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964826 | Proof of AM-GM Inequality (setting [imath]a_n[/imath] in the last step)
I have been reading this and this, but I don't understand how one of the steps works. Let [imath]a_n=\frac{a_1+a_2+\cdots+a_{n-1}}{n-1}.[/imath] How do you set [imath]a_n[/imath] to meet certain criteria and not lose generality? | 739804 | How to understand Cauchy's proof of AM-GM inequality(the last step)
The AM-GM inequality: [imath]a_1a_2\cdots a_n\leq\left(\frac{a_1+\cdots + a_n}{n}\right)^n[/imath] the trivial case: [imath]a_1a_2 \leq \left(\frac{a_1+a_2}{2}\right)^2 [/imath] is self-evident. then cauchy use this fact repeatedly. He get: [imath]a_1a_2\cdots a_{2^m} \leq \left(\frac{a_1+a_2+\cdots +a_{2^m}}{2^m}\right)^{2^m}.\tag{$\ast$}[/imath] This is easy to understand, and natural, but next step: he let [imath]b_1 = a_1,\ b_2=a_2,\ \ldots,\ b_n=a_n,\ b_{n+1}=\cdots=b_{2^m}=\frac{a_1+a_2+\cdots a_n}{n}[/imath] replace [imath]a_1, a_2,\ldots ,a_{2^m}[/imath] with [imath]b_1, b_2, \ldots b_{2^m}[/imath] in [imath](\ast)[/imath] and simplified both left and right. He get the answer. I can repeat his proof, but I don't know why Cauchy can got this idea. How does he got it? |
395271 | How can I calculate [imath]\int \frac{\sec x\tan x}{3x+5}\,\mathrm dx[/imath]
How can I calculate [imath] \int{\sec\left(x\right)\tan\left(x\right) \over 3x + 5}\,{\rm d}x [/imath] My Try:: [imath]\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx[/imath] Now Using Integration by Parts:: We Get [imath]\displaystyle = \frac{1}{3x+5}\sec x +\int \frac{3}{(3x+5)^2}\sec x\,\mathrm dx[/imath] Now My Question is How Can I calculate (II) Integral. Please explain this to me. | 618317 | So close yet so far Finding [imath]\int \frac {\sec x \tan x}{3x+5} dx[/imath]
Cruising the old questions I came across juantheron asking for [imath]\int \frac {\sec x\tan x}{3x+5}\,dx[/imath] He tried using [imath](3x+5)^{-1}[/imath] for [imath]U[/imath] and [imath]\sec x \tan x[/imath] for [imath]dv[/imath]while integrating by parts. below is his work. How can I calculate [imath] \int {\sec\left(x\right)\tan\left(x\right) \over 3x + 5}\,{\rm d}x [/imath] My Try:: [imath]\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx[/imath] Now Using Integration by Parts:: We get [imath]= \frac{1}{3x+5}\sec x +\int \frac{3}{(3x+5)^2}\sec x\,\mathrm dx[/imath] Here he hit his road block. I tried the opposite tactic Taking the other approach by parts. let [imath]U= \sec x \tan x[/imath] then[imath] du= \tan^2 x \sec x +\sec^3 x[/imath] and [imath]dv=(3x+5)^{-1}[/imath] then [imath]v=\frac 1 3 \ln(3x+5)[/imath] Thus [imath]\int \frac {\sec x \tan x}{3x+5}\,dx= \frac {\ln(3x+5)\sec x \tan x}{3} - \int \frac {\ln(3x+5) [\tan^2 x \sec x +\sec^3 x]}{3} \,dx[/imath] As you can see I got no further than he did. So how many times do you have to complete integration by parts to get the integral of the original [imath]\frac {\sec x \tan x}{3x+5} \, dx[/imath] or is there a better way? |
964951 | Proof regarding a fixed point
Show that for any strictly increasing function [imath]f:[0,1]\to[0,1][/imath] there is a fixed point such that [imath]f(x)=x[/imath]. ( The function isn't necessarily continuous) . Any ideas ? | 628096 | Fix-Point Theorem Proof.
Firstly, the assignment: Let [imath]a,b \in\mathbb{R}[/imath] and [imath]a < b[/imath]. Furthermore let [imath]f: [a,b] \rightarrow [a,b][/imath] be monotone increasing. Show that if [imath]x:= \mathbf {sup}\{y \in [a,b] \| \ y ≤ f(y)\}[/imath] then [imath]f(x) = x[/imath]. It's good on me that the assignment itself actually tells you what [imath]x[/imath] is so I can prove the theorem by showing that for [imath]\forall x' \not= x: f(x) \not= x[/imath] and I'm done. However, I have not arrived at a contradiction so far. Any tips? |
965561 | Rational number arbitrarily close to a square root of 2
I am trying to prove the proposition by contradiction For all rational [imath]c > 0[/imath], there exists a rational number [imath]x[/imath] such that [imath]x^2 < 2 < (x + c)^2[/imath]. with the negation [imath]x^2 \ge 2 \lor 2 \ge (x+c)^2[/imath]. But I can not prove it. | 233785 | Proof for [imath]\epsilon[/imath] closeness to [imath]\sqrt{2}[/imath] for all values of [imath]\epsilon[/imath]
I am trying to understand the proof above, Proposition 7, given by the author, but I got stuck at some points. My explanation: Now, suppose for contradiction there is an [imath]\epsilon > 0[/imath], for all [imath]x > 0[/imath] such that [imath]NOT (x^2 < 2 < (x+ \epsilon)^2)[/imath], Since [imath]NOT (x^2 < 2 < (x+ \epsilon)^2) \ \ iff \ \ (x^2 \geq 2 \ \ OR \ \ (x+ \epsilon)^2 \leq 2)[/imath], We must show that for some [imath]\epsilon>0[/imath] and for all [imath]x > 0[/imath], the statement [imath]x^2 \geq 2 \ \ OR \ \ (x+ \epsilon)^2 \leq 2[/imath] leads to a contradiction. I think in the proof above, author tries to get a contradiction using right hand side of [imath]OR[/imath] clause, namely [imath](x+ \epsilon)^2 \leq 2[/imath]. But no contradicition is achieved by him for the left hand side side of [imath]OR[/imath] clause namely, [imath]x^2 \geq 2[/imath]. First Question is, am I correct at this point, i.e. the need for a contradiction for the left hand side side of [imath]OR[/imath] clause? Second question is, I don't understand how he derived the following sentence in his proof above, Since [imath]0^2 < 2[/imath], we thus have [imath]\epsilon^2 < 2[/imath], which then implies [imath](2 \epsilon)^2 < 2[/imath] and indeed [imath](n \epsilon)^2 < 2[/imath] Why does he feel a need to multiply [imath]\epsilon[/imath] constantly by [imath]n[/imath] |
965699 | Is there a general formula for the number of elements of order [imath]k[/imath] in [imath]S_n[/imath]?
How many elements of order [imath]n[/imath] does [imath]S_7[/imath] contain? Is there a general formula for to compute how many elements of order [imath]k[/imath] (for a given) in [imath]S_n[/imath]? | 627130 | How many elements of order [imath]k[/imath] are in [imath]S_n[/imath]?
I need to find how many elements of order [imath]k[/imath] are in [imath]S_n[/imath] (where [imath]k \leq n[/imath]). So if [imath]k[/imath] is prime, it's easy: [imath]k[/imath] can't be the [imath]\mathrm{lcm}[/imath] of any integers besides itself and one's (which we're omitting). So the length of the cycle then must be [imath]k[/imath], and the number of elements with [imath]k[/imath]-length cycle representation in [imath]S_n[/imath] is given by [imath]\dbinom{n}{k}(k-1)![/imath] But it gets really tricky when [imath]k[/imath] is not a prime number... I need to consider every set of integers that can divide [imath]k[/imath], and then, for every set, I need to find how many elements of that form are there... Or maybe I'm missing something here? I need help figuring that out. thanks in advanced! |
965651 | Let [imath]|\mathbb F|=q[/imath] where [imath]\mathbb F[/imath] is a field. Then [imath]|GL_n(\mathbb F)|=(q^n-1)(q^n-q)(q^n-q^2)\ldots(q^n-q^{n-1})[/imath].
Let [imath]|\mathbb F|=q[/imath] where [imath]\mathbb F[/imath] is a field. Now [imath]|GL_n(\mathbb F)|=(q^n-1)(q^n-q)(q^n-q^2)\ldots(q^n-q^{n-1}).[/imath] How it is so? My attempt I just viewed in the matrix such that [imath]\left(\begin{array}{ccccc} a_{11}&a_{12}&\cdots&a_{1n-1}&a_{1n}\\ \vdots&&\vdots&&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn-1}&a_{nn} \end{array}\right)[/imath] For the first element I will have q choices and rest of the elements also have q choices. So there are [imath]q^n[/imath] choices and i should neglect [imath]0[/imath] in all at a time. So I removed one choice from the total so for the first row I will have [imath]q^n-1[/imath] choices. But for the second and third i am having confusion how to see the choices. And I am thinking that this will give the number of distinct basis for [imath]GL_n(\mathbb F)[/imath]. Is it correct. Give me a diagrammatical explanation. | 391817 | Cardinality of [imath]GL_n(K)[/imath] when [imath]K[/imath] is finite
I don't know how to do the last task of an exercise. Let [imath]K[/imath] be a field, [imath]G=GL_n(K)[/imath] and [imath]X=K^n\backslash\{0\}[/imath]. First task: Show that [imath]G \times X \to X[/imath], [imath](A,x)\mapsto Ax[/imath] defines an action of [imath]G[/imath] on [imath]X[/imath]. Done. Second task: find the stabilizer of [imath]\displaystyle x=\left(\begin{eqnarray} 1\\ 0 \\ \vdots \\0 \end{eqnarray}\right)[/imath]. Done. Third task: Show that there is only one orbit. Done. Last task: Let [imath]|K|<\infty[/imath]. Use the last two tasks and induction to find [imath]|G|[/imath]. I know that [imath]|G|=[G:St_G(x)]\cdot|St_G(x)|[/imath] and i know that [imath][G:St_G(x)]=|\bar{x}|=m^n-1[/imath] but i can't figure out what [imath]|St_G(x)|[/imath] is. What should i do next? I appreciate every hint :) |
961976 | Functions for non-negative integers
Let [imath]\Bbb{Z}^+[/imath]be the set of all non-negative integers where [imath]n[/imath] and [imath]k[/imath] are given natural numbers. We consider the following non-decreasing function, [imath]f:\Bbb{Z}^+ \to \Bbb{Z}^+[/imath] such that \begin{align} f\left(\sum_{i=1}^{n} a_{i}^{n}\right) =\frac{1}{k}\sum_{i=1}^{n} \left(f(a_{i})\right)^{n}\\ \end{align} Found all functions for [imath]n=2014[/imath] and [imath]k=2014^{2013}[/imath] Found, how many functions and which satisfy the condition of the problem depending on the values of parameters [imath]n[/imath] and [imath]k[/imath]. PS. I've tried to make some operations with this function and found [imath]f(0) [/imath], but I can't imagine what can I do with it | 965231 | Integers and integer functions
Let [imath]\Bbb{Z}^+[/imath]be the set of all non-negative integers where [imath]n[/imath] and [imath]k[/imath] are given natural numbers. We consider the following non-decreasing function, [imath]f:\Bbb{Z}^+ \to \Bbb{Z}^+[/imath] such that \begin{align} f\left(\sum_{i=1}^n a_i^n\right) =\frac 1 k \sum_{i=1}^n \left(f(a_i)\right)^n \end{align} 1.Find all functions for [imath]n=2014[/imath] and [imath]k=2014^{2013}[/imath] 2.Find, how many functions and which satisfy the condition of the problem depending on the values of parameters [imath]n[/imath] and [imath]k[/imath]. I have some ideas for this problem . I know how to found [imath]f(0)[/imath] (just take [imath]a_1=a_2=\cdots=a_n=0[/imath]) . I also can found [imath]f(1)[/imath] (take [imath]a_1=1[/imath] and [imath]a_2=\cdots=a_n=0[/imath]) Then if [imath]f(0)=0[/imath] I've get that [imath]f(a_n)+f(b_n)=f(a_n+b_n)[/imath]. I also think thas all functions will be as [imath]f(x)=kx[/imath] but I can't prove it . Can you help me with some partical solution or some ideas which can help to find it. And can someone tell me how in first example ([imath]n=2014[/imath] and [imath]k=2014^{2013}[/imath]) found f(2015) or [imath]f(2016)[/imath] using that [imath]f(i^n)=i\cdot f(1)[/imath] ( for [imath]2015>i>1[/imath])and [imath]f(1)=0[/imath] or [imath]f(1)=2014[/imath] |
965070 | Half open intervals in a Borel [imath]\sigma[/imath]-algebra
I am working on the exercise: prove that a right continuous function [imath]\mathbb{R} \to \mathbb{R}[/imath] is Borel measurable. I found that for every [imath]x \in f^{-1}((a,b))[/imath] we must have [imath] [x, x + \delta_x) \subset f^{-1}((a,b)), [/imath] for some [imath]\delta_x > 0[/imath] depending on [imath]x[/imath]. Consequently [imath] \{ [x, x + \delta_x) \mid x \in f^{-1}((a,b)) \} = f^{-1}((a,b)). [/imath] I know that intervals are Borel measurable, but to prove that [imath]f^{-1}((a,b))[/imath] is measurable I would have to prove that [imath]f^{-1}((a,b))[/imath] is a countable union of such intervals. Could anybody give me a hint on how to do this? | 680854 | Follow-up regarding right-continuous [imath]f:\mathbb{R} \to\mathbb{R}[/imath] is Borel measurable
I have a follow-up to another question here on math.stackexchange, Are right continuous functions measurable?. The thread was a couple of years old, so I hope it's okay if I start a new question. The person who answered the question claimed that the preimage of an open set under a right-continuous function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] is a Borel measurable set because it is the countable union of half-open intervals. I was trying to think through the "countable union" part. If we have an open set [imath]U \subseteq \mathbb{R}[/imath], and let [imath]A = f^{-1}(U)[/imath], then certainly by definition if we take [imath]A \cap \mathbb{Q}[/imath], then for every [imath]a \in A \cap \mathbb{Q}[/imath], there is some [imath]\delta > 0[/imath] such that [imath][a,a+\delta) \subseteq A[/imath], and thus [imath]\bigcup_{a \in A \cap \mathbb{Q}} [a,a+\delta) \subseteq A[/imath]. But then, why does the reverse containment hold (or does it hold at all)? I tried to prove that [imath]A \subseteq \bigcup_{a \in A \cap \mathbb{Q}} [a,a+\delta)[/imath] using contradiction, but didn't get anywhere. The best I could come up with was that if [imath]b \in A[/imath] and [imath]b \notin \bigcup_{a \in A \cap \mathbb{Q}} [a,a+\delta)[/imath], then because of the density of the rationals in the reals, we could write [imath]\{ b \} \cup \bigcup_{n \in \mathbb{N}} [a_n,a_n+\delta_n) \subseteq A[/imath], but this clearly didn't get me anywhere... |
966791 | Help with trigonometry identity derivation
Can someone help me with a complicated trigonometry derivation with only basic trigonometric identities: [imath]\csc^2 \bigg(\frac{\pi}{7}\bigg) + \csc^2 \bigg(2\frac{\pi}{7}\bigg) + \csc^2 \bigg(3\frac{\pi}{7}\bigg) = 8[/imath] Thanks... | 309271 | proving [imath]\csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)=8[/imath]
How can I prove the following identity using complex variables [imath] \begin{align*} 1) & \csc^2 \left( \frac{\pi}{7}\right)+\csc^2 \left( \frac{2\pi}{7}\right)+\csc^2 \left( \frac{4\pi}{7}\right)=8 \\ 2) & \tan^2 \left( \frac{\pi}{16}\right) + \tan^2\left( \frac{3\pi}{16}\right) + \tan^2\left( \frac{5\pi}{16}\right)+ \tan^2\left( \frac{7\pi}{16}\right) = 28 \end{align*} [/imath] On earlier problem, I was given, [imath]\displaystyle (z+a)^{2m}-(z-a)^{2m}=4maz \prod_{k=1}^{m-1} \left(z^2+a^2 \cot^2 \left(\frac{k\pi}{2m} \right )\right ) [/imath] for integer [imath]m>1[/imath]. I am not sure if I can use this is helpful. I am stumped please help. |
966836 | Why is this map homeomorphism?
Munkres - Topology p.143 Let [imath]p:X\rightarrow Y[/imath] be a quotient map. Set [imath]x\sim y[/imath] iff [imath]p(x)=p(y)[/imath]. Let [imath]X/\sim[/imath] be the quotient space of [imath]X[/imath] with respect to [imath]\sim[/imath]. Let [imath]\pi:X\rightarrow X/\sim[/imath] be the projection map. Then, there exists a bijective quotient map [imath]\phi:X/\sim\rightarrow Y[/imath] such that [imath]\phi\circ \pi = p[/imath]. Why is this [imath]\phi[/imath] a homeomorphism? | 966744 | Is there an equivalence relation of which the quotient topology is homeomorphic to codomain of a quotient map?
Let [imath]X,Y[/imath] be topological spaces. Let [imath]p:X\rightarrow Y[/imath] be a quotient map. Then, does there exist an equivalence relation [imath]\sim[/imath] on [imath]X[/imath] such that [imath]X/\sim[/imath] is homeomorphic to [imath]Y[/imath]? |
963374 | Beads on the circle
It is placed the [imath]n[/imath] beads on a circle, [imath]n \geq 3[/imath]. They numbered in random order. They are viewed clockwise. Beads for which the number of the previous bead less than the number of a next bead, are painted in white color,and other - in black. Two colourations, that can be compared by a turn are considered identical. How many different colourations can be occured ? | 965292 | Black and white beads on a circle
There are [imath]n[/imath] beads placed on a circle, [imath]n\ge 3[/imath]. They are numbered in random order as viewed clockwise. Beads for which the number of the previous bead is less than the number of a next bead are painted in white color,and others - in black. Two colourations that can be made equal by rotation are considered identical. How many different colourations can occur? I've write a programm and for [imath]n=3...11[/imath] I've got answers [imath]2 , 1 , 6 , 7 , 18 , 25 , 58 , 93 , 186[/imath] |
962903 | A question on p-groups, and order of its commutator subgroup
[imath]\textbf{QUESTION-}[/imath] Let [imath]P[/imath] be a p-group with [imath]|P:Z(P)|\leq p^n[/imath]. Show that [imath]|P'| \leq p^{n(n-1)/2}[/imath]. [imath]\textbf{TRY- }[/imath] If [imath]P=Z(P)[/imath] it is true. Now let [imath]n > 1[/imath], then If I see [imath]P[/imath] as a nilpotent group and construct its upper central series, it will end , so let it be, [imath]e=Z_0<Z_1<Z_2<......<Z_r=P[/imath] Now as [imath]Z_{i+1}/Z_i=Z(P/Z_i)[/imath], so if if I take some [imath]x\in Z_2[/imath]\ [imath]Z_1[/imath] then [imath]N[/imath]={[imath][x,y]|y\in P[/imath]} [imath]\leq Z_1(P)[/imath] and [imath]N \triangleleft P [/imath], so [imath]P/N[/imath] is a group with order [imath]\leq p^{n-1}[/imath]. Now if I let [imath]H=P/N[/imath] then obviously |[imath]H/Z(H)[/imath]|[imath]\leq p^{n-1}[/imath]. Now [imath]H'\cong P'N/N \cong P'/(P' \cap N)[/imath] so from here I could finally bring [imath]P'[/imath] atleast into the picture, now |[imath]P'[/imath]|=[imath]|H'|.|P'\cap N|[/imath] so [imath]|P'|\leq |H'||N|[/imath]. This is where I am [imath]\textbf{STUCK}[/imath] Now , from here how can I calculate or find some power [imath]p[/imath] bounds on [imath]|H'|[/imath] and [imath]|N|[/imath] so i could get my result. | 944948 | A question on [imath]p[/imath]-groups, and order of its commutator subgroup.
[imath]\textbf{QUESTION-}[/imath] Let [imath]P[/imath] be a p-group with [imath]|P:Z(P)|\leq p^n[/imath]. Show that [imath]|P'| \leq p^{n(n-1)/2}[/imath]. If [imath]P=Z(P)[/imath] it is true. Now let [imath]n > 1[/imath], then If I see [imath]P[/imath] as a nilpotent group and construct its upper central series, it will end , so let it be, [imath]e=Z_0<Z_1<Z_2<......<Z_r=P[/imath] Now as [imath]Z_{i+1}/Z_i=Z(P/Z_i)[/imath], so if if I take some [imath]x\in Z_2[/imath]\ [imath]Z_1[/imath] then [imath]N[/imath]={[imath][x,y]|y\in P[/imath]} [imath]\leq Z_1(P)[/imath] and [imath]N \triangleleft P [/imath], so [imath]P/N[/imath] is a group with order [imath]\leq p^{n-1}[/imath]. Now if I let [imath]H=P/N[/imath] then obviously |[imath]H/Z(H)[/imath]|[imath]\leq p^{n-1}[/imath]. Now [imath]H'\cong P'N/N \cong P'/(P' \cap N)[/imath] so from here I could finally bring [imath]P'[/imath] atleast into the picture, now |[imath]P'[/imath]|=[imath]|H'|.|P'\cap N|[/imath] so [imath]|P'|\leq |H'||N|[/imath]. This is where I am [imath]\textbf{STUCK}[/imath] Now , from here how can I calculate or find some power [imath]p[/imath] bounds on [imath]|H'|[/imath] and [imath]|N|[/imath] so i could get my result. |
969255 | Convergence and Limits of Sequences
Let [imath](x_n)[/imath] and [imath](y_n)[/imath] be two sequences recursively defined as follows: [imath]x_1=a\geq0,y_1=b\geq0[/imath] [imath]x_{n+1}=\sqrt{x_ny_n}, y_{n+1}=\frac{x_n+y_n}{2}, n\geq1[/imath] Prove that [imath](x_n)[/imath] and [imath](y_n)[/imath] are convergent and that [imath]\lim{x_n}=\lim{y_n}[/imath] | 687609 | Sequences of arithmetic and geometric mean
I am assuming my teacher wants us to conclude they converge by being bounded and monotone, since part (i) is to show that [imath]a_n[/imath] is monotone increasing and [imath]b_n[/imath] is monotone decreasing. I am confused though because I have never worked with a sequence defined with... another sequence. |
969577 | Prove the following [imath]\neg((A\cap B)\cup (\neg A \cap C)) = (A\cap\neg B)\cup (\neg A\cap \neg C)[/imath]
How can I prove the following statements are equivalent using laws of set theory? [imath]\neg((A\cap B)\cup (\neg A \cap C)) = (A\cap\neg B)\cup (\neg A\cap \neg C)[/imath] Using De Morgans laws to simplify the first statement: [imath](\neg A\cup\neg B)\cap (A \cup \neg C)[/imath] but I have no idea where to go from here. | 969539 | How to prove that [imath]\neg((A\cap B)\cup (\neg A \cap C)) = (A\cap\neg B)\cup (\neg A\cap \neg C)[/imath]
How can I prove the following statements are equivalent using laws of set theory? [imath]\neg((A\cap B)\cup (\neg A \cap C)) = (A\cap\neg B)\cup (\neg A\cap \neg C)[/imath] I managed to use De Morgans laws to simplify the first statement down to: [imath](\neg A\cup\neg B)\cap (A \cup \neg C)[/imath] but I have no idea where to go from here. |
969599 | Let [imath]G[/imath] be a graph of order [imath]n[/imath] with [imath]\kappa (G) \geq 1[/imath]. Prove that [imath]n> \kappa(G)[\operatorname{diam}(G)-1]+2[/imath]
Let [imath]G[/imath] be a graph of order [imath]n[/imath] with [imath]\kappa (G) \geq 1[/imath]. Prove that [imath]n> \kappa(G)[\operatorname{diam}(G)-1]+2[/imath] I know [imath]\kappa(G) \leq n-1[/imath]. I do not know if do we have to prove it by induction or there is another way. | 106604 | Min. number of vertices in graph as function of [imath]\kappa(G)[/imath] and [imath]\operatorname{diam}(G)[/imath]
Question I found in "Introduction to Graph Theory" by Douglas B. West: Let [imath]G[/imath] graph on [imath]n[/imath] vertices with connectivity [imath]\kappa(G)=k \geq 1[/imath]. Prove that [imath]n \geq k(\operatorname{diam}(G)-1)+2[/imath] where [imath]\operatorname{diam}(G)[/imath] is the the maximal (edge) distance between a pair of vertices in [imath]G[/imath] Seems easy and I've seen the answer, too, which goes the same as thought: by taking the vertices the end vertices of the path of length [imath]\operatorname{diam}(G)[/imath], and applying Menger's Theorem. I'm having a hard time understanding one part - it seems like all the paths need to be of the same length for the [imath]n \geq k(\operatorname{diam}(G)-1)+2[/imath] to apply. Am I missing some part here? |
967735 | quotient by contractible space and homotopy equivalent?
Let [imath]X[/imath] be a space and [imath]L[/imath] be a subspace of [imath]X[/imath] that is contractible. Then is [imath]X/L[/imath] homotopy-equivalent to [imath]X[/imath] itself? it doesn't seem trivial to me at al...If so, how to show it rigorously? Could anyone help me? | 568279 | Is the quotient map a homotopy equivalence?
It is well known that, if [imath]A \subset X[/imath] is a reasonable contractible subspace, then the quotient map [imath]X \to X/A[/imath] is a homotopy equivalence ("reasonable" means that the pair [imath](X,A)[/imath] has the homotopy extension property, e.g. if it is a CW pair). For example, it's proposition 0.17 in Hatcher's celebrated Algebraic Topology. It seems to me that this result should be a consequence of the following result, which seems true but for which I've been unable to find a proof or a reference. Proposition? If [imath]X[/imath] is Hausdorff and [imath]\sim[/imath] is an equivalence relation whose classes [imath]C[/imath] are contractible and such that every pair [imath](X, C)[/imath] has the homotopy extension property, then the quotient map [imath]X \to X/\sim[/imath] is a homotopy equivalence. The classical result would of course be a direct corollary of this one. So, is this proposition true? |
969281 | Ideal Generated by Three Elements in Polynomial Ring
How would one prove that the ideal [imath](xy,xz,yz)[/imath] of [imath]k[x,y,z][/imath] for some field [imath]k[/imath], cannot be generated by two polynomials. In other words, prove: [imath](xy,xz,yz) \neq (f,g)\; \forall f,g \in k[x,y,z].[/imath] Furthermore, can somebody abstract this fact and prove it for [imath]k[x_1,x_2,...,x_n][/imath]? Maybe that the ideal of certain combinations of multiplied variables [imath]x_i[/imath] is not generated by less elements. Please ask if my question is unclear. Edit: Specifically, I want to say that the ideal generated by all combinatorial products of [imath]n-1[/imath] variables in [imath]k[x_1,x_2,...,x_n][/imath] cannot be generated by less elements. But if somebody has a different abstraction, please share. | 928275 | Problem on the number of generators of some ideals in [imath]k[x,y,z][/imath]
I have got stuck with two generator problems: The ideal [imath](zx,xy,yz)[/imath] can't be generated by [imath]2[/imath] elements. The ideal [imath](xz-y^2,yz-x^3,z^2-xy)[/imath] can't be generated by [imath]2[/imath] elements. Here the ring is [imath]k[x,y,z][/imath]. Thanks in advance for any help. |
968182 | Net based on finer filter is a subnet?
Let [imath]G[/imath] be a filter finer than the filter [imath]F[/imath]. Is it necessary that the net based on the filter [imath]G[/imath] is a subnet of the net based on [imath]F[/imath]? | 802886 | Subnets and finer filters
Suppose [imath]G[/imath] is a finer filter than [imath]F[/imath] in a topological space [imath]X[/imath]. Is the net base in [imath]G[/imath] a subnet of the net base in [imath]F[/imath]? I am using the definitions of General Topology of Willard: Definition 12.15. If [imath](x_\lambda)[/imath] is a net in [imath]X[/imath], the filter generated by the filter base [imath]\mathscr C[/imath] consisting of the sets [imath]B_{\lambda_0}=\{x_\lambda; \lambda\ge\lambda_0\}[/imath], [imath]\lambda_0\in\Lambda[/imath], is called the filter generated by [imath](x_\lambda)[/imath]. Definition 12.16. If [imath]\mathscr F[/imath] is a filter on [imath]X[/imath], let [imath]\Lambda_{\mathscr F}=\{(x,F); x\in F\in\mathscr F\}[/imath]. Then [imath]\Lambda_{\mathscr F}[/imath] is directed by the relation [imath](x_1,F_1)\le(x_2,F_2)[/imath] iff [imath]F_2\subset F_1[/imath], so the map [imath]P\colon \Lambda_{\mathscr F} \to X[/imath] defined by [imath]P(x,F)=x[/imath] is a net in [imath]X[/imath]. It is called the net based on [imath]\mathscr F[/imath]. The definition of subnet used in Willard is the second one from this post: Different definitions of subnet Thank you |
969974 | Abstract algebra permutation and order
Let [imath]G[/imath] be a group. Define a relation [imath]\sim[/imath] on [imath]G[/imath] by [imath]a\sim b[/imath] if there exists [imath]g\in G[/imath] such that [imath]a=gbg^{-1}[/imath]. Prove that all elements of order [imath]15[/imath] in [imath]S_8[/imath] are related by [imath]\sim[/imath]. I noticed that in order to have order 15 in S8, we need to have a product of 3 cycle and 5 cycle. Then, I am stuck... What should I do? Thank you in advance. | 969381 | show elements of order 15 in [imath]S_8[/imath] satisfy a relation
Let [imath]G[/imath] be a group. Define a relation [imath]\sim[/imath] on [imath]G[/imath] by [imath]a \sim b[/imath] if there exists [imath]g \in G[/imath] such that [imath]a = gbg^{-1}[/imath]. Prove that all elements of order 15 in [imath]S_8[/imath] are related by [imath]\sim[/imath]. I noticed that in order to have order [imath]15[/imath] in [imath]S_8[/imath]. We need a [imath]5[/imath] cycle and a [imath]3[/imath] cycle. Then, I am stuck. Can anyone give me some ideas about what to do? Thank you in advance. |
971247 | Method of Proof (Computer Science)
Prove that [imath]1+r+r^{2}+...+r^{n-1}=\frac{r^{n}-1}{r-1}[/imath], [imath]r[/imath] a positive integer, [imath]r \neq 1[/imath] | 371612 | Geometric representation of the sum of a Geometric Series
Here is a geometric proof of why [imath]\sum ar^{n}=\dfrac{a}{1-r}[/imath] (|r|<1)-Geometric Proof which I find is quite intuitive. Can someone give me a hint to prove [imath]\sum_{k=0}^{n-1} ar^{k}=\dfrac{a(1-r^{n})}{1-r}[/imath] (k is finite,|r|<1) using similar geometric interpretation. |
965259 | If [imath]abc=1[/imath], then [imath]\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} [/imath]
I've been trying to prove the following inequality Assume that a,b,c are positive reals s.t. [imath]abc=1[/imath] , prove that : [imath]S=\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} [/imath] If you apply AM-GM directy you get : [imath] S \geq \frac{3}{\sqrt[3]{(a^n+(n-1)b^n)(b^n+(n-1)c^n)(c^n+(n-1)a^n)}} [/imath] But from here I get stuck ... I think there is probably a change of variable or another trick (convex function + Jensen). We can apply AM-GM again to get this : [imath]\sqrt[3]{(a^n+(n-1)b^n)(b^n+(n-1)c^n)(c^n+(n-1)a^n)} \leq \frac{n}{3} \left(a^ n+ b^n+c^ n \right) [/imath] [imath] S \geq \frac{9}{n(a^n+b^n+c^n)} [/imath] Unfortunately [imath] a^n+b^n+c^n \geq 3 [/imath] by AM-GM so it doesnt help to conclude Thanks for your help. | 963977 | If [imath]abc=1[/imath], then [imath]\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} [/imath]
Let [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] be positive real numbers with [imath]abc=1[/imath]. Prove that [imath] \frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} [/imath] for each integer [imath]n[/imath]. I have used Cauchy-Schwarz inequalities and Jensen inequality. But I am stuck. I need some idea and advice on this problem. Induction would be cruel. |
966222 | A question about the proof of general lifting theorem in Munkres topology p.479
after the lifting [imath]\tilde f[/imath] of [imath]f[/imath] is defined, the book goes on to prove its continuity. And in that part, the book says that replacing [imath]U[/imath] by a smaller neighborhood, [imath]V_0[/imath] can be contained in [imath]N[/imath]. But why is it possible? How to construct a smaller neighborhood such that [imath]V_0[/imath] can be contained in [imath]N[/imath]? | 966589 | seemingly nontrivial question about covering maps and evenly covered open sets
Let [imath]p : E \to B[/imath] be a covering map. And also assume that [imath]B[/imath] is path-connected and locally path-connected. Then, given y in [imath]B[/imath], there exists a neighborhood [imath]U[/imath] of [imath]y_1[/imath] in [imath]B[/imath] that is path-connected and evenly covered by [imath]p[/imath]. Now the inverse image of [imath]U[/imath] by [imath]p[/imath] consists of slices [imath]\{V_a\}[/imath]. Take a point [imath]x[/imath] in [imath]E[/imath] such that [imath]p(x)=y[/imath] and let [imath]V'[/imath] be the slice containing [imath]x[/imath]. Now the questions is, if [imath]N[/imath] is an arbitrary neighborhood of [imath]x[/imath] in [imath]X[/imath], then how to replace [imath]U[/imath] by a smaller neighborhood of [imath]y[/imath] such that the slice [imath]V'[/imath] containing [imath]x[/imath] can be included in the [imath]N[/imath]? It seems very nontrivial to me...In the text I read, there is no detailed explanation about this. Could anyone give me some rigorous proof? |
396889 | How to find the factorial of a fraction?
From what I know, the factorial function is defined as follows: [imath]n! = n(n-1)(n-2) \cdots(3)(2)(1)[/imath] And [imath]0! = 1[/imath]. However, this page seems to be saying that you can take the factorial of a fraction, like, for instance, [imath]\frac{1}{2}![/imath], which they claim is equal to [imath]\frac{1}{2}\sqrt\pi[/imath] due to something called the gamma function. Moreover, they start getting the factorial of negative numbers, like [imath]-\frac{1}{2}! = \sqrt{\pi}[/imath] How is this possible? What is the definition of the factorial of a fraction? What about negative numbers? I tried researching it on Wikipedia and such, but there doesn't seem to be a clear-cut answer. | 1606364 | Fraction Factorial
How do we find factorial of fractions? For eg: [imath]\frac{1!}{2!}=(\frac{\pi}{4})^{\frac{1}{2}}[/imath] Factorials are used in combinatorics and they can only be functioned on integers to give integers.Then how do we get this? |
453198 | show that [imath]\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}[/imath]
show that [imath]\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}[/imath] using different ways thanks for all | 2013921 | integration of [imath]\sin^3(x)/x^3 [/imath] from [imath]0[/imath] to infinity.Can it be solved using laplace or fourier transform?
Integration of [imath]\sin^3(x)/x^3 [/imath] from [imath]0[/imath] to infinity. Can it be solved using Laplace or Fourier transform ? |
231878 | The Maximum possible order for an element [imath]S_n[/imath]
Given the following groups, what is the maximum possible order for an element for (a) [imath]S_5[/imath] (b) [imath]S_6[/imath] (c) [imath]S_7[/imath] (d) [imath]S_{10}[/imath] (e) [imath]S_{15}[/imath] My book justifies the answer as (a) The greatest order is [imath]6[/imath] and comes from a product of disjoint cycles of length 2 and 3 (b) The greatest order is [imath]6[/imath] and comes from a cycle of length [imath]6[/imath] The other answers were justified exactly the same way, that is (c) 12, (d) 30, (e) 105 I do not understand how in (a) we even got the number "6" from [imath]S_5[/imath] and what disjoint cycles they are referring to. Could someone at least justify one for me? | 1124861 | what is the maximom order of an element is [imath]\mathbb S_{15}[/imath]
Let [imath]S_n[/imath] be the permutation group of order [imath]n[/imath]. What is the maximom order of an element on [imath]\mathbb S_{15}[/imath]. Is there any way to find maximom order of an element in [imath]\mathbb S_{15}[/imath] or find the order of an arbitary element in [imath]\mathbb A_n[/imath] and [imath]\mathbb S_n[/imath] |
972134 | A monoid with left identity and right inverses need not be a group
Let [imath]G[/imath] be a set with an operation [imath]\ast:G\times G \rightarrow G[/imath] such that: (1) For all [imath]a,b,c \in G[/imath] we have [imath](a\ast b)\ast c=a\ast (b\ast c)[/imath] (associativity), (2) There is [imath]e \in G[/imath] such that [imath]e\ast a=a[/imath] for all [imath]a\in G[/imath] (left identity), (3) For all [imath]a \in G[/imath] there is [imath]a^{\prime}\in G[/imath] such that [imath]a\ast a^{\prime}=e[/imath] (right inverse). Show that [imath](G,\ast )[/imath] need not be a group. I can't think of a counter example that satisfies this not being a group. | 935277 | Why is a monoid with right identity and left inverse not necessarily a group?
This problem is from Herstein's 'Topics in Algebra'. I've thought about it a bit but haven't come up with much. Let [imath]G[/imath] be a non-empty set with an associative product which also satisfies: [imath]\exists e\in G [/imath] such that [imath]\forall a \in G[/imath], [imath]a \cdot e=a[/imath]. Given [imath]a \in G[/imath], [imath]\exists y(a) \in G[/imath], such that [imath]y(a) \cdot a =e[/imath]. Prove that [imath]G[/imath] need not be a group. I know that [imath]G[/imath] is a group if any one of those multiplications is switched around, i.e, either [imath]a \cdot y(a)=e[/imath] or [imath]e \cdot a=a[/imath]. But I can't quite understand why in this case [imath]G[/imath] is not a group. I'd be much obliged if someone can give me a good explanation of why it must be and I'd also appreciate a counter-case if nothing else. Thanks a lot. Cheers! |
147199 | A maximal ideal among those avoiding a multiplicative set is prime
Let [imath]S[/imath] be a multiplicatively closed subset of a ring [imath]R[/imath], and let [imath]I[/imath] be an ideal of [imath]R[/imath] which is maximal among ideals disjoint from [imath]S[/imath]. Show that [imath]I[/imath] is prime. If [imath]R[/imath] is an integral domain, explain briefly how one may construct a field [imath]F[/imath] together with a ring homomorphism [imath]R\to F[/imath]. Deduce that if [imath]R[/imath] is an arbitrary ring, [imath]I[/imath] an ideal of [imath]R[/imath], and [imath]S[/imath] a multiplicatively closed subset disjoint from [imath]I[/imath], then there exists a ring homomorphism [imath]f\colon R\to F[/imath], where [imath]F[/imath] is a field, such that [imath]f(x)=0[/imath] for all [imath]x\in I[/imath] and [imath]f(y)\neq 0[/imath] for all [imath]y\in S[/imath]. [You may assume that if [imath]T[/imath] is a multiplicatively closed subset of a ring, and [imath]0\notin T[/imath], then there exists an ideal which is maximal among ideals disjoint from [imath]T[/imath].] Here is a question I need to answer. For the first part I can show if [imath]x, y[/imath] are not in [imath]I[/imath] then nor does [imath]xy[/imath]. The second part is just the field of fractions. For the third part, I think I need to find an ideal [imath]J[/imath] containing [imath]I[/imath] such [imath]J[/imath] is prime, so that [imath]R/J[/imath] is an integral domain, and use the second part. To find [imath]J[/imath] prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from [imath]S[/imath] containing [imath]I[/imath], but how can I do that? | 2256638 | Ring theory- 'maximal' ideal that does not contain an element
Let [imath]R[/imath] be a commutative ring and [imath]S \subset R[/imath] a non empty subset with [imath]0_R \notin S[/imath], and such that [imath]s_1, s_2 \in S[/imath] implies [imath]s_1 \cdot s_2 \in S[/imath]. Let [imath]I[/imath] be an ideal of [imath]R[/imath] with [imath]I \cap S = \emptyset[/imath], and such that if [imath]J[/imath] is an ideal which contains [imath]I[/imath] then either [imath]J = I[/imath], or [imath]J \cap S \neq \emptyset[/imath]. I am wondering how to show that [imath]a \cdot b \in I \Rightarrow a \in I[/imath] or [imath]b \in I[/imath]. |
952212 | If [imath]m,n[/imath] are integers [imath]\gt 2[/imath] then [imath]R(m,n) \leq R(m-1,n)+R(m,n-1)[/imath]
THIS IS NOT THE DUPLICATE OF ABOVE BECAUSE:I require pigeonhole principle argument to my doubt which is not stated in the answer to above question... I missed my lecture the day the following theorem for ramsey numbers was proved in the class: If [imath]m,n[/imath] are integers [imath]\gt 2[/imath] then [imath]R(m,n) \leq R(m-1,n)+R(m,n-1)[/imath] Proof: [imath]R(m,n)=[/imath] smallest possible size of group of persons s.t. among these either [imath]\exists \,m[/imath] persons mutual friends or [imath]\exists \,n[/imath] mutual enemies. Let [imath]p=R(m-1,n)\,,q=R(m,n-1)[/imath] and [imath]r=p+q[/imath]. Consider the group [imath]\{1,2\ldots,r\}[/imath] of [imath]r[/imath] persons. Let L=[imath]\{[/imath]set of persons friends with '[imath]1[/imath]'[imath]\}[/imath] and M=[imath]\{[/imath]set of persons enemies with '[imath]1[/imath]'[imath]\}[/imath] Then [imath]L\cap M=\phi[/imath] .So, we have [imath]|L|+|M|=|L\cup M|=r-1=p+q-1[/imath] We cannot have [imath]|L|\leq p-1[/imath] and [imath]|M|\leq q-1[/imath] . [imath]\therefore[/imath] either [imath]|L| \geq p[/imath] or [imath]|M|\geq q[/imath]. This the starting paragraph of the proof which I have stated above. and later the proof is divided into two cases ,(1. when [imath]L[/imath] has atleast [imath]p[/imath] elements. 2.when [imath]M[/imath] has atleast [imath]q[/imath] elements),which I understood... But what I can't understand is the step at begining of proof that: We cannot have [imath]|L|\leq p-1[/imath] and [imath]|M|\leq q-1[/imath] . [imath]\therefore[/imath] either [imath]|L| \geq p[/imath] or [imath]|M|\geq q[/imath]. The above statement makes the use of pigeonhole principle but I don't know how.Please anyone who can help me with this ,as I have no idea for this..... | 939720 | Question about the proof of Ramsey's Theorem
I was reading up on a proof of Ramsey's Theorem and I can't understand this part of the proof: Pick a vertex [imath]v[/imath] from the graph, and partition the remaining vertices into two sets [imath]M[/imath] and [imath]N[/imath], such that for every vertex [imath]w[/imath], [imath]w[/imath] is in [imath]M[/imath] if [imath](v, w)[/imath] is blue, and [imath]w[/imath] is in [imath]N[/imath] if [imath](v, w)[/imath] is red. Because the graph has [imath]R(r - 1, s) + R(r, s - 1) = |M| + |N| + 1[/imath] vertices, it follows that either [imath]|M| \geq R(r − 1, s)[/imath] or [imath]|N| \geq R(r, s − 1)[/imath]. Why is that [imath]|M| \geq R(r − 1, s)[/imath] or [imath]|N| \geq R(r, s − 1)[/imath]? I know from the proof of [imath]R(3,3) = 6[/imath] that given a vertex, at least [imath]3[/imath] are guaranteed to be either red or blue, so what is the pigeonhole argument for the general case? |
973822 | how to solve [imath]\int\frac{1}{1+x^4}dx[/imath]
i want find the answer and metod of solve of [imath]\int\frac{1}{1+x^4}dx[/imath]. I know [imath]\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C[/imath], How I can use this to solve of that integration. | 973801 | How to evaluate [imath]\int_0^\infty \frac{1}{x^n+1} dx[/imath]
Noticed that the integral [imath]\int_0^\infty \frac{1}{x^n+1} dx[/imath] is often approached with partial fraction decomposition, but this gets increasingly ugly as [imath]n[/imath] gets bigger. Is there a neat trick to do these all in one fell swoop? Or a famous name for these integrals that I can look up for more info? |
973655 | A group is abelian
Let [imath]G[/imath] be a finite group. For any two elements [imath]a,b \neq e\in G[/imath] there exits an automorphism [imath]\sigma[/imath] such that [imath]\sigma(a)=b[/imath]. Prove that [imath]G[/imath] is abelian. Only thing that I could conclude about the problem is that all elements have the same order. | 74284 | A condition of abelian groups related to an automorphism
Let [imath]G[/imath] be a group. If for any [imath]a, b\in G\backslash\{1\}[/imath] there exists an automorphism [imath]\sigma[/imath] of [imath]G[/imath] such that [imath]\sigma (a)=b[/imath] then [imath]G[/imath] is abelian. I am not the best of the algebra and I have a problem with the proof of the fact above although I guess it must be very simple.. I will be grateful for hint. |
974014 | Integrate [imath]e^{x^2} dx[/imath]
I tried [imath]u[/imath]-substiution with [imath]u = x^2[/imath] but that leaves me with [imath]x[/imath] values and not a simpler function to integrate.Is there a better way to integrate? | 634975 | How can one prove the impossibility of writing [imath] \int e^{x^{2}} \, \mathrm{d}{x} [/imath] in terms of elementary functions?
Can we express [imath] \displaystyle \int e^{x^{2}} \, \mathrm{d}{x} [/imath] in terms of elementary functions? (Note: Infinite series are not allowed.) If not, then is there a proof that [imath] \displaystyle \int e^{x^{2}} \, \mathrm{d}{x} [/imath] cannot be written in such a manner? |
974264 | Harmonic conjugate extend to boundary
Suppose u is a harmonic function in disc [imath]|z|<1[/imath], and u can be extended continuously to boundary, what about its harmonic conjugate v? Can it also be extended continuous to boundary? I know v can be represented by integral use the boundary value of u, but I don't think this can help. Can anyone prove it or give a counterexample? Thank all of you. | 870574 | Dirichlet problem in the disk: behavior of conjugate function, and the effect of discontinuities
Dirichlet's problem in the unit disk is to construct the harmonic function from the given continuous function on the boundary circle. It is solved by the convolution with the Poisson kernel, and we know that the constructed function has uniform limit which is the given continuous function as [imath]r\to 1[/imath], and we can complexify the solution by taking the Schwarz integral formula. Now (in this very particular case) does the imaginary part has a uniform limit? If it does, why don't we prove the uniqueness of the solution by a simple application of the Cauchy integral formula? Here is the proof that I reckon to be reasonable(which is what I expose here to be criticised) We construct the Schwarz integral, or [imath]\frac{1}{2\pi}\int _{0}^{2\pi}\frac{\exp(it)+z}{\exp{(it)}-z}u(\exp{(it)})\mathrm{d}t[/imath]. This function is an analytic function in the interior unit disk, and is continuous in the closed unit disk(its closure). Let this analytic function denoted by [imath]f[/imath], and another analytic function that has the same uniform limit function on the circle be [imath]v[/imath]. Then [imath]u-v[/imath] is an complex analytic function in the interior of the unit disk, continuous on the closure, with boundary value identically zero. An application of Cauchy's integral formula gives rise to the conclusion that [imath]u-v[/imath] in identically zero on the entire disk. Another question being if the boundary function has finite jump discontinuities, what is the behaviour of the convolution with the Poisson kernel near the discontinuous points? |
974535 | How can I solve this recursion question, I am really stuck.
I am doing a couple of exercise questions, How do I show that if we let [imath]n \geq 1[/imath] be an integer, and if we consider [imath]n[/imath] people [imath]P_1[/imath],[imath]P_2[/imath],...,[imath]P_n[/imath]. If we let [imath]A_n[/imath] be the number of ways these [imath]n[/imath] people can be divided into groups, such that each group consist of either one or two people. How do we determine [imath]A_{1}[/imath], [imath]A_{2}[/imath], and , [imath]A_{3}[/imath] and how do we prove that for each integer [imath]n\geq 3[/imath], [imath]A_{n}[/imath] = [imath]A_{n-1}[/imath] + [imath]({n-1})[/imath] [imath]\cdot[/imath] [imath]A_{n-2}[/imath] | 969709 | How many combinations can I make?
let [imath]n \gt 1[/imath] be an integer, and consider [imath]n[/imath] people; [imath]P_1, P_2,..., P_n[/imath] let [imath]A_n[/imath] be the number of ways these [imath]n[/imath] people can be divided into groups, such that each group have either one or two people determine [imath]A_1, A_2, A_3[/imath] So I have [imath]A_1 = 1[/imath] way [imath]A_2 = 4[/imath] way [imath]A_3 = 12[/imath] way But I am not sure if this is right... Can anyone confirm? edit: My way of getting this is [imath]A_1 = 1 [/imath] because {P1} only has 1 way to sort [imath]A_2 = 4[/imath] because {P1,P2} = {P1,P2},{P2,P1},{P1P2},{P2P1} etc. |
974788 | Hard Mathematical Induction
I have a mathematical induction question and I know what I need to do just not how to do it. The question is: Prove the equality of: [imath](1 + 2 + . . . + n)^2 = 1^3 + 2^3 . . . + n^3[/imath] Base case: [imath](1 + 2)^2 = 1^3 + 2^3\\ (3)^2 = 1 + 8\\ 9 = 9[/imath] and I understand I have to get the sides to equal each other though I'm not sure how to do that: I use this: [imath](1 + . . . + n + (n + 1))^2 = 1^3 + . . . + n^3 + (n + 1)^3[/imath] but i can't seem to factor anything in anyway to figure it out . . . I've tried putting the [imath]S(n)[/imath] in the [imath]S(n + 1)[/imath]: [imath](1 + . . . + n + (n + 1))^2 = (1 + . . . + n)^2 + (n + 1)^3[/imath] but its just getting the [imath]-(n + 1)^3[/imath] on the first side I can't figure out... Any help would be amazing!!!! | 973242 | Sum of cubes proof
Prove that for any natural number n the following equality holds: [imath] (1+2+ \ldots + n)^2 = 1^3 + 2^3 + \ldots + n^3 [/imath] I think it has something to do with induction? |
975767 | An equivalence relation iff G≈H, where G and H are groups
Problem : Let [imath]S[/imath] be the relation G~H iff G is isomorphic to H. Show reflexive, transitivity and symmetric. First show G is automorphism, which will imply G~G. So the identity mapping gives us this automorphism. So G~G. My question is, is this the right process for showing reflexiveness? Now we show symmetry. Assume G is isomorphic to H. Thus there exists a mapping [imath]\phi:G ->H[/imath]. We must show H is isomorphic to G. Since [imath]\phi[/imath] is this mapping, consider [imath]\phi^{-1}.[/imath] This is a bijection since [imath]\phi[/imath] is a bijection. How do I show it is homomorphic? Finally show its transitive. This I have done in a previous problem. This is different than the duplicate as suggested because previously I had done it incorrectly. I know now I must do it this way. | 970414 | Double check [imath]G\sim H[/imath] iff [imath]G≈H[/imath]
Let [imath]S[/imath] be the collection of all groups. Define a relation on [imath]S[/imath] by [imath]G \sim H[/imath] iff [imath]G ≈ H[/imath]. Prove that this is an equivalence relation. So [imath]S[/imath] is partitioned into isomorphism classes. Proof: Let [imath]S[/imath] be a relation on S defined by [imath]G \sim H[/imath] iff [imath]G ≈ H[/imath]. Show reflexive, symmetric, transitive. It is obvious that [imath]G≈G[/imath] so [imath]G\sim G[/imath]. Assume [imath]G\sim H[/imath], thus [imath]G≈H[/imath], which implies that [imath]H≈G[/imath], so [imath]H\sim G[/imath], thus symmetric. Assume [imath]G\sim H[/imath] and [imath]H\sim K[/imath], for [imath]K[/imath] a group in [imath]S[/imath]. Thus [imath]G≈H[/imath] and [imath]H≈K[/imath]. (I have already shown that if [imath]G≈H[/imath] and [imath]H≈K[/imath], then [imath]G≈K[/imath]). From earlier this implies that [imath]G≈K[/imath], so [imath]\sim[/imath] is an equivalence relation. Thus [imath]S[/imath] is partitioned into isomorphism classes. QED. |
975800 | Independent events satisfying [imath]P(A_n)<1[/imath]
So I'm working on my mathematical probability class, and I have the following question: Suppose [imath]\{A_n\}[/imath] are independent events satisfying [imath]P(A_n)<1[/imath] for all [imath]n[/imath]. Show [imath]P(\bigcup _{n=1}^\infty A_n)[/imath]=1 iff [imath]P(A_n i.o.) =1[/imath]. So, the backwards direction was trivial. For the forward direction, what I have so far is [imath]P\left(\bigcup_{n=1}^{\infty}A_{n}\right)^{C}=0=P\left(\bigcap_{n=1}^{\infty}A_{n}^{C}\right)=\prod_{n=1}^{\infty}(1-P\left(A_{n}\right))[/imath] Since the [imath]A_n[/imath]'s are independent. Now, since [imath]P(A_n)<1[/imath], we have [imath]1-P\left(A_{n}\right)>0[/imath]. Hence, for the infinite product to be [imath]0[/imath], we must have an infinite number of [imath]1-P\left(A_{n}\right)<1[/imath], hence for an infinite number of [imath]n[/imath]'s, [imath]P\left(A_{n}\right)>0[/imath]. Now, I'm not sure if that's sufficient. Does that imply the desired conclusion that [imath]P(A_n i.o.)=1[/imath] I feel like it should....(Or did I make another mistake!) | 640832 | Borel-Cantelli lemma problem
Practice problem for exam: Let [imath]{A_n}[/imath] satisfy [imath]\sum\limits_{n=1}^\infty P(A_n \cap A^c_{n+1}) < \infty[/imath] and [imath]\lim\limits_{n\to \infty} P(A_n) = 0[/imath]. Show that [imath]P(\lim \sup A_n) = 0[/imath]. I can see that it is sufficient to show that [imath]P\left(\sum\limits_{n=1}^\infty A_n\right) < \infty[/imath] and then apply the Borel-Cantelli lemma, but I am having trouble showing this from the information given. I see an almost identical problem here: A variation of Borel Cantelli Lemma, however the hint there is not enough for me to make headway, and this problem is also different in which term gets the complement inside the first sum. |
975799 | Proving Multivairble Limit Exists
How do you deal with multivariable limits? We'll use the example [imath]f: \mathbb R ^2 \rightarrow \mathbb R[/imath] [imath]\lim _{(x,y) \rightarrow (0,0)}\frac{\sqrt{|xy|}}{\sqrt{x^2 + y^2}}[/imath] The limit doesn't exist, if [imath]x=y[/imath] we have the value [imath]1/\sqrt{2}[/imath] and if [imath]y = x^3[/imath] we get [imath]0[/imath]. How would we prove it with the [imath]\epsilon - \delta[/imath] proof? Do we even need to prove it through the definition, or does it suffice to show that approaching the point by different paths leads to different answers? Given that there are an infinite amount of paths in which to approach our point, how would you prove that the limit did actually exist? Take for example [imath]f(x,y) = xy[/imath], where we'll take the limit [imath]\lim _{(x,y)\rightarrow (1,2)} f(x,y) = 2 [/imath] This is obvious because the function is continuous at our point of interest, but how do you prove it directly from the definition? The same tricks we used when dealing with a single variable won't apply here due to the fact that we're dealing with a point, rather than a single number. | 316806 | Is there a step by step checklist to check if a multivariable limit exists and find its value?
Do we rely on certain intuition or is there an unofficial general crude checklist I should follow? I had a friend telling me that if the sum of the powers on the numerator is smaller then the denominator, there is a higher chance that it may not exists. And if the sum of powers on the numerator is higher then the denominator, most likely, it exists. Also if there is a sin or cos or e-constant, it most likely exists. How can I know what to do at the first glance given so little time exists for me in exam to ponder? If I spend all my time on figuring out a two-path test when the limit exists, that would be a huge disaster. Is this one of those cases where practice makes perfect? Example: [imath]\lim_{(x,y)\to(0,0)}\frac{(\sin^2x)(e^y-1)}{x^2+3y^2}[/imath] Please give me a hint and where do you get the hint. Example: [imath]\lim_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{4x^4+y^6}\right)[/imath] I need a hint for this too. Common methods I have learnt for reference: Two-Path test, Polar Coordinates, Spherical Coordinates, Mean Value Theorem using inequalities. |
975880 | How can you tile this checkerboard with trominoes?
Ok, so define a tromino as a [imath]1[/imath]x[imath]3[/imath] tile. If a corner is removed from an [imath]8[/imath]x[imath]8[/imath] board, is it possible to tile it with these trominoes? So far, I have concluded that you would need [imath]21[/imath] trominoes to tile the remaining [imath]63[/imath] squares. If you color the squares three different colors, it is possible to have an equal number of each. I alternated the diagonals for each color. Theoretically, a tiling should be possible right? I have been trying to find such a tiling, but each time I end up with three squares that cannot fit the shape of the given tromino. Are there any possible tilings? Thanks. | 944783 | Simple chessboard exercise
If we remove the square in the upper right corner of a [imath]8\times 8[/imath] chessboard. The question is: Is it possible to cover the entire remaining area, with [imath]1\times 3[/imath] chocolate bars? (they can be laid on the chessboard vertically and horizontally aswell) For the answer: We know that there will be [imath]63[/imath] squares left on the chessboard, of which [imath]32[/imath] are white and [imath]31[/imath] are black, since the one in the upper right corner was black. [imath]63[/imath] is divisible by [imath]3[/imath], so it could be possible to cover the area, but that doesn't prove that an arrangement exists to do it. I'm not sure how to continue and am seeking help. Thanks. |
976639 | How to show that the limit of this sequence exists?
We define [imath]a_1=1[/imath] and [imath]a_{n+1}=\dfrac 12\left(a_n+\dfrac{2}{a_n}\right)[/imath]. I'd like to show that this sequence converges (and [imath]a_n\to\sqrt{2}[/imath]). So, I wanted to prove that [imath]a_n[/imath] is monotonic and bounded. I thought that [imath]a_n<\sqrt{2}[/imath], but I was wrong. Maybe this sequence isn't even monotonic. What do you think? I already know that if the limit exists, say [imath]l[/imath], then [imath]l=\dfrac{1}{2}\left(l+\dfrac{2}{l}\right)[/imath]. But we still need to prove that [imath]a_n[/imath] converges. Thanks. | 976600 | Show that [imath]a_n<\sqrt{2}[/imath] for every [imath]n\in\mathbb{N}[/imath].
We define [imath]a_1=1[/imath] and [imath]a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)[/imath]. I want to prove that [imath]a_n[/imath] converges. But first I'd like to show that [imath]a_n<\sqrt{2}[/imath] for every [imath]n\in\mathbb{N}[/imath]. I try to use induction. For a fixed [imath]n[/imath], I suppose that [imath]a_n<\sqrt{2}[/imath]. Any hint to prove that [imath]a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)<\sqrt{2}[/imath]? Thanks. |
976800 | Integrate [imath]\csc(x)^3dx [/imath]
I'm not sure trig identity to use. Would we use [imath]1+\cot^2(x) = \csc^2(x)[/imath]? I know that we break down into [imath]\csc^2(x)[/imath] and [imath]\csc(x)[/imath]. Could I get hints? | 112918 | Integrate [imath]\csc^3{x} \ dx[/imath]
I found these step which explain how to integrate [imath]\csc^3{x} \ dx[/imath]. I understand everything, except the step I highlighted below. How did we go from: [imath]\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%[/imath] to [imath]\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?[/imath] Thank you for your time! [imath] \int \csc^3 x\,dx = \int\csc^2x \csc x\,dx[/imath] To integrate by parts, let [imath]dv = \csc^2x[/imath] and [imath]u=\csc x[/imath]. Then [imath]v=-\cot x[/imath] and [imath]du = -\cot x \csc x \,dx[/imath]. Integrating by parts, we have: [imath]\begin{align*} \int\csc^2 x \csc x \,dx &= -\cot x \csc x - \int(-\cot x)(-\cot x\csc x\,dx)\\ &= -\cot x \csc x - \int \cot^2 x \csc x\,dx\\ &= -\cot x\csc x - \int(\csc^2x - 1)\csc x\,dx &\text{(since }\cot^2 x = \csc^2-1\text{)}\\ &= -\cot x \csc x - \int(\csc^3 x - \csc x)\,dx\\ &= -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx \end{align*}[/imath] From [imath]\int \csc^3 x\,dx = -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx[/imath] we obtain [imath]\begin{align*} \int\csc^3x\,dx + \int\csc^3 x\,dx &= -\cot x \csc x + \int\csc x\,dx\\ 2\int\csc^3 x\,dx &= -\cot x\csc x + \int\csc x\,dx\\ \int\csc^3x\,dx &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\ &=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx\\ &= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx\\ &= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\ &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x - \cot x|+ C \end{align*}[/imath] |
977092 | If I define [imath] I.J=\{ij : i \in I [/imath] & [imath] j \in J \} [/imath]. Then prove that it is not necessrily an ideal, where [imath]I,J[/imath] are ideals in a ring [imath]R[/imath].
If I define [imath] I.J=\{ij : i \in I [/imath] & [imath] j \in J \} [/imath]. Then prove that it is not necessrily an ideal, where [imath]I,J[/imath] are ideals in a ring [imath]R[/imath]. I have found one counter example in [imath]R[x,y,z][/imath] for [imath]I=<x,y>[/imath] & [imath]J=<x,z>[/imath] & [imath]R[/imath] is commutative ring without unity. But can anyone give me any other simple example. Roughly speaking observe what we have show that is [imath] i_1j_1 + i_2j_2 \neq i_0j_0 [/imath]. | 950840 | Which of the following is also an ideal?
If [imath]U,V[/imath] are ideals of a ring [imath]R[/imath], then which of the following is also an ideal of [imath]R[/imath]? [imath]U+V=\{u+v\mid u\in U,v\in V\}[/imath] [imath]U\cdot V=\{u\cdot v\mid u\in U,v\in V\}[/imath] [imath]U\cap V[/imath] My attempt: I have proved [imath]U+V[/imath] and [imath]U\cap V[/imath] are ideals of [imath]R[/imath]. My problem: I have problem about the second option. I think it is not ideal of [imath]R[/imath]. If I am wrong then please prove this otherwise give one counter example to show that [imath]U\cdot V[/imath] is not ideal of [imath]R[/imath]. |
977926 | [imath]\dfrac1a+\dfrac1b=\dfrac1c[/imath], [imath]a, b, c \in \mathbb{N}[/imath] with no common factor, find all solutions
Given [imath]\dfrac1a+\dfrac1b=\dfrac1c[/imath], where [imath]a, b, c \in \mathbb{N}[/imath] with no common factor, find all solutions. Actually, you can think this question as a follow up of this one. Today, I saw this question and thought whether such numbers really exits! And quickly I found some solutions: [imath](2,2,1), (3, 6, 2),[/imath] [imath](4, 12, 3),[/imath] [imath] (5, 20, 4),[/imath] [imath](6, 30, 5), (7, 42, 6), (8, 56, 7), (9, 72, 8) \ldots[/imath] (see my comments). Clearly, [imath](n+1, [/imath] [imath]n(n+1),[/imath] [imath]n)[/imath] for [imath]n \in \mathbb{N}[/imath] is a solution. Interestingly these are the only solutions I found. (Irrelevant after paw88789's comment) So, my question is: Is [imath](n+1, [/imath] [imath]n(n+1),[/imath] [imath]n)[/imath] for [imath]n \in \mathbb{N}[/imath] the characterization of the above problem? Update After paw88789's answer, I realize that there are solutions of other form also. So, I get back to my original question: Find all solutions to the above problem. | 1341162 | Given primitive solution to [imath]\frac{1}{a}+\frac{1}{b}=\frac{1}{c}[/imath], show [imath]a+b[/imath] is a perfect square
If [imath]a,b,c[/imath] are positive integers and [imath]\gcd(a,b,c)[/imath] is [imath]1[/imath]. Given that [imath]\frac{1}{a} + \frac{1}{b} = \frac{1}{c}[/imath] then prove that [imath]a+b[/imath] is a perfect square. I was trying to get something useful from the information given in the question but was unable to get that. |
978082 | Find the next number
I am given a sequence [imath]1, 11, 21, 1211, 111221, 312211, 13112221, \cdots[/imath] what is the next number? | 715981 | What is the next number?
What is the next number in the following set ? [imath]1,11,21,1211,111221, \ldots[/imath] |
978092 | show that [imath](p \to q) \vee (p \to r) \to (q \vee r)[/imath] and [imath]p\vee q\vee r[/imath] are logically equivalent
without using the truth table: Show that [imath](p \to q) \vee (p \to r) \to (q \vee r)[/imath] and [imath]p\vee q\vee r[/imath] are logically equivalent. | 978072 | Prove [imath]\;\big((p\rightarrow q) \lor (p \rightarrow r)\big) \rightarrow (q\lor r)\equiv p \lor q \lor r[/imath] without use of a truth table.
Without using the truth table, I need to prove: [imath]\big((p\rightarrow q) \lor (p \rightarrow r)\big) \rightarrow (q\lor r)\equiv p \lor r \lor q[/imath] Up until now, we've been using truth-tables to verify equivalences. So I'm a bit lost on where to begin without using a truth-table. |
525970 | Norm of an inverse operator: [imath]\|T^{-1}\|=\|T\|^{-1}[/imath]?
I am a beginner of funcional analysis. I have a simple question when I study this subject. Let [imath]L(X)[/imath] denote the Banach algebra of all bounded linear operators on Banach space X, [imath]T\in X[/imath] is invertible, then [imath]||T^{-1}||=||T||^{-1}[/imath]? Is this result correct? | 2263286 | Relation between operator norm of a matrix and norm of inverse
Given a matrix [imath]A[/imath] over [imath]\mathbb{R}[/imath], define the operator norm as [imath]\|A\|: = \sup\{\|A\mathbf{x}\| : \|\mathbf{x}\|=1\}[/imath]. If [imath]A[/imath] is invertible, I realize that in general we have [imath]\|A^{-1}\|\|A\|\geq 1[/imath]. My question: What is a concrete example where this inequality is strict? I can't think of one. |
976655 | Ideal and its number of generators
Consider an ideal [imath]I[/imath] in [imath]\mathbb{C}[x_1, x_2, x_3, x_4][/imath] such that [imath]I[/imath] is generated by [imath]x_1x_3, x_2x_3, x_1x_4,[/imath] and [imath]x_2x_4[/imath]. I think this ideal cannot be generated by two elements, but can't give a rigorous proof. Can anyone suggest a proof? | 702095 | Can [imath](X_1,X_2) \cap (X_3,X_4)[/imath] be generated with two elements from [imath]k[X_1,X_2,X_3,X_4][/imath]?
Can [imath](X_1,X_2) \cap (X_3,X_4)[/imath] be generated with two elements in the ring [imath]R=k[X_1,X_2,X_3,X_4][/imath]? Can it be generated with three elements? (Here [imath]k[/imath] is a field.) Thanks for any help. |
14977 | Supremum and Infimum
Let [imath]A[/imath] be a nonempty set of real numbers which is bounded below. Let [imath]-A[/imath] be the set of all numbers [imath]-x[/imath], where [imath]x \in A[/imath]. Prove that [imath]\inf A = -\sup(-A)[/imath]. We know that [imath]-A[/imath] is bounded above. Hence [imath]\sup(-A)[/imath] exists. So to proceed, one would show that [imath]\inf A \leq -\sup(-A)[/imath] and [imath]\inf A \geq -\sup(-A)[/imath]? Or [imath]-\inf A > \sup(-A)[/imath] and [imath]-\inf A < \sup(-A)[/imath]. I get the sense that one can use the following fact as well: (i) Let [imath]\alpha = \sup S[/imath]. If [imath]\beta < \alpha[/imath] then [imath]\beta[/imath] is not an upper bound for [imath]S[/imath]. | 1324921 | Define [imath]-S = \{-x \mid x \in S\}[/imath] , prove [imath]\sup(-S) = -\inf(S)[/imath] and [imath]\inf(-S) = - \sup(S)[/imath], with [imath]S[/imath] is bounded both sides
Define [imath]-S = \{-x \mid x \in S\}[/imath] Prove [imath]\sup(-S) = -\inf(S)[/imath] and [imath]\inf(-S) = - \sup(S)[/imath], with [imath]S[/imath] bounded both sides Can someone help me with this, i dont get this type of more "complex" inf and sup excersises, detailed explanation would be appreciated, thanks alot ! |
978989 | Why is [imath]i^2[/imath] equal to [imath]-1[/imath]?
In this KhanAcademy link at 2:25, Sal (the narrator) says that [imath]i^2[/imath] is negative 1 and he didn't explain why. Why is this so? What is the intuition behind it? | 144364 | Is the square root of a negative number defined?
I have been in a debate over 9gag with this new comic: "The Origins" And I thought, "haha, that's funny, because I know [imath]i = \sqrt{-1}[/imath]". And then, this comment cast a doubt: There is no such thing as sqrt(-1). The square root function is only defined for positive numbers. Sorry... This wasn't by ignorance of complex numbers. In the short round of arguing that happened there, the guys insisted that negative numbers don't have square roots, and that [imath]i[/imath] is defined such that [imath]-1 = i^2[/imath], which is not the same as [imath]i = \sqrt{-1}[/imath]. In their opinion, too, no respectable math textbook should ever say that [imath]i = \sqrt{-1}[/imath]; which is precisely how Wolfram MathWorld defines [imath]i[/imath]. Is there a consensus over if negative numbers have a square root? Are there some gotchas or disputes with the definitions I should be aware of in the future? |
979689 | Proving a certain sum is 1
I'd like to prove that, for any [imath]M,N\in\mathbb{N}[/imath] with [imath]M\leq N[/imath], and any [imath]n\in\mathbb{N}[/imath] with [imath]n\leq M[/imath], the sum: [imath]\sum\limits_{k=0}^n\frac{\binom{M}{k}\!\!\binom{N-M}{n-k}}{\binom{N}{n}}=1.[/imath] I have proved it for [imath]n=2[/imath] by hand, and got stuck on the induction because trying to extract the previous sum is always unsuccesful. Let me expand on that. I tried twice, and always got stuck. I'll tell you only about the first time, because it is the attempt that tried to go from the induction hypothesis to the thesis, and it lays out all the equalities used in the second attempt, which went the other way, and was also unsuccessful. The induction step is: [imath]\sum\limits_{k=0}^{n-1}\frac{\binom{M}{k}\!\!\binom{N-M}{n-k-1}}{\binom{N}{n-1}}=1.[/imath] Using that the following holds: [imath]\binom{N}{n}=\frac{N!}{n!(N-n)!}=\frac{N-n+1}{n}\binom{N}{n-1}\!\!\implies\!\!\!\!\binom{N}{n-1}=\frac{n}{N-n+1}\binom{N}{n},[/imath] one turns the inductive hypothesis into: [imath]\sum\limits_{k=0}^{n-1}\frac{\binom{M}{k}\!\!\binom{N-M}{n-k-1}}{\binom{N}{n}}\frac{N-n+1}{n}=1.[/imath] This first step is lucky since the extra factor doesn't depend on [imath]k[/imath], so can be extracted from the sum and brought to the right side. The next step, though, is less lucky. One can easily see that: [imath]\binom{N-M}{n-k}=\frac{(N-M)!}{(n-k)!(N-M-n+k)!}=\frac{N-M-n+k+1}{n-k}\binom{N-M}{n-k-1}\!\!\implies[/imath] [imath]\implies\!\!\!\!\binom{N-M}{n-k-1}=\frac{n-k}{N-M-n+k+1}\binom{N-M}{n-k}.[/imath] This shows us how going on like this would give us a factor that can't be taken outside the sum, and can therefore not be isolated to get to the sum of the thesis (Note: "thesis" is what I have to prove, so the statement for [imath]n[/imath]; I'm not sure if that term is used in English with that sense; it is in Italian). I tried using: [imath]\binom{N-M}{n-k}=\!\!\binom{N-M-1}{n-k-1}+\!\binom{N-M-1}{n-k}[/imath] as well, but that didn't get me any further. If we use the second equality above to substitute, we get: [imath]\sum\limits_{k=0}^{n-1}\frac{\binom{M}{k}\binom{N-M}{n-k-1}}{\binom{N}{n}}\frac{N-n+1}{n}\frac{n-k}{N-M-n+k+1}=1.[/imath] Any ideas? | 68564 | Is this binomial coefficient identity known?
I stumbled across this identity involving binomial coefficients this morning: If [imath]n[/imath], [imath]k[/imath], [imath]a[/imath], and [imath]b[/imath] are positive integers and [imath]n=a+b[/imath], then [imath] \binom{n}{k} =\sum_{i=0}^k \binom{a}{k-i}\binom{b}{i}.[/imath] The proof is trivial. (Partition a set [imath]N[/imath] of [imath]n[/imath] things into disjoint sets [imath]A[/imath] of [imath]a[/imath] things and [imath]B[/imath] of [imath]b[/imath] things. Now to pick [imath]k[/imath] items from [imath]N[/imath], either all [imath]k[/imath] are from [imath]A[/imath], or [imath]k-1[/imath] are from [imath]A[/imath] and [imath]1[/imath] is from [imath]B[/imath], or...) Has this identity appeared before (I expect the answer to be "Yes!") and if so, does it have a name? |
979177 | Does [imath]1.0000000000\cdots 1[/imath] with an infinite number of [imath]0[/imath] in it exist?
Does [imath]1.0000000000\cdots 1[/imath] (with an infinite number of [imath]0[/imath] in it) exist? | 1046368 | Is [imath]1.0000\cdots 0001[/imath] equal to [imath]1[/imath]
I was wondering if this is true: [imath]1.\underbrace{0000\dots0000}_{\infty}1 = 1[/imath] I tried expressing this as a limit, but I have no idea how to continue: [imath]\lim_{x\to-\infty}1+10^{x}[/imath] Also, I'm not so experienced with limits, etc. |
961679 | [imath]\binom{p^{\alpha}-1}{k} = (-1)^k\pmod{p}[/imath]?
I need to show that [imath]\binom{p^{\alpha}-1}{k} = (-1)^k\pmod{p}[/imath] for [imath]0 \leq k \leq p^{\alpha}-1[/imath]. Not really sure how to start going about this... how should I transform the term on the left? Thanks. | 961938 | Combinations of sets raised to the power of a prime modulus: $\binom{p^{\alpha}-1}k \equiv (-1)^k \pmod p$
This is a problem out of the text Introduction to the Theory of Numbers by Niven, Zuckerman, and Montogmery and I am having quite a bit of trouble with it. I tried to prove it directly, but that didn't work. The question is: Let [imath]$p$[/imath] be a prime and [imath]$\alpha$[/imath] a positive integer. Show that [imath]$\binom{p^{\alpha}-1}k\equiv(-1)^k \pmod p$[/imath] for [imath] 0 \leq k \leq p^\alpha -1[/imath]. If relevant, I did prove that [imath]${{p^{\alpha}}\choose k}\equiv0 \pmod p$[/imath] for [imath] 0 < k < p^\alpha [/imath] but I can't seem to incorporate it. |
981437 | comparing cardinality of infinte sets
Let's say I have two infinite sets: 1) the set of all functions from [imath]\mathbb{R}[/imath] to {0,1}; 2) the set of all polynomials whose coefficients are in [imath]\mathbb{R}[/imath]. Which is greater? I figure the polynomial set would be considered greater. | 233707 | Comparing the cardinality of sets
An exercise is the following: Compare the cardinality of the following sets: The class of all real numbers [imath]\mathbb{R} =: A[/imath] The class of all polynomials [imath]\mathbb{R}[X] =: B[/imath] The class of all real functions [imath]f:\mathbb{R} \to \{0,1\} =: C[/imath] ... The hint is to use the Cantor-Schröder-Bernstein-Theorem. Is it also possible to show that two sets [imath]A,B[/imath] satisfy [imath]|A| \leq |B|[/imath] and [imath]|A| \neq |B|[/imath]? Obviously there is a injective function [imath]f: A \to B[/imath]. But if you could show that there is no injective function [imath]f: B \to A[/imath] would that imply [imath]|A| \neq |B|[/imath] ? |
980470 | Find the continued fraction of the square root of a given integer
How to find the continued fraction of [imath]\sqrt{n}[/imath], for an integer [imath]n[/imath]? I saw a site where they explained it, but it required a calculator. Is it possible to do it without a calculator? | 265690 | Continued fraction of a square root
If I want to find the continued fraction of [imath]\sqrt{n}[/imath] how do I know which number to use for [imath]a_0[/imath]? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question. If anyone has a good site that answers these questions either, please let me know. Thanks! |
981783 | Proving [imath]\rm{card}(\Bbb Z)=\rm{card}(\Bbb N)[/imath]
So I'm trying to prove that the set of integers has the same cardinality as the set of naturals just using the definition, that is, I'm trying to find a bijective function between the two sets. I found this Why do the rationals, integers and naturals all have the same cardinality? but I couldn't quite find the answer. Thanks for any help. | 74820 | Cardinality and infinite sets: naturals, integers, rationals, bijections
I have alot of questions. Do Infinite sets have the same cardinality when there is a bijection between them? Are [imath]\mathbb{N}[/imath] and [imath]\mathbb{Z}[/imath] infinite sets? I assume they are, but why? Why does there not exist a bijection between them? Why do [imath]\mathbb{N}, \mathbb{Z}, \mathbb{Q}[/imath] have the same cardinality? I assume there exists a bijection between them, but how can I find this function? |
981994 | Bound on difference of two i.i.d. variables
Prove that for every two independent, identically distributed real random varaibles [imath]X,Y[/imath], [imath]Pr(|X-Y|\leq 2)\leq 3\cdot Pr(|X-Y|\leq 1)[/imath] [Source: The probabilistic method, Alon and Spencer] | 15251 | Random Variable Inequality
Doing a little reading over the break (The Probabilistic Method by Alon and Spencer); can't come up with the solution for this seemingly simple (and perhaps even a little surprising?) result: (A-S 1.6.3) Prove that for every two independent, identically distributed real random variables X and Y, [imath]Pr[|X-Y| \leq 2] \leq 3 Pr[|X-Y| \leq 1][/imath]. Thanks for your time. |
982530 | Let [imath]f[/imath] be entire in [imath]\Bbb C[/imath]. If [imath]Re(z)>0[/imath] [imath]\forall z \in \Bbb C[/imath]. Then prove that [imath]f(z)[/imath] is constant.
Let [imath]f[/imath] be entire in [imath]\Bbb C[/imath]. If [imath]Re(z)>0[/imath] [imath]\forall z \in \Bbb C[/imath]. Then prove that [imath]f(z)[/imath] is constant. Please read below before marking duplicate. If [imath]Re(f(z))>M[/imath] then taking [imath]|\frac 1f|<\frac 1M[/imath]. Applying Liouville's theorem we are done. But if [imath]Re(f)>0[/imath] then can we conclude [imath]|\frac 1f|<\infty[/imath] & [imath]\frac 1f[/imath] is bounded so [imath]f[/imath] is constant. If it is so then comment otherwise give me some solution. | 538221 | Let [imath]u:\mathbb{C}\rightarrow\mathbb{R}[/imath] be a harmonic function such that [imath]0\leq f(z)[/imath] for all [imath] z \in \mathbb{C}[/imath]. Prove that [imath]u[/imath] is constant.
Let [imath]u:\mathbb{C}\rightarrow\mathbb{R}[/imath] be a harmonic function such that [imath]0\leq u(z)[/imath] for all [imath] z \in \mathbb{C}[/imath]. Prove that [imath]u[/imath] is constant. I think i should use Liouville's theorem, but how can i do that? Help! |
980481 | no. of elements in [imath]\mathbb Z[i]/\langle 3+i\rangle[/imath].
[imath]\mathbb Z[i]/\langle 3+i\rangle[/imath] can be represented as :[imath]\{a+3b+\langle 3+i\rangle\big|~~a,b\in \mathbb Z\}[/imath] How shall I find the total no. of elements in [imath]\mathbb Z[i]/\langle 3+i\rangle[/imath].. please provide some hint how should I proceed? | 785522 | Number of elements in the ring [imath]\mathbb Z [i]/\langle 2+2i\rangle[/imath]
The question is : Show that [imath]I=\langle 2+2 i\rangle[/imath] is not a prime ideal of [imath]\mathbb Z[i][/imath]. Also find the number of elements in [imath]\mathbb Z[i]/I[/imath] and its characteristic. My try: I started with elements [imath]2, 1+i \in \mathbb Z[i][/imath]. Clearly, the product is [imath]0[/imath] in the coset representation. The part where i am struck at is the number of elements and the characteristic. Well, [imath]2+2i+\langle 2+2i\rangle=0+\langle 2+2i\rangle[/imath] Can we directly write [imath]i^2=-1[/imath] in the coset representation? |
981838 | [imath]X_n[/imath] are r.v.s, is it true that [imath]E[\sum_{n=1}^{\infty} X_n] = \sum_{n=1}^{\infty} E[X_n] [/imath]?
[imath]X_n[/imath] are r.v.s, is it true that [imath]E[\sum_{n=1}^{\infty} X_n] = \sum_{n=1}^{\infty} E[X_n] [/imath]? My feeling is that this is not necessarily true. But cannot come up with an example. Can someone provide a counterexample or give a proof for this statement? | 83721 | When can a sum and integral be interchanged?
Let's say I have [imath]\int_{0}^{\infty}\sum_{n = 0}^{\infty} f_{n}(x)\, dx[/imath] with [imath]f_{n}(x)[/imath] being continuous functions. When can interchange the integral and summation? Is [imath]f_{n}(x) \geq 0[/imath] for all [imath]x[/imath] and for all [imath]n[/imath] sufficient? How about when [imath]\sum f_{n}(x)[/imath] converges absolutely? If so why? |
983198 | Is the number 0.2343434343434.. rational?
Consider the following number: [imath]x=0.23434343434\dots[/imath] My question is whether this number is rational or irrational, and how can I make sure that a specific number is rational if it was written in decimal form. Also, is [imath]0.234[/imath] rational or irrational? | 198810 | Proof that every repeating decimal is rational
Wikipedia claims that every repeating decimal represents a rational number. According to the following definition, how can we prove that fact? Definition: A number is rational if it can be written as [imath]\frac{p}{q}[/imath], where [imath]p[/imath] and [imath]q[/imath] are integers and [imath]q \neq 0[/imath]. |
983370 | T is not compact and orthonormal sequence
I want to show that if [imath]\,T[/imath] is not compact then there exists an orthonormal sequence [imath]x_{n}[/imath] and [imath]R>0[/imath] such that [imath] \forall n\in \mathbb{N}\,\,\,\,\|T(x_{n})\|\geq R[/imath]. It is obvious by the definition of compact operator that we can find a sequence satisfying the above condition but how can we find an orthonormal one? Thank you. | 981486 | bounded operator [imath]T[/imath] is not compact then there exists an orthonormal sequence [imath]e_n[/imath] and [imath]d>0[/imath] such that [imath]\|T(e_n)\|>d[/imath] for all [imath]n\in\Bbb{N}[/imath]?
I want to prove that if a bounded operator [imath]T[/imath] is not compact then there exists an orthonormal sequence [imath]e_n[/imath] and [imath]d>0[/imath] such that [imath]\|T(e_n)\|>d[/imath] for all [imath]n\in\Bbb{N}[/imath]. Could someone helps me? I think that the fact that T is not compact implies that there exists an arbitrary sequence in H that doesn't contain convergent subsequence, not for a orthonormal sequence. Moreover, we know that if T is a compact operator and en an orthonormal sequence then ||T(en)|| converges strongly to 0 but not the inverse. We want to show this. |
982791 | Discrete Mathematics Symmetric Diffirence Proof
I've been trying to find a proof for the following problem but have been unable to come up with anything myself: Say we have A, B, C part of a universe U show that if [imath]A \Delta C = B \Delta C \rightarrow A = B[/imath] I've been able to use set operations to change the form of the problem into a logics one: [imath]{x | X \in A \cup C \vee x \notin A \cap C} [/imath] [imath]{x | x \in A \vee x \in C \vee x \notin (x \in A \wedge x \in C)}[/imath] A second thought was using the definition of symmetric diffirence where [imath]A \Delta C = (A \cup C) - (A \cap C)[/imath] Can anyone provide a step by step proof or point me in the right direction? | 537172 | [imath]A \oplus B = A \oplus C[/imath] imply [imath]B = C[/imath]?
I don't quite yet understand how [imath]\oplus[/imath] (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use [imath]\oplus[/imath]. For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not [imath]A \oplus B = A \oplus C[/imath] implies [imath]B = C[/imath] is true. I know that the venn diagram of [imath]\oplus[/imath] in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this: I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the [imath]\oplus[/imath], but I'm not sure if that's an adequate enough proof. On the other hand, I could be completely wrong about my reasoning. Also just for clarity's sake: Would [imath]A\cup B = A \cup C[/imath] and [imath]A \cap B = A \cap C[/imath] be proven in a similar way to show whether or not the conditions imply [imath]B = C[/imath]? A counterexample/ proof of this would be appreciated as well. |
983930 | How to determine roots of a complex polynomial
Let us consider an equation [imath]x^3+10x^2-100x+1729=0[/imath]. Will this equation have at least one complex root having modulus [imath]>12[/imath]? | 690033 | Show that the equation, [imath]x^3+10x^2-100x+1729=0[/imath] has at least one complex root [imath]z[/imath] such that [imath]|z|>12.[/imath]
Show that the equation, [imath]x^3+10x^2-100x+1729=0[/imath] has at least one complex root [imath]z[/imath] such that [imath]|z|>12.[/imath] |
984171 | About the rationality of [imath]1.1010010001\dots[/imath]
Let's define [imath]\rho=1.1010010001\dots[/imath] which can be expressed by: [imath]\rho=\frac{1}{10^{0}}\underbrace{+\frac{1}{10^{1}}}_{\text{power}+1}\underbrace{+\frac{1}{10^{3}}}_{\text{power}+2}\underbrace{+\frac{1}{10^{6}}}_{\text{power}+3}\underbrace{+\frac{1}{10^{10}}}_{\text{power}+4}+\dots[/imath] That is to say: [imath]\rho=\sum_{i=0}^{\infty}10^{-i(i+1)/2}[/imath] Can we express [imath]\rho[/imath] in a simple form? I think it's not a rational but I don't know how to prove it, any ideas? Thanks. | 778218 | Is [imath]0.1010010001000010000010000001 \ldots[/imath] transcendental?
Does anyone know if this number is algebraic or transcendental, and why? [imath]\sum\limits_{n = 1}^\infty {10}^{ - n(n + 1)/2} = 0.1010010001000010000010000001 \ldots [/imath] |
984721 | Applying the Implicit Function Theorem from R3 to R
Suppose [imath]f:\mathbb{R}^3 \rightarrow \mathbb{R}[/imath] is such that the Implicit Function Theorem applies to [imath]F(x,y,z) = 0[/imath] to determine [imath]z = f(x,y)[/imath], [imath]x=g(y,z)[/imath] and [imath]y=h(x,z)[/imath] in a neighborhood of a point [imath]a \in \mathbb{R}^3[/imath]. Use the Implicit Function Theorem to prove [imath]\frac{\partial f}{\partial x} \frac{\partial g}{\partial y} \frac{\partial h}{\partial z} = -1[/imath] at a I haven't used the implicit function on functions that aren't explicitly defined, so I'm a little thrown off as to how I can show those partial fractions equal -1. I know I can show things like given [imath]F(x, y, f(x,y)) = 0[/imath] I can take the partial with respect to x and get [imath]\frac{\partial F}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial x} +\frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0[/imath] and I can show similar with the other two definitions, but I can't figure out how to use these to show those partials with respect to each of the functions equals -1. | 942457 | understanding [imath]\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1[/imath]
In particular, when [imath]x=y=z[/imath] isn't the above expression equal to 1 and not -1? |
983623 | Let A be a non-empty set, and p an equivalence relation on A . Let a , b be an element of A . Prove that [ a ] = [ b ] is equivalent to apb
the question: a) Let A be a non-empty set, and p an equivalence relation on A . Let a , b be an element of A . Prove that [ a ] = [ b ] is equivalent to [imath]apb[/imath] b) If p is both an equivalence relation and (simultaneously) a partial order on A , describe p. this is currently what I have done for question a. there exists [imath]x[/imath] element of [b] intersect [b] therefore xRa and xRb therefore aRx by symmetry rule therefore aRb by transitive rule therefore by lemma 1 (which wass a proof that "if R is an equivalence relation on a set A, and if a,b element A satisfy aRb, then [a] = [b], that is, the equivalence classes in which a and b lie are identical.") [a] = [b] how can i now show that this is equivalent to [imath]apb[/imath]? furthermore how do i do question b? | 974488 | A question about equivalence relations.
Let [imath]A[/imath] be a non-empty set, and [imath]\rho[/imath] an equivalence relation on [imath]A[/imath]. Let [imath]a, b \in A[/imath]. Prove that [imath][a] = [b] \iff a \mathrel\rho b[/imath]. |
985232 | Is it an open problem about Riemman Hypothesis non-trivial zero?
Let's assume RH was correct, and [imath]1/2+Ki[/imath] is any one of non-trivial zero of [imath]\zeta[/imath], is following problem open? 1) [imath]K[/imath] is irrational number 2) [imath]K[/imath] is transcendental number | 121786 | Rational Roots of Riemann's [imath]\zeta[/imath] Function
A look at the first few zeros [imath]14.134725,21.022040,25.010858,30.424876,32.935062,37.586178,\dots[/imath] is in accord with Numerical evidence suggests that all values of [imath]t[/imath] (the imaginary part of a root of [imath]\zeta[/imath]) corresponding to nontrivial zeros are irrational (e.g., Havil 2003, p. 195; Derbyshire 2004, p. 384). (numbers and quote taken from here). What are the attempts to prove that all values of [imath]t[/imath] are irrational? Would it mean something to the distribution of primes, if one, some or plenty of rational roots [imath]\frac{1}{2}+i\frac{q}{r}[/imath] exist? |
986257 | Does every linearly independent set of n vectors in [imath]R^n[/imath] forms a basis in [imath]R^n[/imath]?
Basically does a vector set that is linearly independent in [imath]R^n[/imath] automatically span [imath]R^n[/imath]? My initial thought is yes, but is there some counterexample that can disprove this? | 810551 | If [imath]n[/imath] vectors are linearly independent, is their span [imath]\mathbb{R}^n[/imath]?
Have [imath]n[/imath] vectors in [imath]\mathbb{R}^n[/imath]. If the [imath]n[/imath] vectors are linearly independent, can we conclude that their span is [imath]\mathbb{R}^n[/imath]? |
986506 | Finding all the values of n, such that [imath] \varphi (n) = 12 [/imath]
I have not broken this down very far. I have come to the conclusion that there are infinitely many values for n where there exists 12 coprimes to n. Since there are infinitely many primes, and primes are coprimes to any number smaller than that prime, I reach that conclusion. Can anyone stop me and tell me where I'm going wrong or how to approach this. I've used the theorem: [imath] \varphi (n) = ((1-\frac{1}{p_{1}})...(1-\frac{1}{p_{k}})) [/imath] But to compute these values would one not need to write a program to run which slots in all these primes? | 101566 | Using Euler's Totient Function, how do I find all values n such that [imath]\phi(n)=12[/imath]?
How do I generalize the equation to be able to plug in any result for [imath]\phi(n)=12[/imath] and find any possible integer that works? |
986143 | Looking for proof of formula in WolframMathWorld article
I came across the formula (24) in the WolframMathWorld article on Web page http://mathworld.wolfram.com/TrigonometryAngles.html where no source of the proof could be identified by me. The formula is [imath]\prod_{k=1}^{\lfloor \frac n2\rfloor}\sin(\frac{k\pi}{n})=\sqrt{\frac{n}{2^{n-1}}}[/imath] I am near a proof but it is rather intricate and maybe awkward and much to complicated. Does anybody know the source of the original proof or is there an elementary homework-style one without deeper necessary experience ? | 70231 | How to prove those "curious identities"?
How to prove [imath] \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}[/imath] and [imath] \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}[/imath] |
986825 | Proving an orthonormal set is an orthonormal basis in Hilbert space
Consider a separable Hilbert space [imath]H[/imath], and [imath]\{g_n\}[/imath] is an orthonormal basis of [imath]H[/imath]. Now there is an orthonormal set [imath]\{f_n\}[/imath] that satisfies [imath]\sum_n\|f_n-g_n\|^2<1[/imath]. Show that [imath]\{f_n\}[/imath] is also an orthonormal basis of [imath]H[/imath]. I am not sure how to use the condition [imath]\sum_n\|f_n-g_n\|^2<1[/imath]. It implies [imath]\|f_n-g_n\|\rightarrow 0[/imath], which looks like [imath]f_n[/imath] and [imath]g_n[/imath] are almost the same when [imath]n[/imath] is large. But I don't know how to utilize that property. Thanks in advance! | 967474 | Proving that if [imath]\sum\|f_n-e_n\|^2< 1[/imath], [imath]\{f_n\}[/imath] is a complete sequence
Let [imath]\{e_n\}[/imath] be a complete orthonormal sequence in an Hilbert space [imath]H[/imath] and let [imath]\{f_n\}[/imath] be an arbitrary sequence of elements in [imath]H[/imath] s.t [imath]\sum_{n=1}^\infty\|f_n-e_n\|^2<1[/imath]Show that [imath]\{f_n\}[/imath]is a complete sequence First I thought simplifying the last inequality:[imath]1>\sum_{n=1}^\infty\|f_n-e_n\|^2=\sum_{n,k=1}^\infty|(e_n-f_n,e_k)|^2=\sum_{n,k=1}^\infty|\delta_{n,k}-(f_n,e_k)|^2[/imath]but I was stuck there. Incorrect Solution: Another option I thought about was using parsevall's generalized identity on [imath]f_n[/imath] means: [imath]0=(f_n,f)=\sum(f_n,e_n)\overline{(f,e_n)}[/imath] and since the inner product is non-negative, if [imath](f,e_n)=0[/imath] then [imath]f\equiv 0[/imath] and we are done and otherwise [imath](f_n,e_n)=0[/imath] which means that [imath]\forall n\in\mathbb N,f_n=0[/imath] and then by definition of inner product [imath]\{0\}[/imath] is indeed a complete sequence. What's incorrect in the 'incorrect solution' and how can I use the fact about the inequality? |
987159 | How to differentiate this negative power?
I'm reading the book "Calculus made easy" and I'm stuck with a step of a derivative with a negative power. Here is what is in the book: Case of a negative power. Let [imath]y=x^{-2}[/imath]. Then proceed as before [imath]\eqalign{y+dy&=(x+dx)^{-2}\\&=x^{-2}\left(1+\dfrac{dx}x\right)^{-2}.}[/imath] I don't understand what I have to do to go from [imath] (x+dx)^{-2} [/imath] to [imath] x^{-2} \cdot (1+dx/x)^{-2} [/imath] Any help will be much appreciated! | 927906 | Differentiation - simple case
In the book calculus made easy, page 22 the case of the negative power for [imath]y=x^{-2}[/imath] [imath]\begin{align} y+dy & =(x+dx)^{-2}\tag{1}\\ \\ & = x^{-2}\left(1+\frac{dx}{x}\right)^{-2}\tag{2} \end{align}[/imath] How do they get from [imath](1)[/imath] to [imath](2)[/imath]? And why isn't the binomial function applied to the second line but applied on the third line (the line that comes after line [imath](2))[/imath]? |
985619 | Real analytic methods for the following integral
A few days back, the following integral was posted [imath]\int_0^1 x^x(1-x)^{1-x}\sin(\pi x)\,dx=\frac{\pi e}{24}[/imath] The integral was answered using complex analysis tools but I am interested in other methods which do not use residue calculus. I am unable to come up with anything for this. I noticed that the integrand is symmetric about [imath]x=1/2[/imath] but this doesn't help much. Help is appreciated. Thanks! | 958624 | Prove that [imath]\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} [/imath]
I've found here the following integral. [imath]I = \int_{0}^{1}\sin{(\pi (1-x))}x^x(1-x)^{1-x}\,dx=\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\frac{\pi e}{24}[/imath] I've never seen it before and I also didn't find the evaluation on math.se. How could we verify it? If it is a well-known integral, then could you give a reference? |
988293 | How do I prove that this is or isn't isomorphic?
[imath]\mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_6[/imath]? How can I show that the groups are isomorphic? (Or not?) | 795919 | [imath]\mathbb Z_{mn}[/imath] isomorphic to [imath]\mathbb Z_m\times\mathbb Z_n[/imath] whenever [imath]m[/imath] and [imath]n[/imath] are coprime
How to show that [imath]\mathbb Z_{mn}[/imath] is isomorphic to [imath]\mathbb Z_m \times\mathbb Z_n[/imath] when [imath]m[/imath] and [imath]n[/imath] are coprime? It is easy to show that the natural map from [imath]\mathbb Z_{mn}[/imath] to [imath]\mathbb Z_m \times\mathbb Z_n[/imath] is a ring homomorphism. How to show that it is bijective? |
988622 | Is local compactness preserved by continuous closed onto functions?
I've just shown for a homework problem that if [imath]f[/imath] is an open continuous function from [imath]X[/imath] onto a [imath]T_2[/imath]-space [imath]Y[/imath], and [imath]X[/imath] is locally compact, then [imath]Y[/imath] is locally compact. I wonder, does this hold for closed continuous functions too? Intuitively, I want to say no, but I'm not sure if I can find a counter example. Am I correct? How can I prove / disprove it? | 118005 | Closed image of locally compact space
If [imath]X[/imath] is locally compact and [imath]f : X \rightarrow Y[/imath] is continuous closed and surjective, must [imath]Y[/imath] be locally compact? This seems like it should be relatively simply to answer, but I am unable to find either a proof or a counterexample. Any ideas? |
988879 | Why [imath]L^1[/imath] is not reflexive
We already known that [imath] (L^p(\Omega))^* = L^q(\Omega), [/imath] for all [imath]1\le p < \infty [/imath] and [imath]q[/imath] is the exponent conjugate to [imath]p[/imath]. So that, [imath]L^p(\Omega)[/imath] is reflexive with [imath]1<p<\infty[/imath]. However, [imath]L^1(\Omega)[/imath] is not, due to [imath] (L^\infty(\Omega))^* \supset L^1(\Omega), [/imath] but I don't know why. Can we find a element [imath]f\in (L^\infty(\Omega))^*[/imath] and [imath]f \notin L^1(\Omega)[/imath]? | 866312 | [imath]L^1(μ)[/imath] is finite dimensional if it is reflexive
If [imath](X,\Omega,\mu)[/imath] is a [imath]\sigma -[/imath] finite measure space, show that if [imath]L^1(X,\Omega,\mu)[/imath] is reflexive then it is finite dimensional. My attempt: I want to show there is a copy of [imath]\ell^1[/imath] in [imath]L^1(X,\Omega,\mu)[/imath]. For this suppose [imath]L^1(X,\Omega,\mu)[/imath] is infinite dimensional. there is a sequence [imath]\{x_n\}[/imath] of disjoint points of [imath]X[/imath]. For every n, Put [imath]\chi_n:=\chi(x_n)[/imath] (characteristic function ). define [imath]\phi:\ell^1\to L^1(X,\Omega,\mu)[/imath] such that [imath]\phi(\{a_n\})=\Sigma a_n \chi_n[/imath]. But I can not show that [imath]\phi[/imath] is an isometry. I think I did not in a correct way. Please help me. Thanks in advance. |
114800 | Infinite Series: Fibonacci/ [imath]2^n[/imath]
I presented the following problem to some of my students recently (from Senior Mathematical Challenge- edited by Gardiner) In the Fibonacci sequence [imath]1, 1, 2, 3, 5, 8, 13, 21, 34, 55,\ldots[/imath] each term after the first two is the sum of the two previous terms. What is the sum to infinity of the series: [imath]\frac{1}{2} + \frac{1}{4}+ \frac{2}{8} + \frac{3}{16} + \frac{5}{32} +\frac{8}{64} + \frac{13}{128} +\frac{21}{256} +\frac{34}{512}+ \frac{55}{1024} + \cdots[/imath] Now, I solved this using an infinite geometric matrix series (incorporating the matrix version of the relation [imath]a_n= \frac{a_{n-1}}{2}+ \frac{a_{n-2}}{4}[/imath]), and my students, after much hinting on my part, googled the necessary string to stumble across Binet's formula (which allows one to split the series into two simple, if rather messy, geometrics). Both of these are good methods, but neither really seems plausible for a challenge set for 15-18 year olds under exam conditions. So how is one supposed to do it? | 1634222 | How to prove this series about Fibonacci number: [imath]\sum_{n=1}^{\infty }\frac{F_{n}}{2^{n}}=2[/imath]?
How to prove this series result: [imath]\sum_{n=1}^{\infty }\frac{F_{n}}{2^{n}}=2[/imath] where [imath]F_{1}=1,~F_{2}=1,~F_n=F_{n-1}+F_{n-2},~~n\geq 3[/imath]. I have no idea where to start. |
989220 | Is 1^2^3 = [imath]1^{2^3}[/imath] or [imath](1^2)^3[/imath]
Caret ^ signs can be used to describe the power of numbers. Is [imath]1[/imath]^[imath]2[/imath]^[imath]3 = 1^{(2^3)}[/imath] or [imath](1^2)^3[/imath] How do you calculate it? Do you start with [imath]2^3[/imath] and then do [imath]1^8[/imath] or do you start with [imath]1^2[/imath] and then do [imath]1^3[/imath]? | 439075 | Incorrect notation in math?
Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, [imath]{{2^2}^2}^3[/imath] (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this? a radical expression with the root being a radical expression? [imath]\sqrt[\sqrt{2^3}]{2}[/imath] |
989171 | [imath]H,H'[/imath] normal in [imath]G,G'[/imath] respectively and [imath]G \cong G' [/imath] and [imath]H \cong H' [/imath] [imath]\implies[/imath] [imath]G/H \cong G'/H'[/imath] ?
Let [imath]H[/imath] be a normal subgroup of [imath]G[/imath] and [imath]H'[/imath] be a normal subgroup of [imath]G'[/imath] such that [imath]G \cong G' [/imath] and [imath]H \cong H' [/imath] , then is it true that [imath]G/H \cong G'/H'[/imath] ? Denoting by [imath]f[/imath] the isomorphism between [imath]G , G'[/imath] I tried the map [imath]g(xH)=f(x)H'[/imath] , but this didn't work , I know I somehow also have to incorporate the isomorphism between [imath]H,H'[/imath] . Please help | 966100 | Isomorphism of factor groups
I came across an intuitive statement on factor groups that I realized I did not know how to prove. Let [imath]G[/imath] be a group, and let [imath]H \unlhd G, K \unlhd G[/imath] such that [imath]H \cong K[/imath]. I want to show that [imath]G/H \cong G/K[/imath]. I'm not sure how exactly to construct such an isomorphism from an isomorphism [imath]\varphi: H \to K[/imath]. Is that the correct approach? By extending such an isomorphism to an automorphism on [imath]\varphi^{\uparrow}:G \to G[/imath], then we could apply the First Isomorphism Theorem to the composition of the natural homomorphism [imath]\mu: G \to G/K[/imath] and that automorphism (for [imath]\ker \mu\varphi^{\uparrow} = H[/imath], if such a [imath]\varphi^\uparrow[/imath] exists). Edit: Hamou points out this is false in the infinite case: can it still be salvaged in the finite case? Are there restrictions we can put on [imath]G, H,[/imath] and [imath]K[/imath] such that this is true? |
528034 | Fixed field of automorphism [imath]t\mapsto t+1[/imath] of [imath]k(t)[/imath]
I'm working on the following problem: Determine the fixed field of the automorphism [imath]t\mapsto t+1[/imath] of [imath]k(t)[/imath]. (Ex 7, section 14.1, Abstract Algebra by Dummit & Foote). Here is my attempt of a solution: Let be [imath]\sigma[/imath] the automorphism determined by [imath]t\mapsto t+1[/imath] and fixing [imath]k[/imath]. An arbitrary element of [imath]\alpha(t)\in k(t)[/imath] has the form [imath]\alpha(t)=\dfrac{p(t)}{q(t)}[/imath] with [imath]p,q\in k\left[ t\right][/imath]. We can take [imath]p,q[/imath] such that [imath](p,q)=1[/imath] and [imath]q[/imath] monic. If [imath]\alpha[/imath] is a fixed element by [imath]\sigma[/imath], then we have [imath]\alpha(t)=\alpha(t+1)[/imath], so that [imath]\dfrac{p(t)}{q(t)}=\dfrac{p(t+1)}{q(t+1)}[/imath] and finally [imath]p(t)q(t+1)=p(t+1)q(t)[/imath]. As [imath](p,q)=1[/imath], [imath]q(t)[/imath] must divide [imath]q(t+1)[/imath]. By the same reason, [imath]q(t)[/imath] must divide [imath]q(t+1)[/imath], and as we took [imath]q[/imath] monic, we conclude that [imath]q(t+1)=q(t)[/imath], ie, [imath]\sigma(q(t))=q(t)[/imath]. Now we note that, as [imath]\sigma[/imath] fixes both [imath]\alpha(t)[/imath] and [imath]q(t)[/imath], it must also fix [imath]p(t)[/imath]. Now I'm getting stuck. If the characteristic of the field [imath]k[/imath] is [imath]0[/imath], I can say both [imath]p[/imath] and [imath]q[/imath] have an infinite number of roots (if they're not constant polynomials), so both have to be constant. But if the characteristic is non-zero, I can't conclude in the same way. What can I say about the polynomials? Thanks in advance! | 86949 | What exactly is the fixed field of the map [imath]t\mapsto t+1[/imath] in [imath]k(t)[/imath]?
Suppose [imath]k[/imath] is a field, and [imath]k(t)[/imath] is the rational function field. If [imath]f(t)=P(t)/Q(t)[/imath] for some polynomials [imath]P(t)[/imath] and [imath]Q(t)\neq 0[/imath], then the map [imath]t\mapsto t+1[/imath] sends [imath]f(t)[/imath] to [imath]f(t+1)[/imath]. So the fixed field for this map the set of rational functions [imath]P(t)/Q(t)[/imath] such that [imath] \frac{P(t)}{Q(t)}=\frac{P(t+1)}{Q(t+1)}\text{ or }P(t)Q(t+1)=P(t+1)Q(t). [/imath] This does not seem awfully descriptive. Is there a more explicit description of what the fixed field looks like, (perhaps in terms of the form of [imath]P[/imath] and [imath]Q[/imath]?), in order to get a better handle on it? Later: With the wonderful help of Professor Suárez-Alvarez, I understand that the [imath]\sigma[/imath]-invariant polynomials in [imath]k[t][/imath] have all factors of form [imath]\phi_\alpha[/imath] where [imath]\phi_\alpha=\prod_{i=1}^{p-1} (t-\alpha-i)[/imath] for [imath]\alpha[/imath] a root of [imath]f[/imath] in [imath]k[/imath] when [imath]k[/imath] is algebraically closed. What happens when [imath]k[/imath] is not algebraically closed? I think you can still factor out polynomials of form [imath]\phi_\alpha[/imath] for [imath]\alpha[/imath] roots of [imath]f[/imath] in [imath]k[/imath]. But if [imath]f[/imath] does not split completely then there will be some [imath]\sigma[/imath]-invariant factor with not roots in [imath]k[/imath] in the factorization of [imath]f[/imath]. Is there anything more one can say to describe this remaining part? Or is that about the best we can do? Thank you for your time. |
990103 | Proving [imath]a^ab^bc^c\ge(abc)^{(a+b+c)/3}[/imath] for positive real numbers.
Prove that [imath]a^ab^bc^c\ge(abc)^{(a+b+c)/3}[/imath] where [imath]a,b,c\in\mathbb{R^+}[/imath] I tried using powered AM-GM but didn't get anything. please give me a hint to solve it. | 109783 | Prove [imath]a^ab^bc^c\ge (abc)^{\frac{a+b+c}3}[/imath] for positive numbers.
Prove that the following inequality holds [imath]a^a b^b c^c\ge (abc)^{\frac{a+b+c}{3}}[/imath] if [imath]a,b,c[/imath] are positive. I'm not sure how to handle these kinds of powers. Are there any "famous" but not so advanced inequalities that involve this kinds of expressions? I was trying to do something like ordering them without lost of generality because it's symmetric, but I get nowhere and I just make the thing even more complicated. I'm asking for solution rather than hints since this is my first time encountering these kinds of problems. |
988923 | [imath]2 \times 3 = 5+1[/imath] and [imath]2+3 = 5 \times 1[/imath]. When else can we switch the operators like this?
I noticed the following: [imath]2 \times 3 = 5+1[/imath] If you switch the operators, it is still true: [imath]2+3 = 5 \times 1[/imath] There is another obvious/trivial example where you can swap the operators: [imath]2\times 2 = 2+2[/imath] I think these are the only solutions (for positive integers). Can anyone give an elegant proof? | 219762 | [imath]a+b=c \times d[/imath] and [imath]a\times b = c + d[/imath]
There is a 'nice' relationship between the integers (1,5) and (2,3) as [imath]1+5=2 \times 3;[/imath] [imath]1\times 5 = 2 + 3.[/imath] So I tried to find all positive integers pairs [imath](a, b)[/imath] and [imath](c, d)[/imath] such that [imath]a+b=c \times d;[/imath] [imath]a\times b = c + d.[/imath] To find this, [imath]a, b, c, d[/imath] must satisfy [imath](a+1)(b+1)=(c+1)(d+1).[/imath] However, this condition is only necessary but not sufficient. Any idea? |
990766 | If the inner product of Ax with x is 0 for all x, then A=0.
Given matrix [imath]A\in M_{n}(\mathbb{C})[/imath], if [imath]\left<Ax,x\right>=0[/imath] for all [imath]x\in \mathbb{C^n}[/imath], then [imath]A=0_{n}[/imath]. (Here [imath]\left<a,b\right> = b^{\ast}a[/imath] where "*" is the conjugate transpose.) Can anyone help me prove this? By Schur's theorem, A is similar to a strictly upper-triangular matrix, which is nilpotent. I'm not sure if this will help me. ETA: In the comments I linked to an answer in which the author states "for [imath]\lambda\neq 0[/imath] we get [imath]\left<Ax,y\right>=-\frac{\bar{\lambda}}{\lambda}\left<Ay,x\right>[/imath]. For fixed [imath]x,y[/imath] we can vary [imath]\frac{\bar{\lambda}}{\lambda}[/imath], so that [imath]\left<Ax,y\right>=0[/imath]. Can someone help me understand that last statement, i.e. how we can get [imath]\frac{\bar{\lambda}}{\lambda}[/imath] to be [imath]0[/imath]? | 67244 | Question on equivalence of inner product
How to show that if [imath]A, B \in M_{n}(\mathbb{C})[/imath], and [imath]x,y\in \mathbb{C}^{n}[/imath] then the following two conditions are equivalent, 1.) [imath]\langle Ax,y\rangle =\langle Bx,y\rangle[/imath] 2.) [imath]\langle Ax,x \rangle= \langle Bx,x \rangle[/imath] for all [imath]x,y[/imath]? |
991191 | Entire function f(z) bounded above by 1/|Im(z)|
How do I show that an entire function [imath]f(z)[/imath] that satisfies [imath] |f(z)|\leq\frac{1}{|Im(z)|} \quad \forall z\in\mathbb C [/imath] is a constant function? | 377782 | Application of Liouville's Theorem
Let [imath]f(z)[/imath] be an entire function such that [imath]|f(z)|<\frac{1}{|\text{Im}(z)|},\qquad z\in\Bbb C-\Bbb R.[/imath] The question asked me to prove that [imath]f(z)=0[/imath]. At least looking at it, it really seems to have an application of Liouville's theorem lurking around somewhere, but I haven't found it. My thoughts first led me to think about doing this by contradiction and using Picard's little theorem. So, I've considered a strip containing the real axis (say of width [imath]2[/imath] for simplicity). This will imply the complement maps into the unit disk, which implies that this strip has to map almost everywhere else on the complex plane. Now, on the imaginary axis, I know that [imath]f(z)[/imath] will vanish as [imath]z\to\infty[/imath], and this seems like it might be useful. I'm sort of under the impression [imath]f(z)[/imath] might have an essential singularity at [imath]\infty[/imath], which would mean [imath]f(1/z)[/imath] has an essential singularity at [imath]0[/imath]. Either way, it of course can't have a pole at [imath]\infty[/imath] because of [imath]f(z)[/imath] vanishing on the imaginary axis, and if it is a removable singularity, it must be [imath]0[/imath], which still gives the solution. Therefore, I think it must have something to do with [imath]f(z)[/imath] having an essential singularity at [imath]\infty[/imath]. I was hoping to combine this with the above inequality and deduce a contradiction, but I haven't thought of one yet. Any hints or suggestions? I'm studying for a prelim. exam, so this isn't homework. Update: So, in the spirit of searching methods using the essential singularity at [imath]z=0[/imath], we have [imath]\left|f\left(\frac{1}{z}\right)\right|\leq\frac{1}{\left|\text{Im}\left(\frac{1}{z}\right)\right|}=\frac{|z|^2}{|y|}.[/imath] Thus, for [imath]|y|>1[/imath], we have [imath]\left|f\left(\frac{1}{z}\right)\right|<|y|\left|f\left(\frac{1}{z}\right)\right|<|z|^2.[/imath] Thus, if I can show that [imath]|f(1/z)|[/imath] is a polynomial, we'll know it can't have an essential singularity at the origin, concluding my proof. |
990536 | Product of bounded and convergent to [imath]0[/imath] sequence is a convergent to [imath]0[/imath] sequence
Let [imath]a_n[/imath] be a null sequence and let [imath]b_n[/imath] be a bounded sequence. Prove that [imath]a_n \cdot b_n[/imath] is a null sequence. I tried using the product rule of sequences but cannot because [imath]b_n[/imath] is not necessarily convergent and may not have a limit. How do I go about answering this? | 2666857 | The product of two sequences where one is bounded and the other is convergent to zero
I'm trying to prove, given only that one sequence [imath]B[/imath] is bounded and the other sequence [imath]A[/imath] is convergent to zero, that their product also converges. My work probably wouldn't be considered a formal proof but I'm curious if my strategy/thought process is on the right track. My work: Since [imath]A[/imath] converges to zero, there exists a point in this sequence where all terms are between [imath]-1[/imath] and [imath]1[/imath]. In this case, whatever behavior [imath]B[/imath] has will be dictated by the terms of [imath]A[/imath]. If [imath]B=0[/imath], then the conclusion is trivial. Thus, [imath]AB[/imath] converges to zero. |
985686 | Integral Contest
Before you answer this OP, please read all the terms and conditions below. Thank you... Today I hold an unofficial little contest on brilliant.org. Now, I will hold it here on Math S.E. It's just for fun guys. (>‿◠)✌ Before we start the contest, here are the rules of my little contest that you should obey as a contestant (the one who post an answer to this OP during contest period): The contest will be started on Sunday, October [imath]26[/imath], [imath]2014[/imath] at [imath]12.00[/imath] p.m. noon (UTC). So, until D-Day, please do not make any attempts (posting comments or answers) that can lead users here to answer the problem contest. You may post your answer after this OP is marked as a bounty question by me. I will give a bounty to this OP at least [imath]250[/imath] reps. The bounty will be awarded to the winner. The winner of this contest is the highest voted answer. So, the jury of this contest is not only me, but all of us. Each contestant can only post one answer and the answer can be posted anytime during the contest period. Once the answer is posted, it cannot be edited during the contest period. Since you have at least [imath]4[/imath] days to prepare the answer to this OP, please make sure you create a zero error answer. Do not post this contest problem on other sites for asking a hint or an answer. The contest will be ended on Sunday, November [imath]2[/imath], [imath]2014[/imath] (after bounty period over). The contestant will be disqualified for violating the contest rules above. The term disqualified means the contestant who violates the rules but she/ he owns the highest voted answer at the end of the contest will not be declared as the winner. Therefore, the bounty will not be awarded to her/ him. If necessary, the rules may be changed accordingly. To be fair, I will not answer this OP nor give any hints to anyone. I really need your cooperation in order to make this contest succeed. Thank you. Here is the contest problem: Prove\begin{equation}\large{\int_0^{\Large\frac{\pi}{2}}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta=\frac{\pi^2}{4}\end{equation} Feel free to use any methods to solve this problem (real or complex analysis approach). Okay, happy problem solving and best of luck! ٩(˘◡˘)۶ | 990337 | To determine a definite integral
I have been trying to solve the following integral [imath]\int_{0}^{\frac {\pi}{2}} \ln\left (\frac {\ln^2 (\sin x)}{\pi^2+\ln^2 (\sin x)}\right) \frac {\ln \cos x}{\tan x} dx[/imath] I tried substituting for [imath]\frac {\ln^2 (\sin x)}{\pi^2+\ln^2 (\sin x)}[/imath] and using the properties of definite integrals, but I am not able to proceed with the integral as the [imath]\ln \cos x[/imath] term doesn't get substituted. Is there any other trick that I can employ? |
991538 | Proof that [imath]||x||_0 = supp(x)[/imath]
I was reading a book on compressive sensing wherein they mentioned that in the limit, the zero'th norm of a vector is the number of non-zero elements in that vector. That is, [imath]\lim_{p\rightarrow 0} ||x||_p = ||x||_0 = | \text{supp}(x) |[/imath] where [imath]\text{supp}(x)[/imath] is the number of non-zero elements in [imath]x[/imath]. I could not prove this. Any help would be greatly appreciated. | 13125 | What is the 0-norm?
On [imath]\mathbb{R}^n[/imath] and [imath]p\ge 1[/imath] the [imath]p[/imath]-norm is defined as [imath]\|x\|_p=\left ( \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}[/imath] and there is the [imath]\infty[/imath]-norm which is [imath]\|x\|_\infty=\max _j |x_j|[/imath]. It's called the [imath]\infty[/imath] norm because it is the limit of [imath]\|\cdot\|_p[/imath] for [imath]p\to \infty[/imath]. Now we can use the definition above for [imath]p<1[/imath] as well and define a [imath]p[/imath]-"norm" for these [imath]p[/imath]. The triangle inequality is not satisfied, but I will use the term "norm" nonetheless. For [imath]p\to 0[/imath] the limit of [imath]\|x\|_p[/imath] is obviously [imath]\infty[/imath] if there are at least two nonzero entries in [imath]x[/imath], but if we use the following modified definition [imath]\|x\|_p=\left ( \frac{1}{n} \sum _{j=1} ^n |x_j| ^p \right ) ^{1/p}[/imath] then this should have a limit for [imath]p\to 0[/imath], which should be called 0-norm. What is this limit? |
991983 | Proof for divisions that in include prime number.
How do I prove, that if [imath]m^2[/imath] can be divided [imath]p[/imath] (where [imath]m[/imath] is a whole number and [imath]p[/imath] is a prime number) then also m can be divided by [imath]p[/imath]? | 768955 | How to prove if [imath]n[/imath] is prime and [imath]n | a^2[/imath] then [imath]n | a[/imath]?
My professor assigned this for homework but I don't understand how to connect the dots. He suggested using the fact that [imath]\gcd (x,y) \cdot \operatorname{lcm} (x,y) = xy[/imath] but I'm not sure how that's relevant. |
992230 | Limit of the reciprocal of the mean harmonic
Suppose [imath](x_n)[/imath] is a convergent sequence such that [imath]x_n>0 \forall n \in \mathbb{N}[/imath]. Set [imath]y_n=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}, n\in\mathbb{N}[/imath] Prove [imath]\lim_{n\to\infty}{x_n}=\lim_{n\to\infty}{y_n}[/imath] <- How should I approach this? Also prove that if [imath]x_n[/imath] diverges to [imath]+\infty[/imath], [imath](y_n)[/imath] diverges to [imath]+\infty[/imath] <- I suppose I should be able to do this if I prove the above statement? | 989242 | Proving one limit is equal to another
Let ([imath]x_s[/imath]) be a convergent sequence, where [imath]x_s>0[/imath] for all s, and [imath]y_s[/imath] be a sequence such that [imath]\displaystyle \Large y_s=\frac{s}{\frac{1}{x_1}+...+\frac{1}{x_s}}[/imath] Prove [imath]\displaystyle \lim x_s[/imath]=[imath]\lim y_s[/imath] So I know things like [imath]x_s[/imath] is bounded, and that it's Cauchy, but I can't think of anything to solve it. Any help is appreciated. |
982409 | Given [imath]n \sim r \iff n \equiv r \pmod d[/imath], prove [imath]\sim[/imath] is an equivalence relation.
It is given that n belongs to Z and d belongs to N. How do I prove that n=r mod d defines equivalence relation? I know I have to prove it is reflexive, symmetric and transitive. But how do I do that? | 306429 | prove that [imath]a \equiv b \mod m[/imath] is an equivalence relation on the integers
prove that [imath]a \equiv b \mod m[/imath] is an equivalence relation on the integers I believe there are 3 properties that it must meet to prove and equivalence relationship. Any reference material would be greatly appreciated |
978338 | Product notation [imath]\prod[/imath] when product does not commute
This is kind of a dubious question, but is the product notation [imath]\prod[/imath] often used in noncommutative rings? For example, if [imath]M_i[/imath] are matrices, I guess the common definition of [imath]\prod[/imath] is [imath]\prod_i M_i=M_1\,M_2\dots[/imath] but is it also sometimes used as [imath]\prod_i M_i=\dots M_2\,M_1\text{ ?}[/imath] If not, is there a common notation for the latter product? | 149398 | Notation for infinite product in reverse order
This question is related to notation of infinite product. We know that, [imath] \prod_{i=1}^{\infty}x_{i}=x_{1}x_{2}x_{3}\cdots [/imath] How do I denote [imath] \cdots x_{3}x_{2}x_{1} ? [/imath] One approach could be [imath] \prod_{i=\infty}^{1}x_{i}=\cdots x_{3}x_{2}x_{1} [/imath] I need to use this expression in a bigger expression so I need a good notation for this. Thank you in advance for your help. |
992932 | Expressing an arbitrary natural number as the difference of squarefree ones
Is it true that [imath]\forall n \in \mathbb{N}, \exists x, y \in \mathbb{N}: x,y[/imath] are squarefree, & [imath]n = x-y[/imath]? I think so, but I am not sure that I am quite proving it. Here is my line of thought: Let [imath]S \subset \mathbb{N}[/imath] be the set of squarefree natural numbers. Represent upper density by [imath]D[/imath]. Then we know that [imath]D(S) = \frac{6}{\pi^2} > 0[/imath]. There is a proposition that says that, since [imath]S[/imath] has positive upper density, [imath]\exists n \in \mathbb{N}: D(S \cap (S-n)) > 0[/imath] too. Then [imath]\exists x,y \in S: y = x-n[/imath]. Therefore [imath]\exists x,y \in S: n = x-y[/imath]. So, I think that I have proven existence of at least one such [imath]n[/imath], but I am not sure that I have universality. The statement is supposed to hold for all [imath]n \in \mathbb{N}[/imath]. | 482496 | Two problems about Squarefree numbers
(1) For every [imath]n\in \Bbb N[/imath], there exist squarefree numbers [imath]x[/imath] and [imath]y[/imath] such that [imath]x+y=n[/imath], i.e., every natural number can be written as the sum of two squarefree numbers; (2) For every [imath]n\in \Bbb N[/imath], there exist squarefree numbers [imath]k[/imath] and [imath]l[/imath] such that [imath]k-l=n[/imath], i.e., every natural number can be written as the difference of two squarefree numbers. For the squarefree numbers, I only know the definition of it, and I really did not quite understand how to prove these two from the definition. Any help, please. Thanks in advance. |
991280 | The ideal [imath](x_1,x_2,x_3,.........)[/imath] (with infinitely many variables) in the ring [imath]K[x_1,x_2,x_3,....][/imath] is not finitely generated.
How can I show that the ideal [imath](x_1,x_2,x_3,...)[/imath] (with infinitely many variables) in the ring [imath]K[x_1,x_2,x_3,...][/imath] is not finitely generated. I cannot complete the arguments as I thought if the ideal is generated by finitely many polynomials say [imath]f_1,f_2,........f_n[/imath] then these polys will contain only finitely many variables. Let [imath]x_k[/imath] be the variable with highest subscript which will be involved in [imath]f_i 's[/imath]. So [imath]x_{k+1}[/imath] will not appear in any of the [imath]f_i's[/imath]. My target is to show that [imath]x_{k+1}[/imath] cannot be written as [imath]K[x_1,x_2,x_3,.....][/imath] linear combination of [imath]f_i's[/imath]. If possible suppose [imath]x_{k+1}[/imath]=[imath]g_1.f_1 +g_2.f_2+.........+g_n.f_n[/imath]. I am trying by putting [imath]x_1=x_2=...=x_k[/imath]=[imath]1[/imath] but then I can't argue further. I want this to be completed in this way. Help me. Thank You. | 978361 | An ideal which is not finitely generated
Let [imath]K[/imath] be a field and [imath]A=K[x_1,x_2,x_3,...][/imath]. Prove that the ideal [imath]I:=\langle x_i: i \in \mathbb N\rangle[/imath] is not finitely generated as [imath]A[/imath]-module. I have no idea what can I do here, I mean, suppose [imath]I[/imath] is finitely generated, then there is a subset [imath]S \subset I[/imath] with [imath]S=\{x_{i_1},...,x_{i_n}\}[/imath] such that [imath]\langle\{x_{i_1},...,x_{i_n}\}\rangle=I[/imath]. How can I arrive to a contradiction? Any suggestions would be appreciated. |
983259 | The ideal [imath]\langle x,y \rangle[/imath] in [imath]F[x,y][/imath] is not principal.
Let [imath]F[/imath] be a field. Apparently we know that [imath]\langle x,y \rangle \neq \langle g(x,y) \rangle[/imath] for any [imath]g \in F[x,y][/imath]. Why is this the case? | 110708 | The ideal [imath]I= \langle x,y \rangle\subset k[x,y][/imath] is not principal
The ideal [imath]I= \langle x,y \rangle\subset k[x,y][/imath] is not a principal ideal. I don't know how to consider it. Any suggestions? |
995071 | How to calculate decimal factorials, like [imath]0.78![/imath]
When I enter [imath]0.78![/imath] in Google, it gives me [imath]0.926227306[/imath]. I do understand that [imath]n! = 1\cdot2\cdot3 \cdots(n-1)\cdot n[/imath], but not when [imath]n[/imath] is a decimal. I also have seen that [imath]0.5!=\frac12\sqrt{\pi}[/imath]. Can anyone explain me how to calculate decimal factorials? Thanks a lot! | 140926 | What is the definition of [imath]2.5![/imath]? (2.5 factorial)
I was messing around with my TI-84 Plus Silver Edition calculator and discovered that it will actually give me values when taking the factorial of any number [imath]n/2[/imath] where [imath]n[/imath] is any integer greater than [imath]-2[/imath]. Why does this happen? I thought factorials were only defined for positive integers and [imath]0[/imath], so what is my calculator doing to get the answer [imath]3.32335097[/imath] when I enter [imath]2.5![/imath]? Is there actually a definition of [imath]2.5![/imath] or is my calculator just being weird? How is the factorial function implemented? I understand the binary implications of [imath]2.5[/imath], so that could possibly have something to do with it. I get a domain error when trying to take the factorial of [imath]-1, 2.3, e[/imath], and any number that is not of the form [imath]n/2[/imath] where [imath]n[/imath] is any integer greater than [imath]-2[/imath]. |
995285 | Is the following series converging [imath]\sum_{n=2}\dfrac{1}{(\ln n)^{\ln n}}[/imath]
Is the following series converging [imath]\sum_{n=2}\dfrac{1}{(\ln n)^{\ln n}}[/imath] I am not able to compare this with anything, can some show the way | 652345 | Test for convergence of the series [imath]\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}[/imath]
Could I have a hint for testing the convergence of the following series please? [imath]\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}[/imath] Edit The integral test does not work because [imath]\int_1^n\frac{1}{(\ln x)^{\ln x}}dx[/imath] has not an elementary primitive. Thank You. |
995969 | Does this version of Schröder-Bernstein-Cantor imply choice?
Consider the following statement: [imath](*)[/imath] For all sets [imath]A[/imath],[imath]B[/imath] and surjections [imath]f\colon A \rightarrow B[/imath], [imath]g\colon B \rightarrow A[/imath] there is a bijection [imath]h\colon A \rightarrow B[/imath] Given choice, this follows from the Bernstein-Schröder-Cantor Theorem. Is it also true that choice holds in any given model of [imath]ZF+(*)[/imath]? Maybe I shouldn't ask this question right now (considering how tired I am), but somehow it appears to me that [imath](*)[/imath] might be weaker than choice and I wonder if there is a well-known equivalent axiom over [imath]ZF[/imath]. | 176972 | Is there a Cantor-Schroder-Bernstein statement about surjective maps?
Let [imath]A,B[/imath] be two sets. The Cantor-Schroder-Bernstein states that if there is an injection [imath]f\colon A\to B[/imath] and an injection [imath]g\colon B\to A[/imath], then there exists a bijection [imath]h\colon A\to B[/imath]. I was wondering whether the following statements are true (maybe by using the AC if necessary): Suppose [imath]f \colon A\to B[/imath] and [imath]g\colon B\to A[/imath] are both surjective, does this imply that there is a bijection between [imath]A[/imath] and [imath]B[/imath]. Suppose either [imath]f\colon A\to B[/imath] or [imath]g\colon A\to B[/imath] is surjective and the other one injective, does this imply that there is a bijection between [imath]A[/imath] and [imath]B[/imath]. |
996182 | If [imath]f : D(0,1) \rightarrow \mathbb{C}[/imath] is a function, [imath]f^2[/imath] is holomorphic, and [imath]f^3[/imath] is holomorphic, then prove that [imath]f[/imath] is holomorphic.
If [imath]f : D(0,1) \rightarrow \mathbb{C}[/imath] is a function, [imath]f^2[/imath] is holomorphic, and [imath]f^3[/imath] is holomorphic, then prove that [imath]f[/imath] is holomorphic. MY ATTEMPT SO FAR: If [imath]f^3[/imath] is holomorphic, then we can say that [imath]\frac{\partial}{\partial z}(f^3)[/imath] must be holomorphic. Since the chain-rule for differentiation still applies with complex-valued functions, we can say that [imath]\frac{\partial}{\partial z}(f^3) = 3f^2\left(\frac{\partial f}{\partial z}\right) \implies 3f^2\left(\frac{\partial f}{\partial z}\right) \textrm{ is holomorphic}[/imath] Now it may be unclear whether this implies that [imath]\frac{\partial f}{\partial z}[/imath] is holomorphic even knowing that [imath]f^2[/imath] is holomorphic. Certainly the product of two holomorphic functions must be holomorphic, but the case in which one is holomorphic and the other isn't but the product of those two function is holomorphic isn't as clear, so it will be shown. Suppose we have two functions, [imath]g[/imath] and [imath]h[/imath]. Suppose [imath]g[/imath] is holomorphic on some open subset [imath]\Omega \subset \mathbb{C}[/imath]. Now suppose their product, defined as [imath]\eta = g\cdot h[/imath], is holomorphic. This means that [imath]\eta \cdot (1/g)[/imath] is holomorphic except at those points in [imath]\Omega[/imath] where [imath]g[/imath] is zero. Thus, [imath]h[/imath] must be also holomorphic at all points except those in which [imath]g[/imath] is zero, so [imath]h[/imath] is at least meromorphic. Using that, we can see that [imath]\frac{\partial f}{\partial z}[/imath] must be at least meromorphic, being holomorphic at all points with no guarantee as to whether it has singularities at those points where [imath]f^2[/imath] is zero. And that's where I'm stuck. I know that [imath]\frac{\partial}{\partial z}(f^3)[/imath] and [imath]3f^2[/imath] are related, but as long as the possibility that [imath]\frac{\partial f}{\partial z}[/imath] blows up at those points where [imath]f^2\rightarrow 0 [/imath] has to be considered, I'll stay confused. Is my approach alright, or should I completely change what I'm doing? Have I gone wrong anywhere? Should I be more explicit anywhere? Could you give me a small minor hint as to which direction I should go? | 783550 | [imath]f^2[/imath] and [imath]f^3[/imath] are holomorphic implies [imath]f[/imath] is holomorphic.
Suppose [imath]f[/imath] is a continuous complex valued function on a domain [imath]\Omega[/imath]. Suppose [imath]f^2[/imath] and [imath]f^3[/imath] are holomorphic in [imath]\Omega[/imath]. Show that [imath]f[/imath] is also holomorphic in [imath]\Omega[/imath]. Assume [imath]f=u+iv[/imath]. I see that if [imath]u,v[/imath] are in [imath]C^1[/imath] then [imath]f^2[/imath] is holomorphic can imply [imath]f[/imath] is holomorphic considering Cauchy-Riemann equations. But I don't know how to get [imath]u,v[/imath] are in [imath]C^1[/imath] by the adding condition [imath]f^3[/imath] is holomorphic. (I can't get [imath]u,v[/imath] from some combinations of real and imaginary part of [imath]f^2[/imath] and [imath]f^3[/imath]). Do someone know how to do this? |
995775 | Syntax of an epsilon delta proof/why is this version incorrect
So we have the regular [imath]\delta[/imath]-[imath]\epsilon[/imath] definition of continuity as: (1) For all [imath]\epsilon>0[/imath], there exists a [imath]\delta>0[/imath] such that, if [imath]|x-a|<\delta[/imath], then [imath]|f(x)-f(a)|<\epsilon[/imath]. My question is why is the following definition incorrect? (2) There exists a [imath]\delta>0[/imath] such that, if [imath]|x-a|<\delta[/imath], then [imath]|f(x)-f(a)|<\epsilon[/imath] for all [imath]\epsilon>0[/imath]. The obvious response is that [imath]\forall x[/imath] [imath]\exists y[/imath] [imath]\neq[/imath] [imath]\exists y[/imath] [imath]\forall x[/imath] (or rather, they are not always equal), but look harder at the grammar: is that necessarily what is going on? Let us define [imath]p:=[/imath] "There exists a [imath]\delta>0[/imath] such that, if [imath]|x-a|<\delta[/imath], then [imath]|f(x)-f(a)|<\epsilon[/imath]" So we now have the statement: "[imath]p[/imath] is true for all [imath]\epsilon>0[/imath]" Isn't that sentence identical to the English sentence: "for all [imath]\epsilon>0[/imath], [imath]p[/imath] is true"? In which case you would have (1): For all [imath]\epsilon>0[/imath], there exists a [imath]\delta>0[/imath] such that, if [imath]|x-a|<\delta[/imath], then [imath]|f(x)-f(a)|<\epsilon[/imath]. The counterargument that [imath]\forall x[/imath] [imath]\exists y[/imath] [imath]\neq[/imath] [imath]\exists y[/imath] [imath]\forall x[/imath] makes sense if you look at it immediately from a logical view, but because of the way English sentences work (and their vagueness, in a case like this), the "for all [imath]\epsilon>0[/imath]" clause can be placed anywhere without changing the meaning of the statement in English. To illustrate better what I'm talking about, let us imagine that we write our definition as follows: There exists a [imath]\delta>0[/imath] such that, if [imath]|x-a|<\delta[/imath], then [imath]|f(x)-f(a)|<\epsilon[/imath] (for all [imath]\epsilon>0[/imath]). Or, maybe: There exists a [imath]\delta>0[/imath] such that, if [imath]|x-a|<\delta[/imath], then [imath]|f(x)-f(a)|<\epsilon[/imath] (N.B. we're talking about all [imath]\epsilon>0[/imath] here!). Aren't these parentheticals sort of "overlying" the whole statement? Is the fault in my reasoning that [imath]\epsilon[/imath] is being "bound" to an if-then statement before it has been defined? Or am I just blatantly incorrect and this is a case of [imath]\forall x[/imath] [imath]\exists y[/imath] [imath]\neq[/imath] [imath]\exists y[/imath] [imath]\forall x[/imath]. The essential point of all of this is that, if you have the statement: [imath] \forall A \exists B : {B \subseteq A} [/imath] Then in English, this is equivalent to saying, both "For all A, there exists a B such that if B then A", and, "There exists a B such that if B then A, for all A". | 865871 | [imath]\forall[/imath] At the beginning or at the end?
I have a set of real numbers [imath]x_1, x_2, \ldots, x_n[/imath] and two functions [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] and [imath]g:\mathbb{R} \rightarrow \mathbb{R}[/imath]. What are the differences between the following statements? [imath]\forall v \in \{1, \ldots, n\} ~f(x_v) = 0 \vee g(x_v) = 0[/imath] [imath]f(x_v) = 0 \vee g(x_v) = 0 ~\forall v \in \{1, \ldots, n\} ~[/imath] |
996464 | Formal explanation for this change of integration
Formally, why is true that [imath]\int_0^\infty \int_{x}^\infty f(x,y)dy dx =\int_0^\infty \int_{0}^y f(x,y)dx dy [/imath] ? I know and understand perfectly the geometric interpretation, and with that, I´m well satisfied that this is true, but, what about the formal explanation? is a merely use of change variables theorem? or Fubinni´s Theorem? or a use of both? | 852127 | Integral change that I don't understand
If [imath]f(t)[/imath] is a Probability density function of a positive RV. [imath]\int_0^\infty\int_x^{\infty}f(t)dtdx[/imath] Using fubini theorem should become [imath]\int_0^\infty\int_0^{t}f(t)dxdt[/imath] But why? Surely the answer is very simple but I don't see it. Fubini theorem allows the change of the order of integration. So it should become [imath]\int_0^\infty\int_x^{\infty}f(t)dxdt[/imath] but then I don't understand what meaning could have [imath]\int_x^{\infty}dx[/imath] |
992189 | Newton's Method Convergence
Let r be the zero of multiplicity 2 of the polynomial p(x), how do I prove that [imath]x_{n}[/imath] converges quadratically to r? I only know the basic Newton's Method which is [imath]x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}[/imath], I can't think other way to prove that from this. I tried to googled, but couldn't find any reference for this. Thanks in advance. | 312819 | Rate of convergence of modified Newton's method for multiple roots
I've got a problem with a modified Newton's method. We've got a function [imath]f \in C^{(k+1)}[/imath] and [imath]r[/imath] which is it's multiple root of multiciplity [imath]k[/imath]. Also [imath]f^{(k)}(r) \neq 0[/imath] and [imath]f'(x) \neq 0 [/imath] in the neighbourhood of [imath]r[/imath]. The modified Newton's method is [imath]x_{n+1} = x_n - k\frac{f(x_n)}{f'(x_n)}[/imath] How to prove that [imath]x_n[/imath] converges to [imath]r[/imath] quadratically? I found one method using a function [imath]G(x) = (x-r) f'(x) - kf(x)[/imath] but then it's said that [imath] G^{(k+1)}(r) \neq 0[/imath] but it comes from the fact that [imath]f^{(k+1)}(r) \neq 0[/imath], but it doesn't necessarily have to be true, am I right? We know that [imath]f^{(k)}(r) \neq 0[/imath], but we have no information about the [imath]k+1[/imath]st derivative. I would appreciate if somebody explained to me precisely, why it quadratically converges. |
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