck. Can anyone help me? Thanks. |
849327 | SMO 2013 Junior First Round Q35.
In a competition that i have recently taken part in, one of this questions popped out: [imath]2^{29}[/imath] has nine distinct digits. Find the digit that is not in the sequence. My answer will be placed as a 'Answer-your-question'. However, my main question is: are there other methods? | 819412 | The number [imath]2^{29}[/imath] has exactly [imath]9[/imath] distinct digits. Which digit is missing?
The number [imath]2^{29}[/imath] has exactly [imath]9[/imath] distinct digits. Which digit is missing? I came across this question in a math competition and I am looking for how to solve this question without working it out manually. Thanks. |
849274 | The convergence of [imath]\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}[/imath]
I would like to know if this sequence converges [imath]\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}[/imath]. I know this sequence is increasing monotone, but I couldn't prove it's bounded. Thanks | 555778 | Show that [imath]\sqrt{2+\sqrt{2+\sqrt{2...}}}[/imath] converges to 2
Consider the sequence defined by [imath]a_1 = \sqrt{2}[/imath], [imath]a_2 = \sqrt{2 + \sqrt{2}}[/imath], so that in general, [imath]a_n = \sqrt{2 + a_{n - 1}}[/imath] for [imath]n > 1[/imath]. I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least upper bound of this sequence. But how? |
849366 | Prove that [imath]1+\tan^2 x=\sec^2 x[/imath]
I have no idea how to prove this. Does anyone know where to start? We're allowed to use other trigonometric identities but i'm not sure why these are useful. | 607269 | Prove [imath]1 + \tan^2\theta = \sec^2\theta[/imath]
Prove the following trigonometric identity: [imath]1 + \tan^2\theta = \sec^2\theta[/imath] I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant. |
849322 | Show that if [imath]B \subseteq C[/imath], then [imath]\mathcal{P}(B) \subseteq \mathcal{P}(C)[/imath]
Can someone please verify this? Show that if [imath]B \subseteq C[/imath], then [imath]\mathcal{P}(B) \subseteq \mathcal{P}(C)[/imath] let [imath]x \in \mathcal{P}(B)[/imath]. Then, [imath]x \subseteq B[/imath] This implies that [imath]\forall a \in x, a \in B[/imath] This further implies that [imath]\forall a \in x, a \in C[/imath] Therefore, [imath]x \subseteq C[/imath] This implies that [imath]x \in \mathcal{P}(C)[/imath] | 857277 | Prove this result about power sets
I have to prove this result: If [imath]P[/imath] be the power set, and [imath]B[/imath] and [imath]C[/imath] are two sets, then if [imath]B \subseteq C[/imath] prove that [imath]P(B) \subseteq P(C)[/imath]. Now, it seems obvious to me that since all the elements of B are in C, all possible combinations of the elements of C must include those that of B. But other than this straightforward argument, I'm at a loss to see what kind of proof can be given for statements like these. Can anyone suggest something more "proper"? |
767267 | About partitions, majorization, and conjugates
I am trying to solve a question of a property of conjugation. What I am trying to show is that conjugation reverses the order of majorization. Let [imath]\lambda[/imath] and [imath]\mu[/imath] are partitions of n and [imath]\lambda[/imath] is majorized by [imath]\mu[/imath]. [imath]\lambda:n=\lambda_1+\lambda_2+\lambda_3+ \cdots + \lambda_k[/imath] [imath]\mu:n=\mu_1+\mu_2+\mu_3+ \cdots + \mu_k[/imath] '[imath]\lambda[/imath] is majorized by [imath]\mu[/imath]' means that [imath]\lambda_1+\lambda_2+\lambda_3+ \cdots + \lambda_i \leq \mu_1+\mu_2+\mu_3+ \cdots + \mu_i[/imath] for [imath]i=1,2,\cdots,k[/imath] Then, showing that [imath]\mu^*\leq\lambda^*[/imath] would be enough to say that conjugation reverses the order of majorization. But, I am stuck from the beginning after writing the definitions of majorization and conjugation. Please help me. Thanks | 108559 | Proof of the duality of the dominance order on partitions
Could anyone provide me with a nice proof that the dominance order [imath]\leq[/imath] on partitions of an integer [imath]n[/imath] satisfies the following: if [imath]\lambda, \tau[/imath] are 2 partitions of [imath]n[/imath], then [imath]\lambda \leq \tau \Longleftrightarrow \tau ' \leq \lambda '[/imath], where [imath]\lambda'[/imath] is the conjugate partition of [imath]\lambda[/imath] (i.e. the transpose of the set of 'dots' which [imath]\lambda[/imath] represents). I feel like there must be a nice clever and concise/intuitive proof of this, but all I can come up with is an ugly brute force approach based on the definition of sums of the components [imath]\lambda_i[/imath]. Could anyone suggest a particularly nice way to obtain this result? Many thanks. |
849839 | Answered: Converting between Schwarz-Christoffel formulas for disk and half-plane
Is there a way to conveniently use change of variables (e.g. with a MΓΆbius transformation) in order to convert from a Schwarz-Christoffel integral of the form [imath] C_1 + C_2 \int _0^w \prod _{k=1}^n (\zeta - w_k)^{-\beta_k} \ d\zeta [/imath] that maps the closed unit disk onto a polygonal region to an analogous integral of the form [imath] C_3 + C_4 \int ^w \prod _{k=1}^{n-1} (\zeta - \xi_k)^{-\beta_k} \ d\zeta [/imath] that does the same for the upper half plane? | 446092 | Modification of Schwarz-Christoffel integral
I found two different formulations of the Schwarz-Christoffel formula (e.g. Link1, p.20 and Link2, p. 9). The first is \begin{align*} z=w(\zeta)=&A+C\int\limits^{z}\prod\limits_{k=1}^n\left(\zeta-z_k\right)^{\alpha_k-1}d\zeta \end{align*} The second is \begin{align*} z=w(\zeta)=&A+C\int\limits^{z}\prod\limits_{k=1}^n\left(1-\frac{\zeta}{z_k}\right)^{\alpha_k-1}d\zeta \end{align*} In both equations [imath]A[/imath] and [imath]C[/imath] are complex constants, [imath]z_k[/imath] are the coordinates of point [imath]k[/imath] on the unit circle corresponding to the vertex [imath]k[/imath] of the rectangle, [imath]\alpha_k[/imath] are the interior angles of the vertices by means of multiples of [imath]\pi[/imath] and [imath]\zeta[/imath] is a point outside the unit circle such that [imath]|\zeta| > 1[/imath]. In the second link on page 20 it is said that: "Composing the first equation with standard conformal maps leads to variations of the Schwarz-Christoffel formula for mapping from other fundamental domains [...]. The simplest such modification has the unit disk as domain. The vertices then lie in counterclockwise order on the unit circle and equation one can be transformed to equation two." Why is this transformation possible for the upper complex half-plane and how is it carried out? |
850006 | Help with improper integral
I need help solving this integral: [imath]\int_0 ^\infty \frac{\sin(x)}{x} dx[/imath] I have a help that says that try to calculate the integral of [imath]\frac{e^{iz}}{z}[/imath] for a "proper path"... but I don't know how to use that, help. | 294490 | A Complex approach to sine integral
this integral: [imath]\int_0^{+\infty}\frac{\sin x}{x}\text{d}x=\frac{\pi}{2}[/imath] is very famous and had been discussed in the past days in this forum. and I have learned some elegant way to computer it. for example: using the identity: [imath]\int_0^{+\infty}e^{-xy}\sin x\text{d}x=\frac{1}{1+y^2}[/imath] and [imath]\int_0^{\infty}\int_0^{\infty}e^{-xy}\sin x\text{d}y\text{d}x[/imath] and Fubini theorem. the link is here:Post concern with sine integral In this post, I want to discuss another way to computer it. since[imath]\int_0^{+\infty}\frac{\sin x}{x}\text{d}x=\frac{1}{2i}\int_{-\infty}^{+\infty}\frac{e^{ix}-1}{x}\text{d}x[/imath] this fact inspire me to consider the complex integral:[imath]\int_{\Gamma}\frac{e^{iz}-1}{z}\text{d}z[/imath] and [imath]\Gamma[/imath] is the red path in the above figure, with counter-clockwise orientation, by Cauchy's theorem, we have [imath]\int_{\Gamma}\frac{e^{iz}-1}{z}\text{d}z=0[/imath] the above integral can be written as:[imath]\int_{-R}^{-\epsilon}\frac{e^{ix}-1}{x}\text{d}x+\int_{\Gamma_{\epsilon}}\frac{e^{iz}-1}{z}\text{d}z+\int_{\epsilon}^{R}\frac{e^{ix}-1}{x}\text{d}x+\int_{\Gamma_{R}}\frac{e^{iz}-1}{z}\text{d}z[/imath] Let [imath]R\rightarrow +\infty[/imath] and [imath]\epsilon \rightarrow 0[/imath], we have: [imath]\int_{-R}^{-\epsilon}\frac{e^{ix}-1}{x}\text{d}x+\int_{\epsilon}^{R}\frac{e^{ix}-1}{x}\text{d}x \rightarrow \int_{-\infty}^{+\infty}\frac{e^{ix}-1}{x}\text{d}x=2i\int_0^{+\infty}\frac{\sin x}{x}\text{d}x[/imath] and [imath]\int_{\Gamma_{\epsilon}}\frac{e^{iz}-1}{z}\text{d}z=\int_\pi^0\frac{e^{i\epsilon e^{i\theta}}-1}{\epsilon e^{i\theta}}i\epsilon e^{i\theta}\text{d}\theta=i\int_\pi^0(\cos(\epsilon e^{i\theta})+i\sin(\epsilon e^{i\theta})-1)\text{d}\theta \rightarrow 0[/imath] as [imath]\epsilon \rightarrow 0[/imath] so I am expecting that:[imath]\int_{\Gamma_{R}}\frac{e^{iz}-1}{z}\text{d}z=-i\pi[/imath] when [imath]R \rightarrow +\infty[/imath] but I can't find it. Could you help me? Thanks very much. |
460296 | If [imath]\forall \mathcal F(\bigcup \mathcal F = A \Rightarrow A \in \mathcal F)[/imath] then A has exactly one element
While working through Velleman's How To Prove It, I came across the following problem. Is my proof correct? Suppose [imath]A[/imath] is a set, and for every family of sets [imath]\mathcal F[/imath], if [imath]\bigcup \mathcal F = A[/imath] then [imath]A \in \mathcal F[/imath]. Prove that [imath]A[/imath] has exactly one element. Proof. Existence: Suppose [imath]A = \emptyset[/imath]. Then in particular, taking [imath]\mathcal F = \emptyset[/imath], we can conclude that if [imath]\bigcup \emptyset = \emptyset[/imath] then [imath]A \in \emptyset[/imath]. Since [imath]\bigcup \emptyset = \emptyset[/imath], it follows that [imath]A \in \emptyset[/imath], which is clearly a contradiction. Thus [imath]A \neq \emptyset[/imath]. Uniqueness: Now suppose there are [imath]x \in A[/imath] and [imath]y \in A[/imath] such that [imath]x \neq y[/imath]. Let [imath]\mathcal G = \{ \{x\}, \{y\}, A \setminus \{x,y\} \}[/imath]. Then in particular, taking [imath]\mathcal F = \mathcal G[/imath], we can conclude that if [imath]\bigcup \mathcal G = A[/imath] then [imath]A \in \mathcal G[/imath]. Clearly [imath]\bigcup \mathcal G = A[/imath], so [imath]A \in \mathcal G[/imath]. But this is a contradiction since [imath]A \neq \{x\}[/imath], [imath]A \neq \{y\}[/imath], and [imath]A \neq A \setminus \{x,y\}[/imath]. Thus for all [imath]x \in A[/imath] and [imath]y \in A[/imath], [imath]x = y[/imath]. | 359550 | If [imath]\cup \mathcal{F}=A[/imath] then [imath]A \in \mathcal{F}[/imath]. Prove that [imath]A[/imath] has exactly one element.
I'm reading through How to Prove It by Velleman and I'm having trouble with this exercise in the section about Existence and Uniqueness proofs. Here is the exercise: Suppose [imath]A[/imath] is a set and for every family of sets [imath]\mathcal{F}[/imath], if [imath]\cup \mathcal{F}=A[/imath] then [imath]A \in \mathcal{F}[/imath]. Prove that [imath]A[/imath] has exactly one element. He hints that for both the existence and uniqueness parts of the proof it would be a good idea to use contradiction. I've been playing around with this proof for a while but I can't seem to make any substantial progress. I current idea is considering some cases where for some family of sets [imath]\mathcal{G}[/imath], [imath]A \in \mathcal{G}[/imath] and [imath]A \notin \mathcal{G}[/imath]. I thought if I could show that if [imath]A= \varnothing[/imath] lead to contradictions, I could at least say that there is something in [imath]A[/imath], and try to prove that it is unique from there. I haven't been able to make any progress with this though. Any help with this problem would be greatly appreciated! |
850992 | Show that [imath]\sum\limits_n1/x_{n}^{2} = 1/10[/imath] where [imath]x_{n}[/imath] is the [imath]n^{\text{th}}[/imath] positive root of [imath]\tan x = x[/imath]
Recently I encountered this problem Show that [imath]\sum_{n = 1}^{\infty}\frac{1}{x_{n}^{2}} = \frac{1}{10}[/imath] where [imath]x_{n}[/imath] is the [imath]n^{\text{th}}[/imath] positive root of [imath]\tan x = x[/imath]. I found many threads on MSE regarding solutions to [imath]\tan x = x[/imath] like this and this which suggest that the series [imath]\sum 1/x_{n}^{2}[/imath] is convergent if we compare it with [imath]\sum 1/n^{2}[/imath]. But at the same time I have no idea how we can sum this up. Please help with suggestions or an answer. Update: I further tried to relate it with the sum [imath]\sum 1/n^{2} = \pi^{2}/6[/imath] and we can rewrite it as [imath]\sum 1/(n^{2}\pi^{2}) = 1/6[/imath] so that if [imath]y_{n}[/imath] is [imath]n^{\text{th}}[/imath] positive root of [imath]\sin x = 0[/imath] then [imath]\sum 1/y_{n}^{2} = 1/6[/imath]. I believe this has got to do with the product [imath]\frac{\sin x}{x} = \prod_{n = 1}^{\infty}\left(1 - \frac{x^{2}}{n^{2}\pi^{2}}\right)[/imath] where we can see the factors corresponding to actual roots. We can now compare the coefficients of [imath]x^{2}[/imath] on both sides to get [imath]\sum 1/y_{n}^{2} = 1/6[/imath]. I don't see any product related to equation [imath]\tan x = x[/imath]. | 75206 | Sum of the squares of the reciprocals of the fixed points of the tangent function
The sum of the squares of the reciprocals of the positive fixed points of the tangent function is [imath]1/10[/imath]. I've seen this proved by means of residues, but I don't remember the details. I've also heard it asserted that it that it can be done by means of Green's functions. What proofs of this fact are published or otherwise known? PS: Maybe it is of interest to note that the positive fixed points of the tangent function are also the abscissas of the extreme values of the function [imath]x\mapsto\dfrac{\sin x} x.[/imath] |
851297 | If [imath]a+\frac{1}{b}, b+\frac{1}{c}, c+\frac{1}{a}\in\mathbb{Z}[/imath], find [imath]a+b+c[/imath].
Let [imath]a,b,c[/imath] be positive rational numbers such that [imath]a+\frac{1}{b}, b+\frac{1}{c}, c+\frac{1}{a}[/imath] are all integers. Find all the possible values of [imath]a+b+c[/imath]. it would be too complicate to solve by quadratic equation(the discriminant is 6 degree polynomial of 3 variables...). Either, no thought at all⦠| 538905 | Making [imath]x+\frac1y[/imath], [imath]y+\frac1z[/imath], and [imath]z+\frac1x[/imath] all integer.
Let [imath]x,y,z\in\Bbb R[/imath], and at least one is be a postive integer, and such that [imath]x+\frac{1}{y},\;y+\frac{1}{z},\;z+\frac{1}{x}\in\Bbb Z[/imath] Find the value of [imath]x,y,z[/imath]. My try: [imath]\left(x+\dfrac{1}{y}\right)\left(y+\dfrac{1}{z}\right)\left(z+\dfrac{1}{x}\right)=\dfrac{(xy+1)(yz+1)(xz+1)}{xyz}\in Z[/imath] Then I can't. Thank you. I konw [imath](x,y,z)=(1,1,1),(-1,-1,1),(2,0.5,1)[/imath] is such condition,But How prove it? |
851373 | Factoring over ring of power series
How would we factor [imath]6+x[/imath] over [imath]\mathbb{Z}[[x]][/imath], the ring of formal power series with integer coefficients? Proving things are prime is easy, but factoring a nonprime is difficult. Thanks in advance. | 124357 | Show [imath]100+t[/imath] is reducible in [imath]\mathbb{Z}[[t]][/imath]
[imath]\mathbb{Z}[[t]][/imath] is the ring of formal power series with integer coefficients, and the problem asks to show that [imath]100+t[/imath] is reducible. This means I need to find two power series, [imath]\sum_{i=0}^\infty a_i t^i[/imath] and [imath]\sum_{j=0}^\infty b_j t^j[/imath] with [imath]a_i,b_j\in\mathbb{Z}[/imath], that when multiplied give [imath]100+t[/imath]. Since the integers are an integral domain, I know that neither of these power series may be finite, but I'm having a really hard time figuring out how to come up with two series that will telescope properly. Any suggestions? |
43519 | Squares in arithmetic progression
It is easy to find 3 squares (of integers) in arithmetic progression. For example, [imath]1^2,5^2,7^2[/imath]. I've been told Fermat proved that there are no progressions of length 4 in the squares. Do you know of a proof of this result? (Additionally, are there similar results for cubes, 4th powers, etc? If so, what would be a good reference for this type of material?) Edit, March 30, 2012: The following question in MO is related and may be useful to people interested in the question I posted here. | 1246493 | [imath]4[/imath] different square numbers can't create an arithmetic sequence.
I have the following task: Prove, that four different square numbers can't create an arithmetic sequence(obviously, the [imath]0,0,0,0[/imath] case is forbidden). How can I prove this statement? I tried to write down the numbers in their form: [imath]a<b<c<d \rightarrow a^2 , b^2, c^2, d^2[/imath] is my sequence, and this can't occur: [imath]b^2-a^2=D, c^2-b^2=D, d^2-c^2= D, D \in \mathbb{N^+}[/imath] is a fixed number. What should I do after this? Summing the equations didn't really help me to go further. Thanks in advance for the help. |
852296 | [imath]A \setminus B \cup C = A \setminus (B \cup C)[/imath]?
[imath]A \setminus B \cup C[/imath] or [imath]A \setminus (B \cup C)[/imath]? Sorry as this is a very soft question, but I couldn't find the answer anywhere. Are these two things generally considered the same? | 266182 | Order of precedence of set operators
I don't know how to evaluate this expression: [imath]A-B\cap C[/imath] Is it true that: [imath]A-(B\cap C)=(A-B)\cap C[/imath] or, in other words, that the associative property applies to these operations? |
851030 | How prove this [imath]|A||M|=A_{11}A_{nn}-A_{1n}A_{n1}[/imath]
Question: let the matrix [imath]A=(a_{ij})_{n\times n},i=1,2,\cdots,n,j=1,2,\cdots,n[/imath], and the matrix [imath]M=(a_{ij})_{(n-2)\times (n-2)},[/imath] mean that [imath]A=\begin{bmatrix} a_{11}&\cdots&a_{1n}\\ \vdots& M&\vdots\\ a_{n1}&\cdots&a_{nn} \end{bmatrix}[/imath] show that [imath]\det|A|\cdot \det |M|=A_{11}A_{nn}-A_{1n}A_{n1}[/imath] where [imath]A_{ij}[/imath] is cofactor with the matrix [imath]A[/imath]. This problem is from linear problem book ,and this problem I can't deal it. because this value [imath]|A||M|[/imath] I can't choose something to deal it? | 816392 | Determinant identity: [imath]\det M \det N = \det M_{ii} \det M_{jj} - \det M_{ij}\det M_{ji}[/imath]
Let [imath]M[/imath] be a (real) [imath]n \times n[/imath] matrix. For [imath]1 \leq i, j \leq n[/imath] we denote by [imath]M_{ij}[/imath] the [imath](n-1) \times (n-1)[/imath] matrix that we get when the [imath]i[/imath]th row and [imath]j[/imath]th column of [imath]M[/imath] are removed. Now, consider fixed [imath]i[/imath] and [imath]j[/imath] with [imath]i\neq j[/imath]. Let [imath]N[/imath] be the [imath](n-2) \times (n-2)[/imath] matrix that we get when removing both the [imath]i[/imath]th and [imath]j[/imath]th row and the [imath]i[/imath]th and [imath]j[/imath]th column from [imath]M[/imath]. Then the following identity holds: [imath] \det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}. [/imath] We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of [imath]n![/imath] terms (for instance, terms containing [imath]a_{ji}a_{ij}[/imath] are counted both [imath]1[/imath] time in the LHS and [imath]1[/imath] time in the RHS). We are looking for a slick proof that does not involve writing out the determinant. Any suggestions or approaches to this problem are appreciated! |
852086 | Finding the remainder when a polynomial is divided by another polynomial.
Find the remainder when [imath]x^{100}[/imath] is divided by [imath]x^2 - 3x + 2[/imath]. I tried solving it by first calculating the zeroes of [imath]x^2 - 3x + 2[/imath], which came out to be 1 and 2. So then, using the Remainder Theorem, I put both their values, and so the remainder came out to be [imath]1 + 2^{100}[/imath]. But the correct answer is [imath](2^{100} - 1)x + (2 - 2^{100})[/imath]. Can you please explain the exact process to reach the solution? Thanks in advance. :) | 842715 | how to find the remainder when a polynomial [imath]p(x)[/imath] is divided my another polynomial [imath]q(x)[/imath]
i was solving the question from the book IIT FOUNDATION AND OLYMPIAD - X and i was solving the problems of polynomials-III. so on the page number 136, there is a question (question 17) given below: The remainder when [imath]x[/imath]^100 is divided by [imath]x^2-3x+2[/imath] is: a) [imath](2[/imath]^100[imath]-1)x + (-2[/imath]^100[imath] +2) [/imath] b) [imath](2[/imath]^100[imath]+1)x + (-2[/imath]^100[imath] -2) [/imath] c) [imath](2[/imath]^100[imath]-1)x + (-2[/imath]^100[imath] -2) [/imath] d) none as far as i tried to find the remainder, i tried long division method but it was getting more and more complicated, then i used systematic method of division but i can't get the corret option what is the correct option. please explain me how did you find the remainder. thanks and yes its answer is option (a) |
852416 | Structure Theorem for Finite Commutative Rings with unity
The Structure Theorem for Finite Commutative Rings with unity state that: A finite commutative ring [imath]R[/imath] with multiplication identity is isomorphic to a direct sum of local rings. Suppose all the maximal ideals of [imath]R[/imath] are [imath]M_1, M_2, ..., M_n[/imath], and let [imath]J[/imath] be the Jacobson radical of [imath]R[/imath]. There exists a positive integer [imath]k[/imath] such that [imath]J^k=0[/imath]. There is an assertion in the proof which says that [imath]M_i[/imath] is the unique maximal ideal such that [imath]M_i^k\subseteq M_i[/imath], why? The complete proof of the theorem see p. 9, lem. 9 in the thesis http://bfhaha.blogspot.tw/2014/06/thesis-classification-of-finite-rings.html p.95, thm. (VI.2) in the book "Finite rings with identity", Bernard R. McDonald p.40, thm. 3.1.4 in the article "Finite commutative rings and their applications", Gilberto Bini and Flamino Flamini | 746111 | Is quotient of a ring by a power of a maximal ideal local?
Say I have a commutative ring [imath]R[/imath] with a maximal ideal [imath]m[/imath]. Then [imath]m/m^k[/imath] is a maximal ideal in [imath]R/m^k[/imath] for any [imath]k[/imath]. Is it the only maximal ideal, i.e. is [imath]R/m^k[/imath] a local ring? This is a well known result for [imath]k = 1[/imath], as [imath]R/m[/imath] is a field. It seems to be true in other cases, e.g. [imath]p\mathbb{Z}\subset \mathbb{Z}[/imath] for prime [imath]p[/imath] and for [imath](x,y) \subset \mathbb{F}[x,y][/imath], the maximal ideal generated by [imath]x[/imath] and [imath]y[/imath] in a polynomial ring over the field [imath]\mathbb{F}[/imath]. Equivalently, if an element [imath]x\notin m[/imath], is [imath]x + m^k[/imath] an invertible element in [imath]R/m^k[/imath]? |
852520 | Calculus Problem general polynomial limit to infinity
I have to solve the following problem for homework for a calculus class. I really have no idea where to start, does anyone have any hints?: let n be a positive integer greater than 0. Let P(x) be a polynomial s.t.: [imath]P(x) = x^n+b_1x^{n-1}+b_2x^{n-2}+...+b_{n-1}+b_n[/imath], Show that [imath]lim_{x\to\infty}([P(x)]^{1/n}-x)=b_1/n[/imath]. We're not allowed to use induction. I was thinking maybe somehow apply the pinching theorem? | 530218 | Evaluating limit making it [imath]\frac{\infty}{\infty}[/imath] and using L'Hopital Rule
Let [imath]P(x)=x^n+\displaystyle\sum\limits_{k=0}^{n-1}a_kx^k[/imath]. Find [imath] \lim_{x \to +\infty} ([P(x)]^{1/n}-x) [/imath] I know that in order to solve this problem I need to multiply it by something that will make it [imath]\frac{\infty}{\infty}[/imath] and then use L'Hopital Rule. I also know that the answer should be [imath]1/n[/imath] if I am not mistaken. I have tried multiplying the expression by [imath]\frac{e^x}{e^x}[/imath] and then using L'H. Rule but with not much success. Any suggestions on how I should proceed? Thank you for the help. |
852690 | What's wrong with having the same variable in the integrand as in the limits?
To give you a bit of context, if I'm trying to calculate the work done by a force from a distance [imath]x_0[/imath] (given constant) to [imath]x[/imath] (variable), I've been told (in lectures) to write the work done as [imath]W=\int_{x_0}^x F(\xi)d\xi \tag{1},[/imath] rather than [imath]W=\int_{x_0}^xF(x)dx \tag{2},[/imath] the reason being that "we've got the same variable in two places". But what exactly is wrong with this? Is [imath](2)[/imath] wrong? | 109105 | Limit of integration can't be the same as variable of integration?
I am told that an expression like [imath] \int_a^x f(x)dx [/imath] is not well formed, i.e. it should be [imath] \int_a^xf(t)dt [/imath] or similar. Why is it that the limits of integration can't depend on the variable of integration? |
852955 | For vector spaces [imath]V,W[/imath], probe that [imath]\Lambda (V)\otimes\Lambda (W) \cong \Lambda (V\oplus W)[/imath]
For vector spaces [imath]V,W[/imath], probe that [imath]\Lambda (V)\otimes\Lambda (W) \cong \Lambda (V\oplus W)[/imath]. I have tried to use the universal property, but I can not create the necessary linear transformation. | 822470 | Exterior power "commutes" with direct sum
I know that for vector spaces [imath]V, W[/imath] over a field [imath]K[/imath], we have the following identity : [imath] \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) [/imath] which holds in a canonical way : for [imath]0 \le k \le n[/imath], the maps [imath] \Lambda^k(V) \times \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W), \\ (v_1 \wedge\cdots \wedge v_k,w_{k+1} \wedge \cdots \wedge w_n) \mapsto (v_1 \wedge \cdots \wedge v_k) \wedge (w_{k+1} \wedge \cdots \wedge w_n) [/imath] factors through the tensor product's universal properties, which gives us a map [imath]\varphi_k : \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W)[/imath], and since the category of [imath]K[/imath]-modules has biproducts, we get a canonical map [imath] \varphi = \bigoplus_{k=0}^n \varphi_k : \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \longrightarrow \Lambda^n(V \oplus W) [/imath] What if I replace [imath]K[/imath] by a (commutative ring with [imath]1[/imath]) ring [imath]A[/imath]? Does this map have a kernel? I know it is surjective (it suffices to "look at it", i.e generators of the codomain are obviously all hit by [imath]\varphi[/imath]). I don't know if it helps, but I am working in integral domains. In the field case, this is obvious if [imath]V[/imath] and [imath]W[/imath] are finitely generated, since one can use dimension arguments. I wouldn't mind if the above identity held only in the finitely generated case either. If it worked in general it would be cool though. Note that when trying to compute the inverse map with the universal property, there seems to be a problem... my guess would have been [imath] (v_1 + w_1,\cdots,v_n + w_n) \mapsto \bigoplus_{k=0}^n (v_1 \wedge \cdots \wedge v_k) \otimes (w_{k+1} \wedge \cdots \wedge w_n) [/imath] and this map is obviously [imath]A[/imath]-multilinear, but there is a problem with the alternating property ; I get some terms left. I'm guessing this is the wrong map... if there is one. |
286947 | Number of zero digits in factorials
Here is a riddle someone has been asked in a job interview: How many zero digits are there in [imath]100![/imath]? Well, I found the first [imath]24[/imath] quite fast by counting how many times five divides [imath]100![/imath] ([imath]5[/imath] divides [imath]20[/imath] times and [imath]25[/imath] divides it [imath]4[/imath] times). However, there are more zero digits in the middle of the number (these can be found by hand, by typing factorial(100) in sage). My question is whether there is a smart way to determine the number of zero digits in [imath]100![/imath], and more generally in [imath]n![/imath]. By the way, this will not affect the job interview as it was finished some time ago. | 1604179 | How many trailing zeroes does 4617! contain?
I am getting [imath]1151[/imath] as answer on continuous division by [imath]5[/imath]. Is it right? On each division by 5, some remainder is generated...doesn't that count? Example: 4617/5 + 923/5 + 184/5 + 36/5 + 7/5 Divisor: 923 + 184 + 36 + 7 + 1 = 1151 Remainder: 2 , 3 , 4 , 1 , 2 |
853175 | [imath]p[/imath] and [imath]q[/imath] are primes. Prove [imath]\forall n,k\in \mathbb N, (p^n\mid q^kβp=q)[/imath]
I'm having trouble answering this question, can anyone help explain a full solution of this problem? I will be very grateful. Thanks! | 853071 | Question regardles primes and the fundamental theorem of arithmetic
I have been reading through my book of practice proofs and came across this particular question which has stumped me. [imath]p[/imath] and [imath]q[/imath] are primes. Prove [imath]\forall p \in \mathbb{Z}, \forall k \in \mathbb{Z}, (p^2\mid q^2) \Rightarrow p=q[/imath]. I assume that the proof starts with stating that [imath]p^n[/imath] and [imath]q^k[/imath] in their product of primes form but I am unsure on how division in these forms work or if I'm even correct in assuming that is the first step. Can anyone give me an explanation of the answer so that even a 16 year old can understand? Thanks! |
557975 | Solving [imath]\cos x=x[/imath]
I would like to know how can we solve the equation [imath]\cos x = x[/imath], without graphing. I know that there would only be one solution, that is obvious, that too in between [imath]0[/imath] and [imath]\frac{\pi}{2}[/imath]. Is there any real expression in finite terms [perhaps we call that closed form, I am not sure] that could give [imath]x[/imath] or [imath]\cos x[/imath]. Although I have not studied Taylor series, I know that it only gives an infinite series, which is not what I want. I suspect that it could not be done, but can anyone explain me why? Just for completeness, Wolfram Alpha, gives the approximate answer [imath]x = 0.7390851332151606416553120876738734040134[/imath], but fails to give a exact solution. | 853825 | How to solve [imath]x = \arccos(x)[/imath]
How does one solve [imath] x = \arccos(x) [/imath] for [imath]x[/imath]? Is there an exact solution achievable by hand? I tried wolfram|alpha, but it only spits out the solution [imath] x = \text{ root of }\quad x-\cos^{-1}(x) \quad \text{ near } x = 0.739085 [/imath] |
853844 | Help with a proof with combination: [imath]\binom{n}{k}\binom{k}{m} = \binom{n}{m} \binom{n-m}{k-m}[/imath]
I have to prove that: [imath]\binom{n}{k}\binom{k}{m} = \binom{n}{m} \binom{n-m}{k-m}[/imath] I'm not really sure how to approach this problem. I know the formula definition of combination but am unsure how to apply it to this question | 720206 | Prove [imath]\binom{n}{m}\binom{m}{k} = \binom{n}{k}\binom{n-k}{m-k}[/imath]
I need to prove the following: If [imath]n,m,k\in \mathbb{N}[/imath] and [imath]k\leq m \leq n[/imath], then [imath]\binom{n}{m}\binom{m}{k} = \binom{n}{k}\binom{n-k}{m-k}[/imath]. I did the following steps: \begin{align} \require{cancel} \binom{n}{m}\binom{m}{k} &= \binom{n}{k}\binom{n-k}{m-k} \\ \frac{n!}{m!(n-m)!}\cdot \frac{m!}{k!(m-k)!} &= \frac{n!}{k!(n-k)!}\cdot \frac{(n-k)!}{(m-k)!(n-k-m+k)!}\\ \frac{n!}{\cancel{m!}(n-m)!}\cdot \frac{\cancel{m!}}{k!(m-k)!} &= \frac{n!}{k!\cancel{(n-k)}!}\cdot \frac{\cancel{(n-k)}!}{(m-k)!(n-k-m+k)!} \\ \frac{n!}{k!(n-m)!(m-k)!} &= \frac{n!}{k!(n-m)!(m-k)!} \end{align} The question is: is my proof correct? Are all my steps valid? Thanks |
850963 | Evaluate the [imath]L^p(R^2)[/imath] norm of [imath]f(x_1,x_2) \times \chi_{ \{(x_1,x_2):x_1^2+x_2^2 \leq 1\} }[/imath]
Let [imath]f(x_1,x_2)=\frac{1}{(x_1+ix_2)^2} \times \chi_{(-1,1) \times (-1,1)}(x_1,x_2)[/imath]. Evaluate the [imath]L^p(R^2)[/imath] norm of [imath]f(x_1,x_2) \times \chi_{ \{(x_1,x_2):x_1^2+x_2^2 \leq 1\} }[/imath] for every [imath]1 \leq p \leq \infty[/imath]. This last function is =f inside the disk of radius 1 and is =0 outside the disk. [imath]||f||_{L^p(R^2)}=(\int_{ \{(x_1,x_2):x_1^2+x_2^2 \leq 1 \} } |f(x_1,x_2)|^p dx_1dx_2)^{1/p} =(\int_{ \{(x_1,x_2):x_1^2+x_2^2 \leq 1 \} } \frac{dx_1dx_2}{(x_1+ix_2)^{2p}})^{1/p} =(\int_0^{2 \pi} \int_0^1 \frac{rdrd \theta}{(rcos \theta+irsin \theta)^{2p}})^{1/p} =(\int_0^{2 \pi} \int_0^1 \frac{rdrd \theta}{(rcos \theta+irsin \theta)^{2p}})^{1/p} =(\int_0^{2 \pi} \int_0^1 \frac{drd \theta}{r^{2p-1}e^{i \theta 2p}})^{1/p} =(\int_0^{2 \pi} e^{-i \theta 2p} d \theta \times \int_0^1 r^{1-2p}dr)^{1/p} =(\frac{e^{-i \theta 2p}}{-i2p}]_0^{2 \pi} \times r^{-2p}]_0^1)^{1/p} =[(\frac{e^{-i4 \pi p}}{-i2p}+\frac{1}{i2p} \times (1-?)]^{1/p}[/imath] Where I have the question mark is my problem can anyone help me finish solving the problem? | 853899 | How can apply the [imath]L^p[/imath] norm in a circle to [imath]L^2[/imath] norm in a square?
Let [imath]f(x_1,x_2)=\frac{1}{(x_1+ix_2)^2} \chi_{ \{ (-1,1) \times (-1,1) \}}[/imath] Part A: Evaluate the [imath]L^p(R^2)[/imath] norm of [imath]f(x_1,x_2)\chi_{ \{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}}[/imath] for every [imath]1 \leq p \leq \infty[/imath] (this function is =f inside the disk of radius 1 and is =0 outside the disk) [imath]||f||_{L^p(R^2)}=(\int\int_{\{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}} |f(x_1,x_2)|^p dx_1dx_2)^{1/p} =(\int\int_{\{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}} (\sqrt{x_1^2+x_2^2})^{-2p} dx_1dx_2)^{1/p} =(\int_0^{2 \pi}\int_0^1 r^{-2p} rdr d \theta)^{1/p} =(\int_0^{2 \pi}d \theta \times \int_0^1 r^{-2p+1}dr)^{1/p} =(2 \pi \times r^{-2p+2}]_0^1)^{1/p} =(2 \pi \times \frac{1}{2-2p})^{1/p} =(\frac{\pi}{1-p})^{1/p}[/imath] Please check that the work is correct. Part B: Evaluate [imath]L^2(R^2)[/imath] norm of [imath]f(x_1,x_2)[/imath] [imath]||f||_{L^2(R^2)}=(\int\int_{\{ (-1,1) \times (-1,1) \}} |f(x_1,x_2)|^2 dx_1dx_2)^{1/2} =(\int_{-1}^!\int_{-1}^1 (\sqrt{x_1^2+x_2^2})^{-4} dx_1dx_2)^{1/2} =(\int_{-1}^!\int_{-1}^1 \frac{dx_1dx_2}{(x_1^2+x_2^2)^2} dx_1dx_2)^{1/2}[/imath] If I continue down this path I end up using multiple trig and u substitutions, how can I use Part A to solve this problem? |
854055 | Without using the value for [imath]\sin 18^Β°[/imath] or [imath]\cos 36^Β°[/imath], prove that [imath]4\sin 36^Β°\cos 18^Β°=\sqrt 5[/imath]
Without using the value for [imath]\sin 18^Β°[/imath] or [imath]\cos 36^Β°[/imath], prove that [imath]4\sin 36^Β°\cos 18^Β°=\sqrt 5[/imath] I tried to square it use some trigonometric formulas but could solve it. | 816195 | [imath]4 \sin 72^\circ \sin 36^\circ = \sqrt 5[/imath]
How do I establish this and similar values of trigonometric functions? [imath] 4 \sin 72^\circ \sin 36^\circ = \sqrt 5 [/imath] |
854068 | Locally unital ideals
Let [imath]R[/imath] be a ring with unity not necessarily commutative and [imath]I[/imath] an ideal of [imath]R[/imath]. Let for every element [imath]a \in I[/imath] there exists an element [imath]c\in I[/imath] such that [imath]ac=a[/imath]. Note that [imath]c[/imath] is related to [imath]a[/imath]. Now we have the following question: Can we say that for every element [imath]a\in I[/imath] there exists an idempotent element [imath]c\in I[/imath] such that [imath]ac=c[/imath]? Off course we have many examples such that the answer is true for them but in general we don't know. | 857416 | idempotents acting as local identities
Let [imath]R[/imath] be a ring with unity (not necessarily commutative) and [imath]I[/imath] an ideal of [imath]R[/imath]. Suppose that for every element [imath]a \in I[/imath] there exists an element [imath]c\in I[/imath] such that [imath]ac=a[/imath]. Note that [imath]c[/imath] is related to [imath]a[/imath]. Now we have the following question: Can we say that for every element [imath]a\in I[/imath] there exists an idempotent element [imath]c\in I[/imath] such that [imath]ac=a[/imath]? Of course we have many examples such that the answer is true for them but in general we don't know. |
854061 | Prove that [imath] 1+ \cos A + \cos B + \cos C = 0[/imath].
If [imath]A+B+C=180^\circ[/imath] and [imath]\tan \left[\dfrac{A+B-C} 4 \right] \tan \left[ \dfrac{-A+B+C} 4\right] \tan\left[\dfrac{A-B+C} 4 \right] =1[/imath] then prove that [imath] 1+ \cos A + \cos B + \cos C = 0[/imath] I tried to use the formula of [imath]\tan(A+B+C)[/imath], but couldn't solve it. | 624189 | Prove that [imath]1 + \cos\alpha + \cos\beta + \cos\gamma = 0[/imath]
If [imath]\alpha + \beta + \gamma = \pi [/imath] and [imath]\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1[/imath] Then prove that: [imath]1 + \cos\alpha + \cos\beta + \cos\gamma = 0[/imath]. I have no idea how to go about this. Please help. |
854447 | How to evaluate the sum [imath]\sum_{k = 0}^{n}2^k {{n}\choose {k}}[/imath]
How do I evaluate the sum: [imath]\sum_{k = 0}^{n}2^k {{n}\choose {k}}[/imath] I know that [imath]2^k = {n \choose 0} + {n \choose 1} + {n \choose 2} + {n \choose 3}... {n \choose k}[/imath], but I don't know how to proceed from this. | 69281 | Combinatorial proof help
I have to prove the identity using a combinatorial proof: [imath]\displaystyle\sum\limits_{k=0}^n 2^k \binom{n}{k} = 3^n[/imath] I think this should be my combinatorial proof: We want to form a committee of [imath]k[/imath] people from a total of [imath]n[/imath] people. There are two ways of counting this committee. 1) Go through each member from the [imath]n[/imath] total people, and decide if they should be added to the committee or not, until we have reached [imath]k[/imath] people. This gives us the LHS. ...For the RHS, however, I am not sure how to form it. I think it should be something like forming subsets of [imath]3[/imath] people and choosing from that, but I'm not sure how that will form the same committee as the LHS. EDIT: Okay, I also had the idea of forming a ternary string, and I could get the RHS this way. But I was not sure about the LHS. But the first answer gave me the right idea. Thanks a lot. |
188972 | Why complete measure spaces?
This is a longstanding confusion of mine without a clear answer. Why do we complete measure spaces? Certainly, it is nice to have the property that if some set is of measure zero, then a smaller set also should be so. However, when I looked up the mathematical literature, I am unable to find a single theorem that works better for complete measures. So, the only reasons in support of completing the measure seem to be: 1)., it agrees with our intuition of what can be neglected, 2)., the Caratheodory extension process automatically gives a complete measure. I was also surprised to realize that in probablity theory, very little use is made of the Lebesgue [imath]\sigma[/imath]-algebra. Almost always the Borel [imath]\sigma[/imath]-algebra is used. This prominence of Borel [imath]\sigma[/imath]-algebras also seems to be the case in ergodic theory, where one considers for instance the space of all probability measures on a given measurable space(usually equipped with a Borel [imath]\sigma[/imath]-algebra), and the null sets might differ from one measure to other(for example, consider Dirac measures concentrated at different points). So, what are some better mathematical reasons to argue for complete measures? Are there some theorems that I do not know, which are true only for complete measures? There ought to be some, to give credence to all W. Rudin's going-on about the process of completing a measure being as significant in analysis, if not more than, the process of completing the rationals into the real numbers. One can readily find a number of theorems in analysis that would break without the least upper bound property of real numbers. Such does not seem to be the case with completing the measure space, with a superficial look. | 147381 | Why do we essentially need complete measure space?
While reading the motivation of complete measure space on Wikipedia, what I concluded was, completeness is not really necessary when we define on one measure space and it is necessary when we want to measure on product of measure spaces (is it true ?). If [imath]\lambda[/imath] is measure on [imath]X[/imath] and [imath]Y[/imath] then is it true that [imath]\lambda^2[/imath] is measure of [imath]A[/imath]x[imath]B[/imath] and how ? I am not able to understand that [imath]\lambda^2(A\times B)=\lambda(A)\times\lambda(B)[/imath] ? Essentially what is the flaw in the measure without being complete ? Waiting for response. Thanks! |
855132 | Integration help containing exp and square root
Integrate this [imath] \int_{-\pi}^{\pi} \frac{e^{-inx}}{\sqrt{x^2+a^2}} \, \mathrm{d}x [/imath] Here [imath]a[/imath] is a constant. | 849218 | Integration of exponential and square root function
I need to solve this [imath]\int_{-\pi}^{\pi} \frac{e^{ixn}}{\sqrt{x^2+a^2}}\,dx,[/imath] where [imath]i^2=-1[/imath] and [imath]a[/imath] is a constant. |
855493 | Is [imath]2^{\alpha} < 2^{\beta}[/imath], where [imath]\alpha[/imath] and [imath]\beta[/imath] are cardinal numbers, such that [imath]\alpha < \beta[/imath]?
Let [imath]\alpha[/imath] and [imath]\beta[/imath] be cardinal numbers such that [imath]\alpha < \beta[/imath]. Isn't it always true that [imath]2^{\alpha} < 2^{\beta}[/imath] ? Because if I am not wrong, [imath]2^{\alpha}[/imath] denotes the immediate successor of [imath]\alpha[/imath] by GCH. Can anyone please suggest me some good book which can clarify my concepts on cardinals, ordinals and cofinality. Thank you in advance. | 244870 | Why continuum function isn't strictly increasing?
Is there any example that for cardinal numbers [imath]\kappa < \lambda[/imath], we have [imath]2^\kappa = 2^\lambda[/imath]? My guess is that it only depends on whether GCH holds. Is it true? |
855755 | For a real valued function [imath]f(x,y)[/imath] on [imath]\mathbb{R}^2[/imath] which is in [imath]L_2[/imath], show that [imath]f(x+Ξ΅,y+Ξ΅) β f(x,y)[/imath] in [imath]L_2[/imath] when [imath]Ξ΅ β 0.[/imath]
For a real valued function [imath]f(x,y)[/imath] on [imath]\mathbb{R}^2[/imath] which is in [imath]L_2[/imath], show that [imath]f(x+Ξ΅,y+Ξ΅) β f(x,y)[/imath] in [imath]L_2[/imath] when [imath]Ξ΅ β 0.[/imath] Not sure how to go about this problem. I tried Fubini. But that didn't seem to work well. I tried doing t straight from the definition. It seems the point is that it epsilon is changing the [imath]x[/imath] and [imath]y[/imath] coordinate at the same speed. | 488393 | How can I show that if [imath]f\in L^p(a, b)[/imath] then [imath]\lim_{t\to 0^{+}}\int_{a}^b |f(x+t)-f(x)|^p\ dx=0[/imath]..
can anyone help me show that if [imath]f\in L^p(a, b)[/imath] then [imath] \lim_{t\to 0^{+}}\int_{a}^b|f(x+t)-f(x)|^p\ dx=0.[/imath] Thanks, any help will be useful.. |
603833 | Conformal Mappings of Parabola?
Seems like it should be simple enough, but I haven't made any real progress.... Construct a bijective conformal mapping from the region below the parabola [imath]y=x^2[/imath] to the upper half plane. My first thoughts are to map [imath]y=x^2[/imath] or looking around online it seems smart to map to the disk, which is something I am familiar with.... but how would I go about doing this? Any ideas? I also thought to convert to polar coordinates and work with those, but It really just got uglier from there. Much Thanks! | 463436 | On a conformal mapping
I was asked to find a one-to-one analytic map [imath]f[/imath] of unit disc [imath]\mathbb{D}\subset \mathbb{C}[/imath] so that [imath]\mathbb{D}[/imath] is mapped to [imath]\{(x,y):y<x^2\}[/imath]. I thought the core procedure could be done by certain one-to-one holomorphic function, say [imath]g(z)=z^2[/imath] on [imath](-1,1)\times(0,\infty) \subset \mathbb{C}[/imath](which is one-to-one), but this maps to some field of parabolic that does not contain an interval: [imath]\{(x,y);x\leq 1-y^2/4\}\setminus [0,1][/imath]. I found it difficult to overcome the wield displacement of [imath][0,1][/imath], or was I going ahead in a wrong way? Thanks for the comment below it suffices to choose domain such as [imath]H:=\{\Re z>1\}[/imath] we can get the desired conformal map. But now I am concerned if the target set is the other side of the parabola, say, [imath]\{(x,y):y>x^2\}[/imath]? It seems function [imath]z \to z^2[/imath] does not send simple domain to the other side of the parabola, so how should I modify my conformal mapping at this time? |
856092 | Prove that if the quadratic form [imath]x^{T}Ax[/imath] is positive definite, so is [imath]x^{T}A^{-1}x[/imath]
Is it because the values in the inverse matrix will have the same positive/negative values? | 211453 | Inverse of a Positive Definite
Let K be nonsingular symmetric matrix, prove that if K is a positive definite so is [imath]K^{-1}[/imath] . My attempt: I have that [imath]K = K^T[/imath] so [imath]x^TKx = x^TK^Tx = (xK)^Tx = (xIK)^Tx[/imath] and then I don't know what to do next. |
180002 | Legendre symbol, second supplementary law
[imath]\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8}[/imath] how did they get the exponent. May be from Gauss lemma, but how. Suppose we have a = 2 and p = 11. Then n = 3 (6,8,10), but not [imath]15 = (11^2-1)/8[/imath] n is a way to compute Legendre symbols from Gauss lemma: [imath]\left(\frac{a}{p}\right) = (-1)^n[/imath] | 2189669 | Figuring out when [imath](\frac{2}{p}) = 1[/imath] for [imath]p[/imath] an odd prime
First I want to define the Legendre symbol [imath](\frac{a}{p})[/imath]. [imath](\frac{a}{p}) = 1[/imath] when a is a square mod p, and [imath](\frac{a}{p}) = -1[/imath] otherwise. I want to figure out when [imath](\frac{2}{p}) = 1, -1[/imath]. According to the answers, [imath](\frac{2}{p}) = 1[/imath] for [imath]p = 1, -1 (mod 8)[/imath] and [imath](\frac{2}{p}) = -1[/imath] for [imath]p = 3, -3 (mod 8)[/imath]. What I can't figure out is how to get to the answer. What I do know is that for any odd prime [imath]p[/imath], [imath]p = 1, 3 (mod 4)[/imath] and intuitively it seems that this is related to [imath]mod 8[/imath] somehow, but I'm not sure how to make the connection. Any help would be appreciated! (And apologies in advance for my horrible formatting here.) |
856396 | Show that [imath]d^T Z\sim N(d^T\mu, d^TVd)[/imath]
Consider [imath]Z=(Z_1,\ldots,Z_n)^T\sim N(\mu,V)[/imath] with [imath]\mu=(\mu_1,\ldots,\mu_n)^T[/imath] and [imath]V=\text{Cov}(Z)[/imath]. Show that for [imath]d\in\mathbb{R}^n[/imath] it is [imath] d^TZ\sim N(d^T\mu,d^TVd). [/imath] For me it is not clear how to show that [imath]d^TZ[/imath] is multivariate normal distributed. If that would be the case then because of [imath] E(d^TZ)=d^TE(Z)=d^T\mu [/imath] and [imath] \text{Cov}(d^TZ)=d^T\text{Cov}(Z)(d^T)^T=d^T\text{Cov}(Z)d [/imath] everything would be shown. I've read that [imath]d^TZ[/imath] is multivariate normal distributed exactly then when for each [imath]b\in\mathbb{R}[/imath] [imath] bd^TZ [/imath] is univariate normal distributed. So I choose an arbitrary [imath]b\in\mathbb{R}[/imath] and would have to show that [imath] bd^TZ=\sum_{i=1}^{n}bd_iZ_i~~~(*) [/imath] is univariate normal distributed. If the [imath]Z_i[/imath] were independent then it would be clear because a linear combination of independent univariate normal distributed random variables is itself univariate normal distributed. But here the [imath]Z_i[/imath] are not independent, what to do now to show that (*) is univariate normal distributed? | 803089 | Show that [imath]Y\sim N(a+A\mu,AVA^T)[/imath]
Consider [imath]Z=(Z_1,\ldots,Z_n)^T\sim N(\mu,V)[/imath]. Show: If [imath]a\in\mathbb{R}^m[/imath] and [imath]A[/imath] is a [imath](m\times n)[/imath]-matrix with [imath]\text{rang}(A)=m[/imath] then [imath] Y=a+AZ\sim N(a+A\mu,AVA^T). [/imath] My questions is, what do I have to show for this? I think I have to show that [imath]Y[/imath] has the density function [imath] f(Y)=(2\pi)^{-m/2}\text{det}(AVA^T)^{-1/2}\exp\left\{-\frac{1}{2}(Y-(a+A\mu))^T(AVA^T)^{-1}(Y-(a+A\mu)\right\}. [/imath] But how can I show that? Don't know how to start. |
856699 | proof question about an infinite sums
I once saw a place one interesting question, the question is: proof that if [imath]\displaystyle\sum_{n=1}^{+\infty}a_n[/imath] diverges, then [imath]\displaystyle\sum_{n=1}^{+\infty}\frac{a_n}{a_1+\cdots+a_n}[/imath] diverges. I have no idea what to do. | 845278 | Series of sequence with terms over series.
Prove or give a counterexample to disprove the following statement about series of real numbers: If [imath]a_k > 0, s_k = a_1 + \cdots + a_k[/imath] and [imath]b_k = a_k/s_k[/imath] then [imath]\sum a_k[/imath] and [imath]\sum b_k[/imath] either both converge or both diverge. |
856723 | Infinite product [imath]1+1/k^3[/imath]
Ramanujan's notebooks contain the result [imath]\prod_{k=1}^{\infty} \Big( 1 + \frac{1}{k^3}\Big) = \frac{1}{\pi} \mathrm{cosh}\Big( \frac{\pi \sqrt{3}}{2}\Big).[/imath] It doesn't seem like this is proved there but I guess it is well-known. How could you show this? | 679373 | Evaluate these infinite products [imath]\prod_{n\geq 2}(1-\frac{1}{n^3})[/imath] and [imath]\prod_{n\geq 1}(1+\frac{1}{n^3})[/imath]
What is [imath]\prod\limits_{n\geq 2}(1-\frac{1}{n^3})=?[/imath] [imath]\prod\limits_{n\geq 1}(1+\frac{1}{n^3})=?[/imath] I am sure about their convergence. But don't know about exact values. Know some bounds as well. For example first one is in interval (2/3,1) and second one is in (2,3). |
856965 | Infinite product: [imath]\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}[/imath]
I am trying to find [imath]\lim_{n\rightarrow \infty}\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}.[/imath]I have no idea how I can start. Please help me. Thanks! | 1595730 | To Find value of this infinite product [imath]P=\frac{7}{9} \frac{26}{28} \frac{63}{65}...\frac{n^3-1}{n^3+1}...[/imath]
Given [imath]P=\frac{7}{9} \frac{26}{28} \frac{63}{65}...\frac{n^3-1}{n^3+1}...[/imath] I have no idea how to do this Thanks |
857314 | Prove A or (A and B) is equivalent to A
Prove [imath]A \lor (A \land B) \Leftrightarrow A[/imath] without using truth table. The proof may involve expanding [imath]B[/imath] into [imath]B \land B[/imath] or possibly [imath]B \lor B[/imath]. I am stuck after playing with distributive law several times. Thanks for any help. | 844164 | Showing that [imath]\lnot Q \lor (\lnot Q \land R) = \lnot Q[/imath] without a truth table
I've done a truth table after reducing it to this and it seems to be equal to [imath]\neg Q[/imath]: [imath]\lnot Q \lor (\lnot Q \land R) = \lnot Q[/imath] But when I try to show it without a truth table (with just transformations), I end up in a loop between that and: [imath]\lnot Q \land (\lnot Q \lor R)[/imath] Is there a way to show this is true without using a truth table? What am I missing?! Thanks in advance! |
857265 | Prove that a sequence of positive integers [imath]d_1, d_2, \ldots, d_n[/imath] is a degree sequence of a tree if and only if [imath]d_1+d_2+\ldots+d_n = 2(n-1)[/imath]
Prove that a sequence of positive integers [imath]d_1, d_2, \ldots, d_n[/imath] is a degree sequence of a tree if and only if [imath]d_1+d_2+\ldots+d_n = 2(n-1)[/imath] [imath](\Rightarrow)[/imath] This follows from the handshaking lemma. [imath](\Leftarrow)[/imath] Suppose that for all [imath]k < n[/imath], the statement holds. Now, let [imath]d_n \geq d_{n-1} \geq \ldots \geq d_1[/imath], with [imath]d_1+d_2+\ldots+d_n = 2(n-1)[/imath]. Note that [imath]d_1 = 1[/imath]. Otherwise, [imath]\sum\limits_{k=1}^n d_k > 2(n-1)[/imath]. I don't really know how to proceed from here. It isn't the case that a simple graph with this property is a tree, and there are too many cases and caveats to take care of in the inductive step. Does anyone know a better/cleaner way to solve the problem? | 14100 | Condition on degrees for existence of a tree
Here is what I need to prove: Let [imath]d_1,d_2,...,d_n[/imath] be a sequence of natural numbers (>0). Show that [imath]d_i[/imath] is a degree sequence of some tree if and only if [imath]\sum d_i = 2(n-1)[/imath]. I know that: 1. for any graph [imath]\sum_{v \in V}\ deg(v) = 2e[/imath]; 2. for any tree [imath]e=v-1[/imath]. From 1 and 2 it follows that for any tree [imath]\sum_{v \in V}\ deg(v) = 2(v-1)[/imath]. If I understand it correctly, this is only a half of the proof ([imath]\rightarrow[/imath]), isn't it? Any hints on how to prove it the other way? Edit (induction attempt): [imath]n=1[/imath], we have [imath]d_1 = 2(1-1) = 0[/imath] and [imath]d_1[/imath] is a degree sequence of a tree. Let's assume the theorem holds for all [imath]k<n[/imath]. So we know that [imath]d_1 + d_2 + ... + d_{n-1} = 2(n-2) [/imath] and that it's a degree sequence of a tree. In order to add one vertex to this tree, so that it remains a tree we only can add a vertex of degree one (we connect it to any existing vertex with a single edge). By doing this our new sequence is [imath]1,d_1,...,d_j+1,...,d_{n-1}[/imath], where [imath]j[/imath] is the vertex to which we attached the new vertex (it's degree increments). Clearly this sums to [imath]2(n-1)[/imath]. Is this proof correct or am I missing the point? |
629630 | Simple proof EulerβMascheroni [imath]\gamma[/imath] constant
I'm searching for a really simple and beautiful proof that the sequence [imath](u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)[/imath] converges. At first I want to know if my answer is OK. My try: [imath]\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) = \lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} \log(k)-\log(k+1) = \lim\limits_{n\to\infty} \frac{1}{n} + \sum\limits_{k=1}^{n-1} \log(\frac{k}{k+1})+\frac{1}{k}[/imath] [imath] = \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})[/imath] Now we prove that the last sum converges by the comparison test: [imath]\frac{1}{k}-\log(\frac{k+1}{k}) < \frac{1}{k^2} \Leftrightarrow k<k^2\log(\frac{k+1}{k})+1[/imath] which surely holds for [imath]k\geqslant 1[/imath] As [imath] \sum\limits_{k=1}^{\infty} \frac{1}{k^2}[/imath] converges [imath] \Rightarrow \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})[/imath] converges and we name this limit [imath]\gamma[/imath] q.e.d | 1554114 | How to prove [imath]x_n[/imath] converges as [imath]n \to \infty[/imath].
Let [imath]x_n = \sum_{k=1}^{n} \frac{1}{k} - \log n[/imath]. Prove that [imath]x_n[/imath] converges as [imath]n \to\infty[/imath]. |
857693 | What is the sum of this series: [imath]\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}[/imath]
I tried getting it into a closed form but failed. Could someone help me out? [imath]\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}[/imath] | 3016886 | Calculate [imath]\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}[/imath]
[imath]\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}[/imath] should be calculated using complex numbers I think, the Wolfram answer is : [imath] \frac{1}{3} (e^x + 2 e^{-x/2} \cos(\frac{\sqrt{3}x}{2})) [/imath] How to approach this problem? |
858118 | Is [imath]\infty = \frac{1}{0}[/imath]?
Is [imath]\; \infty = \frac{1}{0}[/imath]? My teacher says no but he wouldn't explain it. My question is why [imath]\; \infty \neq \frac{1}{0}\;?[/imath] My thinking: Let [imath]\frac{1}{x}=p[/imath] Now as [imath]x[/imath] becomes smaller [imath]p[/imath] gets bigger. Ultimately when [imath]x[/imath] is smallest in magnitude then [imath]p[/imath] is largest which is what infinity is. Can anyone help me out? PS: I have just started Calculus, therefore please try to give answers according to the level of my understanding | 57890 | Why is [imath]\frac{1}{\infty } \approx 0 [/imath] and [imath] \frac{1}{0} = {\infty}[/imath]?
First I have checked the search option but found nothing relevant to my problem and also level of math. I just started learning the language of mathematics, on my own and I have trouble understanding why [imath] \frac{1}{\infty } \approx 0 [/imath] and why [imath]\frac{1}{0} = {\infty}[/imath]? I want to know how exactly do I get to these answers. It might sound strange but it's hard for me as a beginner to picture things like this on my own. |
858240 | [imath]B=\{f \in C[0,1] : \sup_{x\in [0,1]} |f(x)| \leq 1\}[/imath], [imath]\Gamma \in C^{*}[/imath] but [imath]\Gamma(B)[/imath] is open
Let [imath]C[/imath] be the Banach space of all complex continuous functions on [imath][0,1][/imath], with the supremum norm. Let [imath]B[/imath] be the closed unit ball of [imath]C[/imath]. Show that there exists continuous linear functionals [imath]\Gamma[/imath] on [imath]C[/imath] for which [imath]\Gamma(B)[/imath] is an open subset of the complex plane ; in particular, [imath]|\Gamma|[/imath] attains no maximum on [imath]B[/imath]. This is the problem in the book 'Rudin's fuctional analysis', p[imath]87[/imath] problem [imath]8[/imath]. By duality, [imath]\Gamma[/imath] can be expressed as follows. [imath]\Gamma(f)=\int_{[0,1]}f(x) \, d\mu(x)[/imath] for some regular Borel measure [imath]\mu[/imath]. The typical example is a point mass measure, but it doesn't work. I have no idea. (I'm calculating the case that [imath]\mu(1/n)=(1/2)^n(-1)^n, n=1,2,3,4,...[/imath]) Thank you in advance | 467402 | Separation theorem on the space of all complex continuous functions
Let [imath]C[/imath] be the Banach space of all complex continuous functions on [imath][0,1][/imath] endowed with the supremum norm. Let [imath]B[/imath] be the closed unit ball of [imath]C[/imath]. Does there exist a [imath]\Lambda\in C^*[/imath] for which [imath]\Lambda (B)[/imath] is open? Note that [imath]C[/imath] is a locally convex space and [imath]B[/imath] is convex, balanced and closed set, by the separation theorem of convex sets,we know there exists [imath]\Lambda\in C^*[/imath] such that [imath]|\Lambda f|\leq 1[/imath] for all [imath]f\in B[/imath] and [imath]|\Lambda f|>1[/imath] for [imath]f\in B^c[/imath]. But how about this when we consider some concrete space above? |
854013 | find all functions [imath] f : \mathbb{R} \rightarrow \mathbb{R} [/imath] such that : [imath] f(f(x))=x^2-2 [/imath]
This is a very hard functional equation. the problem is this : find all functions [imath] f : \mathbb{R} \rightarrow \mathbb{R} [/imath] such that : [imath] f(f(x))=x^2-2 [/imath] to solve it i have no idea! can we solve it with highschool olympiad education? please help : ) | 481017 | Find [imath]f(x)[/imath] such that [imath]f(f(x)) = x^2 - 2[/imath]
Find all [imath]f(x)[/imath] satisfying [imath]f(f(x)) = x^2 - 2[/imath]. Presumably [imath]f(x)[/imath] is supposed to be a function from [imath]\mathbb R[/imath] to [imath]\mathbb R[/imath] with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further. Possibly Helpful Links: Information on similar problems can be found here and here. Source: This question. It is about to be closed for containing too many problems in one question. I'm posting each problem separately. |
858624 | Functional Equation involving derivatives and time-steps
I am attempting to solve the equation [imath]f(x + 1) = f'(x)[/imath] for distributions [imath]C \rightarrow C: f(x)[/imath] My first guess to exploit the fact that this seems similar to identity [imath]\sin\left( \frac{\pi}{2} - x\right) = \cos(x) = \frac{d}{dx} [\sin(x)] [/imath] I therefore assume my answer takes on the form: [imath]Ce^{ax}[/imath] And now attempt to solve: [imath]Ce^{a(x+1)} = Cae^{ax}[/imath] Yielding [imath]e^{a}e^{ax} = ae^{ax}[/imath] Yielding [imath]e^a = a[/imath] How do I extract a single solution, let alone the infinitely many complex solutions for [imath]a[/imath] that satisfy this equation? | 199691 | Find [imath]f[/imath] where [imath]f'(x) = f(1+x)[/imath]
Let [imath]f \colon \mathbb{R} \rightarrow \mathbb{R}[/imath] be a smooth function such that [imath]f'(x) = f(1+x)[/imath] How can we find the general form of [imath]f[/imath]? I thought of some differential equations, but not sure how to use them here. Thanks. |
858075 | Graph theory: graph coloring quesiton
[imath]G_1[/imath] is graph on the set of vertices [imath]\{1,2,3,4,5,6,7,8\}[/imath], [imath]G_1[/imath] vertice chromatic number is 5. [imath]G_2[/imath] is graph on the set of vertices [imath]\{7,8,9,10,11,12,13,14,15,17,18,19,20\}[/imath], [imath]G_2[/imath] vertice chromatic number is 7. we know [imath]G_1[/imath] and [imath]G_2[/imath] have an edge between vertice 7 and vertice 8. (we already used 2 differnt colors on them both). [imath]G[/imath] is graph on the set of vertices [imath]\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}[/imath], that it's set of edges is the union of the edges set of [imath]G_1[/imath] with the edges set of [imath]G_2[/imath]. the vertice chromatic number of [imath]G[/imath] is: 5 / 7 / 12 / we can't know without further information prove the answer. As I understand, no vertice of [imath]G_1[/imath] can be connected to a vertice of [imath]G_2[/imath], except vertices 7 and 8, if we use 7 different colors on [imath]G_2[/imath], with vertice 7 and vertice 8 counted as 2 of them, and we use 5 colors on [imath]G_1[/imath] (with vertice 7 and vertice 8, counted as the same colors we used to color [imath]G_2[/imath] ) [imath]G[/imath] must have a vertice chromatic number of 7. too simple in my opinion, I'm already waiting for my mistake. | 858820 | graph vertex chromatic number in a union of 2 sub-graphs
[imath]G_1[/imath] is graph on the set of vertices [imath]\{1,2,3,4,5,6,7,8\}[/imath], [imath]G_1[/imath] vertice chromatic number is 5. [imath]G_2[/imath] is graph on the set of vertices [imath]\{7,8,9,10,11,12,13,14,15,17,18,19,20\}[/imath], [imath]G_2[/imath] vertice chromatic number is 7. we know [imath]G_1[/imath] and [imath]G_2[/imath] have an edge between vertice 7 and vertice 8. (we already used 2 differnt colors on them both). [imath]G[/imath] is graph on the set of vertices [imath]\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}[/imath], that it's set of edges is the union of the edges set of [imath]G_1[/imath] with the edges set of [imath]G_2[/imath]. the vertice chromatic number of [imath]G[/imath] is: 5 / 7 / 12 / we can't know without further information prove the answer. I think I got it, can someone approve? after the union to [imath]G[/imath], we copy the colors of [imath]G_2[/imath] to 7,8,9,10,11,12,13,14,15,16,17,18,19,20, obviously, [imath]G[/imath] vertex chromatic number is 7 and can't be less. we go through the vertices 1,2,3,4,5,6,7,8 we look for a vertice that doesn't connect to a different vertice. (there must be one, because if only one edge was missing in [imath]G_1[/imath], our chromatic number of [imath]G_1[/imath] was 7, but it's 5). once we find a missing edge, we color both it's vertices in the same color. we go through vertices 1-6, and color each vertice that doesn't have a color, with the colors of [imath]G_2[/imath] that we didn't use in [imath]G_1[/imath] (there should be 4-5 of them). |
859193 | Matrix with trace zero
Question is that : Suppose a matrix [imath]A\in M_n(\mathbb{C})[/imath] is a commutator by which i mean [imath]A=BC-CB[/imath] for some [imath]B,C\in M_n(\mathbb{C})[/imath] then we see that Trace of [imath]A[/imath] is [imath]0[/imath] But Suppose a matrix is of trace zero can we say that it is commutator? I have tried seeing this for [imath]2\times 2[/imath] matrices and end up with nothing reasonable.. I would be thankful if some one can suggest me a way to proceed in this case. | 181430 | If [imath]V_0[/imath] is the subspace of matrices of the form [imath]C=AB-BA[/imath] for some [imath]A,B[/imath] in a vector space [imath]V[/imath] then [imath]V_0=\{A\in V|\operatorname{Trace} (A)=0\}[/imath]
If [imath]V_0[/imath] is the subspace consisting of matrices of the form [imath]C=AB-BA[/imath] for some [imath]A,B[/imath] in a vector space [imath]V[/imath] then [imath]V_0=\{A\in V|\operatorname{Trace}(A)=0\}[/imath]. The problem above is one of the past qualifying exam problems. I can prove that [imath]V_0\subset\{A\in V|\operatorname{Trace}(A)=0\}[/imath] but I do not know what to do with converse. Thank you in advance. |
859393 | Tarski's theorem follows from choice
It is known that Tarski's theorem and axiom of choice are equivalent. Implication [imath]\Rightarrow[/imath] follows from considering bijection [imath](A+\aleph(A))^2\rightarrow(A+\aleph(A))[/imath]. Implication [imath]\Leftarrow[/imath] is often cited as "trivial". However, some time ago I started doubting triviality of this implication, and it isn't obvious for me anymore that [imath]|\alpha^2|=|\alpha|[/imath] for every [imath]\alpha\in\text{Ord}[/imath]. It's still clear for me that [imath]|\alpha+1|=|\alpha|[/imath] (because we can apply [imath]\omega+1=\omega[/imath]) and [imath]|\alpha2|=|\alpha|[/imath] (as [imath]\alpha[/imath] can be divided into [imath]\alpha[/imath] pairs, and we can rearrange them into [imath]\alpha2[/imath]), but I don't see such nice argument for [imath]|\alpha^2|=|\alpha|[/imath]. Sorry if this turns out as a silly question, but I would really want this to become clear for me. Thanks! | 54892 | About a paper of Zermelo
This about the famous article Zermelo, E., Beweis, daΓ jede Menge wohlgeordnet werden kann, Math. Ann.Β 59Β (4), 514β516Β (1904), available here. Edit: Springer link to the original (OCR'ed, may be behind a paywall) An English translation can be found in the book Collected Works/Gesammelte Werke by Ernst Zermelo. Alternative source: the book From Frege to GΓΆdel: a source book in mathematical logic, 1879-1931, by Jean Van Heijenoort. [See also this interesting text by Dan Grayson.] I don't understand the paragraph on the last page whose English translation is Accordingly, to every covering [imath]\gamma[/imath] there corresponds a definite well-ordering of the set [imath]M[/imath], even if the well-orderings that correspond to two distinct coverings are not always themselves distinct. There must at any rate exist at least one such well-ordering, and every set for which the totality of subsets, and so on, is meaningful may be regarded as well-ordered and its cardinality as an "aleph". It therefore follows that, for every transfinite cardinality, [imath]\mathfrak m=2\mathfrak m=\aleph_0\,\mathfrak m=\mathfrak m^2,\mbox{and so forth;}[/imath] and any two sets are "comparable"; that is, one of them can always be mapped one-to-one onto the other or one of its parts. It seems to me Zermelo says that the fact that any set can be well-ordered immediately implies that any infinite cardinal equals its square. Is this interpretation correct? If it is, what is the argument? Side question: What is, in a nutshell, the history of the statement that any infinite cardinal equals its square? Where was it stated for the first time? Where was it proved for the first time? (In a similar vein: where was the comparability of any two cardinal numbers proved for the first time?) Edit: German original of the passage in question: Somit entspricht jeder Belegung [imath]\gamma[/imath] eine ganz bestimmte Wohlordnung der Menge [imath]M[/imath], wenn auch nicht zwei verschiedenen Belegungen immer verschiedene. Jedenfalls muΓ es mindestens eine solche Wohlordnung geben, und jede Menge, fΓΌr welche die Gesamtheit der Teilmengen usw. einen Sinn hat, darf als eine wohlgeordnete, ihre MΓ€chtigkeit als ein βAlefβ betrachtet werden. So folgt also fΓΌr jede transfinite MΓ€chtigkeit [imath]\mathfrak m=2\mathfrak m=\aleph_0\,\mathfrak m=\mathfrak m^2\text{ usw.,}[/imath] und je zwei Mengen sind miteinander βvergleichbarβ, d.Β h. es ist immer die eine ein-eindeutig abbildbar auf die andere oder einen ihrer Teile. |
858948 | Describe the equivalence classes in terms of familiar mathematical objects
Consider the equivalence relation [imath]\sim[/imath] on [imath]\mathbb{Z} \times (\mathbb{Z} \setminus \{0\})[/imath] defined by [imath](a,b) \sim (c,d)[/imath] if [imath]a \cdot d = b \cdot c[/imath]. Describe the equivalence classes in terms of familiar mathematical objects? The above is a homework problem. I can see some patterns in the equivalence classes, however I'm not sure how to answer or approach the question given. Thanks! | 544854 | The class of equivalence.
Let [imath]Q[/imath] be the following set of [imath]\mathbb{Z}\times \mathbb{Z}[/imath] \begin{align*} Q=\{(a,b)\in \mathbb{Z}\times \mathbb{Z} | b\neq 0\} \end{align*} Define the relation [imath]\sim[/imath] on [imath]Q[/imath] as \begin{align*} (a,b)\sim (c,d)\Leftrightarrow ad=bc \end{align*} Prove that [imath]\sim[/imath] is a relation of equivalence, and give the class of equivalence of [imath][(2,3)][/imath]. Furthermore the general order of equivalence of [imath][(a,b)][/imath]. Try to give a description [imath]Q/\sim[/imath]. I've asked the same question (I deleted it one hour later) but it didn't look well when I didn't try to add what I tried. Here we go. I have proven that [imath]\sim[/imath] is a relation of equivalence by showin it's i) reflexive, ii) symmetric and iii) transitive. i) [imath]\forall (a,b)\in Q:(a,b) \sim (a,b)[/imath]. [imath]\forall (a,b)\in Q:(a, b) \sim (a, b) \Leftrightarrow ab \stackrel{\surd}= ab[/imath] ii) [imath](a, b) \sim (c, d) \Rightarrow (c, d) \sim (a,b)[/imath] [imath](a, b) \sim (c, d) \Leftrightarrow ad = bc[/imath] and [imath](c, d) \sim (a, b) \Leftrightarrow cb = da[/imath] then [imath]ad = bc \stackrel{\surd}\Rightarrow cb = da[/imath]. iii) [imath][(a, b) \sim (c,d) \wedge (c, d) \sim (e, f)] \Rightarrow (a, b) \sim (e, f)[/imath] [imath](a, b)\sim (c, d) \Leftrightarrow ad = bc[/imath], [imath](c, d) \sim (e, f) \Leftrightarrow cf = de[/imath] and [imath](a, b) \sim (e, f) \Leftrightarrow af = be[/imath] then [imath]((ad = bc) \wedge (cf = de))\stackrel{\surd}\Rightarrow af = be[/imath]. The class of eq. [imath][(2,3)][/imath] is the class, that contains the element [imath](2,3)[/imath], so for every element [imath](a,b)[/imath] in this class satisfies [imath](a,b) \sim (2,3) \Leftrightarrow 3a=2b[/imath], that is \begin{equation} [(2,3)] = \lbrace (a,b) \in \mathbb{Z}\times\mathbb{Z} | b \neq 0 \wedge 3a = 2b \rbrace \end{equation} So now I've shown what I tried to show. Is there something I'm missing or something I should have mentioned? I need some help to answer the rest. Thanks. |
860162 | What is the solution of the following integral?
I tried to solve the following integral using Maple as well as by hand but unable to do so. Can anybody help me in solving the following integral? [imath] \int_{0}^{R} D\pi r^2 (D\pi r^2-1)^B 2\pi \lambda \alpha r e^{-\pi r^2(\alpha \lambda - D ln(Y))} dr [/imath] where [imath]D[/imath], [imath]B[/imath], [imath]Y[/imath], [imath]\alpha[/imath], [imath]\lambda[/imath] are constants | 860098 | Approximate closed form of Integration involving a lot of terms
I tried to solve the following integral using Maple as well as by hand but unable to do so. Can anybody help me in solving the following integral? [imath] \int_{0}^{R} D\pi r^2 (D\pi r^2-1)^B 2\pi \lambda \alpha r e^{-\pi r^2(\alpha \lambda - D ln(Y))} dr [/imath] In the above equation, [imath]D[/imath], [imath]B[/imath], [imath]Y[/imath], [imath]\alpha[/imath], [imath]\lambda[/imath] are constants |
860303 | [imath]\lim \limits_{x \to \infty} \space f({x})=0[/imath], [imath]|f''(x)| < M[/imath]. Prove: [imath]\lim \limits_{x \to \infty} \space f'({x})=0[/imath]
It's known that: [imath]\lim \limits_{x \to \infty} \space f({x})=0[/imath] [imath]|f''(x)| < M[/imath] How do I prove that: [imath]\lim \limits_{x \to \infty} \space f'({x})=0[/imath] I tried using the mean value theorem, but I don't really know how to approach the question. | 772475 | Show that [imath]\lim_{x\to\infty} f(x)=0 \implies \lim_{x\to\infty} f'(x)=0[/imath]
I'm looking for a good proof for the following problem. I think it's pretty intuitive why this is true, but I'm having trouble expressing it rigorously. Suppose that [imath]f[/imath] is twice differentiable and that [imath]f''[/imath] is bounded on [imath](0, \infty)[/imath]. Furthermore, suppose that [imath]\lim_{x\to \infty} f(x)=0[/imath]. Prove that [imath]\lim_{x\to\infty} f'(x)=0[/imath]. I've already tried writing out the Taylor expansion for [imath]f[/imath] and tweaking that, but I haven't had much success. Any insight would be greatly appreciated. |
860201 | [imath]f\in L^1(0,\infty)[/imath] monotone, show [imath]\lim_{x\rightarrow \infty} xf(x) = 0[/imath]
Here is the solution: First [imath]f[/imath] is monotone and integrable on [imath](0,\infty)[/imath], wolg we can assume that [imath]f>0[/imath] and approaches [imath]0[/imath] as [imath]x[/imath] goes to infinity. Observe that [imath]xf(2x) \leq \int_x^{2x} f(t)[/imath] since when [imath]t\in [x,2x][/imath] we have [imath]f(t) \geq f(2x)[/imath] because [imath]f[/imath] is decreasing. And since [imath]f[/imath] is integrable, we have [imath]\lim_{x\rightarrow \infty} \int_x^{2x} f(t) = 0,[/imath] thus [imath]\lim_{x\rightarrow \infty} 2xf(2x) = 0.[/imath] | 119409 | If monotone decreasing and [imath]\int_0^\infty f(x)dx <\infty[/imath] then [imath]\lim_{x\to\infty} xf(x)=0.[/imath]
Let [imath]f:\mathbb{R}_+ \to \mathbb{R}_+[/imath] be a monotone decreasing function defined on the positive real numbers with [imath]\int_0^\infty f(x)dx <\infty.[/imath] Show that [imath]\lim_{x\to\infty} xf(x)=0.[/imath] This is my proof: Suppose not. Then there is [imath]\varepsilon[/imath] such that for any [imath]M>0[/imath] there exists [imath]x\geq M[/imath] such that [imath]xf(x)\geq \varepsilon[/imath]. So we can construct a sequence [imath](x_n)[/imath] such that [imath]x_n \to \infty [/imath] and [imath]x_n f(x_n ) \geq \varepsilon[/imath]. So [imath]\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.[/imath] So we get a contradiction. I feal like I have the correct idea but some details are wrong. Any help would be appreciated. |
860442 | Simplify: [imath]\displaystyle\frac{\frac{a}{x}-\frac{a}{y}}{\frac{a}{x}+\frac{a}{y}}[/imath]
Simplify: [imath]\displaystyle\frac{\frac{a}{x}-\frac{a}{y}}{\frac{a}{x}+\frac{a}{y}}[/imath] I did [imath]\displaystyle\frac{ay}{xy}-\frac{ax}{xy}=y-x[/imath] on top in underside. I did [imath]\displaystyle\frac{ay}{xy}+\frac{ax}{xy}=ay+ax=a(y+x)[/imath]. My final answer is [imath]\displaystyle\frac{y-x}{ay+ax}[/imath]. Is this correct? I posted question before but now just bit different. Please explain in steps so I understand this form and solve my other problems. | 860381 | equation [imath]\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}}[/imath]
Frist I solved this equation [imath]\displaystyle \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{xy}}[/imath] by multiplying [imath]xy \left(\frac{xy}{x}+\frac{xy}{y}\right)=y+x[/imath] The answer is correct. Next equation was: [imath]\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}}[/imath] I made the bottom multiplied and got [imath]x^2 y^2.[/imath] So I made the up also multiply and got [imath]\frac{\frac{y^2}{x^2}-\frac{x^2}{y^2}}{\frac{1}{x^2y^2}}.[/imath] Now I'm stuck. So now I doubt I did the first one the right way. I might have gotten the answer right but not the correct way to get it. Im sorry if Im confusing. Can someone help me with the two were I go wrong? |
860731 | True or False and why ? basic integration
[imath]\int_0^2f(x)\,\mathrm dx=\int_0^2 f(t)\,\mathrm dt.[/imath] I would say true because a&b = a&b and it would not make a change if is x,d,f,q,w,t it will logically be all the same, am I right ? but because it is MATH not logic so everything could be unexpected! | 646238 | How do I convince my students that the choice of variable of integration is irrelevant?
I will be TA this semester for the second course on Calculus, which contains the definite integral. I have thought this since the time I took this course, so how do I convince my students that for a definite integral [imath]\int_a^b f(x)\ dx=\int_a^b f(z)\ dz=\int_a^b f(βΊ)\ dβΊ[/imath] i.e. The choice of variable of integration is irrelevant? I still do not have an answer to this question, so I would really hope someone would guide me along, or share your thoughts. (through comments of course) NEW EDIT: I've found a relevant example from before, that will probably confuse most new students. And also give new insights to this question. Example: If [imath]f[/imath] is continuous, prove that [imath]\int_0^{\pi/2}f(\cos x)\ dx = \int_0^{\pi/2}f(\sin x)\ dx[/imath] And so I start proving... Note that [imath]\cos x=\sin(\frac{\pi}{2} -x)[/imath] and that [imath]f[/imath] is continuous, the integral is well-defined and [imath]\int_0^{\pi/2}f(\cos x)\ dx=\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx [/imath] Applying the substitution [imath]u=\frac{\pi}{2} -x[/imath], we obtain [imath]dx =-du[/imath] and hence [imath]\int_0^{\pi/2}f(\sin(\frac{\pi}{2}-x))\ dx=-\int_{\pi/2}^{0}f(\sin u)\ du=\int_0^{\pi/2}f(\sin u)\ du\color{red}{=\int_0^{\pi/2}f(\sin x)\ dx}[/imath] Where the red part is the replacement of the dummy variable. So now, students, or even some of my peers will ask: [imath]u[/imath] is now dependent on [imath]x[/imath], what now? Why is the replacement still valid? For me, I guess I will still answer according to the best answer here (by Harald), but I would love to hear more comments about this. |
81690 | If a local ring is [imath]\mathfrak{m}[/imath]-adically complete is it also [imath]I[/imath]-adically complete
Suppose [imath](R,\mathfrak{m})[/imath] is a local ring and [imath]I[/imath] a proper ideal. If [imath]R[/imath] is [imath]\mathfrak{m}[/imath]-adically complete is it also [imath]I[/imath]-adically complete. | 17685 | [imath]I[/imath]-adic completion
Let [imath]A[/imath] be a commutative noetherian ring, and suppose that [imath]A[/imath] is [imath]I[/imath]-adically complete with respect to some ideal [imath]I\subseteq A[/imath]. Is it true that for any ideal [imath]J\subseteq I[/imath], the ring [imath]A[/imath] is also [imath]J[/imath]-adically complete? Edit. Recall that a ring [imath]A[/imath] is [imath]I[/imath]-adically complete if the canonical morphism [imath]A\to \varprojlim A/I^n[/imath] is an isomorphism. |
861076 | Fun Tan Question
Using only trig identities, how would you approach the following question? Determine the value of [imath] \prod_{i=1}^{89} \tan i^Β° = \tan 1^Β° \cdot \tan 2^Β° \cdots \tan 89^Β° [/imath] | 805704 | [imath] \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ[/imath]
How can I find the following product using elementary trigonometry? [imath] \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ.[/imath] I have tried using a substitution, but nothing has worked. |
861309 | are there two non-zero functons [imath]f, g[/imath] in [imath]L^1(\mathbb{R})[/imath] such that [imath]f\ast g=0[/imath] a.e
are there two non-zero functions [imath]f, g[/imath] in [imath]L^1(\mathbb{R})[/imath] such that [imath]f\ast g=0[/imath] a.e? | 90608 | Does the Banach algebra [imath]L^1(\mathbb{R})[/imath] have zero divisors?
Assume that the functions [imath]f,g: \mathbb{R}\rightarrow \mathbb{R}[/imath] are integrable and equal to zero on [imath](-\infty,0)[/imath], (i.e [imath]f,g \in L^+[/imath]). Then by Titchmarsh's theorem: [imath]f*g[/imath] is zero almost everywhere iff [imath]f[/imath] or [imath]g[/imath] is zero almost everywhere. Hence the Banach subalgebra [imath]L^+[/imath] of [imath]L^1(\mathbb{R})[/imath] has no zero divisors. Is the same true for arbitrary integrable [imath]f[/imath] and [imath]g[/imath] on [imath]\mathbb{R}[/imath] ? |
861308 | How to prove a levi-civita symbol and kronecker delta relationship
[imath]\displaystyle \sum_{i=1}^3 \sum_{j=1}^3 \epsilon_{ijk} \epsilon_{ijn} = 2 \delta_{kn}[/imath] When I do the calculations of that I get 3 times the answer, I mean this is easy, but IΒ΄m just wrong, Could someone show me the way? | 369659 | Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory
In order to prove the following identity: [imath]\sum_{k}\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}[/imath] Instead of checking this by brute force, Landau writes thr product of Levi-Civita symbols as: [imath]\epsilon_{ijk}\epsilon_{lmn}=\det\left| \begin{array}{cccc} \delta_{il} & \delta_{im} & \delta_{in} \\ \delta_{jl} & \delta_{jm} & \delta_{jn} \\ \delta_{kl} & \delta_{km} & \delta_{kn} \end{array} \right| [/imath] The proof that the equalty holds is quite straightforward if you consider what values the indices can take. But I've been told that there's a much more profound and elegant demonstration based on the representation of the symmetric group. Does anybody know this approach based on group theory? |
861483 | Fixed point equivalence
Let [imath]\Bbb D^n[/imath] be n-dimensional ball and [imath]S^{n-1}[/imath] the [imath]n-1[/imath] dimensional sphere realized as boundary of [imath]\Bbb D^n[/imath]. Prove that following are equivalent. There is no retraction [imath]\Bbb D^n\to S^{n-1}[/imath] Every Continuous map [imath]\Bbb D^n \to \Bbb D^n[/imath] has a fixed point. I can do for the case of [imath]n=2[/imath]. For this case, I don't even need the help of 2. to prove 1. just i will show there is induced homomorphism between them such that [imath]\pi _*(S^{1}) \to \pi _*(\Bbb D^2) \to \pi_*(S^1)[/imath] does not commute. Next direction is also easy for the case of n=2 But I can't do for the case of [imath]n>2[/imath]. could u please do for me for higher dimension. | 767980 | Equivalent statements to fixed-point theorem
I'm trying to show that they're equivalent statements: 1) [imath]1_{S^1}[/imath] is not homotopic to a constant map. 2) [imath]S^1[/imath] is not a retract of [imath]D^2[/imath] ([imath]D^2[/imath] is the closed unit ball). 3) Every continuous map [imath]f:D^2\to D^2[/imath] has a fixed point. 1)[imath]\to[/imath]2) I suppose the opposite, then there exists [imath]r:D^2\to S^1[/imath] continuous such that [imath]r(x)=x[/imath] for every [imath]x\in S^1[/imath]. How can we show that [imath]1_{S^1}[/imath] is homotopic to a constant? 2)[imath]\to[/imath]3) I suppose that there exists [imath]f:D^2\to D^2[/imath] such that [imath]f(x)\neq x[/imath] for every [imath]x\in D^2[/imath], but I don't know how to prove that [imath]S^1[/imath] would be a retract of [imath]D[/imath]. 3)[imath]\to[/imath]1) Of course I tried to suppose that [imath]1_{S^1}[/imath] is homotopic to a constant, and show a map without fixed points. Any hint? Thanks. |
861552 | Test for convergence [imath]\sum_{n = 2}^\infty \frac{1}{(n+1)\ln^2(n+1)}[/imath]
Test for convergence [imath]\sum_{n = 2}^\infty \frac{1}{(n+1)\ln^2(n+1)}[/imath] Here's my attempt! I decided to use the integral test for this. [imath]\frac{1}{(n+1)\ln^2(n+1)} > \frac{1}{(n+1)^2\ln^2(n+1)}[/imath] Set [imath]f(x) = \dfrac{1}{x\ln(x)}[/imath] and let [imath]x = (n + 1)^2[/imath] [imath]\int_9^\infty\frac{1}{x\ln(x)}dx = \int_{\ln(9)}^\infty\frac{du}{u} = \ln(u)_{\ln(9)}^\infty = \infty[/imath] | 655240 | Convergence of series involving in iterated logarithms [imath]\sum \frac{1}{n(\log n)^{\alpha_1}\cdots (\log^k(n))^{\alpha_k} }[/imath]
What is the quickest way to show when [imath] S(\alpha_1,\alpha_2,\cdots,\alpha_k) = \sum\limits_{n=3}^\infty \frac{1}{n (\log n)^{\alpha_1}\cdots (\log^k(n))^{\alpha_k}} [/imath] converges, where [imath]\log^k(n)[/imath] is the [imath]k[/imath]-th iteration of natural logarithm? |
861646 | Local minimum and gradient
But the proof here below is specially elegant. Is there any function [imath]f[/imath] such that [imath]f[/imath] has a local minimum at [imath]x[/imath] but [imath]\nabla f(x) \neq 0[/imath]? Only assumption on [imath]f[/imath] is that it has to be differentiable at [imath]x[/imath] so that I can write [imath]\nabla f(x)\neq 0[/imath] to ask you this question. | 392952 | Proof of: If [imath]x_0\in \mathbb R^n[/imath] is a point of local minimum of [imath]f[/imath], then [imath]\nabla f(x_0) = 0[/imath].
Let [imath]f: \mathbb R ^n\to\mathbb R[/imath] be a differentiable function. If [imath]x_0\in \mathbb R^n[/imath] is a point of local minimum of [imath]f[/imath], then [imath]\nabla f(x_0) = 0[/imath]. Where can I find a proof for this theorem? This is a theorem for max/min in calculus of several variables. @Peter Tamaroff Here is my attempt: Let [imath]x_0[/imath] = [x1,x2,...xn] Let g[imath]_i[/imath](x) = f(x[imath]_0[/imath]+(x-x[imath]_i[/imath])e[imath]_i[/imath]) where e[imath]_i[/imath] is the ith standard basis vector of dimension n. Since f has local min at x[imath]_0[/imath], then g[imath]_i[/imath](x) has local min at x[imath]_i[/imath]. So by Fermat's theorem, g'(x[imath]_i[/imath])= 0 which is equal to f[imath]_{x_i}[/imath]([imath]x_0)[/imath] Therefore f[imath]_{x_i}[/imath]([imath]x_0)[/imath] = 0 which is what you wanted to show. Is this right??? Any comments would be appreciated. |
861660 | Conditions must satisfy [imath]f: (a, b) \to \mathbb{R}[/imath] so that its Taylor series converge to f itself.
I have a doubt. What conditions must satisfy [imath]f: (a, b) \to \mathbb{R}[/imath] so that its Taylor series converge to f itself. | 117714 | Prove Taylor series converges to [imath]f[/imath].
Let [imath]f[/imath] be an infinitely differentiable function on an interval [imath]I[/imath]. If [imath]a \in I[/imath] and there are positive constants [imath]C[/imath], [imath]R[/imath] such that for every [imath]x[/imath] in a neighborhood of [imath]a[/imath] and every [imath]k[/imath] it holds that [imath]|f^{(k)}(x)| \leq C \frac{k!}{R^k}[/imath] then prove that the Taylor series of [imath]f[/imath] about [imath]a[/imath] converges to [imath]f(x)[/imath]. I think a good approach would be to estimate the error term. I'm not sure how to proceed exactly though. Thoughts? |
859663 | Probability: NEED HELP to Understand with the follow
I need help to understand the probability derviation of a paper. Please help me. For the following, please only treat [imath]|h_{R,B}|^2[/imath] and [imath]|h_{A,R}|^2[/imath] as random variables (other parameters can be treated as constants). Those two random varibles are i.i.d. such that they are exponential distributed with parameter [imath]E[|h_{R,B}^2|][/imath] and [imath]E[|h_{A,R}^2|][/imath] respectively for [imath]|h_{R,B}|^2[/imath] and [imath]|h_{A,R}|^2[/imath]. In particular, I don't understand what he did after line 74 (the yellowed part). | 861404 | [Probability]need help to understand the following expression
So assume [imath]Y[/imath] and [imath]X[/imath] are exponentially distributed with parameters [imath]y_1[/imath], and [imath]x_1[/imath] respecitively. assume c is a constant. I am having huge trouble to understand the integration of the following expression. [imath]P(Y<c/u(X))[/imath] [imath]=\int_{t}^{\infty}f_X(x)\int_{0}^{c/u(x)}f_Y(y)dydx +\int_{0}^{t}f_X(x)\int_{c/u(x)}^{\infty}f_Y(y)dydx[/imath] where t is the cross-point that [imath]u(x)[/imath] change sign here [imath]c/u(x)[/imath] is given by the plot below, t is the point crossing the zero: Confusion: I don't understand the second integration of the second term "[imath]\int_{c/u(x)}^{\infty}f_Y(y)dydx[/imath]", this isn't right because Y an X only defined for y>0, and x>0. So it's the first quadrant in this plot. |
538051 | Is there any simple method to calculate [imath]\sqrt x[/imath] without using logarithm
Suppose that we don't know logarithm, then how we would able to calculate [imath]\sqrt x[/imath], where [imath]x[/imath] is a real number? More generally, is there any algorithm to calculate [imath]\sqrt [ n ]{ x } [/imath] without using logarithm? More simple techniques would be nice. Here is a simple technique used to approximate square roots by Persian author Hassan be al-Hossein: For example: [imath]\sqrt {78}\approx 8\frac { 14 }{ 17 } [/imath] , where [imath]8[/imath] is the nearest integer root of [imath]78[/imath], [imath]14 = 78 - 8^2[/imath], [imath]17 = 2 \times 8 + 1[/imath]. if [imath]n=2^k[/imath] we can use the method above. For example, for [imath]k=2[/imath] Lets calculate [imath]\sqrt [ 4 ]{ 136 } [/imath]: [imath]\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 } } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 } } =\sqrt { 11\frac { 15 }{ 23 } } \\ \sqrt { 11\frac { 15 }{ 23 } } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\[/imath] The exact result is[imath] \sqrt [ 4 ]{ 136 } =3.4149\cdots[/imath] The method approximates well, but it is working for only [imath]n=2^k[/imath] as I know. | 861509 | What is the sum that the square root button on calculator does so I can do it without the calculator button
I am not very good when it comes to Maths but the current work I am doing means I need to get better and quick. I have been teaching myself about areas, diagonals and square roots. However I am struggling to understand how the calculator works out the square root. For example I know that a diagonal in a square from corner to corner is approx. [imath]1.414[/imath] of the side distance of the square. so if I have a square that is [imath]a=1 \times b=1[/imath] I can do [imath]a \times \sqrt{2}(1.414)[/imath]. I know that the distance of the diagonal is [imath]d=1.414[/imath]. I want to know how to work out the square root without a calculator? Or am I just getting this whole thing wrong? Sorry I did say I am not very good at Maths. |
164986 | Smallest multiple whose digits are only ones and zeros
I have a collection of typewritten pages that formed the basis of a third year problem solving course offered about 25 years ago at U. Waterloo. I've been slowly working through the problems and have come up with a solution to one of the questions but was wondering if anyone had a better solution. The question is: Find the smallest natural number composed only of ones and zeros which is divisible by 225. My solution: We are looking for a positive integer [imath]k[/imath] so that [imath]N=225k[/imath] has only ones and zeroes in the digits. Since any multiple of 5 has a 0 or 5 for the last digit, then k must be even. But [imath]2\times 225=450[/imath], so it really means that [imath]k[/imath] must be a multiple of 4. Now [imath]N=900K[/imath], where [imath]4K=k[/imath]. Let [imath]100K=\sum_{i=0}^\infty a_i 10^i[/imath], where (potentially) an infinite number of the [imath]a_i[/imath]'s vanish. The solution method is to look at each term [imath]a_i[/imath] starting with [imath]a_0[/imath] and choose the [imath]a_i[/imath] so that the ones digit of the product [imath]9a_i + c_{i-1}[/imath] is a one or zero, where [imath]c_{i-1}[/imath] is the carry from the previous product. This forms a tree, starting with [imath]a_0[/imath] at the root. The smallest [imath]N[/imath] is found from the smallest limb of the tree. Each of the following congruences are [imath]\mod 10[/imath]. We choose [imath]a_0[/imath] so that [imath]9a_0 \equiv 0[/imath] or [imath]9a_0 \equiv 1[/imath]. Only [imath]a_0=9[/imath] works here, so there is a carry of 8 when considering [imath]a_1[/imath]. Then we have the congruences [imath]9a_1+8\equiv 0[/imath] or [imath]9a_1+8\equiv 1[/imath]. This gives two options to consider for [imath]a_1[/imath]: either [imath]a_1=8[/imath] with a carry of 8 or [imath]a_1=7[/imath] with a carry of 7. The first just leads to a larger number for [imath]N[/imath], so we ignore it and move on to [imath]a_2[/imath]. This has four options, but the only one we can use that has smallest [imath]N[/imath] is [imath]a_2=6[/imath] with a carry of 6. Continuing in this way leads to [imath]K=12345679[/imath] and [imath]k=4\times 12345679 = 49072716[/imath], which means that [imath]N=11111111100[/imath]. This was verified by a simple computer program I wrote that checked all of the smaller multiples. Question for the group: For different numbers from 225, this leads to very dense and deep trees when solving for [imath]a_i[/imath]. In fact, I had a originally solved this question using 255 instead of 225 because I misread it, which had a much denser tree because none of the branches could be eliminated at each step. Does anyone know of a cleaner/simpler solution? | 388165 | How to find the smallest number with just [imath]0[/imath] and [imath]1[/imath] which is divided by a given number?
Every positive integer divide some number whose representation (base [imath]10[/imath]) contains only zeroes and ones. One can easily prove that using pigeonhole principle. Some examples: 2 -> 10 3 -> 111 4 -> 100 ... But how to find the smallest answer? Is there any way to find it efficiently? |
652626 | integrability of Hilbert transform of a function
The problem is : Let [imath]\varphi \in \mathcal S(\mathbb R)[/imath] (Schwartz space) with [imath]\int \varphi \ dx = 0[/imath]. Then the Hilbert transform of [imath]\varphi[/imath] belongs to [imath]L^1(\mathbb R)[/imath]. I believe this helps [imath]\lim_{|x| \rightarrow \infty} x H(\varphi) (x) = 0[/imath]. But i dont know what to do.. someone can give me a hint ? Thanks in advance | 142834 | Integrability of the Hilbert transform of a Schwartz function
Given a Schwartz function [imath]f\colon\mathbb{R}\to\mathbb{R}[/imath], define its Hilbert transform by [imath](Hf)(x)=\frac{1}{\pi}\left(\int_{|t|\leq 1}\frac{f(x-t)-f(x)}{t}\,dt + \int_{|t|\geq 1}\frac{f(x-t)}{t}\,dt\right)[/imath] (the first integral is interpreted as an appropriate limit/principal value). It can be shown that [imath]Hf[/imath] is continuous and that [imath]\lim_{|x|\to\infty}x(Hf)(x)=\frac{1}{\pi}\int_{\mathbb{R}}f(t)\,dt=\frac{1}{\pi}\hat{f}(0)[/imath]. Using this, it is easy to prove that, if [imath]Hf[/imath] is absolutely integrable, [imath]\hat{f}(0)[/imath] must be [imath]0[/imath]. I want to prove the converse. So assume [imath]\lim_{|x|\to\infty}x(Hf)(x)=0[/imath]. This by itself doesn't guarantee integrability of [imath]Hf[/imath], since [imath]Hf[/imath] might behave like [imath]\frac{1}{x\log x}[/imath] at infinity. A stronger decay condition for [imath]Hf[/imath] is needed, but I'm not sure where to get it from. The fact that [imath]f[/imath] is Schwartz should be important; does this imply [imath]Hf[/imath] decays faster than any polynomial? Note, since this is a homework question, please don't be overly explicit in your answers. |
863192 | Trigonometry - simplifying a given equation
Question: [imath]\tan 9 - \tan 27 - \tan 63 + \tan 81[/imath] Answer I'm getting : 0 What I did: Well I clubbed together [imath]\tan 9[/imath] and [imath]\tan 81[/imath] and [imath]\tan 27[/imath] and [imath]\tan 63[/imath] (took out negative as common). Then using the identity for [imath]\tan (A+B)[/imath], I rearranged to formula to get what [imath]\tan A + \tan B[/imath] is. With that I'm getting zero multiplied by [imath]\tan 90[/imath]. Since anything multiplied by zero, even infinity, is zero, I guess it should be zero. I'm pretty sure my logic fails me somewhere, please tell me where (probably in the infinity and zero multiplication part) | 823619 | Calculate [imath]\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}[/imath]
Calculate [imath]\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}[/imath]? The correct answer should be 4. |
800464 | What five odd integers have a sum of [imath]30[/imath]?
I've been asked the following question: What five odd integers from the set [imath]\{1, 3, 5, 7, 9, 11, 13, 15\}[/imath] that when summed together equals to [imath]30[/imath]? Note that any integer can be used more than once. If my limited knowledge of maths is correct, there should be no answer, as no odd number of odd integers summed together can give an even number. | 2125228 | Three odd integers from the set [imath]\{1,3,5,7,9,11,13,15\}[/imath] that equals to [imath]30[/imath].
What three odd integers from the set [imath]\{1,3,5,7,9,11,13,15\}[/imath] that when summed together equals to [imath]30[/imath] ? Note: You can also repeat the numbers I found the Malfunction question here. But this question has the answer with five integer numbers. Please help me |
864017 | [imath]y' = \exp(-\frac{y}{x}) + \frac{y}{x}[/imath]
Could you help me to solve this differential equation: [imath]y' = \exp\left(-\frac{y}{x}\right) + \frac{y}{x}[/imath] | 435462 | solution of [imath]y' = \exp \left(-\frac yx\right) + \frac yx[/imath]
Could you help me to solve equation [imath]y' = \exp \left(-\frac yx\right) + \frac yx;\quad y(e) = 0[/imath] I know how to solve 1st order linear de like [imath]y' = \exp \bigl(-\frac 1x\bigr) + \frac yx[/imath] but here I have the dependent variable in the part that usually (in my practice) was free of it. |
864794 | sin x integral qestions
How could the following integral be solved in a good manner? [imath]\int \frac{\sin(x)}{x}\;\mathrm{d}x[/imath] Regards: | 163305 | What is the integral of function [imath]f(x) = (\sin x)/x[/imath]
I want to know if the function sinx/x is integrable and if it is, then what's its integral? My high school book says its a non-integrable function while WolframAlpha says its integral is Si(x) + constant Please shed some light on this topic and explain from basic level like what is Si(x), who used it for the first time etc. |
864980 | Prove that any group of order 15 is cyclic.
Prove that any group of order [imath]15[/imath] is cyclic. I know that if order of group is a prime then the group is cyclic, but how to approach such questions? | 1318939 | If [imath]G[/imath] s.t. [imath]|G|=15[/imath] has only one subgroup of order [imath]3[/imath] and only one of order [imath]5[/imath], then [imath]G[/imath] is cyclic. Generalize to [imath]|G|=pq[/imath], for [imath]p,q[/imath] primes.
Let [imath]G[/imath] be a group where [imath]o(G)=15[/imath]. If [imath]G[/imath] has only one subgroup of order [imath]3[/imath] and only one of order [imath]5[/imath], prove that [imath]G[/imath] is cyclic. Generalize to [imath]o(G)=pq[/imath], where [imath]p[/imath] and [imath]q[/imath] are primes. |
545431 | Fourier transform of [imath]\exp(-t^2)[/imath] using contour integration.
I am calculating the Fourier transform of [imath]\exp(-t^2)[/imath] using contour integration. I am left with the integral [imath]\int_{-\infty}^\infty \exp(i\omega t)\exp(-t^2)[/imath]. Usually I would now use the residue theorem, but I cannot find the singularities. Can someone help me? | 1602218 | Showing that the Gaussian function is a fixed point of the Fourier transformation using Complex Analysis
I want to show that the Gaussian function, defined by: [imath]f(x) = \frac{1}{\sqrt{2Ο}} e^{-\frac{1}{2} x^2}[/imath] is a fixed point of the Fourier transformation [imath]\hat{f}[/imath] (meaning that [imath]\hat{f} = f[/imath]), where, for any (integrable) function [imath]f[/imath], [imath]\hat{f}[/imath] is given by: [imath]\hat{f}(x) = \frac{1}{\sqrt{2Ο}} \int_\mathbb{R} f(y) e^{-i x y} d y[/imath] I'm allowed to use (without proof) that [imath]\int_{-\infty}^\infty f(x)dx = 1[/imath]. Now I want to do this using the means and tools of Complex Analysis. I know that one can solve this by differentiating [imath]f[/imath], which gives a ODE of first order with respect to [imath]f[/imath], and that one can show that [imath]\hat{f}[/imath] also solves this ODE. I'm aware of this solution (and I know that there are other threads where this has been discussed), but I want to do it specifically without using ODE's. (And I haven't found any threads on solving this with Complex Analysis so far.) I think, one way of doing this is to integrate over a large enough semicircle or rectangle in [imath]\mathbb{C}[/imath] (that has the real line as one of it's edges), for which we could maybe utilize the Residue theorem. (For which we first needed to find an isolated singularity, and then choose the semicircle/rectangle accordingly for it to contain one (but ideally only one) isolated singularity.) Given the integral over the edges of the rectangle, I think we then would need to decompose the path parameterizing the edges of the rectangle in order to get the a formula for the section of the edge that goes through the real numer line. This is how I would theoretically solve it; but I honestly don't really know how to concretely do all of that. I'm having difficulties deciding if my approach is even the right one, considering [imath]f(y) e^{- i x y} = \frac{1}{e^{-1/2 y^2 + i x y}}[/imath] doesn't seem to have any singularities at all, or am I forgetting something? |
865131 | Distribution of the sum of two independent normal variables
Given the two variables [imath]A\sim \text{N}(\mu, \phi^2)[/imath] and [imath]B\sim \text{N}(\xi, \omega^2)[/imath] with [imath]\mu, \xi \in R[/imath] and [imath]\phi^2, \omega^2 > 0[/imath] how do I prove that [imath]C := A + B\sim \text{N}(\mu + \xi, \phi^2 + \omega^2)[/imath] ? | 238576 | The sum of [imath]n[/imath] independent normal random variables.
How can I prove that the sum of [imath]X_1, X_2, \ldots,X_n[/imath] random variables, all of which have normal distributions [imath]N(\mu_i, \sigma_i)[/imath], is a random variable that is itself normally distributed with mean [imath]\mu =\sum_{i=1}^n \mu_i[/imath] and variance [imath]\sigma^2 = \sum_{i=1}^n \sigma_i^2[/imath] Edit: I forgot to add that this was with the assumption that all [imath]X_1, X_2,\ldots,X_n[/imath] are independent. |
865619 | limit of a sum of powers of integers
I ran across the following problem in my Advanced Calculus class: For a fixed positive number [imath]\beta[/imath], find [imath]\lim_{n\to \infty} \left[\frac {1^\beta + 2^\beta + \cdots + n^\beta} {n^{\beta + 1}}\right][/imath] I tried manipulating the expression inside the limit but didn't come up with anything useful. I also noted that the numerator can be rewritten as [imath]\sum_{i=1}^{n}i^\beta[/imath] which is a well-known formula with a closed form (Faulhaber's formula) but I don't fully understand that formula and we haven't talked about the Bernoulli numbers at all, so I think the author intended for the problem to be solved a different way. Any suggestions on how to tackle this would be much appreciated. | 478344 | What is the result of [imath] \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} [/imath] and why?
What is the result of the next limit: [imath] \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} [/imath] Why (theorem)? |
865926 | Computing complex number
"Compute [imath](1 + i)^{1000}[/imath]. So far I have: [imath](1+i)^{4 (2^2 5^3)} [/imath] but I am not sure how to proceed. Ideas? | 848740 | How do i find [imath](1+i)^{100}?[/imath]
How do I find [imath](1+i)^{100}[/imath] without expanding [imath](1+i)[/imath] 100 times? Is there a quicker way to do this? The hint was to find the modulus and argument of [imath]1+i[/imath] which I've got as [imath]\sqrt{2}[/imath] and [imath]\pi/4[/imath] but I'm not sure what to do from here. |
866415 | Putnam Exam question
Prove or disprove: if [imath]x[/imath] and [imath]y[/imath] are real numbers with [imath]y\ge 0[/imath] and [imath]y(y+1)\le (x+1)^2[/imath], then [imath]y(y-1)\le x^2[/imath]. How should I approach this proof? The solution starts with assuming [imath]y\ge 0[/imath] and [imath]y\le 1[/imath], but I'm not sure how to arrive at that second assumption or go from there. Thank you in advance! | 363169 | Does [imath]y(y+1) \leq (x+1)^2[/imath] imply [imath]y(y-1) \leq x^2[/imath]?
Can anyone see how to prove the following? If [imath]x[/imath] and [imath]y[/imath] are real numbers with [imath]y\geq 0[/imath] and [imath]y(y+1) \leq (x+1)^2[/imath] then [imath]y(y-1) \leq x^2[/imath]. It seems it is true at least according to Mathematica. |
866519 | Showing that if [imath]a[/imath] is a transitive set then [imath]\cup a[/imath] is also a transitive set.
I am working on a problem in Enderton's text on set-theory that appears to be deceptively easy. It is likely that I making a mistake somewhere so if someone can comment it would be much appreciated. Question: Show that if [imath]a[/imath] is a transitive set then [imath]\cup a[/imath] is also a transitive set. Answer so far: Assume that [imath]a[/imath] is a transitive set. By definition, [imath]\cup a = \{x \in y \ |[/imath] for some [imath]y \in a[/imath]}. Since [imath]a[/imath] is transitive, whenever we have for every [imath]x \in y \in a[/imath], we have [imath]x \in a[/imath]. But these are just the conditions of set [imath]\cup a[/imath]. So [imath]\cup a[/imath] is transitive. | 49221 | A subset of a transitive set is transitive
Given a transitive set[imath]a,[/imath] I can prove that [imath]\bigcup a[/imath] is also transitive, but I don't quite like my method because I must first prove [imath]a \subseteq \mathcal{P}(\bigcup a)[/imath] to get at [imath]\bigcup a\subseteq \mathcal{P}(\bigcup a).[/imath] Could you please show me a smarter proof? |
242286 | Estimated probability looks too large, what I am doing wrong?
A factory produces links for heavy metal chains. The research lab of the factory models the length (in cm) of a link by the random variable [imath]{X}[/imath], with expected value [imath]{E(X) = 5}[/imath] and variance [imath]{Var(X) = 0.04}[/imath]. The length of a link is defined in such a way that the length of a chain is equal to the sum of the lengths of its links. The factory sells chains of 50 meters; to be on the safe side 1002 links are used for such chains. The factory guarantees that the chain is not shorter than 50 meters. If by chance a chain is too short, the customer is reimbursed, and a new chain is given for free. Give an estimate of the probability that for a chain of at least 50 meters more than 1002 links are needed. For what percentage of the chains does the factory have to reimburse clients and provide free chains? Number of links in a chain: [imath]{n = 1002}[/imath] Mean lengh of 1002 links: [imath]{ \mu = 1002 \times 0.05 = 50.1 }[/imath] We know variance, so we can find the standard deviation: [imath]{ \sigma = \sqrt{0.04} = 0.2 }[/imath] We are interested how often chain is shorter then 50 meters: [imath]{ \bar{x} = 50 }[/imath] Thus let define an appropriate test statistic: [imath]{ \displaystyle z = \frac{ \mu - \bar{x} }{ \sigma } = \frac{50.1 - 50}{0.2} = \frac{0.1}{0.2} = 0.5 }[/imath] Probability of getting value in a lower tail is: [imath]{ \displaystyle P(z < 0.5) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz }[/imath] By making the substitution [imath]{ z = \sqrt{2} x }[/imath], we get: [imath]{ \displaystyle \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz = \frac{1}{\sqrt{\pi}} \int\limits_{0}^{\frac{0.5}{\sqrt{2}}} e^{-x^2} dx }[/imath] As we can not solve integral above, let's consider [imath]{ e^{-x^2} }[/imath] as a Taylor series: [imath]{ \displaystyle \sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} }[/imath] Then the approximate probability is: [imath]{ \displaystyle P(z < 0.5) = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \sum\limits_{n=0}^{\infty} \frac{(-1)^n (\frac{c}{\sqrt{2}})^{2n+1} }{ n! (2n+1) } \simeq 0.691462 }[/imath] But this probability looks too large. What I am doing wrong? | 241296 | Estimated probability looks too large, what I am doing wrong?
A factory produces links for heavy metal chains. The research lab of the factory models the length (in cm) of a link by the random variable [imath]{X}[/imath], with expected value [imath]{E(X) = 5}[/imath] and variance [imath]{Var(X) = 0.04}[/imath]. The length of a link is defined in such a way that the length of a chain is equal to the sum of the lengths of its links. The factory sells chains of 50 meters; to be on the safe side 1002 links are used for such chains. The factory guarantees that the chain is not shorter than 50 meters. If by chance a chain is too short, the customer is reimbursed, and a new chain is given for free. Give an estimate of the probability that for a chain of at least 50 meters more than 1002 links are needed. For what percentage of the chains does the factory have to reimburse clients and provide free chains? Number of links in a chain: [imath]{n = 1002}[/imath] Mean lengh of 1002 links: [imath]{ \mu = 1002 \times 0.05 = 50.1 }[/imath] We know variance, so we can find the standard deviation: [imath]{ \sigma = \sqrt{0.04} = 0.2 }[/imath] We are interested how often chain is shorter then 50 meters: [imath]{ \bar{x} = 50 }[/imath] Thus let define an appropriate test statistic: [imath]{ \displaystyle z = \frac{ \mu - \bar{x} }{ \sigma } = \frac{50.1 - 50}{0.2} = \frac{0.1}{0.2} = 0.5 }[/imath] Probability of getting value in a lower tail is: [imath]{ \displaystyle P(z < 0.5) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz }[/imath] By making the substitution [imath]{ z = \sqrt{2} x }[/imath], we get: [imath]{ \displaystyle \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz = \frac{1}{\sqrt{\pi}} \int\limits_{0}^{\frac{0.5}{\sqrt{2}}} e^{-x^2} dx }[/imath] As we can not solve integral above, let's consider [imath]{ e^{-x^2} }[/imath] as a Taylor series: [imath]{ \displaystyle \sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} }[/imath] Then the approximate probability is: [imath]{ \displaystyle P(z < 0.5) = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \sum\limits_{n=0}^{\infty} \frac{(-1)^n (\frac{c}{\sqrt{2}})^{2n+1} }{ n! (2n+1) } \simeq 0.691462 }[/imath] But this probability looks too large. What I am doing wrong? |
867481 | Does taking power sets conserve inequality?
Given infinite sets [imath]X[/imath] and [imath]Y[/imath] such that [imath]X[/imath] has strictly smaller cardinality than [imath]Y[/imath], is it the case that the power set of [imath]X[/imath] has strictly lower cardinality than the power set of [imath]Y[/imath]? | 629912 | If two sets have the same cardinality, then so do their power sets. Converse can't be answered?
The following is my rewrite of this proof for the following assertion : For infinite sets [imath]A, B[/imath], [imath]|A| = |B| \Longrightarrow \require{cancel} \cancel{\Longleftarrow} |P(A)| = |P(B)|[/imath]. [imath]\bbox[2px,border:1px solid black]{\text{ Proof Strategem : }}[/imath] We are given that [imath]f :A \rightarrow B[/imath], is a bijection. How do we construct a bijection between subsets of A and subsets of B? We need a rule whereby, if we take one subset of A, then we can use the rule to uniquely construct a unique subset of B. My intuition on this is to take all the elements of the subset of A, and map them under [imath]f[/imath] to elements of B, which will form a subset of B. [imath]\bbox[2px,border:1px solid black]{\text{ Proof : }}[/imath] In mathematical notation, we define [imath]g[/imath] as: [imath]g : P(A) \rightarrow P(B)[/imath] by means of [imath]\color{#009900}{g(S) = \{f(x) : x β S\}} \; \forall S \subseteq A[/imath]. [imath]\bbox[2px,border:5px solid grey]{\text{ [/imath]g[imath] onto ? }}[/imath] For all [imath]T \subseteq B[/imath], does there exist [imath]S \subseteq A[/imath] such that [imath]\color{#009900}{g(S) = \{f(x) : x β S\}} \; = T[/imath]? Since [imath]f^{-1}[/imath] exists, define [imath]S = \{f^{-1}(x) : x β T\}.[/imath] We must prove [imath]\color{#009900}{\{f(x) : x β S\}} = T[/imath]. [imath]\bbox[5px,border:1px solid grey]{T \subseteq g(S)}[/imath] Take [imath]y \in T \iff f^{-1}(y) β S \iff \color{ #FF4F00}{f}[f^{-1}(y)] β \color{ #FF4F00}g[S] \iff y \in g[S]. [/imath] [imath]\bbox[5px,border:1px solid grey]{{g(S)} \subseteq T}[/imath] Take [imath]y \in g(S)[/imath] which means there exists [imath]x \in S \ni f(\color{#C154C1}{x}) = y [/imath]. Since [imath]x \in S[/imath] means there exists [imath]t \in T \ni \color{#C154C1}{f^{-1}(t) = x}[/imath], substitute this into the previous equation: [imath]f(\color{#C154C1}{f^{-1}(t)}) = y \implies t = y [/imath]. Since [imath]t \in T[/imath], thus [imath]y \in T[/imath]. [imath]\bbox[2px,border:5px solid grey]{\text{ [/imath]g[imath] 1-1 ? }}[/imath] Suppose we have [imath]G,H β A[/imath] such that: [imath]g(G) = g(H)[/imath] which means [imath]\{f(x) : x β G\} = \{f(x) : x β H\}[/imath]. [imath]\bbox[5px,border:1px solid grey]{\{f(x) : x β G\} \subseteq \{f(x) : x β H\}}[/imath] Take [imath]g \in G \implies f(g) β \quad g(G) = g(H) \quad \implies f(g) β g(H)[/imath]. Thus there exists a [imath]h β H[/imath] such that [imath]f(g) = f(h)[/imath] [imath]\implies g = h[/imath] because [imath]f[/imath] is a bijection. Since [imath]g β G[/imath], thus [imath]g β H[/imath]. We can prove the other direction by the same argument, just with [imath]G[/imath] and [imath]H[/imath] swapped around. [imath]1.[/imath] In the proof for [imath]g[/imath] onto, how would you (fore)know (ie: divine or presage) to define [imath]S = \{f^{-1}(x) : x β T\}[/imath]? [imath]2.[/imath] Would someone please explain the step [imath]f^{-1}(y) β S \iff \color{ #FF4F00}{f}[f^{-1}(y)] β \color{ #FF4F00}g[S] [/imath]? [imath]3.[/imath] Are there easier proofs? The result seems intuitive but the proof is complex. [imath]4.[/imath] Are there any pictures? [imath]5.[/imath] My course doesn't include [imath]ZFC[/imath]. So what's the intuition why the converse is false? |
867506 | How to prove there is no surjection [imath]f\colon X \rightarrow 2^X[/imath]
This is the following problem: Let [imath]X[/imath] be a set. Prove that there is not a surjection from [imath]X \rightarrow 2^X[/imath] (Hint: Assume to the contrary that [imath]f\colon X \rightarrow 2^X[/imath] is a surjection and consider the set [imath]M=\{ x \in X | x \notin f(x) \}[/imath]). Use these facts to conclude that there is an infinite number of sizes of infinity cardinals. My answer to the question is: This function is only bijective if its inverse is bijective and since [imath]f(x)=\log(x)[/imath] isn't bijective then the original function [imath]f(x)=2^x[/imath] isn't bijective. Since this function isn't bijective, then it isn't surjective. As for the cardinality, I'm not sure what that really means. Thank you in advance for any hints or help! | 214807 | Non-existence of a Surjective Function from a Set to Its Subsets (Cantor's theorem)
Show that: Let A be a set and let [imath]P(A)[/imath] be the set of all subsets of [imath]A[/imath]. Then there is no surjection [imath]f: AβP(A)[/imath]. Here is what I thought: if [imath]A=\{a,b\}[/imath] then it has only two elements where [imath]P(A)=\{β
,\{a\},\{b\},\{a,b\}\}[/imath] has 4 elements. Therefore [imath]f:AβP(A)[/imath] cannot be surjective. But I have some problems: 1) How is it possible that any [imath]f[/imath] function to take [imath]\{a\}[/imath] from set [imath]A[/imath] to [imath]\{a,b\}[/imath]? Maybe because I am thinking mainly about functions with real values like [imath]f(x)=2x[/imath], I find it a little bit strange that a function to take an element of a set to another set which has more elements. Is it possible? edit: Now I thought that if [imath]f(x)[/imath] is [imath]\sqrt{x}[/imath], then [imath]f(4)=Β±2[/imath] which means it took an element from a set to a set which has 2 elements. But still I find it kind of strange to denote [imath]f(\{a\})=\{a,b,c,...\}[/imath] 2) How can I construct a explicit proof for this question? Regards |
867690 | CDF for non-homogeneous Poisson process
I am trying to understand the inverse transform method for simulating random processes and have managed to completely confuse myself. Consider a Poisson process whose conditional intensity is [imath]\lambda(t) = \alpha e^{-t}[/imath] starting at time [imath]t=0[/imath] for some parameter [imath]\alpha>0[/imath]. To apply the inverse transform method I need to start with the cdf (called [imath]F[/imath] in the notes that I have linked to) and then compute its inverse. What is the relevant cdf [imath]F[/imath] for this process? My overall goal is to be able to simulate the arrival times from this process. | 867125 | Simulate simple non-homogeneous Poisson proces
Consider a Poisson process whose conditional intensity is [imath]\lambda(t) = \alpha e^{-t}[/imath] starting at time [imath]t=0[/imath] for some parameter [imath]\alpha>0[/imath]. I would like to simulate arrival/event/failure times as efficiently as possible. (I am not sure what the standard term is for the times you get from a Poisson process.) Is there a fast way to do this? |
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