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532625 | Prove that [imath]f\colon \Bbb R^2\to \Bbb R[/imath] is continuous
Let [imath]f\colon \mathbb{R}^2 \to \mathbb{R}[/imath] be function, such that for every fixed [imath]x_0[/imath] [imath]f(x_0,y): \mathbb{R} \to \mathbb{R}[/imath] is monotone and continuous function and for every fixed [imath]y_0[/imath] [imath]f(x,y_0): \mathbb{R} \to \mathbb{R}[/imath] is continuous function. Prove that [imath]f[/imath] is continuous. I am trying to check the continuity at [imath](x',y')[/imath], So let [imath]\epsilon > 0[/imath] and now [imath]|f(x,y)-f(x',y')| \le |f(x,y)-f(x,y')|+|f(x,y')-f(x',y')|[/imath] by triangle inequality (I'm trying to move colloterally to axises). Now I think I should use the continuity of [imath]f(x_0,y)[/imath] and [imath]f(x_0,y)[/imath] in apriopriate way, but I don't se how. | 524631 | Continuous Function
Let [imath] f: \Bbb R^2 \to \Bbb R [/imath] such that : [imath] \forall _{y_0 \in \Bbb R\ }: [/imath] function [imath] x \to f(x,y_0) [/imath] continuous function and increasing [imath] \forall _{x_0 \in \Bbb R\ }: [/imath] function [imath] y \to f(x_0,y) [/imath] continuous function I mean the continuity of one variable. Prove the continuity of a function [imath]f: \Bbb R^2 \to \Bbb R [/imath] I know definition, but I can not do. I wanted to use the definitions of Cauchy. |
533352 | Proof [imath]\displaystyle \text{lcm}(x,y)=\frac{|x\cdot y|}{\text{gcd}(x,y)}[/imath]
Prove: [imath] \text{lcm}(x,y)=\frac{|x\cdot y|}{\text{gcd}(x,y)}[/imath] I used many ways to do it, all failed. One of them was to represent [imath]|x\cdot y|[/imath] as a sum of primes then [imath]\text{gcd}(x,y)[/imath] as a sum of primes and do the operation but I ended up with a false result since some prime factors were left. So what is the simplest proof for the gcd to lcm relation? | 44835 | For integers [imath]a[/imath] and [imath]b[/imath], [imath]ab=\text{lcm}(a,b)\cdot\text{hcf}(a,b)[/imath]
I was reading a text book and came across the following: Important Results (This comes immediately after LCM:) If 2 [integers] [imath]a[/imath] and [imath]b[/imath] are given, and their [imath]LCM[/imath] and [imath]HCF[/imath] are [imath]L[/imath] and [imath]H[/imath] respectively, then [imath]L \times H = a \times b[/imath] Can some please help me understand why the above result is true? Thanks in advance. |
533490 | Group isomorphism between rationals with sum and rationals with multiplication
It is well known that the exponential function induces an isomorphism between the additive group of real numbers and the multiplicative group [imath]\mathbb{R}_{>0}[/imath]. I was wondering if there exists an isomorphism between the additive group of rationals and the multiplicative group [imath]\mathbb{Q}_{>0}[/imath]. | 33607 | Group of positive rationals under multiplication not isomorphic to group of rationals
A question that may sound very trivial, apologies beforehand. I am wondering why [imath]( \mathbb{Q}_{>0} , \times )[/imath] is not isomorphic to [imath]( \mathbb{Q} , + )[/imath]. I can see for the case when [imath]( \mathbb{Q} , \times )[/imath], not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here? |
492109 | What is wrong with this funny proof that 2 = 4 using infinite exponentiation?
Out of boredom, I decided to recall the following equation: [imath]x^{x^{x\cdots}} = 2.[/imath] Which, I simply rewrote like this: [imath]x^2 = 2[/imath], and therefore [imath]x = \sqrt{2}[/imath]. Then I took a look at the more general form: [imath]x^{x^{x\cdots}} = p.[/imath] I then concluded that [imath]x = (p)^{\frac{1}{p}}[/imath], and this was the solution set for all [imath]p[/imath]. Then I thought for a while and determined that there are multiple values where this function equals the square root of [imath]2[/imath], so its not injective. However, this led to an awkward statemend based on the fact that [imath]\sqrt{2} = (4)^{\frac{1}{4}}[/imath]. We deduce the following: The equation [imath]x^{x^{x\cdots}} = 2[/imath] and the equation [imath]x^{x^{x\cdots}} = 4[/imath] have the same solution, namely [imath]x = \sqrt{2}[/imath]. So we conclude that [imath]2 = 4[/imath]. I showed this to friend of mine, and he conjectured that there are no valid solutions for all [imath]p > e[/imath]. To prove this, I tried writing the following: I tried differentiating [imath]f(x) = (x)^{(1/x)}[/imath] and setting it equal to [imath]0[/imath] to find the maximum. I determined that there was a maximum at the point where [imath]x = e[/imath]. Does this mean that the maximum possible solution [imath]x[/imath] for the infinite power is [imath]e^{(1/e)}[/imath] therefore showing that the maximum value for [imath]p[/imath] is indeed [imath]p = e[/imath]? | 1595641 | Extra root of hyperpower equations
Consider the hyperpower equation [imath]x^{x^{x^{x...}}}=2[/imath] We will use the method that Let [imath]y=x^{x^{x^{x...}}}[/imath] so [imath]x^y=x^{x^{x^{x...}}}=2[/imath] and [imath]x^2=2[/imath] to give the solution [imath]x=\sqrt2[/imath] However consider another equation such [imath]x^{x^{x^{x...}}}=4[/imath] Use the similar method so we have [imath]y=x^{x^{x^{x...}}}[/imath] to give [imath]x^y=x^4=4[/imath] , this also makes [imath]x=\sqrt2[/imath] as our solution But the [imath]\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2...}}}[/imath] have to equal to something, not 2 and 4 at the same time. If there is only one solution [imath]x=\sqrt2[/imath], then what is wrong with the second equation? |
529767 | real analysis Basic real analysis proof limit point
[imath]A_n = (-1)^n \left(\frac{\sqrt 2 \;n}{n+1}\right),\;\; n = 1, 2, 3, \ldots[/imath] My question is to find the limit point of this equation I'm still not very good concepted with limited point. This equation is alternating -1 and 1 and converge to [imath]\sqrt 2[/imath]. So can you say [imath]\sqrt 2[/imath] is the limited point? If it is not what is the difference between limited point and limit? How to find limit points of [imath]A_n[/imath] ? | 529794 | Is [imath](-\infty,3]\cap(0,1)^c\cap[-1,\infty)[/imath] closed? Open? What are interior points?
[imath]\text{Let }\;\;A_1 = (-\infty, 3]\cap (0, 1)^c \cap [-1, \infty), \quad A_2 = \left\{(-1)^n \frac {\sqrt 2\,n}{n+1}: n = 1, 2, \ldots\right\}[/imath] Is [imath]A_1[/imath] open? Is [imath]A_1[/imath] closed? Find interior points of [imath]A_1[/imath]. Find limit points of [imath]A_2[/imath]. 1) I think first answer is not opened and closed, and I'm not sure how to find interior points. 2) [imath]A_2[/imath] converges to [imath]\sqrt2[/imath], so can you say [imath]\sqrt2[/imath] is the limit point of [imath]A_2[/imath]? What is the difference between "limit point" and "limit"? And how to find limit points of [imath]A_2[/imath]? |
316297 | Is the result still true if we drop completeness?
I know how to prove the following exercise ( from Folland) : If [imath]X[/imath], [imath]Y[/imath] are Banach spaces. [imath]T:X\rightarrow Y[/imath] is a linear map such that [imath]f\circ T\in\operatorname{dual}(X)[/imath] whenever [imath]f\in \operatorname{dual}(Y)[/imath], then [imath]T[/imath] is bounded. Now I want to know whether the result is still true if [imath]X[/imath] and [imath]Y[/imath] are just normed vector spaces instead of Banach spaces. Thank you so much. | 532272 | Continuous iff composition with every linear functional is continuous
Let [imath]X[/imath] and [imath]Y[/imath] be normed spaces, [imath]T:X \rightarrow Y[/imath] a linear operator. Show that [imath]T[/imath] is continuous if [imath]y' \circ T[/imath] is continuous for every [imath]y' \in Y'[/imath]. My idea is the following: Suppose [imath]T[/imath] is not continuous, then there exists a sequence [imath]x_n[/imath] with [imath]\|x_n\|=1[/imath] such that [imath]\|Tx_n\|>n[/imath]. If I could assure there is a subsequence [imath]x_{n_k}[/imath] such that [imath]Tx_{n_k}[/imath] are linearly independent I think I could construct an element [imath]z'[/imath] of [imath]Y'[/imath] such that [imath]|(z' \circ T)x_{n_k}|>n_k[/imath]. The first thing is, can I assure the existence of that subsequence? And if I can, how do I conclude applying H-B? I'd like a solution with this same approach, but if there is a simpler proof i'd like to see it too. |
534261 | Product of complex numbers goes to 0.
Can someone help me with this assignment: Let [imath]\Re z <0[/imath]. Prove that [imath]z_n=(1+z) \left(1+\frac{z}{2} \right) ... \left( 1+ \frac{z}{n} \right) \to 0[/imath]. I was trying to look for a sequence [imath]a_n[/imath] which goes to [imath]0[/imath] as [imath]n[/imath] goes to [imath]\infty[/imath] such that [imath]|z_n|^2 \le a_n[/imath] but I've failed... | 532038 | [imath]\prod_{k=1}^n (1+ \frac{z}{k})[/imath] converges to [imath]0[/imath] when [imath]\Re (z)<0[/imath]
We have a complex number [imath]z[/imath] such that [imath]\Re (z)<0[/imath] and the sequence [imath] z_n = \prod_{k=1}^n (1+ \frac{z}{k})[/imath]. Prove that [imath]\lim_{n \rightarrow \infty} z_n = 0[/imath]. How to do it? I guess it will be easier to prove that [imath]\lim_{n \rightarrow \infty} |z_n|^2 = 0[/imath], it's a trick that usually helps, but even though I thought about if for a longer while, I don't know how to prove it. May somebody help? |
534796 | For sets [imath]C, D \subset Q[/imath] why is [imath]|CD|=\dfrac{|C||D|}{|C\cap D|}[/imath]
Could someone please clearly explain why the following set cardinality is true? For subgroups [imath]C, D \subset Q[/imath] why is [imath]|CD|=\dfrac{|C||D|}{|C\cap D|}[/imath] Where [imath]CD= \{cd|c\in C, d\in D\}[/imath] and [imath]C,D[/imath] are subgroups of [imath]Q[/imath]. I think the trouble for me is that I can't really see what [imath]CD[/imath] is. I know that every element in [imath]CD \in Q[/imath], but that is about it. Thanks for the help. | 518838 | Needing help picturing the group idea.
I have a question; X and Y are subgroups of a group G. If [imath]|X|, |Y| < \infty[/imath], show that [imath]|XY| = \frac{|X||Y|}{|X \cap Y|}[/imath] but I can't really picture what it is talking about to even get started. Is it to do with looking for different ways so combine the elements {x, y, xy} ? Thanks |
534939 | Is this function Lebesgue integrable?
I have to decice if the following function is Lebesgue-integrable on [imath][0,1][/imath]: [imath]g(x)=\frac{1}x\cos\left(\frac{1}x\right) [/imath] where [imath]x\in[0,1][/imath]. [imath]g(x)[/imath] is Lebesgue integrable if and only if the integral of [imath]|g(x)|[/imath] is finite So, [imath]\int_{[0,1]} |\frac{1}x\cos\left(\frac{1}x\right)| dm < infinite???[/imath] I don't know how to proof that, I have thought about using the monotone convergence theorem but I don't have any idea of how to define [imath]f_n[/imath] a sequence of measurable functions. | 529927 | Is [imath]\frac{1}x\cos\left(\frac{1}x\right)[/imath] Lebesgue integrable on [imath][0,1][/imath]?
I have to decice if the following function is Lebesgue-integrable on [imath][0,1][/imath]: [imath]g(x)=\frac{1}x\cos\left(\frac{1}x\right) [/imath] where [imath]x\in[0,1][/imath]. |
13734 | How to find shortest distance between two skew lines in 3D?
If given 2 lines [imath]\alpha[/imath] and [imath]\beta[/imath], that are created by 2 points: A and B 2 plane intersection I want to find shortest distance between them. [imath]\left\{\begin{array}{c} P_1=x_1X+y_1Y+z_1Z+C=0 \\ P_2=x_2X+y_2Y+z_2Z+C=0\end{array}\right.[/imath] [imath]A=\left(x_3;y_3;z_3\right)[/imath] [imath]B=\left(x_4;y_4;z_4\right)[/imath] [imath]\alpha =n_1\times n_2=\left(\left|\begin{array}{cc} y_1 & z_1 \\ y_2 & z_2\end{array}\right|;-\left|\begin{array}{cc} x_1 & z_1 \\ x_2 & z_2\end{array}\right|;\left|\begin{array}{cc} x_1 & y_1 \\ x_2 & y_2\end{array}\right|\right)[/imath] [imath]\beta =[/imath] From here I tried: The question of "shortest distance" is only interesting in the skew case. Let's say [imath]p_0[/imath] and [imath]p_1[/imath] are points on the lines [imath]L_0[/imath] and [imath]L_1[/imath], respectively. Also [imath]d_0[/imath] and [imath]d_1[/imath] are the direction vectors of [imath]L_0[/imath] and [imath]L_1[/imath], respectively. The shortest distance is [imath](p_0 - p_1)[/imath] * , in which * is dot product, and is the normalized cross product. The point on [imath]L_0[/imath] that is nearest to [imath]L_1[/imath] is [imath]p_0 + d_0(((p_1 - p_0) * k) / (d_0 * k))[/imath], in which [imath]k[/imath] is [imath]d_1 \times d_0 \times d_1[/imath]. Read more: http://wiki.answers.com/Q/If_the_shortest_distance_between_two_points_is_a_straight_line_what_is_the_shortest_distance_between_two_straight_lines#ixzz17fAWKFst I tried, but failed. | 1508463 | How to find shortest distance between nonparallel lines in 3D
[imath]\begin{bmatrix}x\\y\\z\\\end{bmatrix}=\begin{bmatrix}1\\-1\\0\end{bmatrix}+s\begin{bmatrix}1\\1\\1\end{bmatrix}; [/imath] [imath]\begin{bmatrix}x\\y\\z\\\end{bmatrix}=\begin{bmatrix}2\\-1\\3\end{bmatrix}+t\begin{bmatrix}3\\1\\0\end{bmatrix}[/imath] How do you find the shortest distance between the two unparalleled lines and also find the points on the lines that are closest together? |
535985 | Why is dividing by zero = undefine.
I know that [imath] a\cdot\left(\frac1a\right)=1[/imath] as long as [imath]a[/imath] is not [imath]= 0[/imath], but when we divide [imath](\frac a0)[/imath] is [imath]=[/imath] we say it's undefined. why is it really undefined? what's the big conspiracy here. | 11161 | Why does [imath]x[/imath] divided by zero not equal [imath]x[/imath]?
Why does [imath]x[/imath] divided by zero not equal [imath]x[/imath]? After all, [imath]x[/imath] is not being divided by anything. |
536286 | Bases for a topology
I have the following topology : [imath]\tau= \Bigl\{U\subseteq \mathbb{R}^2: (\forall(a,b) \in U) (\exists \epsilon >0) \bigl([a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq U\bigr)\Bigr\}[/imath] Are these a basis for the previous topology: [imath]\beta_1= \{[a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}[/imath] [imath]\beta_2= \{[a,a+\epsilon) \times [b-\epsilon, b+\epsilon)\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}[/imath] I have asked this question before here:https://math.stackexchange.com/posts/536105/edit And I got stuck for [imath]\beta_2[/imath], considering e [imath]a,b=0,\epsilon=1[/imath] this set is not open in our topology. They gave me the advice to look at the point [imath](0,0)[/imath].What is the problem here? Thank you | 536089 | Are these bases for a topology?
I have the following topology : [imath]\tau= \Bigl\{U\subseteq \mathbb{R}^2: (\forall(a,b) \in U) (\exists \epsilon >0) \bigl([a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq U\bigr)\Bigr\}[/imath] Are these a basis for the previous topology: [imath]\beta_1= \{[a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}[/imath] [imath]\beta_2= \{[a,a+\epsilon) \times [b-\epsilon, b+\epsilon)\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}[/imath] The first one is obviously a basis for [imath]\tau[/imath] because of the definition of [imath]\tau[/imath] and I would say that the second is also a basis, because [imath][a,a+\epsilon) \times [b-\epsilon, b+\epsilon) \subseteq [a,a+\epsilon] \times [b-\epsilon, b+\epsilon] [/imath] Is it correct? what do you think? |
536374 | Volume forms under diffeomorphisms
I have an exercise and I have no idea how to do it: Let be [imath]U[/imath] and [imath]V[/imath] open sets of [imath]\mathbb{R}^{n}[/imath] and [imath]f:U\rightarrow V[/imath] an orientation-preserving diffeomorphism, then [imath]f^{*}(\operatorname{vol}_{V}) = \sqrt{\det(g_{ij}(x))} \operatorname{vol}_{U}[/imath] Where [imath]\operatorname{vol}_{V}[/imath] and [imath]\operatorname{vol}_{U}[/imath] are the volume forms of [imath]U[/imath] and [imath]V[/imath] [imath]f^{*}[/imath] the pullback of [imath]f[/imath], and [imath]g_{ij}(x)=D_xf(e_i)\cdot D_xf(e_j)[/imath] | 536361 | Pullback of a Volume Form Under a Diffeomorphism.
I have an exercise here, which I have no idea how to do. Problem: Let [imath] U [/imath] and [imath] V [/imath] be open sets in [imath] \mathbb{R}^{n} [/imath] and [imath] f: U \to V [/imath] an orientation-preserving diffeomorphism. Then show that [imath] {f^{*}}(\text{vol}_{V}) = \sqrt{\det \! \left( \left[ \left\langle {\partial_{i} f}(\bullet),{\partial_{j} f}(\bullet) \right\rangle \right]_{i,j = 1}^{n} \right)} \cdot \text{vol}_{U}. [/imath] Notation: [imath] \text{vol}_{U} [/imath] and [imath] \text{vol}_{V} [/imath] denote the volume forms on [imath] U [/imath] and [imath] V [/imath] respectively. [imath] f^{*}: {\Lambda^{n}}(T^{*} V) \to {\Lambda^{n}}(T^{*} U) [/imath] denotes the pullback operation on differential [imath] n [/imath]-forms corresponding to [imath] f [/imath]. |
536465 | Probability question on expected values
If when [imath]P(X ≥ n)[/imath], we have [imath]E(X)=\sum_{n=1}^{\infty}P(X ≥ n)=\sum_{n=1}^{\infty}(1-p)^{n-1}=\frac{1}{1-(1-p)}=\frac{1}{p}[/imath] how would i work out [imath]P(X\leq n)[/imath] Any help appreciated, Many thanks | 536383 | Tossing a biased coin until heads comes up, Find [imath]P(X=n)[/imath] and [imath]P(X\leq n)[/imath]
A biased coin that comes up heads with probability p, is tossed until the first head appears. Let random variable X be the number of tosses. For a fixed [imath]n \in N[/imath] find [imath]P(X=n)[/imath] and [imath]P(X\leq n)[/imath]. My attempt: Completely stuck on finding [imath]P(X=n)[/imath] but i have a vague idea of [imath]P(X\leq n)[/imath]: So i know that the probability of not getting heads (getting tails) is [imath]p-1[/imath] thus [imath]P(X\leq n)=(1-p)^{n+1}[/imath] so [imath]E(x)=\sum_{n=1}^{\infty}P(X\leq n)=\sum_{n=1}^{\infty}(1-p)^{n+1}=\frac{1}{1-(1-p)}=\frac{1}{p}[/imath] so the expected number of tosses until a head appears on a biased coin is [imath]\frac{1}{p}[/imath] Is this answer correct or have i gone wrong? Many thanks |
536919 | If [imath]\lim\limits_{x\to\infty} xf(x) = L[/imath], then [imath]\lim\limits_{x\to\infty} f(x) =0[/imath]
Show that if [imath]f: (a,\infty) \rightarrow \mathbb R[/imath] such that [imath]\lim_{x\to \infty} xf(x) = L[/imath] where [imath]L \in \mathbb R, [/imath] then [imath] \lim_{x\to \infty} f(x) = 0. [/imath] | 427541 | Showing that if [imath]\lim_{x\to\infty}xf(x)=l,[/imath] then [imath]\lim_{x\to\infty}f(x)=0.[/imath]
Let [imath]l \in\mathbb{R} [/imath] and [imath] f:(0,\infty) \to \Bbb R [/imath] be a function such that [imath] \lim_{x\to \infty} xf(x)=l [/imath] . Prove that [imath] \lim_{x\to \infty} f(x)=0 [/imath]. Any help would be appreciated - I found this difficult to prove |
536750 | Showing that [imath] U(2^n) [/imath] is not a cyclic group for [imath] n \geq 3 [/imath]
Could anyone please explain to me why [imath] U(2^n) [/imath] is not a cyclic group for [imath] n \geq 3 [/imath]? I need help on this because I have an algebra exam tomorrow. Thanks! | 478082 | [imath](\mathbb{Z}/2^n \mathbb{Z})^*[/imath] is not cyclic Group for [imath]n\geq 3[/imath]
Question is to Prove that [imath](\mathbb{Z}/2^n \mathbb{Z})^*[/imath] is not cyclic Group for [imath]n\geq 3[/imath]. Hint : Find two subgroups of order [imath]2[/imath]. I somehow feel that a cyclic group can not have two distinct groups of same order. but, I am not sure about the proof. I have no idea how to proceed for this. any hint would be appreciated. Thank You. |
537291 | Basis of functions from N to R
Show that the space of functions [imath]f: \mathbb{N} \to \mathbb{R}[/imath] does not have a countable basis. This is a problem I found a few weeks ago and I'm really stuck on it... I haven't thought of a good way to do it so could someone please provide a solution? | 536956 | Countable basis of function spaces
Show that the space of functions [imath]f: \mathbb{N} \to \mathbb{R}[/imath] does not have a countable basis. I really don't know where to start with this one! Could anyone help me? Thanks |
537498 | Hint for real analysis question
Suppose that [imath]f[/imath] is one-to-one and continuous on [[imath]a,b[/imath]]. Prove that [imath]f[/imath] is either strictly increasing or strictly decreasing on [[imath]a,b[/imath]]. | 170147 | A continuous, injective function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] is either strictly increasing or strictly decreasing.
I would like to prove the statement in the title. Proof: We prove that if [imath]f[/imath] is not strictly decreasing, then it must be strictly increasing. So suppose [imath]x < y[/imath]. And that's pretty much how far I got. Help will be appreciated. |
162443 | The set of all finite sequences of members of a countable set is also countable
While I was reading Enderton's "A mathematical introduction to Logic", I came across the proof of the following sentence: "The set of all finite sequences of members of the countable set A is also countable". Proof: The set S of all such finite sequences can be characterized by the equation [imath]S=\bigcup_{n \in N} A^{n+1}[/imath] Since A is countable, we have a function f mapping A one-to-one into N. The basic idea is to map S one-to-one into N by assigning to [imath](a_0,a_1,...,a_m)[/imath] the number [imath]2^{f(a_0)+1}3^{f(a_1)+1}\cdot ... \cdot p_m^{f(a_m)+1}[/imath], where [imath]p_m[/imath] is the [imath](m+1)[/imath]st prime. This suffers from the defect that this assignment might not be well-defined. For conceivably there could be [imath](a_0,a_1,...,a_m)=(b_0,b_1,...,b_n)[/imath], with [imath]a_i[/imath] and [imath]b_j[/imath] in A but with [imath]m\neq n[/imath]. But this is not serious; just assign to each member of S the smallest number obtainable in the above fashion. This gives us a well-defined map; it is easy to see that it is one-to-one. Note: P is a finite sequence of members of A iff for some positive integer [imath]n[/imath], we have [imath]P=(x_1,...,x_n)[/imath], where each [imath]x_i \in A[/imath]. First of all, I cannot understand why the former assignment might not be well-defined and the latter assignment is well-defined. Secondly, I cannot understand what Enderton means by "just assign to each member of S the smallest number obtainable in the above fashion". By the way, is [imath](a,b,c,d) = ((a,b),(c,d))[/imath] true? Also, in which cases can I omit/add parentheses in a tuple so as to have an equal tuple? | 2134838 | Is [imath]\bigcup_{i \in \mathbb{N}} \mathbb{N}^i[/imath] countable?
I was wondering if [imath]\bigcup_{i \in \mathbb{N}} \mathbb{N}^i[/imath] were countable or not, where [imath]\mathbb{N}^{i}[/imath] means the direct product of [imath]i[/imath] copies of [imath]\mathbb{N}[/imath]. I may have read that this is countable, but I think I have never seen a proof of the countability (or uncountability). The fact is that I know a countable union of countable set is countable, and I know how to prove that by using the axion of countable choice. The problem is that [imath]\prod_{1}^{\infty} \mathbb{N}[/imath] is not countable, and I don't know how to get rid of the problem. |
537938 | Doubt on proof showing matrices of certain rank form a submanifold.
I have gone through two proofs to show that matrices of rank [imath]k[/imath], where [imath]0 \leq k \leq \min(m,n)[/imath] form a submanifold of the set [imath] M(m \times n ,\mathbb{R}) [/imath]. Both proofs involve writing down an arbitrary matrix in the set considered, as : [imath] X = \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix}[/imath] where [imath]A[/imath] is assumed to be of a minor of order [imath] k \times k[/imath] that has non-zero determinant,(which I understand) and then they ask us to consider either the matrix : [imath] M_X= \begin{bmatrix} A^{-1} & -A^{-1}B \\ 0 & I_{n-k} \\ \end{bmatrix}[/imath] or alternatively: [imath]M_X= \begin{bmatrix} I & -A^{-1}B\\ 0 & I \end{bmatrix}[/imath] This is followed by multiplication of [imath]X[/imath] with [imath]M_X[/imath]. I understand what follows next. But this particular choice of matrix seemed to be picked out of the air as I viewed it.Could anyone please help motivate this idea?? Ref: The proof is found in J.Lee's Introduction to Smooth Manifolds, pg 117, ex 5.30. | 518202 | What is the codimension of matrices of rank [imath]r[/imath] as a manifold?
I'm reading through G&P's Differential Topology book, but I hit a wall at the end of section 4. There is a result stating The set [imath]X=\{A\in M_{m\times n}(\mathbb{R}):\mathrm{rk}(A)=r\}[/imath] is a submanifold of [imath]\mathbb{R}^{m\times n}[/imath] with codimension [imath](m-r)(n-r)[/imath]. There is a suggestion: Let [imath]A\in M_{m\times n}(\mathbb{R})[/imath] have form [imath] A=\begin{pmatrix} B & C \\ D & E\end{pmatrix} [/imath] where [imath]B[/imath] is an invertible [imath]r\times r[/imath] matrix. Then right multiply by [imath] \begin{pmatrix} I & -BC^{-1} \\ 0 & I \end{pmatrix} [/imath] and show [imath]\mathrm{rk}(A)=r[/imath] iff [imath]E-DB^{-1}C=0[/imath]. I multiplied out and got the matrix [imath] M:=\begin{pmatrix} B & 0 \\ D & E-DB^{-1}C\end{pmatrix}. [/imath] Since I multiplied by a nonsingular matrix, I know that [imath]\mathrm{rk}(A)=\mathrm{rk}(M)[/imath]. If [imath]E-DB^{-1}C=0[/imath], then [imath] M=\begin{pmatrix} B & 0 \\ D & 0 \end{pmatrix} [/imath] has rank [imath]r[/imath], so [imath]A[/imath] has rank [imath]r[/imath]. For the converse, if [imath]A[/imath] has rank [imath]r[/imath], then [imath]M[/imath] has rank [imath]r[/imath], so by performing row operations, [imath]M[/imath] is row equivalent to a matrix of the form [imath] \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}. [/imath] This would imply [imath]E-DB^{-1}C[/imath] is row equivalent to [imath]0[/imath], and I think this implies [imath]E-DB^{-1}C=0[/imath]. My main concern is then, how does this approach imply [imath]\mathrm{codim}(X)=(m-r)(n-r)[/imath]? Is there some special map I can apply the Preimage Theorem to? Thank you. |
534335 | Random variable iff constant on each element of a partition
Let [imath]A=\{A_1,\ldots,A_n\}[/imath] be a partition of a set [imath]\Omega[/imath] and let [imath]F=\sigma(A)[/imath]. Prove that [imath]X : \Omega \to \Bbb R[/imath] is a random variable if and only if it is constant on each element of the partition. I am trying to do the proof by contradiction but it seems the wrong way. Any hint? Thanks in advance for your help. | 381986 | Prove that it is a random variable iff it is constant on each partition
Let [imath]\mathcal{G} = \{A_1, \ldots, A_n\}[/imath] be a partition of a set [imath]\Omega[/imath], [imath]\mathcal{F} = \sigma(\mathcal{G})[/imath]. Prove that [imath]X : \Omega\to\mathbb{R}[/imath] is a random variable if and only if it is constant on each of the partition elements [imath]A_j[/imath]. I have attempted both directions but faced some difficulties to continue writing the proof. I have also consulted some sources that have a not so clear proof. Here is the proof: We know that [imath]\sigma(\mathcal{G}) = \left\{\sum_{i\in I}A_i : I \subset \{1,\ldots,n\}\right\}.[/imath] ([imath]\Leftarrow[/imath]) Let [imath]B\in \mathcal{B}(\mathbb{R})[/imath]. In this case [imath]X(\omega) = a_i[/imath], [imath]\forall\omega\in A_i[/imath] for some constants [imath]a_i[/imath], (then I am not sure how to proceed to show that it is a random variable), the proof says that [imath]X = \sum_{i\in I}^n a_i \mathcal{{1_A}_i}[/imath] and [imath]\{X \in B\} = \sum_{i : a_i\in \mathcal{B}}A_i\in \mathcal{F}[/imath], so [imath]X[/imath] is a random variable. My question is why are we concluding or defining [imath]X = \sum_{i\in I}^n a_i \mathcal{{1_A}_i}[/imath] (what is the purpose), in other words I don't understand it's flow of reasoning. ([imath]\Rightarrow[/imath]) We use proof by contradiction: If [imath]X[/imath] is not constant on [imath]A_i[/imath], then [imath]\exists i_0[/imath] such that for some [imath]\omega_1, \omega_2 \in A_{i_0}[/imath] one has [imath]a:= X(\omega_1) \not= X(\omega_2)[/imath]. Then I am not sure how to proceed from here. Many thanks in advance. |
456372 | Proof that [imath]a\mid b \land b\mid c \Rightarrow a\mid c [/imath]
I am trying to prove: [imath]a\mid b \land b\mid c \Rightarrow a\mid c [/imath] [imath]a\mid b[/imath] means that a divides b if there is an integer k, that [imath]b=k\cdot a[/imath] Please give me a hint on how to start, because I have no idea. | 1727445 | proof that if a|b and b|c then a|c
Just wanted some feed back on the following proof "if [imath]a[/imath] divides [imath]b[/imath] and [imath]b[/imath] divides [imath]c[/imath] then [imath]a[/imath] divides [imath]c[/imath]" I came up with this: If [imath]a|b[/imath] then there exist some [imath]x[/imath] that [imath]a * x = b[/imath] and if [imath]b|c[/imath] there exist some integer [imath]y[/imath] that [imath]b * y =c[/imath] therefore if [imath]a|c=z[/imath] and [imath]z=xy[/imath] then [imath](a|b)(b|c)=a|c )[/imath]. Please let me know if I got this correct and I also wanted to know if there are more than one way to prove this statement, Thank you. |
519118 | Natural Deduction: [imath]p \to (\neg q \leftrightarrow (r \lor s)), \neg s \vdash (p \land \neg q) \to r[/imath]
I have the following formula and need to prove it with natural deduction: [imath]p \to (\neg q \leftrightarrow (r \lor s)), \neg s \vdash (p \land \neg q) \to r[/imath] I was able to get the below finished but can't fill in what is missing. 1. p -> (~q <-> (r v s)) 2. ~s 3. | (p ^ ~q) (assumption) 4. | p (elim ^) 3 5. | ~q (elim ^) 3 6. | (~q <-> (r v s)) (elim ->) 1. 7. | (r v s) (elim ->) 6. n-1.| r n. (p ^ ~q) -> r | 519711 | Natural Deduction: [imath](p \lor q)\to r\vdash p \to r[/imath] and [imath]p\to\neg q\leftrightarrow(r\lor s),\neg s\vdash (p\land \neg q)\to r[/imath]
I am having trouble applying Natural Deduction rule and solving these two questions. How do I start this. \begin{align} \{(p\lor q)\to r\}&\vdash_{\sf ND} p\to r\\ \{p\to\neg q\leftrightarrow(r\lor s),\neg s\}&\vdash_{\sf ND} (p\land\neg q)\to r \end{align} |
538471 | Family with two boys
You call randomly a family with two kids, and ask if there is a kid called Tom. The answer is yes. Then what is the probably that the family has two boys. So we want that [imath]P(2 \text{ boys } | \text{ one of the children is called Tom })[/imath]. We can have: \begin{array}{c} Tom, B \\ B, Tom \\ Tom, G \\ G, Tom \\ \end{array} So then since we know one of the kids is Tom then the probability must be [imath]\frac{1}{2}[/imath]. Now if the question were to ask if we are given we know there is one boy and then we are asked to find the probability of two boys then we have: \begin{array}{c} B (known), B \\ B(known), G \\ G, G \\ \end{array} So then this would be [imath]P ( 2 \text{ boys } | \text{ one of them is a boy} ) = \frac{1}{3}[/imath] Are these reasonings correct? | 436718 | Conditional probability with Bayes' Rule
On a practice exam from statistics I encountered a very difficult exercise I couldn't manage to solve: In the tent next to you there is a family with two children. Early in the morning you see a boy coming out of the tent. What is the probability that the other child is a girl? Use Bayes' Rule My approach to the solution was the following: We assume [imath]P(GIRL)[/imath] = 0.5 and similarly [imath]P(BOY)[/imath] = 0.5. We have to compute the following conditional probability: [imath]P([/imath]One child is a girl| One child is a boy). By applying Bayes' rule we should be able to compute this probability. Bayes Rule: [imath]P(A|B)[/imath] [imath]=[/imath] [imath]\frac{P(B|A)*P(A)}{P(B|A)*P(A) + P(B|A^c)*P(A^c)}[/imath] Could anyone please help me with this, I tried many things but nothing worked out.. |
538783 | How to tell an ideal of integer polynomial ring is principal?
I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, [imath]\mathbb{C}[x][/imath]. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map [imath]\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}[/imath] sending [imath]x \mapsto i[/imath] is the principal ideal of [imath]\mathbb{Z}[x][/imath] generated by [imath]f=x^2+1[/imath]" without any justification. How to show this is true, please? | 538711 | How to tell an ideal of integer polynomial ring is principal?
I understand that uni-variate polynomial rings with coefficients in a field only have principal ideals. For example, [imath]\mathbb{C}[x][/imath]. But how can I tell if an ideal of integer polynomial ring is principal, please? For example, a textbook claims that "the kernel of the map [imath]\mathbb{Z[x]} \rightarrow \mathbb{Z[i]}[/imath] sending [imath]x \mapsto i[/imath] is the principal ideal of [imath]\mathbb{Z}[x][/imath] generated by [imath]f=x^2+1[/imath]" without any justification. How to show this is true, please? |
539009 | Which one is bigger? [imath]2011![/imath] or [imath]1006^{2011}[/imath]
I want to know Which one is bigger ? [imath] 2011! \text{ or } 1006^{2011} [/imath]. Any proof with logic is most appreciable. | 523529 | Proving that [imath]n!≤((n+1)/2)^n[/imath] by induction
I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this: [imath]V(1): 1≤1 \text{ true}[/imath] [imath]V(n): n!≤((n+1)/2)^n[/imath] [imath]V(n+1): (n+1)!≤((n+2)/2)^{(n+1)}[/imath] and I've got : [imath](((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}[/imath] [imath]((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)[/imath] |
539109 | Topology on cartesian product and product topology
Let X and Y be sets and let [imath]\tau[/imath] be topology on X[imath]\times[/imath]Y. Then there are [imath]\tau_1[/imath] and [imath]\tau_2[/imath] are topologies on X and Y , respectively such that [imath]\mathcal B[/imath]={U[imath]\times[/imath]V : U$\in[imath]\tau_1$ and V$\in[/imath]\tau_2$} is basis for [imath]\tau[/imath] or not ? If it is true then I want to proof. But if not then I want to example. | 539073 | Topology on cartesian product and product topology.
Let X and Y be sets. Does it have every topology on cartesian product X[imath]\times[/imath]Y must be product topology ? |
538158 | Solution to differential equation is defined on [imath]\mathbb R[/imath]
Let [imath]y(x,\xi)[/imath] be the value at point [imath]x[/imath] of the maximal solution that satisfies the initial condition [imath]y(0)=\xi[/imath]. Prove that the domain of [imath]y(\cdot, y(s, \xi))[/imath] is [imath]I-s[/imath], where [imath]I[/imath] is the domain of [imath]y(\cdot ,\xi)[/imath]. Prove that for all [imath]s, t[/imath] such that [imath]y(s, \xi)[/imath] and [imath]y(t+s, \xi)[/imath] exist, then [imath]y(t,y(s,\xi))[/imath] also exists and [imath]y(t,y(s,\xi))=y(t+s, \xi)[/imath]. If [imath]y[/imath] is a maximal solution and there exists [imath]T>0[/imath] such that [imath]y(0)=y(t)[/imath] and [imath]f(y(0))\neq 0[/imath], then [imath]y[/imath] is a periodic solution and not constant. In the above problem, I'm missing only that in 3., [imath]y[/imath] is defined on [imath]\mathbb R[/imath]. I asked about this problem here. The user Artem has provided some help in the comments regarding this, but I can't transform it into an answer. I have a theorem that says that if for all [imath]a,b\in I[/imath] such that [imath]y[/imath] is defined on [imath](a,b)[/imath], there exists [imath]K[/imath] such that [imath]|y(x)|\leq K[/imath] for all [imath]x\in (a,b)[/imath], then [imath]y[/imath] is defined on [imath]I[/imath]. How can I use the above to solve the problem? Why is [imath]y[/imath] bounded? Otherwise, how I can use the 1. and 2. to prove that [imath]y[/imath] is defined on [imath]\Bbb R[/imath]? Some detail is appreciated, not because I'm lazy but because I feel like I need it. | 528329 | Differential equation: autonomous system
This isn't homework. I have no idea what theorems I should be looking at to solve this. Guidance, partial and total solutions are all welcomed. Let [imath]f[/imath] be a locally lipschitz function in an open set [imath]G\subseteq \Bbb R^n[/imath]. Consider the autonomous system [imath]y'=f(y)[/imath]. Let [imath]y(x,\xi)[/imath] be the value at point [imath]x[/imath] of the maximal solution that satisfies the initial condition [imath]y(0)=\xi[/imath]. Prove that the domain of [imath]y(\cdot, y(s, \xi))[/imath] is [imath]I-s[/imath], where [imath]I[/imath] is the domain of [imath]y(\cdot ,\xi)[/imath]. Prove that for all [imath]s, t[/imath] such that [imath]y(s, \xi)[/imath] and [imath]y(t+s, \xi)[/imath] exist, then [imath]y(t,y(s,\xi))[/imath] also exists and [imath]y(t,y(s,\xi))=y(t+s, \xi)[/imath]. If [imath]y[/imath] is a maximal solution and there exists [imath]T>0[/imath] such that [imath]y(0)=y(T)[/imath] and [imath]f(y(0))\neq 0[/imath], then [imath]y[/imath] is a periodic solution and not constant. If [imath]y[/imath] is a solution whose domain is [imath](a,+\infty)[/imath], if [imath]\eta:=\lim _{x\to +\infty}y(x)[/imath] and [imath]\eta \in G[/imath], then [imath]f(\eta)=0[/imath]. EDIT I found an alternative solution for 3. Please check my proof and give feedback in comments: let [imath]u[/imath] be the restriction of [imath]y[/imath] to [imath][0,T][/imath], now let [imath]\overline u[/imath] be the periodic extension of [imath]u[/imath] to [imath]\mathbb R[/imath]. It is easy to see that [imath]\overline u[/imath] is a solution to the given differential equation. But since [imath]f[/imath] is locally lipschitz, [imath]\overline u[/imath] must coincide with [imath]y[/imath] wherever they are both defined. Since [imath]y[/imath] is a maximal solution, it must be [imath]\overline u[/imath], so [imath]y[/imath] is defined on [imath]\mathbb R[/imath]. |
539266 | Prove [imath]\sqrt{2} + \sqrt{5}[/imath] is irrational
How do you prove that [imath]\sqrt{2} + \sqrt{5}[/imath] is irrational? I tried to prove it by contradiction and got this equation: [imath]a^2/b^2 = \sqrt{40}[/imath]. | 788775 | Prove that [imath]\sqrt{7}+\sqrt{3}[/imath] is irrational
Is there a method by which we can prove that [imath]\sqrt{3}+\sqrt{7}[/imath] is irrational. It's obviously an irrational number, but I want to prove that mathematically. |
539522 | On the prime in the ring of integers of infinite Galois extensions
I need help to solve the following exercise: Let [imath]K[/imath] be an infinite algebraic extension of [imath]\mathbb Q[/imath] and let [imath]O_K[/imath] denote the ring of algebraic integers in [imath]K.[/imath] If [imath]\frak P[/imath] is a prime ideal of [imath]O_K[/imath] then [imath]{\frak P}\cap{\mathbb Z}[/imath] is a prime ideal of [imath]{\mathbb Z}.[/imath] Moreover, if [imath]\frak P[/imath] is non-zero, then so is [imath]{\frak P}\cap{\mathbb Z}.[/imath] Thus [imath]{\frak P}\cap{\mathbb Z}=p{\mathbb Z}[/imath] for some prime [imath]p.[/imath] 1) Show that [imath]O_K/{\frak P}[/imath] is a field, and is an algebraic extension of [imath]{\mathbb F}_p[/imath]; then show that it is Galois over [imath]{\mathbb F}_p[/imath] with abelian Galois group. 2) Now suppose that [imath]K/F[/imath] is an infinite Galois extension. Let [imath]\frak P,[/imath] [imath]O_K[/imath] be as before and put [imath]{\frak p}={\frak P}\cap O_F[/imath], a prime ideal of [imath]O_F.[/imath] Show that the group [imath]\text{Gal}(K/F)[/imath] acts transitively on the set of prime ideals of [imath]O_K[/imath] above [imath]\frak p.[/imath] Thanks in advance. | 537808 | Infinite algebraic extensions of [imath]\mathbb{Q}[/imath]
I need help to solve the following exercise: If [imath]K[/imath] is an algebraic extension of [imath]\mathbb{Q}[/imath] (finite or infinite), then we let [imath]O_{K}[/imath] denote the ring of algebraic integers in [imath]K.[/imath] If [imath]\frak{P}[/imath] is a prime ideal of [imath]O_{K}[/imath] then [imath]\frak{P}∩\mathbb{Z}[/imath] is a prime ideal of [imath]\mathbb{Z}.[/imath] Moreover, if [imath]\frak{P}[/imath] is non-zero, then so is [imath]\frak{P}∩\mathbb{Z}[/imath] . Thus $\frak{P}∩\mathbb{Z}=[imath] p\mathbb{Z}$ for some prime $p.$ Show that $O_{K}/\frak{P}$ is a field, and is an algebraic extension of $\mathbb{F}_{p}$; then show that it is Galois over $\mathbb{F}_{p}$ with abelian Galois group.[/imath] Now suppose that $K/F$ is a (finite or infinite) Galois extension. Let $\frak{P},$ $O_{K}$ be as before and put $\frak{p}=\frak{P} \cap O_{F}$, a prime ideal of [imath]O_{F}.[/imath] Show that [imath]\mathrm{Gal}(K/F)[/imath] acts transitively on the set of prime ideals of [imath]O_{K}[/imath] above [imath]\frak{p}.[/imath] Thanks in advance |
540061 | Integration by parts, error on Primitive of [imath]\sqrt{1-x^{2}}[/imath]
Problem: Integrate [imath]\sqrt{1-x^{2}}[/imath] Attempt: | 533082 | Integral of [imath]\sqrt{1-x^2}[/imath] using integration by parts
I was asked to solve this indefinite integral using Integration by parts. [imath]\int \sqrt{1-x^2} dx[/imath] I know how to solve if use the substitution [imath]x=\sin(t)[/imath] but I'm looking for the Integration by parts way. any help would be very appreciated. |
540278 | convergence of [imath]\sum _{n=2}^{\infty } \frac{(-1)^n}{\sqrt{n}+(-1)^n}[/imath]
Does the infinite sum [imath]\sum _{n=2}^{\infty } \frac{(-1)^n}{\sqrt{n}+(-1)^n}[/imath] converge ? Can anyone help ? | 455315 | Does [imath]\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}[/imath] converge conditionally?
Discuss the convergence of the sum: [imath]\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}.[/imath] My answer so far: It does not converge absolutely since [imath]\left | \frac{(-1)^k}{\sqrt{k}+(-1)^k} \right | \geq \frac {1} {\sqrt{k} + 1},[/imath] the sum of whose terms diverges. Intuitively, I think it should converge conditionally because [imath] \sum \frac{(-1)^k}{\sqrt{k}}[/imath] does, and the [imath](-1)^k[/imath] in the denominator should not make a big difference. However, I am having trouble demonstrating its conditional convergence. Evidently, the problem is that [imath]\frac{1}{\sqrt{k}+(-1)^k}[/imath] is not monotone. Let's call this sequence [imath]a_n[/imath]. If we can show that [imath]\sum_{k=2}^\infty |a_k - a_{k-1}| < \infty,[/imath] then we're done by Dirichlet's test. But this seems to be [imath]O(1/k)[/imath]. Any ideas? |
15766 | Proving that the set of limit points of a set is closed
From Rudin's Principles of Mathematical Analysis (Chapter 2, Exercise 6) Let [imath]E'[/imath] be the set of all limit points of a set [imath]E[/imath]. Prove that [imath]E'[/imath] is closed. I think I got it but my argument is a bit hand wavy: If [imath]x[/imath] is a limit point of [imath]E'[/imath], then every neighborhood of [imath]x[/imath] contains some [imath]y\in E'[/imath], and every neighborhood of [imath]y[/imath] contains some [imath]z\in E[/imath]. Therefore every neighborhood of [imath]x[/imath] contains some [imath]z\in E[/imath], and so [imath]x[/imath] is a limit point of [imath]E[/imath]. Then [imath]x\in E'[/imath], so [imath]E'[/imath] is closed. The thing that's bugging me is the leap from one neighborhood to another. Is this formally correct? | 597386 | limit point set and closed condition
Here is the question Prove that the set of limit points of a set is closed. I'm not even understand the question fully. Is question meaning [imath]\{E= {N_\epsilon(p) }\}[/imath]? if you said [imath]p[/imath] is the limit point? How do you approach this question? |
234062 | Probability of tossing a fair coin with at least [imath]k[/imath] consecutive heads
Tossing a fair coin for [imath]N[/imath] times and we get a result series as [imath]HTHTHHTT\dots~[/imath], Here '[imath]H[/imath]' denotes 'head' and '[imath]T[/imath]' denotes 'tail' for a specific tossing each time. What is the probability that the length of the longest streak of consecutive heads is greater than or equal to [imath]k[/imath]? (that is we have a [imath]HHHH\dots~[/imath], which is the substring of our tossing result, and whose length is greater than or equal to [imath]k[/imath]) I came up with a recursive solution (though not quite sure), but cannot find a closed form solution. Here is my solution. Denote [imath]P(N,k)[/imath] as the probability for tossing the coin [imath]N[/imath] times, and the longest continuous heads is greater or equal than [imath]k[/imath]. Then (For [imath]N>k[/imath]) [imath] P(N,k)=P(N-1,k)+\Big(1-P(N-k-1,k)\Big)\left(\frac{1}{2}\right)^{k+1} [/imath] | 970853 | Four consecutive heads from ten coin flips
If we flip a fair coin [imath]10[/imath] times, what is the probability we get [imath]\ge 4[/imath] consecutive heads? An approach would be to consider the probability of all consecutive heads, cut off by tails, e.g., HHHHTxxxxx THHHHTxxxx ... HHHHHTxxxx THHHHHTxxxx ... but this is complicated, and also has a problem that some patterns overlap (e.g., HHHHTHHHHT). Is there a simpler way to do this? |
541073 | Proof that [imath]\lim{a_n}=L[/imath] when [imath]n[/imath] goes to infinity, then [imath]\{a_n\}[/imath] its a Cauchy Sequence
Proof that [imath]\lim{a_n}=L[/imath] when [imath]n[/imath] goes to infinity, then [imath]\{a_n\}[/imath] its a Cauchy Sequence I start with this hypothesis [imath]\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}>0}[/imath] [imath]\exists{N}\in{\mathbb{N}}[/imath] such that [imath]\forall{n}>N \left |{a_n-L}\right |< \epsilon[/imath] The same its true fot any [imath]m>N,\left |{a_m-L}\right |=\left |{L-a_m}\right |< \epsilon[/imath] I will sum the inequalities but I havent any idea how do it... Help me please!!! | 427169 | Metric Space & Cauchy Sequence
Question: Consider a metric space [imath](X,d)[/imath] 1) Show that if [imath]x[/imath] is a convergent sequence in [imath](X,d)[/imath], then it is a Cauchy sequence in [imath](X,d)[/imath]. Suppose [imath](X,d)[/imath] is complete. Let [imath]f\colon X \to X[/imath] be a contraction, i.e., there exits a [imath]\beta\in(0,1)[/imath] such that, for all [imath]x,y \in X[/imath], we have [imath]d( f(x) , f(y) )\le \beta d(x,y)[/imath]. Let [imath]x_0\in X[/imath]. Define the sequence [imath]x[/imath] inductively by the formula [imath]x_n=f(x_{n-1})[/imath] for [imath]n\in\mathbb{N}[/imath]. 2) Show that [imath]x[/imath] is a Cauchy sequence, and therefore convergent. 3) Show that a limit point of [imath]x[/imath], say [imath]\alpha[/imath], is a fixed point of [imath]f[/imath], i.e., [imath]\alpha=f(\alpha)[/imath]. 4)Show that [imath]f[/imath] has a unique fixed point. My try for part 1: If [imath]x[/imath] is a convergent sequence in [imath](X,d)[/imath] then there exists [imath]x_0 \in X[/imath] and [imath]M \in \mathbb{R}[/imath] such that for all [imath]x \in X[/imath], [imath]d(x,x_0) \le M[/imath] This implies [imath]|x-x_0| \le M[/imath] . This proves it is Cauchy sequence. |
541178 | Area between [imath]y = 5+x-x^2[/imath] and [imath]y = x+4[/imath]
I have the following 2 equations and I have drawn their graphs. [imath]\begin{cases}y=5+x-x^2\\ y=x+4\end{cases}[/imath] I have found intercepts as well. The question is asking to find area between these 2 curves. I am not sure if I should take as a whole triangle? | 332940 | Area enclosed between the curves [imath]y=x^2[/imath] and [imath]y=60-7x[/imath]
Find the area enclosed between the curves [imath]y=x^2[/imath] and [imath]y=60-7x[/imath]. I am completely new at this but I have tried and I believe that it should be between the numbers [imath]0[/imath] and [imath]60[/imath] is this right? |
541576 | How to prove that the intersection of [imath]L^1(\mathbb{R})[/imath] and [imath]L^2(\mathbb{R})[/imath] is dense in [imath]L^2(\mathbb{R})[/imath]
How to prove that the intersection of [imath]L^1(\mathbb{R})[/imath] space and [imath]L^2(\mathbb{R})[/imath] space is dense in [imath]L^2(\mathbb{R})[/imath] space? | 266049 | Why is [imath]L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)[/imath] dense in [imath] L^2(\mathbb{R}^n)[/imath]?
In Lieb and Loss's Analysis, I saw that they mentioned [imath]L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)[/imath] dense in [imath] L^2(\mathbb{R}^n)[/imath] (dense wrt the [imath]L^2[/imath] norm, I think). But I didn't find its proof in the book. So I wonder why it is? Does this conclusion hold for [imath]L^1(\Omega, \mathcal{F}, \mu) \cap L^2(\Omega, \mathcal{F}, \mu)[/imath] for any measure space [imath](\Omega, \mathcal{F}, \mu)[/imath]? Are there similar statements if replace [imath]L^1[/imath] with [imath]L^p[/imath], and [imath] L^2[/imath] with [imath]L^q[/imath] for [imath]p \leq q \in (0, \infty ][/imath] or [imath]\in [1, \infty][/imath]? Thanks and regards! |
541758 | Where am I going wrong on this second derivative?
If a given first derivative is: [imath]\ {dy \over dx} = {-48x \over (x^2+12)^2} [/imath] What are the steps using the quotient rule to derive the second derivative: [imath]\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} [/imath] My Steps: [imath] {{d^2y \over dx^2} = {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} =-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} = -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} = -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} = ???} [/imath] Once I get to this point I am unsure how to derive the second derivative shown. | 541696 | How to derive this second derivative using the quotient rule?
If a given first derivative is: [imath]\ {dy \over dx} = {-48x \over (x^2+12)^2} [/imath] What are the steps using the quotient rule to derive the second derivative: [imath]\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} [/imath] My Steps: \begin{align*} {d^2y \over dx^2} &= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\ &=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\ &= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\ &= -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\ &= {???} \end{align*} |
539894 | Argument principle on fourth-order equation
How many roots of the equation [imath]z^4+8z^3+3z^2+8z+3=0[/imath] lie in the right half plane? The hint is to apply the argument principle. I am not sure how that will go. I can take [imath]f(z)=z^4+8z^3+3z^2+8z+3[/imath], so that for any closed curve [imath]C[/imath], [imath]\int_C\frac{f'(z)}{f(z)}dz=2\pi i(N-P)[/imath] where [imath]N[/imath] and [imath]P[/imath] denote respectively the number of zeros and poles of [imath]f(z)[/imath] inside the contour [imath]C[/imath]. But here we are looking at the right half plane. I can't see how to apply the result from the argument principle. | 31651 | roots of [imath]f(z)=z^4+8z^3+3z^2+8z+3=0[/imath] in the right half plane
This is a question in Ahlfors in the section on the argument principle: How many roots of the equation [imath]f(z)=z^4+8z^3+3z^2+8z+3=0[/imath] lie in the right half plane? He gives a hint that we should "sketch the image of the imaginary axis and apply the argument principle to a large half disk." Since [imath]f[/imath] is an entire function, I think I understand that the argument principle tells us that for any closed curve [imath]\gamma[/imath] in [imath]\mathbb{C}[/imath], the winding number of [imath]f(\gamma)[/imath] around 0 is equal to the number of zeros of [imath]f[/imath] contained inside [imath]\gamma[/imath]. How would you go about actually applying the hint though? I am having trouble figuring out what the image of a large half disk under [imath]f[/imath] would look like. |
541991 | Linear transformation diagonalization
I encountered this question and I need some assistance to solve it. [imath]T[/imath] is a linear transformation from [imath]V \to V[/imath] We are given that [imath]T^2 = T[/imath] Show that [imath]T[/imath] can be diagonalized. | 73862 | Diagonalization of a projection
If I have a projection [imath]T[/imath] on a finite dimensional vector space [imath]V[/imath], how do I show that [imath]T[/imath] is diagonalizable? |
531442 | Idempotent of closure and other properties
I have to prove: [imath]\mbox{ClCl}A = \mbox{Cl}A[/imath] [imath]\mbox{diamCl}A = \mbox{diam}A[/imath] If [imath]G[/imath] is open set then for any [imath]A[/imath] we have [imath] \mbox{Cl}(G \cap A) = \mbox{Cl}(G \cap \mbox{Cl}A)[/imath] My attempt: [imath]\mbox{Cl}A \subseteq \mbox{ClCl}A[/imath] is quite easy by fact that [imath]A \subseteq \mbox{Cl}A [/imath]. But I don't have idea how can I prove that [imath] \mbox{ClCl}A \subseteq \mbox{Cl}A[/imath] But the point 2. and 3. is for me too difficult. Could you give me some hints? Thanks in advance! | 2943419 | Proof of topological space
My solution: 1) [imath] cl(A) \subseteq cl(cl(A))[/imath] let [imath]x \in cl(cl(A))[/imath] by definition [imath]U_x \cap cl(A) \neq \varnothing [/imath] [imath] \forall U_x \in Tau[/imath] such that [imath]x\in U_x [/imath] .Thus [imath]x \in cl(cl(A))[/imath] [imath]cl(A) \subseteq cl(cl(A))[/imath] Is it true? And how prove other side. |
542312 | Do there exist functions such that [imath]f(f(x)) = -x[/imath]?
I am wondering about this. A well-known class of functions are the "involutive functions" or "involutions", which have that [imath]f(f(x)) = x[/imath], or, equivalently, [imath]f(x) = f^{-1}(x)[/imath] (with [imath]f[/imath] bijective). Now, consider the "anti-involution" equation [imath]f(f(x)) = -x[/imath]. It is possible for a function [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath] to have [imath]f(f(z)) = -z[/imath]. Take [imath]f(z) = iz[/imath]. But what about this functional equation [imath]f(f(x)) = -x[/imath] for functions [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath], instead of [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath]? Do such functions exist? If so, can any be described or constructed? What about in the more general case of functions on arbitrary groups [imath]G[/imath] where [imath]f(f(x)) = x^{-1}[/imath] (or [imath]-x[/imath] for abelian groups), [imath]f: G \rightarrow G[/imath]? Can we always get such [imath]f[/imath]? If not, what conditions must there be on the group [imath]G[/imath] for such [imath]f[/imath] to exist? | 312385 | Find a real function [imath]f:\mathbb{R}\to\mathbb{R}[/imath] such that [imath]f(f(x)) = -x[/imath]?
I've been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success: Find a function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] satisfying [imath]f(f(x)) = -x[/imath] for all [imath]x \in \mathbb{R}[/imath]. I am also wondering, can we find [imath]f[/imath] so that is continuous? I was thinking of letting [imath]f[/imath] be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist. Source: http://www.halfaya.org/Casti/CalculusTheory2/challenge.pdf |
542567 | [imath]f : [a, ∞) → R[/imath] is a continuous function. If [imath]\lim_{x→∞} f (x) = L[/imath], prove that [imath]f[/imath] is uniformly continuous on [imath][a, ∞)[/imath].
Suppose that [imath]f : [a, ∞) → R[/imath] is a continuous function. If [imath]\lim\limits_{x→∞} f (x) = L[/imath], prove that [imath]f[/imath] is uniformly continuous on [imath][a, ∞)[/imath]. My attempt at the proof: Well since I have to use both facts, I think my proof needs to be divided into two parts: 1- I have to prove that [imath]f[/imath] is uniformly continuous on [imath](N,∞)[/imath] where [imath]N>a[/imath] (by using the limit definition somehow) 2- Using compactness, I can easily show that [imath]f[/imath] is uniformly continuous on [imath][a,N][/imath] I am having trouble with [imath](1)[/imath] because I can't find an [imath]\delta>0[/imath] that works for all [imath]\epsilon>0[/imath] Just need a hint in the right direction, thank you!! | 75491 | How does the existence of a limit imply that a function is uniformly continuous
I am working on a homework problem from Avner Friedman's Advanced Calculus (#1 page 68) which asks Suppose that [imath]f(x)[/imath] is a continuous function on the interval [imath][0,\infty)[/imath]. Prove that if [imath]\lim_{x\to\infty} f(x)[/imath] exists (as a real number), then [imath]f(x)[/imath] is uniformly continuous on this interval. Intuitively, this argument makes sense to me. Since the limit of [imath]f(x)[/imath] exists on [imath][0,\infty)[/imath], we will be able to find a [imath]\delta > |x_0 - x_1|[/imath] and this implies that, for any [imath]\epsilon>0[/imath], we have [imath]\epsilon > |f(x_0) - f(x_1)|[/imath] (independent of the points chosen). I am aware that the condition of uniform continuity requires that [imath]\delta[/imath] can only be a function of [imath]\epsilon[/imath], not [imath]x[/imath]. What information does the existence of a real-valued limit provide that implies [imath]f(x)[/imath] is uniformly continuous on this interval? |
543408 | Manually computing a galois field element
[imath]F = GF(2^6)[/imath] modulo the primitive polynomial [imath]h(x) = 1 + x^2 + x^3 + x^5 + x^6[/imath] and [imath]\alpha[/imath] is the class of [imath]x[/imath]: [imath]GF(2^6) = \{0,1,\alpha, \alpha^2...\alpha^{62}\}[/imath] How do I manually compute [imath]\alpha^{53}[/imath] (or any other large power of [imath]\alpha[/imath])? It is straightforward to compute e.g. [imath]\alpha^{13}[/imath] - I can just divide [imath]x^{13}[/imath] by [imath]h(x)[/imath] and take the reminder. I came across such a problem in a problem set and I suspect there should be an efficient technique to compute the remainder of [imath]x^{53}[/imath] divided by [imath]h(x)[/imath] by hand. Perhaps I should take into account that [imath]53 = -11 (mod 64)[/imath]? | 539632 | Finding element in binary representation of [imath]GF(2^6)[/imath]
I got the following task: Let [imath]F = GF(2^6 )[/imath] be K[x] modulo the primitive polynomial [imath]h(x) = 1 +x ^2 +x ^3 +x ^5 +x ^6[/imath] , and let [imath]\alpha[/imath] be the class of x. I have a table with the binary representation of the elements of [imath]GF(2^6 ) = \{0; 1; \alpha^1; \alpha^2 ; ...; \alpha^{62}\}[/imath]. Is there a quicker way to check [imath]\alpha^{52}[/imath] than performing a long division of [imath]x^{52}[/imath] by [imath]h(x)[/imath] ? It has [imath]2^6[/imath] elements so after [imath]\alpha^{62}[/imath] it will be back to 1 again, but maybe there is some trick to find it faster rather than performing such a long division? I have also [imath]\alpha^{13}[/imath] so it's [imath]\alpha^{13^4}[/imath], but no idea how to use it |
544246 | Number of digits of [imath]2^{1000}[/imath]
A friend asks me to find the number of digits of [imath]2^{1000}[/imath]. I tried to look for a pattern by calculating the first powers of [imath]2[/imath] but I didn't find it. How should I proceed? Thanks. | 119913 | How many digits does [imath]2^{1000}[/imath] contain?
I tried this way, I only need to know if this is correct or if there are better ways to solve this: [imath]2^{1000}[/imath] does not have a factor of [imath]5[/imath] obviously therefore we can assume [imath] 10^{m} < 2^{1000} < 10^{m+1}[/imath] for some [imath]m[/imath] Assume [imath] k = 2^{1000}[/imath], then take log on both sides [imath]\log k = 1000 \log 2 \approx 301.02999 > 301[/imath] Therefore [imath]2^{1000}[/imath] has [imath]302[/imath] digits. |
544127 | Exercise about stationary processes
I need some help with this exercise: [imath]\bullet[/imath]If [imath]X(t)[/imath] is a mean square differentiable wide-sense stationary stochastical process then the processes [imath]X(t)[/imath] and [imath]X' (t)[/imath] are orthogonal. [imath]\bullet[/imath]If [imath]X(t)[/imath] is a twice mean square differentiable, stationary and Gaussian stochastical process, such that [imath]E[X(t)] = 0[/imath], then [imath]X(t)[/imath] is independent of [imath]X' (t)[/imath] but not independent of [imath]X''(t)[/imath] I was given the hint that I should use the formula [imath]\Gamma_{X^{(n)},X^{(m)}}(t,s)=(-1)^m\frac{d^{(n+m)}\Gamma_X(\tau)}{d\tau^{(n+m)}}[/imath], where [imath]\tau=t-s[/imath]. For the first part, I have this: Denoting [imath]s=E[X(t)][/imath] and [imath]r=E[X'(t)][/imath], using that X is ms differentiable, we have: [imath]h(t)=\Gamma_{X-s,X'-r}=E[(X(t)-s(t))*(X'(t)-r(t))]=E[(X(t)-s(t))*(X'(t)-r'(t))]=E[(X(t)-s(t))*(X(t)-s(t))']=\Gamma_{(X-s(t)),(X-s(t))'}(t)=-\frac{d}{dt}\Gamma_{X-s(t)}(t)=-\frac{E[(X-s)*(X-s)]}{dt}=-\frac{dE[(X-s(t))*(X-s(t))]}{dt}=-E[\frac{d((X-s(t))*(X-s(t)))}{dt}]=-2h(t)[/imath] So h must be zero, which means the processes are orthogonal. Am I correct? For the second part, I need some hint, because I don't know how to start. Computing with the same formula seems to get me nowhere. Thank you very much for any help. | 540055 | About stationary and wide-sense stationary processes
I have just started with stochastical calculus, and I need some help with a pair of problems: [imath]\bullet[/imath]If [imath]X(t)[/imath] is a mean square differentiable wide-sense stationary stochastical process then the processes [imath]X(t)[/imath] and [imath]X' (t)[/imath] are orthogonal. [imath]\bullet[/imath]If [imath]X(t)[/imath] is a twice mean square differentiable, stationary and Gaussian stochastical process, such that [imath]E[X(t)] = 0[/imath], then [imath]X(t)[/imath] is independent of [imath]X' (t)[/imath] but not independent of [imath]X''(t)[/imath] I was given the hint that I should use the formula [imath]\Gamma_{X^{(n)},X^{(m)}}(t,s)=(-1)^m\frac{d^{(n+m)}\Gamma_X(\tau)}{d\tau^{(n+m)}}[/imath], where [imath]\tau=t-s[/imath], but I can't get anything from it. I'm stuck. Any ideas? Thanks a lot for any help. |
544404 | Trigonometry: Isosceles Triangle
I saw the following problem on Facebook (figure not drawn to scale): [imath]\triangle ABC[/imath] is an isosceles triangle with [imath]AB\equiv BC[/imath], [imath]AC\equiv BD[/imath], and [imath]\angle B=20^\circ[/imath]. Find [imath]\angle CAD[/imath]. What I did was let [imath]BC\equiv x[/imath] and divide the triangle in two by bisecting [imath]\angle B[/imath]. This yielded two right triangles with base [imath]x\cos80^\circ[/imath]. Therefore, the base of [imath]\triangle ABC[/imath] has length [imath]2x\cos80^\circ[/imath]. Clearly, [imath]DC=x-2x\cos80^\circ[/imath], and [imath]\angle C=80^\circ[/imath]. Now, to find [imath]AD\equiv y[/imath], I used the law of cosines: [imath] y=\sqrt{(2x\cos80^\circ)^2+(x-2x\cos80^\circ)^2-2(2x\cos80^\circ)(x-2x\cos80^\circ)\cos80^\circ}. [/imath] Finally, to find [imath]\angle CAD\equiv\theta[/imath], I used the law of sines: [imath]\begin{align} \frac{\sin80^\circ}{y}&=\frac{\sin\theta}{x-2x\cos80^\circ}\\ \Longrightarrow\theta&=70^\circ. \end{align}[/imath] Is my result correct? Is there an easier way to solve this problem? | 515684 | Find an angle of an isosceles triangle
[imath]\triangle ABC[/imath] is an isosceles triangle such that [imath]AB=AC[/imath] and [imath]\angle BAC[/imath]=[imath]20^\circ[/imath]. And a point D is on [imath]\overline{AC}[/imath] so that AD=BC, , How to find [imath]\angle{DBC}[/imath]? I could not get how to use the condition [imath]AD=BC[/imath] , How do I use the condition to find [imath]\angle{DBC}[/imath]? EDIT 1: With MvG's observation, we can prove the following fact. If we set on a point [imath]O[/imath] in [imath]\triangle{ABC}[/imath] such that [imath]\triangle{OBC}[/imath] is a regular triangle, then [imath]O[/imath] is the circumcenter of [imath]\triangle{BCD}[/imath]. First, we will show if we set a point [imath]E[/imath] on the segment [imath]AC[/imath] such that [imath]OE=OB=OC=BC[/imath], then [imath]D=E[/imath]. Becuase [imath]\triangle{ABC}[/imath] is a isosceles triangle, the point [imath]O[/imath] is on the bisecting line of [imath]\angle{BAC}[/imath]. [imath]\angle{OAE}=20^\circ/2=10^\circ[/imath]. And because [imath]OE=OC[/imath], [imath]\angle{OCE}=\angle{OEC}=20^\circ[/imath], [imath]\angle{EOA}=20^\circ-10^\circ=10^\circ=\angle{EAO}[/imath]. Therefore [imath]\triangle{AOE}[/imath] is an isosceles triangle such that [imath]EA=EO[/imath]. so [imath]AD=BC=AE[/imath], [imath]D=E[/imath]. Now we can see the point [imath]O[/imath] is a circumcenter of the [imath]\triangle{DBC}[/imath] because [imath]OB=OC=OD.[/imath] By using this fact, we can find [imath]\angle{DBC}=70^\circ[/imath], |
544802 | Gaussian curvature and mean curvature of an ellipsoid
I need help with this differential geometry problem dealing with Gaussian curvature and mean curvature Given an ellipsoid with parametric representation [imath](a\cos u\cos v, b\cos u \sin v, c\sin u)[/imath] compute its Gaussian curvature and mean curvature. (Hint: Compute [imath]E,F,G[/imath] first and [imath]L, M, N[/imath] second then Weingarten map (Shape operator). The Gaussian curvature and the mean curvature are the determinant and the trace of the Weingarten map (Shape operator). | 540710 | gaussian and mean curvatures
I am trying to review, and learn about how to compute and gaussian and mean curvature. Given [imath]\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1[/imath], how can I compute the gaussian and mean curvatures? This is what I have so far, [imath]K(u, v) = \frac{a^2 b^2 c^2}{[c^2 \sin^2(v) (a^2 \sin^2(u)+b^2 \cos^2(u))+a^2 b^2 \cos^2(v)]^2}[/imath] Please help me out. |
535981 | If [imath]\displaystyle \lim _{x\to +\infty}y(x)\in \mathbb R[/imath], then [imath]\lim _{x\to +\infty}y'(x)=0[/imath]
Not homework. I need this (or something similar) to solve 4. in this question. Let [imath]y:(a ,+\infty)\to \mathbb R[/imath] be [imath]C^1[/imath]. Prove that [imath]\lim_{x\to +\infty}y(x)=\eta\text{ for some }\eta\in \mathbb R\implies\text{the following limit exists and } \lim_{x\to +\infty}y'(x)=0[/imath] Intuitively this is true because [imath]\displaystyle \lim_{x\to +\infty}y(x)=\eta[/imath] means that [imath]y[/imath] almost stops increasing or decreasing, so [imath]\displaystyle \lim_{x\to +\infty}y'(x)=0[/imath]. But how to prove it? I tried [imath]\lim_{x\to +\infty}y'(x)=0=\lim_{x\to +\infty} \lim_{h\to 0}\dfrac{y(x+h)-y(x)}h= \lim_{h\to 0}\lim_{x\to +\infty}\dfrac{y(x+h)-y(x)}h= \lim_{h\to 0}\dfrac{a-a}h=0,[/imath] but why I can change the order of the limits? If I can't even do that, how can I prove this? After reading the threads Tyler provided, I now just need to prove that [imath]\displaystyle \lim_{x\to +\infty}y'(x)[/imath] exists? Please consider a suitable adaption to the linked question. | 162078 | If a function has a finite limit at infinity, does that imply its derivative goes to zero?
I've been thinking about this problem: Let [imath]f: (a, +\infty) \to \mathbb{R}[/imath] be a differentiable function such that [imath]\lim\limits_{x \to +\infty} f(x) = L < \infty[/imath]. Then must it be the case that [imath]\lim\limits_{x\to +\infty}f'(x) = 0[/imath]? It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy: [imath] \begin{align} \lim_{x \to +\infty} f'(x) &= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\ &= \lim_{h \to 0} \frac1{h}(L-L) \\ &= \lim_{h \to 0} \frac{0}{h} \\ &= 0 \end{align} [/imath] In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is a similar question already, but I think this is different in two aspects. First, that question assumes that [imath]\lim\limits_{x \to +\infty}f'(x)[/imath] exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case. |
545200 | There are [imath]12[/imath] stations between A and B, in how many ways you can select 4 stations for a halt in such a way that no two stations are consecutive
How do we solve this question with permutation and combination? Can we solve this by Fibonacci Series? Is the answer (12-2)C4 {10 combination 4}? | 153030 | Counting train stops using combinatorics
There are 12 intermediate stations on a railway line between 2 stations. Find the number of ways a train can be made to stop at 4 of these so that no two stopping stations are consecutive. My attempt: Initially I found the maximum allowed stop number for the first stop that satisfies the consecutive station condition. [imath]A \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7 \quad 8 \quad 9 \quad 10 \quad 11 \quad 12 \quad B[/imath] I can have a stop at stations 8, 10, 12. Hence the train can travel a maximum of 6 stations before coming to its first stop, so number of ways [imath]= 6 \cdot 1 \cdot 1 \cdot 1 = 6[/imath]. Then I shift the first stop to station number 5. Now I have 5 options for stop 1 and 2 possible options for any of the next three stops. Hence number of ways [imath]= 5 \cdot 2 \cdot 3 \cdot 1 = 30[/imath]. Again, I shift the first stop to station number 4. Now I have 4 options for stop 1 and 3 possible options for any of the next 3 stops. Hence number of ways [imath]= 4\cdot 3\cdot 3 = 36[/imath]. Continuing the same logic, I arrive at an answer of 156. But the answer I have with me is 126. Help appreciated. Thanks in advance. |
545325 | Find a example in [imath]L^2[/imath], nonempty closed but contains no element of smallest norm.
As we know in Hilbert Space, a nonempty closed convex set always contains a element of smallest norm. I wanna find a example to show the convexity in some way is necessary. No loss of generosity, for simplification, let the domain be [imath][0,1]\subset R[/imath]. Thanks. | 92497 | Minimum principle in Hilbert space
Minimum principle is following: Let [imath]M[/imath] be a closed convex nonempty subset of Hilbert space. Then there exists [imath]x\in M[/imath] which have a minimum norm. Assume that [imath]M[/imath] is not convex subset. What is a counterexample, when there have not the element with minimum norm in space [imath]\ell^{2}[/imath]? |
545357 | Is [imath]\left\{ \frac{1}{n}: n \in \mathbb{N} \right\} \cup \left\{ 0\right\}[/imath] closed set?
Is it true that [imath]\left\{ \frac{1}{n}: n \in \mathbb{N} \right\} \cup \left\{ 0\right\}[/imath] is closed set? I suppose that yes, but I have no idea how can I prove it. | 394570 | Prove that [imath] S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)[/imath] is a compact set in [imath]\mathbb{R}[/imath].
Prove that [imath] S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)[/imath] is a compact set in [imath]\mathbb{R}[/imath], but [imath]\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}[/imath] is not a compact set. (Can we use definition of compact set for this problem? ) |
377073 | Calculate the Fourier transform of [imath]b(x) =\frac{1}{x^2 +a^2}[/imath]
I need help to calculate the Fourier transform of this funcion [imath]b(x) =\frac{1}{x^2 +a^2}\,,\qquad a > 0[/imath] Thanks. | 2375324 | Inverse Fourier transform of [imath]\frac{1}{k^2+a^2}[/imath]
I am trying to take the inverse Fourier transform of [imath]\frac{1}{k^2+a^2}[/imath]. I know what the answer is (something like [imath]e^{-|x|}[/imath]), but I'm having trouble actually doing the integral in the inverse Fourier transform. First I noted that [imath]\frac{1}{k^2+a^2}[/imath] is even in [imath]k[/imath], so it's inverse Fourier transform is [imath]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{1}{k^2+a^2} \cos(kx) dk[/imath] Where do I go from here? I tried integration by parts, but that didn't lead anywhere. And looking at the Fourier transform of [imath]e^{-|x|}[/imath] didn't really help. Ideas? |
545467 | Induction: [imath]\sqrt{2\sqrt{3\sqrt{\cdots\sqrt n}}} < 3[/imath]
I'm trying to prove that [imath] \sqrt{2\sqrt{3\sqrt{4\cdots\sqrt{n}}}} < 3 [/imath] for any [imath]n[/imath] and have decided to use strong induction and instead just show that [imath] \sqrt{k\sqrt{(k + 1)\cdots\sqrt{n}}} < k + 1 [/imath] for any choice of [imath]k[/imath]. Does anyone have an idea on where to proceed from here? | 472945 | Prove [imath]\sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}}<3[/imath] by induction.
Problem: prove [imath]\sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}}<3[/imath] by induction. I tried some, but stopped in [imath]\sqrt[2^n]{n+1}[/imath]. Also tried with [imath]2\sqrt{3\cdots}<3^2[/imath] and so on. |
545447 | A linear operator [imath]T: V \rightarrow V[/imath] commuting with all linear operators is a scalar multiple of the identity.
Let [imath]\mathbb{K}[/imath] a field, [imath]V[/imath] a vector space over [imath]\mathbb{K}[/imath]. If [imath]T:V\to V[/imath] commutes with all other linear operators [imath]V \to V[/imath], then there exists [imath]\lambda \in \mathbb{K}[/imath] such that [imath]T= \lambda I[/imath], where [imath]I[/imath] is the identity V. | 27808 | A linear operator commuting with all such operators is a scalar multiple of the identity.
The question is from Axler's "Linear Algebra Done Right", which I'm using for self-study. We are given a linear operator [imath]T[/imath] over a finite dimensional vector space [imath]V[/imath]. We have to show that [imath]T[/imath] is a scalar multiple of the identity iff [imath]\forall S \in {\cal L}(V), TS = ST[/imath]. Here, [imath]{\cal L}(V)[/imath] denotes the set of all linear operators over [imath]V[/imath]. One direction is easy to prove. If [imath]T[/imath] is a scalar multiple of the identity, then there exists a scalar [imath]a[/imath] such that [imath]Tv = av[/imath], [imath]\forall v \in V[/imath]. Hence, given an arbitrary vector [imath]w[/imath], [imath]TS(w) = T(Sw) = a(Sw) = S(aw) = S(Tw) = ST(w)[/imath] where the third equality is possible because [imath]S[/imath] is a linear operator. Then, it follows that [imath]TS = ST[/imath], as required. I am, however, at a loss as to how to tackle the other direction. I thought that a proof by contradiction, ultimately constructing a linear operator [imath]S[/imath] for which [imath]TS \neq ST[/imath], might be the way to go, but haven't made much progress. Thanks in advance! |
545614 | how to integrate sqrt(sin(x)).without using elliptic integration .
How do you solve [imath] \int\sqrt{\sin(t)}dt [/imath] using the substitution method. | 177709 | Indefinite Integral of [imath]\sqrt{\sin x}[/imath]
[imath]\int \sqrt{\sin x} ~dx.[/imath] Does there exist a simple antiderivative of [imath]\sqrt{\sin x}[/imath]? How do I integrate it? |
545539 | Prove, square of quadrilateral is the sum of squares of 4 triangles
Let [imath]A_1[/imath], [imath]B_1[/imath], [imath]C_1[/imath], and [imath]D_1[/imath] - midpoints of the sides [imath]AB[/imath], [imath]BC[/imath], [imath]CD[/imath] and [imath]DA[/imath] convex quadrilateral [imath]AВСD[/imath]. Directs [imath]AC_1[/imath], [imath]ВD_1[/imath], [imath]CA_1[/imath] and [imath]DВ_1[/imath] - divide it by [imath]5[/imath] quadrilaterals and [imath]4[/imath] triangles. Prove that the area of the central quadrilateral is the sum of squares of [imath]4[/imath] triangles. I think, it should be proved in such a way: the square of quadrilateral is finding out by such formula: [imath]S=(p-a)(p-b)(p-c)-abcd\cos^2\left(\frac{A+B}{2}\right)[/imath]. If we will prove that the sum of big triangles are the same as the square of [imath]ABCD[/imath], it will be automatically proved area of the central quadrilateral is the sum of areas of small triangles. | 538258 | Area of quadrangle
In the quadrangle [imath]ABCD[/imath], the points [imath]E,F,G,H[/imath] are the midpoints of respectively [imath]AB, BC, CD, DA[/imath]. We know that area [imath]\triangle AHL=a[/imath], [imath]\triangle DIG=b[/imath], [imath]\triangle FJC=c[/imath], [imath]\triangle EBK=d[/imath]. Prove that area [imath]IJKL=a+b+c+d[/imath]. |
545879 | How to prove that [imath]\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n)[/imath]
I know that [imath]\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n),[/imath] but I cannot find a way how to prove it. I tried induction but it did not work. On wiki they say that I should use differentiation but I do not know how to apply it to binomial coefficient. Thanks for any response. (Presumptive) Source: Theoretical Exercise 1.12(b), P18, A First Course in Pr, 8th Ed, by S Ross | 1431112 | What is the sum [imath]\sum_{k=0}^{n}k^2\binom{n}{k}[/imath]?
What should be the strategy to find [imath]\sum_{k=0}^{n}k^2\binom{n}{k}[/imath] Can this be done by making a series of [imath]x[/imath] and integrating? |
546386 | Show that [imath] \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{n^2+k^2}= \frac{\pi}{4}[/imath]
Show that: [imath] \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{n^2+k^2}= \frac{\pi}{4}[/imath] My idea: I thought that this could be rewritten as an integral, then use trig substitution (perhaps tangent) and then solve to get the answer, but I keep getting stuck. I think I am not setting the problem up correctly. Help would be appreciated. Thanks! :) | 546298 | Show that [imath]\lim_{n\to\infty} \sum_{k=1}^{n} \frac{n}{n^2+k^2}=\frac{\pi}{4}[/imath]
Show that [imath]\lim_{n\to\infty} \sum\limits_{k=1}^{n} \frac{n}{n^2+k^2}=\frac{\pi}{4}[/imath] Using real analysis techniques. |
397978 | Every subsequence of [imath]x_n[/imath] has a further subsequence which converges to [imath]x[/imath]. Then the sequence [imath]x_n[/imath] converges to [imath]x[/imath].
Is the following true? Let [imath]x_n[/imath] be a sequence with the following property: Every subsequence of [imath]x_n[/imath] has a further subsequence which converges to [imath]x[/imath]. Then the sequence [imath]x_n[/imath] converges to [imath]x[/imath]. I guess that it is true but I am not sure how to prove this. | 1029438 | Convergence implied by partial sequences.
I am not sure how to deal with this Question: Show, that the sequence [imath](a_{n})_{n \in \mathbb{N}}[/imath] converges towards the limit [imath]a \in \mathbb{R}[/imath], exactly when every subsequence [imath](a_{n_{k}})_{k \in \mathbb{N}}[/imath] of [imath](a_{n})_{n \in \mathbb{N}}[/imath] owns itself a converging subsequence [imath](a_{n_{k_{j}}})_{j \in \mathbb{N}}[/imath] towards [imath]a[/imath]. To show: [imath]PS(a_{n}):=[/imath] the set of all subsequences of a given sequence. [imath]a_{n}\rightarrow a \Leftrightarrow \forall a_{n_{k}} \in PS(a_{n})\; \exists \; a_{n_{k_{j}}} \in PS(a_{n_{k}}): a_{n_{k_{j}}}\rightarrow a[/imath] It is easy to show the "[imath]\Rightarrow[/imath]" direction, as every subsequence of a given converging sequence, converges towards the same limit. But what about "[imath]\Leftarrow[/imath]"? |
546624 | Find [imath]\int_0^1 \mathrm{\frac{x-1}{ln(x)}}\,\mathrm{d}x[/imath]
Find [imath]\int_0^1 \mathrm{\frac{x-1}{ln(x)}}\,\mathrm{d}x[/imath] I tryed this: [imath]\int_0^1 \mathrm{\frac{x-1}{ln(x)}} = \int_0^1 \mathrm{\frac{x}{ln(x)}} - \int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x[/imath] To [imath]\int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x[/imath] Let [imath]t=lnx [/imath] then [imath]\frac{dt}{dx}=\frac{1}{x}[/imath] and [imath]dx=e^tdt[/imath] [imath]\int \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x \equiv \int \mathrm{\frac{e^t}{t}}[/imath][imath]\mathrm{d}t[/imath] and well [imath]e^t=\sum _{n=0}^\infty \frac{t^n}{n!}[/imath] [imath]\frac{e^t}{t}=\sum_{n=0}^\infty \frac{t^{n-1}}{n!}[/imath] [imath]\int \mathrm{\frac{e^t}{t}}\,\mathrm{d}t=\int \mathrm \sum_{n=0}^\infty{\frac{t^{n-1}}{n!}}\,\mathrm{d}t=\sum_{n=0}^\infty\frac{t^n}{n*(n)!}[/imath] How can I solve [imath]\int_0^1 \mathrm{\frac{x}{ln(x)}}[/imath] ? Thanks for your help :) have a nice day | 461245 | Proving [imath]\int_{0}^1\frac{x-1}{\log x}dx=\log 2[/imath] and [imath]\int_{0}^1\frac{\log x}{x-1}dx=\frac{\pi^2}{6}[/imath]
I would like to know how to prove the following two definite integrals. A: [imath]\int_{0}^1\frac{x-1}{\log x}dx=\log 2[/imath] B:[imath]\int_{0}^1\frac{\log x}{x-1}dx=\frac{\pi^2}{6}[/imath] I found these two by using wolfram-alpha, but I can't prove them. I suspect that the following relations might be used. [imath]\log 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}[/imath]and[imath]\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\sum_{n=1}^\infty\frac{1}{n^2}[/imath] I need your help. |
545957 | New Question: Proving Divergence of a Sequence
After I negated the definition of a convergent sequence, I ended up with the following mathematical statement: [imath]\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon[/imath] Is this correct? I'd like clarification. Anyway, I'm now asked to use my negation of the definition of convergence to prove that [imath](a_n)=((-1)^nn)[/imath] is divergent. Can someone show me how? | 545738 | Proof of Divergence for a Sequence
Previous related question. After I negated the definition of a convergent sequence, I ended up with the following mathematical statement: [imath]\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon[/imath] Is this correct? I'd like clarification... Anyway, I'm now asked to use my negation of the definition of convergence to prove that [imath](a_n)=((-1)^nn)[/imath] is divergent... Am I right in assuming it's enough to prove that there exists at least one [imath]\varepsilon[/imath] such that [imath]|(-1)^nn -\mathscr l|\ge\varepsilon[/imath]? Can I set [imath]\mathscr l[/imath] to be any real number I please? |
540705 | Evaluating [imath] \lim_{x\rightarrow \infty}e^{-x } + 2\cos(3x)[/imath]
Find the limit or prove that it does not exist by [imath]\varepsilon-\delta[/imath] approach: [imath] \lim_{x\rightarrow \infty}e^{-x } + 2\cos(3x)[/imath] Note:I found this question when I was doing exercise from the book Calculus:Early Transcendentals. The book just need me to show it does not exist, but I think it would be interesting to strictly prove it by [imath]\varepsilon-\delta[/imath] language. Update: This question is not duplicate to the other question at all, since I am asking a strictly proof by [imath]\varepsilon-\delta[/imath] approach here rather than just show it does not exist. I also emphasized the requirement in my Note when I post this question. Although those two questions hold the same functions, but they have different requirements.Hence, It actually totally different from another one. | 299094 | Find the limit [imath]\lim\limits_{x\to \infty}(e^{-x}+2\cos3x)[/imath], or show that it does not exist.
I have calculus homework question that to be quite frank I don't even know how to begin to solve. Here is my attempt on how to to solve the question; however I'm not sure if I'm tackling this correctly. I'm hoping that someone here can tell me if I'm doing this correctly and if my answer would be correct. Here is the question. Find the limit or show that it does not exist: [imath]\lim_{x\to \infty}(e^{-x}+2\cos3x)[/imath] My attempt to the solution is DNE or does not exist simply because cosine oscillates between -1 and 1 and it never approaches any one number to reach a horizontal asymptote. Is this answer correct? Or how should I approach this problem if it's not? Or should it be [imath]-\infty[/imath] since that would be the the value of [imath]e^{-x}\quad ?[/imath]Please forgive my stupidity if I'm incorrect in all fronts. |
417582 | Evaluating [imath]\lim_{n \to \infty} (1 + 1/n)^{n}[/imath]
I was recently thinking about how I could evaluate the famous limit of 'e' as I haven't ever seen a proof. I can't really find anything online so I've tried to evaluate the limit myself. And I was also thinking it would be nonsensical to use L'Hopital's rule, am I right? So I did the following: [imath]\lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = \exp \left(\lim_{n \to \infty} n \cdot \ln \left(1 + \frac{1}{n} \right) \right)[/imath] [imath]=\lim_{n \to \infty} \exp \left( n \cdot \left( \frac{1}{n} - \frac{(1/n)^{2}}{2} + \cdots \right) \right)[/imath] [imath]= \lim_{n \to \infty} \exp \left( 1 - \frac{(1/n)}{2} + \cdots \right)[/imath] [imath]= e[/imath] I am not sure, is my logic correct or does it create circularity by taking logarithms and assuming [imath]f(x) = \exp \left( \ln f(x) \right)[/imath] ? | 96606 | How to prove [imath]\lim\limits_{n \to \infty} (1+\frac1n)^n = e[/imath]?
How to prove the following limit? [imath]\lim_{n \to \infty} (1+1/n)^n = e[/imath] I can only observe that the limit should be a very large number! Thanks. |
546954 | Deriving ¬R from {R↔(R∨(P∧¬P)),R↔¬P,¬P→(P↔(Q→Q)),P→Q}
Give a derivation of [imath]\neg R[/imath] from the following premises: [imath]\{R\leftrightarrow(R\lor(P\land \neg P)), R\leftrightarrow\neg P, \neg P\to(P\leftrightarrow(Q\to Q)), P\to Q\}[/imath] using the rules found here. My assignment is due today, if someone can help me out. ty Is this correct? I know some dependence columns are wrong. 1 1) R↔(R/lor(P∧¬P)) P 2 2) R↔¬P P 3 3) ¬P→(P↔(Q→Q)) P 4 4)P→Q P 5 5)R P 2 6)R→¬P 2,BE 2,5 7)¬P 5,6,MP 2,3 8)P↔(Q→Q) 3,7,MP 2,3 9)(Q→Q)→P 8,BE 2,3,5 10)¬(Q→Q) 7,9,MT 2,3,5 11)Q∧¬Q 10,NC 2,3,5 12)Q 11,simp 2,3,5 12)¬Q 11,simp 2,3 14)¬R 12,13,RAA 2,15)¬R 5,14,RAA | 546585 | Deriving [imath]\neg R[/imath] from [imath]\{R↔(R∨(P∧¬P)), R↔¬P, ¬P→(P↔(Q→Q)), P→Q\}[/imath]
Give a derivation of [imath]\neg R[/imath] from the following premises: [imath]\{R\leftrightarrow(R\lor(P\land \neg P)), R\leftrightarrow\neg P, \neg P\to(P\leftrightarrow(Q\to Q)), P\to Q\}[/imath] using the rules found here. I don't know where to start... Edit: 1 1) R↔(R/lor(P∧¬P)) P 2 2) R↔¬P P 3 3) ¬P→(P↔(Q→Q)) P 4 4)P→Q P 5 5)R P 2 6)R→¬P 2,BE 2,5 7)¬P 5,6,MP 2,3 8)P↔(Q→Q) 3,7,MP 2,3 9)(Q→Q)→P 8,BE 2,3,5 10)¬(Q→Q) 7,9,MT 2,3,5 11)Q∧¬Q 10,NC 2,3,5 12)Q 11,simp 2,3,5 12)¬Q 11,simp 2,3 14)¬R 12,13,RAA 2,15)¬R 5,14,RAA |
547129 | Is a left invertible element of a ring necessarily right invertible?
Let [imath]R[/imath] be a unitary ring and let [imath]a,b \in R[/imath] such that [imath]ab=1[/imath]. Does it imply that [imath]a[/imath] is invertible? The definition of invertible element requires that [imath]ab = ba = 1[/imath], so I guess it doesn't imply that a is invertible, but I didn't manage to find a unitary ring in which this happens. | 70777 | A ring element with a left inverse but no right inverse?
Can I have a hint on how to construct a ring [imath]A[/imath] such that there are [imath]a, b \in A[/imath] for which [imath]ab = 1[/imath] but [imath]ba \neq 1[/imath], please? It seems that square matrices over a field are out of question because of the determinants, and that implies that no faithful finite-dimensional representation must exist, and my imagination seems to have given up on me :) |
239364 | Surjective endomorphisms of finitely generated modules are isomorphisms
My Problem: Let [imath]M[/imath] be a finitely generated [imath]A[/imath]-module and [imath]T[/imath] an endomorphism. I want to show that if [imath]T[/imath] is surjective then it is invertible. My attempt: Let [imath]m_1,...,m_n[/imath] be the generators of [imath]M[/imath] over [imath]A[/imath]. For every [imath]b = b_1 m_1 + ... + b_n m_n[/imath] with [imath]b_i \in A[/imath] there is [imath]a = a_1 m_1 + ... + a_n m_n[/imath] with [imath]a_i \in A[/imath] such that [imath] T(a)=b [/imath] or in matrix-vector notation [imath] T \vec{a} = \vec{b} [/imath] where [imath]\vec{x}[/imath] is the column vector of [imath]x_1,...,x_n[/imath] where [imath]x = x_1 m_1 + ... + x_n m_n[/imath]. I multiply by the adjugate matrix to get [imath] \mathrm{adj}(T) \vec{b} = \mathrm{adj}(T) T \vec{a} = \det(T) I_n \vec{a} = \det(T) \vec{a} \ . [/imath] Now take [imath]\vec{b}=0[/imath]. Then [imath]\vec{0} = \det(T) \vec{a}[/imath] and thus [imath]T[/imath] is injective if and only if [imath]\det(T)[/imath] is not a zero divisor. If I prove that [imath]T[/imath] is injective, then I'll get it is invertible. For that, I think the way is to prove that [imath]\det(T)[/imath] is not a zero divisor. The importance of finitely generated condition: Let [imath]M = A^{\aleph_0} =\{ ( a_1 , a_2 , ... ) \mid a_i \in A \}[/imath] be a not finitely generated [imath]A[/imath]-module. Let [imath]T : M \to M[/imath] defined by [imath] T(a_1, a_2, a_3, ... ) = (a_2, a_3, ... ) \ . [/imath] Then clearly [imath]T[/imath] is surjective but not injective ([imath]\ker T = \{ ( a , 0 , 0 , ... ) \mid a \in A \}[/imath]), and thus not invertible. The importance of surjective and not injective condition: Need to find a counter-example. | 269483 | Suppose R is a commutative ring, [imath]A \in R_n[/imath], and the homomorphism [imath]f: R^n \to R^n[/imath] defined by [imath]f(b) = Ab[/imath] is surjective. Show [imath]f[/imath] is an isomorphism.
How would I go about showing this? Suppose [imath]R[/imath] is a commutative ring, [imath]A \in R_n[/imath], and the homomorphism [imath]f: R^n \to R^n[/imath] defined by [imath]f(b) = Ab[/imath] is surjective. Show [imath]f[/imath] is an isomorphism. (Edit: [imath]R_n[/imath] is the ring of [imath]n\times n[/imath] matrices.) I'm thinking that since [imath]f[/imath] is surjective on [imath]R^n \to R^n[/imath], for every [imath]x \in R^n[/imath], [imath]x = Ab[/imath], so that [imath]f^{-1}(x) = A^{-1}x[/imath]. And since the inverse of any element in a ring is nonzero, [imath]\ker(f) = 0[/imath]. I'm not terribly confident with my answer though. Thanks! |
546081 | Find all real roots of [imath]x^5+10x^3+20x - 4=0[/imath]
Find all real roots of the following fifth-degree equation [imath]x^5+10x^3+20x-4=0.[/imath] In fact, the only one root is [imath]\sqrt[5]8-\sqrt[5]4[/imath]. How to get that? | 545651 | How solve this equation
Find the equation [imath]x^5+10x^3+20x-4=0[/imath] My try:I think this equation maybe take Trigonometric functions Now I have solution:let [imath]x=t-\dfrac{2}{t}[/imath],then [imath]x^5+10x^3+20x-4=(t-\dfrac{2}{t})^5+10(t-\dfrac{2}{t})^3+20(t-\dfrac{2}{t})-4=0[/imath] so [imath]t^5-4-\dfrac{32}{t^5}=0[/imath] My question: why we take [imath]x=t-\dfrac{2}{t}[/imath] and This problem have other solution? Thank you |
547480 | Lp optimality proof
i have a general question. if there is a general LP problem [imath]c^Tx[/imath] s.t [imath]A\cdot x \le b[/imath], and [imath]x \ge 0[/imath] and assuming that the components of [imath]c[/imath] are non-zero entries then how can I prove that when [imath]x[/imath] satisfies [imath]a\cdot x < b[/imath] (notice the strict inequality here) and [imath]x > 0[/imath], then [imath]x[/imath] cannot be an optimal solution? | 544659 | Lp optimal solution question
i have a general question. if there is a general LP problem [imath]c^Tx[/imath] s.t [imath]A\cdot x \le b[/imath], and [imath]x \ge 0[/imath] and assuming that the components of [imath]c[/imath] are non-zero entries then how can I prove that when [imath]x[/imath] satisfies [imath]a\cdot x < b[/imath] (notice the strict inequality here) and [imath]x > 0[/imath], then [imath]x[/imath] cannot be an optimal solution? |
547578 | Showing the simplifying steps of this equality ...
Can someone please show me how these are equivalent in steps[imath]\frac{(h_2^2 - h_3^2 )}{\dfrac{1}{h_3}-\dfrac{1}{h_2}}=h_2 h_3 (h_2+h_3)[/imath] I thought it simplifies to [imath](h_2^2-h_3^2)(h_3-h_2)[/imath]This would be much appreciated, I cant wrap my head around it. | 547586 | simplifying [imath] (h^2_2-h^2_3)\over ({1\over h_3})-({1\over h_2})[/imath]
Can someone please show me how these are equivalent in steps [imath]\frac{(h_2^2 - h_3^2 )}{\dfrac{1}{h_3}-\dfrac{1}{h_2}}=h_2 h_3 (h_2+h_3)[/imath] I thought it simplifies to [imath](h_2^2-h_3^2)(h_3-h_2)[/imath]This would be much appreciated, I cant wrap my head around it. |
547600 | Automorphism on [imath]\mathbb{Q}_p[/imath]
How many automorphism are there in the field of [imath]p[/imath]-adic? % I suspect that there's only one automorphism, the Identity. But I stuck with the continuously of the automorphism in [imath]\mathbb{Q}_p[/imath]. | 449424 | An automorphism of the field of [imath]p[/imath]-adic numbers
Is an automorphism of the field [imath]\mathbb{Q}_p[/imath] of [imath]p[/imath]-adic numbers the identity map? If yes, how can we prove it? Note:We don't assume an automorphism of [imath]\mathbb{Q}_p[/imath] is continuous. |
547118 | Existence of an injection
Let [imath]A[/imath] and [imath]B[/imath] be two sets. Prove the existence of an injection from [imath]A[/imath] to [imath]B[/imath] or an injection from [imath]B[/imath] to [imath]A[/imath]. I don't know how to proceed, since I don't have any information on [imath]A[/imath] or [imath]B[/imath] to begin with. Does anybody have a hint ? | 421638 | Proving [imath](A\le B)\vee (B\le A)[/imath] for sets [imath]A[/imath] and [imath]B[/imath]
For any pair of sets [imath]A[/imath] and [imath]B[/imath], we can define [imath]A\le B[/imath] iff there exists injection [imath]f\colon A\rightarrow B[/imath]. I am trying prove that [imath](A\le B)\vee (B\le A).[/imath] I have tried assuming [imath]\neg (A\le B)[/imath], then proving [imath]B\le A[/imath] by constructing the required injection, but I haven't been able to make any progress. Any hints, etc. would be appreciated. EDIT Assuming [imath]\neg (A\le B)[/imath], can you prove there exists a surjection [imath]f: A\rightarrow B[/imath]? Then it would be easy, by applying AC, to construct an injection [imath]g: B\rightarrow A[/imath] |
547771 | Induction: [imath]2^n = \sum_{v=0}^{n} \binom{n}{v}[/imath]
I have to prove the following identity for [imath]n \in \mathbb{N}[/imath]: [imath]\displaystyle 2^n = \sum_{v=0}^{n} \binom{n}{v}[/imath] Is there a way to show it through induction? Or is there a easier way? My steps so far: [imath]\displaystyle n=2: 2^2=\sum_{v=0}^{2} \binom{n}{v} \Rightarrow 2 \cdot 2 = \binom{0}{2} + \binom{1}{2} + \binom{2}{2} \Rightarrow 4 = 1 + 2 +1 \checkmark[/imath] Now we suppose that the idenity is true for [imath]n \in \mathbb{N}[/imath], then it has to be true for [imath]n+1 \in \mathbb{N}[/imath], too: [imath]\displaystyle n \to n+1: 2^{n+1} = \sum_{v=0}^{n+1} \binom{n}{v} \Rightarrow 2^n \cdot 2 = \sum_{v=0}^{n} \binom{n}{v}+\binom{n+1}{v} \Rightarrow \ldots ?[/imath] | 295802 | Induction proof of [imath]\sum_{k=1}^{n} \binom n k = 2^n -1 [/imath]
Prove by induction: [imath]\sum_{k=1}^{n} \binom n k = 2^n -1 [/imath] for all [imath]n\in \mathbb{N}[/imath]. Today I wrote calculus exam, I had this problem given. I have the feeling that I will get [imath]0[/imath] points for my solution, because I did this: Base Case: [imath]n=1[/imath] [imath]\sum_{k=1}^{1} \binom 1 1 = 1 = 2^1 -1 .[/imath] Induction Hypothesis: for all [imath]n \in \mathbb{N}[/imath]: [imath]\sum_{k=1}^{n} \binom n k = 2^n -1 [/imath] Induction Step: [imath]n \rightarrow n+1[/imath] [imath]\sum_{k=1}^{n+1} \binom {n+1} {k} = \sum_{k=1}^{n} \binom {n+1} {k} + \binom{n+1}{n+1} = 2^{n+1} -1[/imath] Please show me my mistake because next time is my last chance in this class. |
547648 | Proving a group is cyclic
Let [imath]G[/imath] be a group of order [imath]pq[/imath], where [imath]p,q[/imath] are primes, [imath]p < q[/imath] and [imath]q≢1[/imath] (mod [imath]p[/imath]); how do we prove that [imath]G[/imath] is cyclic ? (I have no idea) | 67129 | Groups of order [imath]pq[/imath] without using Sylow theorems
If [imath]|G| = pq[/imath], [imath]p,q[/imath] primes, [imath]p \gt q, q \nmid p-1 [/imath], then how do I prove [imath]G[/imath] is cyclic without using Sylow's theorems? |
28940 | Why is [imath]\infty \cdot 0[/imath] not clearly equal to [imath]0[/imath]?
I did a bit of math at school and it seems like an easy one - what am I missing? [imath]n\times m = \underbrace{n+n+\cdots +n}_{m\text{ times}}[/imath] [imath]\quad n\times 0 = \underbrace{0 + 0 + \cdots+ 0}_{n\text{ times}} = 0[/imath] (i.e add [imath]0[/imath] to [imath]0[/imath] as many times as you like, result is [imath]0[/imath]) So I thought an infinite number of [imath]0[/imath]'s cannot be anything but [imath]0[/imath]? But someone claims different but couldn't offer a reasonable explanation why. Google results seemed a bit iffy on the subject - hopefully this question will change that. | 698690 | When 0 is multiplied with infinity, what is the result?
Any number multiplied by [imath]0[/imath] is [imath]0[/imath]. Any number multiply by infinity is infinity or indeterminate. [imath]0[/imath] multiplied by infinity is the question. Answer with proof required. |
549586 | Mathematical Induction Factorials, sum r(r!) =(n+1)! -1
How do I prove that [imath]\sum\limits_{r=1}^{n} r(r!) = (n+1)!-1[/imath] I was able to get to factor: [imath]LHS = k(k!) + (k+1)(k+1)![/imath] [imath]\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\, RHS = (k+2)! -1[/imath] | 301615 | Prove by Mathematical Induction: [imath]1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1[/imath]
Prove by Mathematical Induction . . . [imath]1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1[/imath] I tried solving it, but I got stuck near the end . . . a. Basis Step: [imath](1)(1!) = (1+1)!-1[/imath] [imath]1 = (2\cdot1)-1[/imath] [imath]1 = 1 \checkmark[/imath] b. Inductive Hypothesis [imath]1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1[/imath] Prove k+1 is true. [imath]1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1[/imath] [imath]\big[RHS\big][/imath] [imath](k+2)!-1 = (k+2)(k+1)k!-1[/imath] [imath]\big[LHS\big][/imath] [imath]=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)![/imath] (Explicit Last Step) [imath]= \underbrace{(k+1)!-1}+(k+1)(k+1)![/imath] (Inductive Hypothesis Substitution) [imath]= (k+1)!-1 + (k+1)(k+1)k![/imath] [imath]= (k+1)k!-1 + (k+1)^{2}k![/imath] My [LHS] looks nothing like my [RHS] did I do something wrong? EDIT: [imath] = (k+1)k! + (k+1)^2k! -1 [/imath] [imath] = (k+1)(k!)(1 + (k+1))-1[/imath] [imath] = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1[/imath] |
550151 | Show that [imath]f(E-F) \not\subseteq f(E)-f(F)[/imath]
Can someone help explain this question to me? Let [imath]f : X \to Y[/imath] be a function and [imath]E \subseteq X[/imath] and [imath]F \subseteq X[/imath]. Show that in general: 1) [imath]f(E-F) \not\subseteq f(E)-f(F)[/imath] 2) [imath]f(E \bigtriangleup F) \not\subseteq f(E)\bigtriangleup f(F)[/imath] | 548594 | Let [imath]f : X \to Y[/imath] be a function and [imath]E \subseteq X[/imath] and [imath]F \subseteq X[/imath]. Show that in general
Let [imath]f:X\to Y[/imath] be a function and [imath]E\subseteq X[/imath] and [imath]F\subseteq X[/imath]. Show that in general [imath]f(E − F)\nsubseteq f(E) − f(F)[/imath]. I have no idea about how to prove this; and could anyone please explain the basic theory of functions(by relating to this question. Thank you. |
550256 | Stirling numbers of second type
How can I do a combinatoric proof that for Stirling number of second type the equality if true: [imath]{n\brace k} = \frac{1}{k!}\sum_{i=0}^{k}{k \choose i}i^n(-1)^{k-i}[/imath] | 436719 | Stirling numbers combinatorial proof
This is a Homework Question. I am required to give a Combinatorial proof for the following. [imath]S(m,n)=\frac 1{n!} \sum_{k=0}^{n} (-1)^k\binom nk (n-k)^m[/imath] Hint given is : Show that [imath]n!S(m,n)[/imath] equals the number of onto functions [imath]f\colon A \rightarrow B[/imath] when [imath] |A|=m[/imath] and [imath]|B|=n. [/imath] There were some other combinatorial proof questions on the assignment which I found easier to do but this one not so much. Could use help. |
550571 | How do I prove that [imath]x^2\equiv-1\pmod{p}[/imath]
How do I prove that [imath]x^2\equiv-1\pmod{p}[/imath] iff [imath]p[/imath] is prime at form: [imath]p=4n+1[/imath]. I have to use Wilson theroem... (I'm asking this, becuase I didn't understand it from my prev. Q, how it proves that there is a solution...). Thank you! | 547191 | Proving something with Wilson theorem
I need to prove that [imath]x^2\equiv -1\pmod p[/imath] if [imath]p=4n+1[/imath]. ([imath]p[/imath] is prime of course...) I need to use Wilson theorem. |
536753 | Showing the Irreducibility of a Polynomial and "Splitting" the Field
The following problem is part of my Algebra problem set: Let [imath]\alpha= \sqrt[3]{2}+\omega[/imath] ,where [imath]\omega=e ^{2\pi i /3} = -\frac{1}{2} + \frac{\sqrt{-3}}{2}[/imath] is a primitive cube root of [imath]1[/imath]. [imath]\alpha[/imath] is a zero of [imath]f(x) = 9 + 9x +3x^3 + 6x^4 + 3x^5 + x^6[/imath]. Show that [imath]f(x)[/imath] is irreducible. ** Eisenstein's Criterion fails here I realize that Gauss's Lemma is the best method of proving this although I am unclear as to whether proving the irreducibility of a polynomial in a subfield is sufficient to prove that it is irreducible. ** Please Also note that we have not had any Galois Theory. Show that [imath]\mathbb{Q}(\alpha)[/imath] is the smallest field which contains all three zeros of [imath]x^3 -2 [/imath] (this is called the splitting field of [imath]x^3 -2[/imath]). Thank you in advance for your help! | 536985 | Show that [imath]9+9x+3x^3+6x^4+3x^5+x^6[/imath] is irreducible given one of its roots
Given a polynomial [imath]f(x)=9+9x+3x^3+6x^4+3x^5+x^6[/imath] and one of its roots [imath]\alpha=2^{1/3}+e^{2\pi i/3}[/imath]. Show that [imath]f(x)[/imath] is irreducible in [imath]\mathbb Q[x][/imath]. Eisenstein's criterion fails, it also didn't use the fact that [imath]\alpha[/imath] is a root of [imath]f(x)[/imath]. How should I approach? Thanks |
550891 | A set dense in [imath]L^2(\mathbb{R})[/imath]
I am curious about the following problem: Can we show that the set [imath]\{p(x)\cdot\exp(-x^2/2)\}\subseteq L^2(\mathbb{R})[/imath], where [imath]p(x)[/imath] is any real-domained polynomial function (i.e., functions in the form of the product between a polynomial function and the exponential function [imath]\exp(-x^2/2)[/imath]), is dense in [imath]L^2(\mathbb{R})[/imath]? I am confident in this claim, but fail to prove it. The barrier to me is how to properly use the Stone-Weierstrass theorem (if we need to use it), as it only works on a compact set. Can anyone give me some hint or suggestions? Many thanks! | 355901 | proof for a basis in [imath]L^2[/imath]
I know, correct me if I am wrong, that the functions [imath]H_n(x)\exp(-x^2/2)[/imath] form a complete basis in [imath]L^2(\mathbb{R},dx)[/imath], where [imath]H_n(x)[/imath] is the [imath]n[/imath]th Hermite polynomial. This must be true also for [imath]x^n\exp(-x^2/2)[/imath] with [imath]n\in\mathbb{N}_0[/imath]. Does someone know a proof of the latter or can give me a reference? Since I am not a mathematician, I will really appreciate it if the proof contains all the details that might puzzle a non-mathematician. I also have the problem to choose functions [imath]f_k[/imath] such that \begin{equation} \int_{-\infty}^{+\infty}f_k(x)x^n\exp(-x^2)dx = \delta_{kn} \mbox{.} \end{equation} Does anyone know a method to construct the [imath]f_k[/imath]s? |
551807 | [imath]Y^\varnothing[/imath] has one element; if [imath]X \ne \varnothing[/imath], then [imath]\varnothing^X = \varnothing[/imath]
From Section 8 of Halmos' Naive Set Theory. (i) Show that [imath]Y^\emptyset[/imath] has exactly one element, namely [imath]\emptyset[/imath]. (ii) Show that if [imath]X\neq\emptyset[/imath], then [imath]\emptyset^X=\emptyset[/imath]. I'm not completely sure I understand what this means nor how to prove it. For (i), we want to show that the set of functions [imath]f:\emptyset\to X[/imath] is exactly [imath]\{\emptyset\}[/imath]. To me, this means that there is one "map", the "empty map", which associates nothing to something. I'm not really sure how to show this. We know that [imath]Y^\emptyset\subseteq\mathcal P(\emptyset \times Y)=\{\emptyset\}[/imath]. Then we can show that [imath]\{\emptyset\}\subseteq Y^\emptyset[/imath] to obtain the equality. So this is vacuously true? For (ii), since [imath]X\neq\emptyset[/imath], to me this exercise means that we're showing that there are no maps which associate something to nothing. But how to show that there is no function [imath]X\to\emptyset[/imath]? Is this just immediate from the definition of function? | 505379 | If [imath]A \ne \varnothing[/imath], then [imath]\varnothing^A=\varnothing[/imath]
Prove that if [imath]A\ne\varnothing[/imath], then [imath]\varnothing^A=\varnothing[/imath]. Here, [imath]\varnothing^A=\{f\in \mathcal P(A\times \varnothing): f:A \to \varnothing\}[/imath] is the set of functions from [imath]A[/imath] to [imath]\varnothing[/imath]. What if [imath]A = \varnothing[/imath]? Proof: Assume [imath]f \in \varnothing^A[/imath] Then [imath]f \subseteq A \times \varnothing = \varnothing[/imath] Thus [imath]f = \varnothing[/imath] Now we are trying to show that [imath]\varnothing \in \varnothing^A[/imath] [imath]\varnothing \subseteq A \times \varnothing[/imath]. Thus [imath]\varnothing \in \mathcal P(A \times \varnothing)[/imath]. and [imath]\operatorname{dom}(\varnothing)=\varnothing=A\times \varnothing[/imath] And [imath]\varnothing[/imath] is a function, since if not then there exist [imath]x,y,z[/imath] such that [imath](x,y), (x,z)\in \varnothing[/imath] and [imath]y\ne z[/imath]. But this contradicts that [imath]\varnothing[/imath] has no elements. Thus [imath]\varnothing[/imath] is a function. [imath]\varnothing^\varnothing=\{\varnothing\}[/imath] Am I doing it right? |
552547 | How prove that [imath]\sum_{n=1}^{\infty}\frac{\pi{(n)}}{n^2}<\infty[/imath] for some bijective [imath]\pi(n)[/imath]
Let [imath]\mathbb{N}=\{1,2,\cdots\}[/imath]. Does there exist a bijective function [imath]\pi:\mathbb{N} \to \mathbb{N}[/imath] such that [imath]\sum_{n=1}^{\infty}\dfrac{\pi{(n)}}{n^2}<\infty ?[/imath] My try: note [imath]\sum_{n=1}^{\infty}\dfrac{1}{n^2} [/imath] is convergent, and then I can't. Thank you. | 2120 | Does there exist a bijective [imath]f:\mathbb{N} \to \mathbb{N}[/imath] such that [imath]\sum f(n)/n^2[/imath] converges?
We know that [imath]\displaystyle\zeta(2)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}[/imath] and it converges. Does there exists a bijective map [imath]f:\mathbb{N} \to \mathbb{N}[/imath] such that the sum [imath]\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^2}[/imath] converges. If our [imath]s=2[/imath] was not fixed, then can we have a function such that [imath]\displaystyle \zeta(s)=\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^s}[/imath] converges |
552620 | Showing the limit of a sup
Let [imath]f: \mathbb{R}\to\mathbb{R}[/imath] be a bounded function and suppose that [imath]g(x)=\sup \left(f(t)\right)[/imath] for [imath]t>x[/imath]. Show that [imath]\lim_{x\to a^+} g(x) = g(a)[/imath] for all real [imath]a[/imath]. I'm not sure how to start this. Do we want to show that [imath]\limsup f(t) = g(a)[/imath]? If so, is it possible to say that since we are evaluating a right sided limit, we then want to find a [imath]\sup f(a^+)[/imath] since [imath]a^+>a[/imath] which is then [imath]g(a)[/imath] since [imath]g(x)=\sup \left(f(t)\right)[/imath]? How can I write this more concretely? | 552538 | Show that [imath]\lim_{x\to a^{+}} g(x) =g(a)[/imath]
Let [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] be a bounded function and suppose that [imath]g(x)=\sup_{t>x}f(t)[/imath]. Show that [imath]\lim_{x\to a^{+}}g(x) = g(a)[/imath] for all real [imath]a[/imath]. I have a hard time with these kind of proofs because in high school, I would simply replace [imath]x[/imath] by [imath]a[/imath] in [imath]g(x)[/imath] to get that [imath]g(x) \rightarrow g(a)[/imath]. Obviously, Real Analysis requires a more formal proof. I can always apply the definition of a limit of a function : Given [imath]\varepsilon>0[/imath], there exists [imath]\delta>0[/imath] such that if [imath]x\in A[/imath] and [imath]0<|x-c|<\delta[/imath], then [imath]|f(x)-L| < \varepsilon[/imath]. In particular, we have that if [imath] |x-c| = |x-a^{+}|<\delta[/imath] then [imath]|g(x)-g(a)| < \varepsilon[/imath] How can I find a relation between [imath]f[/imath] and [imath]g[/imath], the sup of [imath]f(t)[/imath] and between [imath]x[/imath] and [imath]a^{+}[/imath]? |
552588 | Help in a question of Hartshorne's algebraic geometry book
I'm trying to solve this question: Let [imath]\psi:\mathbb P^1\times \mathbb P^1\to \mathbb P^3[/imath] be the Segre embedding given by [imath]\psi([a_0:a_1],[b_0:b_1])\to [a_0b_0:a_0b_1:a_1b_0:a_1b_1][/imath]. This corresponding to the ring homomorphism [imath]\theta:k[z_{00},z_{01},z_{10},z_{11}]\to k[x_0,x_1,y_0,y_1][/imath] which maps [imath]z_{ij}[/imath] to [imath]x_iy_j[/imath] (see question 2.14 Hartshorne). I need help to prove that [imath]\ker\theta=\langle z_{00}z_{11}-z_{01}z_{10}\rangle[/imath]. If I prove that, I think I solve the question. Thanks a lot. | 353846 | Hartshorne Problem 1.2.14 on Segre Embedding
This is a problem in Hartshorne concerning showing that the image of [imath]\Bbb{P}^n \times \Bbb{P}^m[/imath] under the Segre embedding [imath]\psi[/imath] is actually irreducible. Now I have shown with some effort that [imath]\psi(\Bbb{P}^n \times \Bbb{P}^m)[/imath] is actually equal to [imath]V(\mathfrak{a})[/imath] where [imath]\mathfrak{a}[/imath] is the ideal generated by the set of all monomials [imath]\Big\{z_{ij}z_{kl} - z_{il}z_{kj} \hspace{1mm} \Big| \hspace{1mm} i,k = 0,\ldots, n; \hspace{2mm} j,l = 0,\ldots,m\Big\}.[/imath] My main problem now is in showing that [imath]\mathfrak{a}[/imath] is actually equal to the kernel of the ring homomorphism [imath]\varphi : k[z_{ij}] \to k[x_0,\ldots,x_n,y_0,\ldots,y_m][/imath] that sends [imath]z_{ij}[/imath] to [imath]x_iy_j[/imath]. I have spent quite a few hours playing around with monomial orderings and trying to show that [imath]\mathfrak{a} \supseteq \ker \varphi[/imath] but to no avail. Of course the other inclusion is immediate. Is there anything I can do apart from playing around with monomial orderings to try and show that the kernel of [imath]\varphi[/imath] is equal to [imath]\mathfrak{a}[/imath]? Perhaps maybe something along the lines of inducting on [imath]n[/imath], those this does not look promising. Note: Please do not close this question; my question is different from the other questions on this site concerning the Segre embedding. Also, I can show my work concerning how I arrived at the conclusion that [imath]V(\mathfrak{a}) = \psi(\Bbb{P}^n \times \Bbb{P}^n)[/imath]. |
551986 | Prove that if [imath]\chi(G) > k \wedge |\{\operatorname{ad}(H): \text{[/imath]H[imath] is an induced subgraph of [/imath]G[imath]}\}|[/imath] then [imath]G[/imath] has a [imath]k-regular[/imath] induced subgraph?
I have found an interesting exercise in my introduction to graphs workbook: Let [imath]\operatorname{ad}(G) = \frac{2\vert E(G)\vert}{\vert V(G) \vert}[/imath] and [imath]\operatorname{mad}(G) = \max\{\operatorname{ad}(H): \text{[/imath]H[imath] is an induced subgraph of [/imath]G[imath]}\}[/imath]. Given a graph [imath]G[/imath] and a number [imath]k \geq 3[/imath], prove that if [imath]\chi(G) > k \wedge \operatorname{mad}(G) \leq k \Rightarrow \exists A \subseteq V(G)[/imath] such that [imath][A]_{G}[/imath] is k - regular. I believe this can proved by contradiction: Given a [imath]G[/imath] and [imath]k[/imath] for which the properties hold, we assume that [imath]G[/imath] has no induced subgraph [imath]H[/imath] that is k - regular. We then prove that if [imath]\nexists H[/imath] then at least one of the property is false and thus we contradict ourselves however I can't quite figure out the connection between [imath]H[/imath] and the mentioned properties. How would you go about solving this? | 549962 | Regular graph (Homework)
Let [imath]G = (V, E)[/imath] be a graph and [imath]ad(G) = \frac{2|E|}{|V|}[/imath] the average degree of [imath]G[/imath]. [imath] mad(G) = max ( ad(H) : H \le G ) \text{ the maximum average degree of a subgraph of $G$} [/imath] We know that [imath]\chi(G)\gt k[/imath]. Proof that [imath]G[/imath] has a [imath]k-regular[/imath] subgraph. I've been chewing this for quite a while. Any pointers that may lead me to a rigorous proof ? EDIT : [imath]mad(G) \le k[/imath] and [imath]k \ge 3[/imath]. I left it out by accident. |
552982 | Newton method to find [imath]\frac{1}{\sqrt{a}}[/imath]
What function should I use to find [imath]\frac{1}{\sqrt{a}}[/imath] without using division? | 550177 | Using Newton's method calculate [imath] \frac{1}{\sqrt{a}} [/imath] without division
Suggest algorithm for the numerical calculation [imath] \frac{1}{\sqrt{a}} \ a > 0 [/imath] without division, use Newton's method. My idea is: [imath] \frac{1}{\sqrt{a}} = (\sqrt{a})^{-1} = a^{-\frac{1}{2}}[/imath] [imath] x = a^{-\frac{1}{2}} \Rightarrow f(x)=x^2 - a^{-1}[/imath] And next step is calulate [imath] f'(x)=2x [/imath], but after substitution I got: [imath] x_{n+1} = \frac{x_n^2+a^{-1}}{2x_n}[/imath] And know I still have divion, I guess. Maybe you've better idea on this task? |
553138 | how [imath]\pi[/imath] is irrational if it is a ratio
How can [imath]\pi[/imath] be an irrational number if it is a ratio of the circumference over the diameter? Thanks! | 184675 | Why is [imath]\pi[/imath] irrational if it is represented as [imath]c/d[/imath]?
[imath]\pi[/imath] can be represented as [imath]C/D[/imath], and [imath]C/D[/imath] is a fraction, and the definition of an irrational number is that it cannot be represented as a fraction. Then why is [imath]\pi[/imath] an irrational number? |
553202 | Proof of statement
I want to prove this statement: [imath](A_1 \cup A_2)^c = {A_1}^c \cap {A_2}^c[/imath] What I have realized so far, is that [imath](A_1 \cup A_2)^c \implies x \not\in A_1 \textrm{ and } x \not\in A_2 \implies x \not\in (A_1^c \cap A_2^c)[/imath] (Though I'm not sure if my approach is correct since I'm stuck) but I am not sure how to proceed. Any help would be greatly appreciated. | 551994 | prove a statement (complements of unions)
I want to prove this statement: [imath](A_1 \cup A_2)^c = {A_1}^c \cup {A_2}^c[/imath] where the [imath]c[/imath] means the complement. Any help would be greatly appreciated. |
218512 | Show that a function that is locally increasing is increasing?
A function [imath]f : \mathbb{R} \to \mathbb{R}[/imath] is locally increasing at a point [imath]x[/imath] if there is a [imath]\delta > 0[/imath] such that [imath]f(s) < f(x) < f(t)[/imath] whenever [imath]x-\delta < s < x < t < x+\delta[/imath]. Show that a function that is locally increasing at every point in [imath]\mathbb{R}[/imath] must be increasing, i.e., [imath]f(x) < f(y)[/imath] for all [imath]x < y[/imath]. | 1821774 | To prove that if a function [imath]f : \mathbb R \to \mathbb R[/imath] is locally increasing at every point then the function is increasing
I am trying to prove that if [imath]f : \mathbb R \to \mathbb R[/imath] is locally increasing at every point , i.e. if for every [imath]x\in \mathbb R , \exists r_x >0[/imath] such that [imath]f(a)\ge f(b) , [/imath] whenever [imath] x+r_x >a>b>x-r_x [/imath] , then [imath]f[/imath] is increasing everywhere . I have proceeded as : Let [imath]a,b \in \mathbb R , a<b[/imath] ; we want to show [imath]f(a) \le f(b)[/imath] ; now for each [imath]p \in [a,b][/imath] , there is an open n.b.d. of [imath]p[/imath] in which [imath]f[/imath] is increasing , so by these open intervals , we get an open cover of [imath][a,b][/imath] and since [imath][a,b][/imath] is compact we can get a finite subcover ; but then I am stuck ; can the proof be finished in this way ? Or is there any better way ? Please help . Thanks in advance |
553341 | Let p be an odd prime number. Prove that
[imath]\left(\frac{1\cdot2}{p}\right) + \left(\frac{2\cdot3}{p}\right) + \left(\frac{3\cdot4}{p}\right) +\ldots+ \left(\frac{(p-2)(p-1)}{p}\right) = -1[/imath] Note: [imath]\left(\frac{a}{b}\right)[/imath] represents the Legendre Symbol. I have tried using this method. For each [imath]k[/imath] between [imath]1[/imath] and [imath]p-2[/imath] denote by [imath]k'[/imath] its multiplicative inverse [imath]\mod p[/imath]. To estimate the sum of all Legendre symbols [imath]\left(\frac{k(k+1)}p\right)[/imath] show first that [imath]\left(\frac{k(k+1)}p\right) = \left(\frac{1+k'}p\right),[/imath] then estimate the sum of [imath]\left(\frac{1+k'}p\right).[/imath] | 552521 | [imath]\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1[/imath]
Let [imath]p[/imath] be an odd prime number. Prove that [imath]\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \left( \frac{3 \cdot 4}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1[/imath] where [imath]\left( \frac{a}{p}\right)[/imath] is the Legendre symbol. This seems to be a tricky one! I've tried using the property [imath]\left( \frac{ab}{p} \right)=\left( \frac{a}{p}\right) \left( \frac{b}{p} \right)[/imath] and prime factoring all the non-primes but to no avail. I had a quick thought of induction haha, but that was silly. I tried factoring the common Legendre symbols like [imath]\left( \frac{3}{p}\right) \left[ \left( \frac{2}{p} \right) + \left( \frac{4}{p} \right) \right][/imath] but that didn't bring anything either. And I've been looking for pairwise cancellation with [imath]1[/imath] term leftover, but that does not seem to work. Can you help? |
553843 | Practical significance of [imath]e[/imath]
We know, for example, the constant [imath]\pi[/imath] is the perimeter of a circle with diameter [imath]1[/imath] unit. In the similar manner how would we explain the constant [imath]e[/imath]. I have searched a lot for it. But I couldn't comprehend it in a practical way. Can anybody help me to understand the constant [imath]e[/imath]? | 177681 | What are the practical uses of [imath]e[/imath]?
How can [imath]e[/imath] be used for practical mathematics? This is for a presentation on (among other numbers) [imath]e[/imath], aimed at people between the ages of 10 and 15. To clarify what I want: Not wanted: [imath]e^{i\pi}+1=0[/imath] is cool, but (as far as I know) it can't be used for practical applications outside a classroom. What I do want: [imath]e[/imath] I think is used in calculations regarding compound interest. I'd like a simple explanation of how it is used (or links to simple explanations), and more examples like this. |
209198 | Show that [imath]f[/imath] vanishes identically.
Let [imath]f[/imath] be an entire function. Assume that [imath]\mid f(1/n)\mid\le e^{-n}[/imath] for all [imath]n\in \mathbb{N}[/imath]. Show that [imath]f[/imath] vanishes identically. | 2820673 | holomorophic function on unit disk
Let [imath]f[/imath] holomorophic function on [imath]\{|z|<1\}[/imath]. Suppose [imath]|f(\frac{1}{n})|\leq e^{-n}[/imath] for [imath]n=2,3,\dots [/imath] ,then [imath]f=0[/imath]. If [imath]f(\frac{1}{n})=0[/imath], by identity theorem, [imath]f=0[/imath].But I cannot prove it. |
553094 | Continuous homomorphisms from [imath]\mathbb{S}^{1} \to \mathbb{S}^{1}[/imath]
I think I read somewhere that all continuous homomorphisms from [imath]\mathbb{S}^{1}[/imath] to itself are of the form [imath]z \mapsto z^{n}[/imath] for some [imath]n \in \mathbb{Z}[/imath]. Is this result true? I am not able to prove it. Nor am I able to find any counterexample. What I have done so far is: If for some [imath]\theta \in [0,2\pi)[/imath], [imath]e^{\iota\theta}[/imath] goes to [imath]e^{\iota\phi}[/imath] for some [imath]\phi \in [0,2\pi)[/imath], then [imath]e^{\iota n\theta}[/imath] goes to [imath]e^{\iota n \phi}[/imath] for all [imath]n \in \mathbb{Z}[/imath]. I thought I could take [imath]n[/imath] to be any rational here and then extend to reals by continuity; but that is not true as a multiple of [imath]2\pi[/imath] may creep in. Next I tried something using [imath]n^{th}[/imath] roots of unity; but again apart from a number of cases, I could not go very far. Could someone point me in the right direction? | 123588 | [imath]\varphi[/imath] in [imath]\operatorname{Hom}{(S^1, S^1)}[/imath] are of the form [imath]z^n[/imath]
I'd like to see a proof why [imath]\varphi \in \operatorname{Hom}{(S^1, S^1)}[/imath] looks like [imath]z^n[/imath] for an integer [imath]n[/imath]. At first I thought I could argue that if I have a homomorphism that maps [imath]e^{ix}[/imath] to some [imath]e^{iy}[/imath] for [imath]x,y \in \mathbb R[/imath] then [imath]z = rx[/imath] for some real number [imath]r[/imath]. But on second thought I'm not sure why I can't have [imath]\varphi (e^{ix} ) = e^{g(x)i}[/imath] for a [imath]g[/imath] other than [imath]g(x) = \lambda x[/imath]. I'm also interested in seeing different proofs. I'm sure there are several ways to prove this. Thanks for your help. |
62852 | In set theory, how are real numbers represented as sets?
In set theory, if natural numbers are represented by nested sets that include the empty set, how are the rest of the real numbers represented as sets? Thanks for the answers. Several answers basically said for irrational numbers that A Dedekind cut is a pair of sets of rational numbers [imath]\{L, R\}[/imath]. The set of real numbers is defined to be the set of all Dedekind cuts, where a Dedekind cut is a pair of sets of rational numbers [imath]\{L, R\}[/imath] which have no elements in common, and where all the elements of [imath]L[/imath] are less than any element of [imath]R[/imath]. Each Dedekind cut is a real number. This is where I have a problem - surely that can’t be correct. The set [imath]L[/imath] is a set of all rationals, and there must be a rational in the set [imath]L[/imath] that is greater than all other rationals in that set, even if we have no method of determining it. And similarly, there must be a rational in the set [imath]R[/imath] that is less than all other rationals in that set, even if we have no method of determining it. If every irrational number has a corresponding set [imath]L[/imath], then each irrational number has some such corresponding largest element of that set [imath]L[/imath], and then each irrational number has some corresponding rational number. And that would mean that the irrational numbers are countable. So, with Dedekind cuts, the only conclusion is that there must be irrational numbers [imath]x[/imath] which are either greater or lesser than some irrational cut [imath]y[/imath] of the rationals, and between [imath]x[/imath] and [imath]y[/imath] there is no rational number. But that is impossible, so that the Dedekind cuts cannot be the correct representation of the real numbers. Surely the problem with Dedekind cuts is in using sets of rationals that include all rationals up to a certain rational. But there is an alternative method of representing irrationals can be defined in terms of infinite sets of rational numbers. For example, in binary notation, the non-integer part of [imath]\pi[/imath] is [imath].00100100\ 00111111\ 01101010\ 10001[/imath]. You define a set by: if the nth digit is a [imath]1[/imath], then the natural number [imath]n[/imath] is in the set. And then we have that, for the real numbers between [imath]0[/imath] and [imath]1[/imath], that the set of real numbers is simply the set of all subsets of natural numbers. Each subset corresponds to some real number between [imath]0[/imath] and [imath]1[/imath]. And in this way, all real numbers can be considered to be some set based only on nested sets of the empty set. But I still haven’t got a satisfactory answer for how negative numbers can be represented in terms only of sets containing the empty set. Any ideas? | 799475 | Integers, rationals and reals as sets?
Natural numbers can be represented as pure sets by defining them to contain every number that is smaller than them. Arithmetic can be performed on them using the Peano axioms. Are there any similar definitions for integers, rationals and reals? For example, I could define a rational to be an ordered pair of dividend and divisor. But that would leave the two rationals [imath]\frac{1}{2}[/imath] and [imath]\frac{2}{4}[/imath] not equal to each other, and it would be based on ordered things rather than pure sets. |
555199 | How do you isolate variables in a trig function?
I wanted to ask how you can solve an equation such as: [imath] \sin x = x + 1 [/imath] How would you solve for x? | 26358 | Solve [imath]\sin x = 1 - x[/imath]
How would you be solve sin x = 1 - x, without drawing the graph and manually measuring the point of intersection? |
555332 | Show that [imath]\pi_1: X \times Y \to X[/imath] is an open map
A map [imath]f: X \to Y[/imath] is said to be an open map if for every open set [imath]U[/imath] of [imath]X[/imath], the set [imath]f(U)[/imath] is open in [imath]Y[/imath]. Show that [imath] \pi_1: X \times Y \to X[/imath] is an open map. I say: Let there exist [imath]\lambda \in X \times Y[/imath] such that [imath]\lambda[/imath] is open and [imath]f(\lambda) \in Y[/imath]. If [imath]\lambda[/imath] is open in [imath] X \times Y[/imath] then it is an element of [imath] U \times V \in X \times Y[/imath] such that [imath] U \in X[/imath] and [imath]V \in Y[/imath] are open sets. Now, if [imath]\lambda[/imath] is [imath]\in U \times V[/imath] then this map must be sending some [imath]\lambda[/imath] to [imath]V[/imath] in [imath]X[/imath]. I can't say why I made the last assumption, it just seemed intuitively right? How do I show it, though? But I suspect that showing that the whole [imath]X[/imath] is an open set (i.e. if we were discussing topologies) would be much easier way to show that this is an open map but I can't for the life of me figure out how to do that. | 247542 | Projection is an open map
Let [imath]X[/imath] and [imath]Y[/imath] be (any) topological spaces. Show that the projection [imath]\pi_1[/imath] : [imath]X\times Y\to X[/imath] is an open map. |
555388 | prove that [imath]T''[/imath] is not injective (difficult computation)
Let [imath]T:c_0 \to c_0[/imath] defined by [imath]T(\{s_j\}_j)=\{s_{j+1}-s_j\}_j[/imath]. Prove that [imath]T''[/imath] is not injective. I tried even knowing that [imath](c_0)' \sim l^1[/imath],in fact if [imath]F\in (c_0)'[/imath] then there exist [imath]s=(s_j)\in l^1[/imath] such that [imath]F=F_s[/imath] and [imath]F_s(t_j)=\sum s_jt_j[/imath] , and similar for [imath](l^1)' \sim l^{\infty}[/imath]. It's easy to prove that [imath]T''[/imath] is injective if and only iff the image of [imath]T'[/imath] is dense on [imath]X'[/imath]. Maybe this is easy to prove. | 555357 | prove that this double transpose [imath]T''[/imath] it's not injective.
Prove that if [imath]T''[/imath] is injective then [imath]T[/imath] it's injective, but the converse it's not true. I proved the first part " [imath]T''[/imath] injective implies [imath]T[/imath] injective". It's easy to prove that [imath]T''[/imath] is injective if and only if the image of [imath]T'[/imath] is dense on [imath]X'[/imath]. So If I prove that the [imath]cl({Ran(T')})[/imath] it's a proper subset of [imath]X'[/imath] then I'm done. But even this it's difficult. |
555513 | Units of [imath]\mathbb Z[\sqrt 3][/imath]
Prove that the number of units of [imath]\mathbb Z[\sqrt{3}][/imath] is infinite, i.e. [imath]a^2 -3b^2=1[/imath] has infinitely many integer solutions. Can you help me? | 87169 | [imath]\mathbb Z[\sqrt 3][/imath] contains infinitely many units
I'm asked to show that there are infinitely many units in the ring [imath]\mathbb Z[\sqrt 3][/imath]. But I don't really see a good approach to this one, so far. Some thoughts: The inverse of [imath]a+\sqrt3 b[/imath] should be given by [imath]\pm(2 - \sqrt 3 b)[/imath], since the norm [imath]N: \mathbb Z[\sqrt{3}] \to \mathbb Z, \qquad N(x+\sqrt{3} y) = x^2 - 3y^2[/imath] is multiplicative. So if [imath](a+\sqrt{3}b)^{-1}=x+\sqrt 3 y[/imath], then [imath]1 = N(1) = N((a+\sqrt{3}b)(x + \sqrt{3}y)) = (a^2 - 3b^2)(x^2 - 3y^2)[/imath] Hence we must have [imath]\pm 1 = (a^2 - 3b^2) = (a+\sqrt{3}b)(a-\sqrt{3}b)[/imath]. Therefore I need to show that there are infinitely many [imath]a,b[/imath] such that [imath]a^2 - 3b^2 = \pm 1[/imath]. Here I don't know how to proceed. Maybe rewriting as [imath]a = \sqrt{3b^2 \pm 1}[/imath], and now trying to prove that there are infinitely many [imath]b[/imath] for which [imath]3b^2 \pm 1[/imath] is a square? A hint would be appreciated! =) Thanks. |
553774 | [imath]\overline{A\cdot B}=\overline{A}\cdot\overline{B}[/imath] for [imath]A,B\subseteq\Bbb R[/imath]?
If [imath]A[/imath] is a closed set in [imath]\Bbb R[/imath] and [imath]B[/imath] a compact set in [imath]\Bbb R[/imath], let [imath]A\cdot B=\{a\cdot b:a \in A,b\in B\}[/imath]. Question: Is [imath]\overline{A\cdot B}=\overline{A}\cdot\overline{B}[/imath]? Note: [imath]\overline{A}[/imath] is the closure of set [imath]A[/imath]. | 553173 | Let [imath]A , B\subseteq\mathbb{R}[/imath]. If [imath]A[/imath] is closed and [imath]B[/imath] is compact, is [imath]A\cdot B[/imath] closed?
Let [imath]A , B\subseteq\mathbb{R}[/imath]. If [imath]A[/imath] is closed and [imath]B[/imath] is compact, is [imath]A\cdot B=\{a. b: a\in A,b\in B\}[/imath] closed? If [imath]p[/imath] is an adherent point in [imath]A\cdot B[/imath], exists a sequence [imath]p_{n}=\alpha_{n} \beta_{n}[/imath], [imath]\alpha_{n} \in A[/imath] and [imath]\beta_{n}\in B[/imath], such that [imath]p_{n}\rightarrow p[/imath]. If [imath]B[/imath] is compact, [imath]\beta_{n}\rightarrow b \in B[/imath]. |
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