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1375212 | How is it that [imath]\pi[/imath] appears in so many formulas that seem to be in no way geometric.
When I first saw: [imath]\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\mp\cdots.[/imath] I was puzzled that such an expression could have anything to do with circles. There are tons more and it seems like a very large amount of summations will end up having to do with [imath]\pi[/imath]. My question is how come [imath]\pi[/imath] finds its way into so many formulas, and could the above equation be thought of geometrically. | 849348 | Geometric intuition for [imath]\pi /4 = 1 - 1/3 + 1/5 - \cdots[/imath]?
Following reading this great post : Interesting and unexpected applications of [imath]\pi[/imath], Vadim's answer reminded me of something an analysis professor had told me when I was an undergrad - that no one had ever given him a satisfactory intuitive explaination for why this series definition should be true (the implied assumption being that it is so simple, there should be some way to look at this to make it intuitive). Now it follows simply from putting [imath]x=1[/imath] in the series expansion [imath] \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots [/imath] but I don't see anything geometrically intuitive about this formula either! |
1375183 | Reorder this series to change its sum
If in the series [imath]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\cdots[/imath] the order of the terms be altered, so that the ratio of the number of positive terms to the number of negative terms in the first [imath]n[/imath] terms is ultimately [imath]a^2[/imath], show that the sum of the series will become [imath]\log 2a[/imath]. It is easy to show that when [imath]a^2=1[/imath], the sum is just the expansion of [imath]\log(1+x)[/imath] when [imath]x=1[/imath]. However I don't know how to prove it in the general case. | 116089 | Evaluating the sum after reordering an infinite series [imath]1-\frac {1} {2}+\frac {1} {3}-\frac {1} {4}+\ldots [/imath]
If in the series [imath]1-\dfrac {1} {2}+\dfrac {1} {3}-\dfrac {1} {4}+\ldots [/imath] the order of the terms be altered, so that the ratio of the number of positive terms to the number of negative terms in the first [imath]n[/imath] terms is ultimately [imath]a^{2}[/imath], show that the sum of the series will become [imath]\log \left( 2a\right) [/imath]. Solution attempt. Let [imath]p[/imath] be the number of positive terms in the first [imath]n[/imath] terms of the reordered series so based on the question we are allowed [imath]a^{2}=\dfrac {p} {n-p}[/imath]. Now solving for p we get [imath]p=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }[/imath] and [imath]n-p=\dfrac {n } {\left( 1+a^{2}\right) }[/imath]. We also observe that in the original series only the odd terms are positive and only the even terms are negative. Let's define [imath]S_{odd}=1+\dfrac {1} {3}+\dfrac {1} {5}+.\ldots +\dfrac {1} {2n-1}[/imath] and [imath]S_{even}=\dfrac {-1} {2}\dfrac {-1} {4}\ldots -\dfrac {1} {2n}[/imath] [imath]S_{Reordered}=S_{odd_{p}}+S_{even_{n-p}}[/imath] [imath]S_{Reordered}=\sum _{t=1}^{t=P}\dfrac {1} {2t-1}+\sum _{t=1}^{t=(n-P)}\dfrac {-1} {2t}[/imath] [imath]S_{Reordered}=\sum _{t=1}^{t=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\dfrac {n } {\left( 1+a^{2}\right) }}\dfrac {-1} {2t}[/imath] In order to extend [imath]S_{Reordered}[/imath] from n terms to an infinite length. I guess i should take the limit of the [imath]S_{Reordered}[/imath] as [imath]n\rightarrow \infty [/imath] which would make the upper values of [imath]t[/imath] (Not sure of technical term here) to be [imath]\infty[/imath] giving me [imath]S_{Reordered}=\sum _{t=1}^{t=\infty}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\infty}\dfrac {-1} {2t}[/imath] and i have lost the [imath]a[/imath] from the expression. I guess i am stuck i need to perform some step to capture a before taking the limit. Any help would be much appreciated. |
785355 | Proof via induction [imath]1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}[/imath]
(b) Prove that for every integer [imath]n \ge 1[/imath], [imath]1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}[/imath] This is the second part of a two part question. Part (a) was the following: Write the sum: [imath]1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2)[/imath] using summation notation. It was simple enough : [imath]\sum k(k+2)[/imath]. For this question, the base case [imath](n=1)[/imath] holds, as [imath]1\cdot(1+2) = 3 = (1\cdot2\cdot9)/6[/imath]. Induction step: Assume the above holds for all [imath]n = k[/imath], prove that it holds for all [imath]n = k+1[/imath] I'm a bit lost from here, help? | 72660 | [imath]1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = n(n+1)(2n+7)/6[/imath] by mathematical induction
I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side. Please guide me how to do it further. [imath]1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)[/imath]. Sol: [imath]P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)[/imath]. [imath]P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)[/imath]. [imath]P(1): 3[/imath]. Hence it is true for [imath]n=n_0 = 1[/imath]. Let it be true for [imath]n=k[/imath]. [imath]P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)[/imath]. We have to prove [imath]P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)[/imath]. Taking LHS: [imath]\begin{align*} 1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\ &= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\ &= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\ &= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\ &= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right]. \end{align*}[/imath] |
1375621 | Size of a triangle using determinant
find the size of a triangle using (determinant) with the following points: [imath](x_1,y_1)=(1,-2)[/imath] [imath](x_2,y_2)=(-4,-2)[/imath] [imath](x_3,y_3)=(-5,-1)[/imath] How should I place those points in the determinant? or firstly in a matrix? | 299352 | Show that the area of a triangle is given by this determinant
I'm not sure how to solve this problem. Can you guys provide some input/hints? Let [imath]A=(x_1,y_1)[/imath], [imath]B=(x_2,y_2)[/imath] and [imath]C=(x_3,y_3)[/imath] be three points in [imath]\mathbb{R}^{2}[/imath]. Show that the area of [imath]\Delta ABC[/imath] is given by [imath]\frac{1}{2}\left| \det(M) \right|[/imath], where [imath]M = \begin{pmatrix} 1 & 1 & 1\\ x_1 & x_2 & x_3\\ y_1 & y_2 & y_3 \end{pmatrix}[/imath] |
1375805 | question on second -> first order systems
I have heard that it is possible to write second order IVP as first order system. What are some strategies to writing [imath]y''=xy^2[/imath], [imath]y(0)=1[/imath], [imath]y'(0)=2[/imath] as a first order system [imath]y'=f(y)[/imath], [imath]y(a)=y_0[/imath]? Or perhaps a good tutorial to help me understand this better? | 501745 | How to reduce higher order linear ODE to a system of first order ODE?
Is there any general and systematic way of reducing the higher order linear ODE to a system of first order ODE? For example, assume we have [imath]a_3x^{(3)}+a_2x^{(2)}+a_1x^{(1)}+a_0x=0[/imath], then how do we convert this into matrix form(a system of first order ODE). And after we solve the system of equation, how to combine them into our final solution [imath]x(t)[/imath]? Thanks for helping me out! |
1375893 | Substitution for limits
How does substitution for limits exactly work? I see often answers that use the substitution [imath]t=\frac1x[/imath], then changing [imath]x\rightarrow\infty[/imath] to [imath]t\rightarrow0^+[/imath]. I have seen this question, this appears to solve my question in the finite case. How does it work when going to infinty? Books I'm using don't cover substitutions, but they seem very useful. | 456029 | Change of Variables in Limits
After answering this question, I was wondering if the following generalization holds true: Claim: If [imath]\lim \limits_{x\to a}g(x)=b[/imath], then [imath]\lim \limits_{x\to a}f(g(x))=\lim \limits_{y\to b}f(y)[/imath]. I've seen some people use this change of variables before when evaluating difficult limits, but I haven't seen this presented as a theorem in a textbook. Is this claim true? If not, is the claim salvageable with additional hypotheses? For example, must [imath]f[/imath] be continuous for the claim to hold? |
1364235 | The space of alternating multilinear maps and existence of a bilinear map
Before, I ask similar this. But here I change question settings since it was incomplete. I hope receive good ideas. Let [imath]E[/imath] be a finite dimensional vector space over field [imath]\mathbb R[/imath] with [imath]E^*[/imath] as the dual space of [imath]E[/imath] and [imath]E^{**}:=(E^*)^*[/imath] is the double duall of [imath]E[/imath]. ([imath]\dim E=n[/imath]) [imath]Alt^P(E):=\{ \alpha\colon \overbrace{E\times\cdots\times E}^{p- times}\rightarrow \mathbb R\ \ , \alpha \text{ is alternating multilinear map}\}[/imath] [imath]Alt^P(E^*):=\{ u\colon \overbrace{E^*\times\cdots\times E^*}^{p- times}\rightarrow \mathbb R\ \ , u \text{ is alternating multilinear map}\}[/imath] I wish to prove that there exist an [imath]\textbf{unique}[/imath] bilinear form [imath]B\colon Alt^p(E^*)\times Alt^p(E)\rightarrow\mathbb R[/imath] such that for all [imath] f_i\in E^* \ , u_j\in E^{**}\stackrel{\scriptsize{isomorph}}\simeq E \qquad i,j=1,\cdots,p[/imath], it has the following rule: [imath]B(u_1\wedge\cdots\wedge u_p,f_1\wedge\cdots\wedge f_p)=\det[f_i(u_j)][/imath]. [imath]\wedge[/imath] is wedge product between alternating maps i.e. [imath]\alpha\wedge\beta(v_1,\cdots,v_{p+p})=\dfrac{1}{p! p!}\displaystyle\sum_{\sigma\in S_{p+p}}sign(\sigma)\alpha(v_{\sigma(1)},\cdots,v_{\sigma(p)})\beta(v_{\sigma(p+1)},\cdots,v_{\sigma(p+p)})[/imath] How can I do this? | 1360424 | Uniqueness in existence of a bilinear form
Let [imath]E[/imath] be a finite dimensional vector space over field [imath]\mathbb R[/imath] with [imath]E^*[/imath] as the dual space of [imath]E[/imath].([imath]\dim E=n[/imath]) [imath]\Omega^P(E):=\{ \alpha\colon \overbrace{E\times\cdots\times E}^{p- times}\rightarrow \mathbb R\ \ , \alpha \text{ is alternating multilinear map}\}[/imath] [imath]\Omega^P(E^*):=\{ u\colon \overbrace{E^*\times\cdots\times E^*}^{p- times}\rightarrow \mathbb R\ \ , u \text{ is alternating multilinear map}\}[/imath] I wish to prove that there exist an [imath]\textbf{unique}[/imath] bilinear form [imath]B\colon \Omega^p(E^*)\times\Omega^p(E)\rightarrow\mathbb R[/imath] such that for all [imath] f_i\in E^* \ , u_j\in E^{**}\simeq E \qquad i,j=1,\cdots,p[/imath], it has the following rule: [imath]B(u_1\wedge\cdots\wedge u_p,f_1\wedge\cdots\wedge f_p)=\det[f_i(u_j)][/imath]. [imath]\wedge[/imath] is wedge product between alternating maps i.e. [imath]\alpha\wedge\beta(v_1,\cdots,v_{p+p})=\dfrac{1}{p! p!}\displaystyle\sum_{\sigma\in S_{p+p}}sign(\sigma)\alpha(v_{\sigma(1)},\cdots,v_{\sigma(p)})\beta(v_{\sigma(p+1)},\cdots,v_{\sigma(p+p)})[/imath] How can I do this? |
1376285 | Show that an irrationally periodic function is also a constant function
Let [imath]f:\mathbb R \to \mathbb R[/imath] be a function such that for any irrational number [imath]r[/imath], and any real number [imath]x[/imath] we have [imath]f(x)=f(x+r)[/imath]. Show that [imath]f[/imath] is a constant function. | 875068 | Functional equation with strange property about irrational numbers
Let [imath]f:\mathbb R \to \mathbb R[/imath] be a function such that for any irrational number r, and any real number x we have [imath]f(x)=f(x+r)[/imath]. Show that f is a constant function. It's easy to see any constant function satisfies the original properti... But I don't see how to show this is the only solution. |
1376353 | Proving that if [imath]0, then a<\sqrt{ab}<\frac{a+b}{2}[/imath]
In proving these inequalities, what I did was to take each of the "<" relations and prove them. Is this a valid way of proving if we have got several inequalities as in this problem? So here is my proof by parts. [imath]a<\sqrt{ab}[/imath]: Let [imath]P[/imath] be the set of all positive numbers. Since [imath]a,b,b-a\in P[/imath], we have [imath]a(b-a)=ab-a^2\in P[/imath], so [imath]a^2<ab[/imath], which implies that [imath]a<\sqrt{ab}[/imath]. (I make a separate proof for this implication, which goes as follows: [imath]a^2<b^2 \iff 0<b^2-a^2=(b-a)(b+a)\in P[/imath], so [imath](b-a)\in P[/imath], thus [imath]a<b[/imath]). So this proves [imath]a<\sqrt{ab}[/imath]. [imath]\displaystyle\sqrt{ab}<\frac{a+b}{2}:[/imath] We have [imath]\displaystyle\sqrt{ab}<\frac{a+b}{2}\iff2\sqrt{ab}<a+b\iff0<a-2\sqrt{ab}+b=(\sqrt{a}-\sqrt{b})^2[/imath]. Since [imath]a,b[/imath] are both non-zero, so [imath]0<(\sqrt{a}-\sqrt{b})^2[/imath] holds, thus, proving [imath]\displaystyle\sqrt{ab}<\frac{a+b}{2}[/imath]. [imath]\displaystyle\frac{a+b}{2}<b[/imath]: By closure under addition, [imath]a<b \implies a+b < b+b = 2b[/imath], dividing by [imath]2[/imath], [imath]\displaystyle\frac{a+b}{2}<b[/imath]. Does this work? If not, where did I go wrong? Is there a much better way to show this? | 1114615 | If [imath]0, prove that a<\sqrt{ab}<\frac{a+b}{2}[/imath]
If [imath]0<a<b[/imath], prove that [imath]a<\sqrt{ab}<\frac{a+b}{2}<b[/imath] So far I've got: [imath]a<b[/imath] [imath]a^2<ba[/imath] [imath]a<\sqrt{ab}[/imath] And: [imath]a<b[/imath] [imath]a+b<2b[/imath] [imath]\frac{a+b}{2}<b[/imath] So I need to prove that [imath]\sqrt{ab}<\frac{a+b}{2}[/imath] How can I do it? |
1376093 | how to calculate the projection of a vector onto a closed convex set?
suppose we have a vector [imath]x \in \mathbb{R}^{n}[/imath], and a closed convex set [imath]C \in \mathbb{R}^{n}[/imath]. [imath]C =\{x|Ax=b\}[/imath] how to calculate the vector [imath]y \in \mathbb{R}^{n}[/imath], which is the projection of [imath]x[/imath] onto set [imath]C[/imath]. The vector [imath]y\in C[/imath] and have the smallest distance to [imath]x[/imath]. | 1320363 | Projection of [imath]z[/imath] onto the affine set [imath]\{x\mid Ax = b\}[/imath]
Suppose [imath]A[/imath] is fat(number of columns > number of rows) and full row rank. The projection of [imath]z[/imath] onto [imath]\{x\mid Ax = b\}[/imath] is (affine) [imath]P(z) = z - A^T(AA^T)^{-1}(Az-b)[/imath] How to show this? Note: [imath]A^T(AA^T)^{-1}[/imath] is the pseudo-inverse of [imath]A[/imath] What I am thinking is from: Least square problem: [imath]\text{min $\left\|\: Ax-b \,\right\|_2$}[/imath] The solution for this is [imath]\hat{x} = A^T(AA^T)^{-1}b[/imath]. It seems [imath](Az - b )[/imath] above is the role of [imath]b[/imath] here. Vector projection of [imath]x[/imath] onto [imath]y[/imath]: [imath]p = \frac{x^Ty}{y^Ty}y[/imath] But I still cannot figure out how to prove the above result. |
599020 | Recurrence relation
Okay, I have the following problem: Find a recurrence relation for the number of ways to climb n stairs if the person climbing the stairs can take one, two, or three stairs at a time? And I basically have no idea how to do it, so I looked up the solution, which is as follows: [imath]a_n[/imath] = [imath]a_{n-1}[/imath] + [imath]a_{n-2}[/imath] for n >= 2, with the initial conditions being [imath]a_1[/imath] = 1 and [imath]a_0[/imath] = 1. The first thing I'm confused about is the fact that one of the initial conditions is [imath]a_0[/imath] = 1. If the only ways to climb the stairs are by climbing one, two, or three stairs at a time, in my mind it seems as there are NO ways (or 0 ways) to climb 0 stairs, which would mean [imath]a_0[/imath] = 0. The second thing is..when I actually try to plug in numbers, for example, n = 3: [imath]a_3[/imath] = [imath]a_{2}[/imath] + [imath]a_{1}[/imath] = 2 + 1 = 3 So there are three ways to climb 3 stairs if you can only climb 1, 2, or 3 steps at a time...and I guess what i'm confused about is..how do you determine a "way" to climb a stair? For example, I thought that the ways to climb three stairs would be as follows: You can take it one step at a time, so 1 step, then another step, then another step, which I'll write as 3(1,1,1). You can take 1 step and then take 2 steps, so 3(1,2) You can take 2 steps first and then take 1 step, so 3(2,1) or you could just take 3 steps 3(3) It seems as though that would mean there are four ways to climb the stairs? Can somebody clear this confusion up for me? | 97902 | a simple recurrence problem
Here's the problem: 1.Find a recurrence relation for the number of ways to climb n stairs if the person climbing the stairs can take one, two, or three steps at the time. 2.Explain how the relation is obtained 3.What are the initial condition (base case) 4.How many ways can this person climb a flight of nine stairs 5.Show that this number is an exponential function of n. That is, find a real constant [imath]c > 1[/imath] such that this number is at least [imath]c^n[/imath]. I know the recurrence formula is [imath]C_n = C_{n-1} + C_{n-2} + C_{n-3}[/imath], but for question five, I used induction to show [imath]C_n >= c^n[/imath] for some constant c, but [imath]C_n >= c^{n-1} + c^{n-2} + c^{n-3} >= c^n * (1/c + 1/c^2 + 1/c^3)[/imath] since [imath]1/c + 1/c^2 + 1/c^3[/imath] is a less than 1 for [imath]c > 1[/imath], I don't know how [imath]C_n[/imath] is guaranteed [imath]>= c^n[/imath] Thanks! |
1376467 | Proving an equivalent definition of the [imath]\lim_{x\to a}f(x)[/imath] exists
Prove that the following statements are equivalent. (a) [imath]\lim_{x\to a}f(x)[/imath] exists (b) Given [imath]\epsilon \gt 0[/imath], there is a [imath]\delta \gt 0[/imath] such that if [imath]0\lt |x-a| \lt \delta, 0\lt |y-a| \lt \delta[/imath], then [imath]|f(x)-f(y)|\lt \epsilon.[/imath] (a) [imath]\to[/imath] (b) is easy. But I'm having trouble showing [imath](b)\to (a)[/imath], actually I'm not even sure if this direction is true. My initial thought was that I can prove this by the contrapositive. For some [imath]\epsilon \gt 0[/imath], suppose the limit does not exist at [imath]a[/imath]. Then there are some sequences [imath]x_n[/imath], [imath]y_n[/imath] such that they both tend to [imath]a[/imath], but are never [imath]a[/imath], however, [imath]\lim f(x_n)\neq \lim f(y_n)[/imath]. Then since [imath]\lim x_n=\lim y_n=a[/imath], we can choose a large enough [imath]N[/imath] such that if [imath]n\ge N[/imath], then [imath]|x_n-a|\lt \delta, |y_n-a|\lt \delta[/imath]. Then we get [imath]|f(x_n)-f(y_n)|\gt 0[/imath]. However, I realized that I cannot prove that such [imath]x_n[/imath] and [imath]y_n[/imath] exist in the first place. From definition, the function [imath]f[/imath] does not have a limit at [imath]a[/imath] if and only if there exists a sequence [imath]x_n[/imath] with [imath]x_n\neq a[/imath] for all [imath]n\in \mathbb{N}[/imath] such that the sequence [imath]x_n[/imath] converges to [imath]a[/imath] but the sequence [imath]f(x_n)[/imath] does not converge in [imath]\mathbb{R}[/imath]. This does not mean that I can find two sequences with different limits from the function values. How can I solve this problem? I would greatly appreciate any help. | 27490 | proof that an alternative definition of limit is equivalent to the usual one
How to prove the following theorem? Let [imath]I \in R[/imath] be an open interval, let [imath]c \in I[/imath], and let [imath]f: I-{c} \rightarrow R[/imath] be a function. Then [imath]\lim \limits_{x \rightarrow c} {f(x)}[/imath] exists if for each [imath]\epsilon > 0[/imath], there is some [imath]\delta > 0[/imath] such that [imath]x,y \in I[/imath] and [imath]\vert x-c \vert < \delta[/imath] and [imath]\vert y-c \vert < \delta[/imath] implies [imath]\vert f(x)-f(y) \vert < \epsilon[/imath]. I think this needs standard limit definition, sup and inf properties to prove. And I came up with a following scratch of proof: (1) Suppose for each [imath]\epsilon >0[/imath], there is some [imath]\delta >0[/imath] such that [imath]\vert x-c \vert < \delta[/imath] and [imath]\vert y-c \vert < \delta[/imath] implies [imath]\vert f(x)-f(y) \vert < \epsilon[/imath]. For each [imath]r>0[/imath], let [imath]A_r[/imath] = [imath]I \bigcap (c-r,c+r)[/imath]. Then, for each [imath]\epsilon>0[/imath], there is some [imath]\delta>0[/imath], such that [imath]x,y \in A_\delta[/imath] implies [imath]\vert f(x)-f(y) \vert < \epsilon[/imath]. Then I want to show there is some [imath]a>0[/imath] such that [imath]f(A_a)[/imath] is bounded. (2) If [imath]f(A_a)[/imath] is bounded, then for each [imath]s \in (0,a)[/imath], [imath]f(A_s) \subseteq f(A_a)[/imath], and thus [imath]f(A_s)[/imath] is bounded. Define [imath]a_s = \mathrm{glb}~f(A_s)[/imath] and [imath]b_s = \mathrm{lub}~f(A_s)[/imath]. Let [imath]A = \{a_s \mid s \in (0,a)\}[/imath] and [imath]B=\{b_s \mid s \in (0,a)\}[/imath]. Then we know that A has a least upper bound and B has a greatest lower bound, and [imath]\mathrm{lub}(A) \leq \mathrm{glb}(B)[/imath]. Now I want to show [imath]\mathrm{lub}(A) = \mathrm{glb}(B)[/imath] (3) If I could show [imath]\mathrm{lub}(A) = \mathrm{glb}(B)[/imath], let [imath]M=\mathrm{lub}(A) = \mathrm{glb}(B)[/imath], I want to show [imath]\lim \limits_{x \rightarrow c} {f(x)} = M[/imath]. Can someone give me some help on how to prove (1) (2) (3), [imath]f(A_a)[/imath] is bounded, [imath]\mathrm{lub}(A) = \mathrm{glb}(B)[/imath], and [imath]\lim \limits_{x \rightarrow c} {f(x)} = M[/imath]? Thanks! |
1368722 | How to solve [imath]z^3 + \overline z = 0[/imath]
I need to solve this: [imath]z^3 + \overline z = 0[/imath] how should I manage the 0? I know that a complex number is in this form: z = a + ib so: [imath]z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace[/imath] [imath]\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace[/imath] but how about the 0? EDIT: ok, following some of your comments/answers this is what I have done: [imath]z^3 = - \overline z[/imath] [imath]\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace[/imath] So [imath] \begin{Bmatrix} \rho^3 = \rho\\ 3\theta = -\theta + 2k\pi \end{Bmatrix}[/imath] [imath] \begin{Bmatrix} \rho^3 = \rho\\ 2\theta = 2k\pi \end{Bmatrix}[/imath] [imath] \begin{Bmatrix} \rho = 0 or \rho = 1\\ \theta = k\frac{\pi}{2} \end{Bmatrix}[/imath] is this the right way? | 145102 | Complex roots of [imath]z^3 + \bar{z} = 0[/imath]
I'm trying to find the complex roots of [imath]z^3 + \bar{z} = 0[/imath] using De Moivre. Some suggested multiplying both sides by z first, but that seems wrong to me as it would add a root ( and I wouldn't know which root was the extra ). I noticed that [imath]z=a+bi[/imath] and there exists [imath]\theta[/imath] such that the trigonometric representation of [imath]z[/imath] is [imath]\left ( \sqrt{a^2+b^2} \right )\left ( \cos \theta + i \sin \theta \right )[/imath] . It seems that [imath]-\bar{z} = -\left ( \sqrt{a^2+b^2} \right )\left ( \cos (-\theta) + i \sin (-\theta) \right )[/imath] However, my trig is pretty rusty and I'm not quite sure where to go from here. |
1376561 | How to find [imath]\int_0^1 \int_x^1 \arctan(\frac{y}{x})dxdy[/imath]?
How to find [imath]\int_0^1 \int_x^1 \arctan \left( \frac{y}{x}\right)~dxdy[/imath] I am not looking for any full solutions just some small hints to get me started would be great. | 1256465 | Find [imath]\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg) \, \, \, dx \, \, dy[/imath]
Find [imath]\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg)dx \, \, dy[/imath] So obviously using cylindrical is the way to go to give [imath]\theta r[/imath] inside the integral (after considering the jacobian term). But how can the limits of [imath]x[/imath] be [imath](x,1)[/imath]. I am guessing it is meant to say [imath]dy \, \, dx[/imath] right so those limits are for [imath]y[/imath]? Please tell me if you disagree. But even if this is the case, it is not possible to compute the integral. After sketching out [imath]x<y<1[/imath] and [imath]0<x<1[/imath], I realised that the region is the triangle bounded by origin, [imath](0,1)[/imath] and [imath](1,0)[/imath]. So the limits of [imath]r[/imath] and [imath]\theta[/imath] are [imath](0,1/\cos\theta)[/imath] and [imath](\pi/4 , \pi /2)[/imath]. So we have [imath]\int \limits_{\pi/4}^{\pi/2} \int \limits_0^{1/\cos\theta} r\theta \, \, \, dr \, \, d\theta[/imath] which gives [imath]\int \limits_{\pi/4}^{\pi/2} \frac 12 \theta \sec^2 \theta \, \, \, d\theta[/imath] The integral of [imath]\sec^2 \theta [/imath] is [imath]ln |\cos \theta |[/imath] and one of the limits is [imath]\pi/2[/imath] which is undefined. Please help. |
1376818 | Repertoire method in solving recurrence
I don't know, how should I start solving this: [imath]a_1 = 2 \\ a_n = 2a_{n-1} +7[/imath] using the repertoire method. Could anyone give me an algorithm or explain, how to use this method in this case? | 1017498 | Mathematical explanation for the Repertoire Method
There are a few questions already about this method, which has stumped me for a long while. The process is explained, for instance, here: Repertoire Method Clarification Required ( Concrete Mathematics ) What I absolutely don't get is: What's the meaning of plugging in simple functions (which are not solutions) in the recurrence and how would that help finding what the right solution really is? As I understand it, we have a function determined by a recurrence relation and we want to find a closed formula for it. The recurrence being linear, we assume this closed form is a linear combination of 3 other functions, with coefficients [imath]\alpha, \beta[/imath] and [imath] \gamma[/imath]. The first method suggested in the book is clear to me: set simple values for the constants, since the function will be the same for all of them, and try to guess A, B and C and prove it by induction. What I don't get is the "dual" repertoire method. [imath]f[/imath] could be anything, but it's something fixed, so I don't see any meaning in this process of plugging in [imath]f(n) = 1[/imath]. [imath]f[/imath] clearly is not 1. What is going on? I expect everything coming down to seeing a vector space / module in two different ways and then somehow picking some basis vectors from each. However, I can't work out the right abstraction. I'd like to ask for a very explicit top down explanation of this repertoire method, if possible with linear algebra vocabulary. |
1377101 | Evaluate [imath]\int_{-2}^{2}\int_{y^2-3}^{5-y^2}dxdy[/imath]
In the black I evaluated the integral and I got 64/3, now I need to evaluate the same integral with [imath]\color{red}{dydx}[/imath] .in the [imath]\color{blue}{\text{blue}}[/imath] color is my attempt, I don't think that my attempt is correct. Can someone please tell me what is my mistake in my attempt (in the blue color)? [imath]\int_{-2}^{2}\int_{y^2-3}^{5-y^2}dxdy=\frac{64}{3}[/imath] [imath]\color{blue}{\int_{-3}^{1}\int_{-\sqrt{x-3}}^{\sqrt{x-3}}dydx+\int_{1}^{5}\int_{-\sqrt{x-5}}^{\sqrt{x-5} }dydx}[/imath] | 1361194 | Evaluate[imath]\int_{-2}^2\int_{y^2-3}^{5-y^2}dxdy[/imath]
I evaluated this integarl: [imath]\int_{-2}^2\int_{y^2-3}^{5-y^2}dxdy=\boxed{\frac{64}{3}}[/imath] Now I need to evaluate it with changing the limits, My attempt: [imath]=\int_{-3}^1 \bigg[\int_0^{\sqrt{x+3}}dy\bigg]dx+\int_{1}^5 \bigg[\int_0^{\sqrt{5-x}}dy\bigg]dx[/imath] [imath]\int_{-3}^1 \sqrt{x+3}dx+\int_{1}^5 \sqrt{5-x}dx=\boxed{\color{red}{\frac{32}{3}}}[/imath] Where am I wrong? |
1377285 | Prove that [imath]\sum_{t \vert n} d^3(t) = (\sum_{t \vert n}d(t))^2[/imath] for all [imath]n \in \mathbb{N}[/imath]
here [imath]d(n)[/imath] counts the number of positive divisors of [imath]n[/imath]. I've tried 2 things: Using Bell series. But then again it just showed me that the bell series of the square of a function is not the square of the bell series of the function. Expand it all and use combinatorics. If: [imath]n = \prod_{i=1}^r p_i^{\alpha_i} [/imath] then if [imath]d \vert n[/imath]: [imath]d = \prod_{i=1}^rp_i^{\beta_i}, 0\leq \beta_i \leq \alpha_i[/imath] so [imath]\sum_{t \vert n}d(t) = \sum \prod_{i=1}^r (\beta_i + 1)[/imath] for all possible combitions of [imath](\beta_1,\beta_2,...,\beta_r)[/imath] but I couldn't develop it very much | 479961 | Prove that if [imath]n \in \mathbb{N}[/imath], then [imath]\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2[/imath] where [imath]d(n)[/imath] is the divisor function.
Prove that if [imath]n \in \mathbb{N}[/imath], then [imath]\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2[/imath] where [imath]d(n)[/imath] is the divisor function. I have know only information is [imath]d(n)=\sum_{d|n}{1}[/imath]. |
1377741 | Which partitions [imath]P[/imath] of [imath]n[/imath] give the row and column sums of some [imath]|P| \times |P|[/imath] [imath](0,1)[/imath]-matrix?
Question: Which partitions [imath]P[/imath] of [imath]n[/imath] give the row and column sums of some [imath]|P| \times |P|[/imath] [imath](0,1)[/imath]-matrix? Someone comes along and gives us the partition [imath]P=\{2,2,3,3,4\}[/imath] of [imath]14[/imath]. How can we determine if there's a [imath]5 \times 5[/imath] matrix whose rows and column sums both give rise to that partition? In the example, it's realized by this matrix: [imath]\begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ \end{bmatrix}[/imath] but I cheated: I generated the example partition from the matrix. Observations: The obvious necessary condition is that [imath]\max P \leq |P|[/imath]. This is equivalent to asking when is [imath]P[/imath] the out-degree sequence of a directed graph where any vertex may have one loop. (So the Erdős–Gallai theorem or Havel–Hakimi algorithm could be applied in some cases, replacing each undirected edge with a bidirected edge; in fact, they would work in my above example.) If we just brute force started filling in [imath]1[/imath]'s, we can get stuck. Here I proceed row-by-row, placing a [imath]1[/imath] wherever possible: [imath] \begin{array}{c|ccccc} & 4 & 3 & 3 & 2 & 2 \\ \hline 4 & 1 & 1 & 1 & 1 & 0 \\ 3 & 1 & 1 & 1 & 0 & 0 \\ 3 & 1 & 1 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 1 & 0 \\ 2 & 0 & 0 & 0 & 0 & ??? \\ \end{array} [/imath] Motivation: This is a first step in an attempt at answering my Mathoverflow question: What is the minimum number of filled cells in a partial Latin rectangle with autotopism group [imath]\cong C_2[/imath] and autoparatopism group [imath]\cong S_3[/imath]?. An answer to this question might help reduce the search space. | 756254 | Construct matrix of ones and zeros based on sequences
We are given ([imath]a_1,a_2,....,a_m[/imath]) and ([imath]b_1, b_2,....,b_n[/imath]) sequences with non-negative integers. Decide whether it's possible and if it is construct a matrix [imath]\Re^{m x n}[/imath] of ones("1") and zeros("0") where the number of ones("1") in row [imath]i[/imath] is [imath]\forall i [/imath] : [imath]a_i[/imath] and in column [imath]j[/imath] is [imath]\forall j[/imath] : [imath]b_j[/imath] Any hint is greatly appreciated. |
452670 | Bilinear Form - Proof
I have to prove that the mapping [imath]f(x,y) = {\displaystyle \sum_{i=1} ^ {n} }{ \displaystyle \sum_{j=1}^{n} }x_iy_j{f}(e_i,e_j)[/imath] is a bilinear form, that is, inter alia, the condition: [imath]f(x+y,z)=f(x,z)+f(y,z)[/imath] I have: [imath]f(x+y,z)=f\left({\displaystyle \sum_{i=1}^{n}}x_{i}e_{i}+{\displaystyle \sum_{k=1}^{n}}y_{k}e_{k},{\displaystyle \sum_{j=1}^{n}}z_{j}e_{j}\right)=f\left({\displaystyle \sum_{i=1}^{n}}{\displaystyle \sum_{k=1}^{n}}\left(x_{i}e_{i}+y_{k}e_{k}\right),{\displaystyle \sum_{j=1}^{n}}z_{j}e_{j}\right)=\ldots[/imath] and what's next? Help! | 452710 | bilinear form - proof
I have to prove that the mapping [imath]f(x,y)={\displaystyle \sum_{i=1}^{n}}{\displaystyle \sum_{j=1}^{n}}x_{i}y_{j}{f}(e_{i},e_{j})[/imath] is a bilinear form, that is, inter alia, the condition: [imath]f(x+y,z)=f(x,z)+f(y,z)[/imath] I have: [imath]f(x+y,z)=f\left({\displaystyle \sum_{i=1}^{n}}x_{i}e_{i}+{\displaystyle \sum_{k=1}^{n}}y_{k}e_{k},{\displaystyle \sum_{j=1}^{n}}z_{j}e_{j}\right)=f\left({\displaystyle \sum_{i=1}^{n}}{\displaystyle \sum_{k=1}^{n}}\left(x_{i}e_{i}+y_{k}e_{k}\right),{\displaystyle \sum_{j=1}^{n}}z_{j}e_{j}\right)=\ldots[/imath] and what's next? Help! |
1378350 | Compute the limit (using LDCT)
Could you please help me with the following? I think it is a Lebesgue Dominated Convergence Theorem but I am not sure. Compute the following: For [imath]1\leq p<\infty[/imath] and [imath]f\in L^p(\mathbb{R})[/imath] [imath]\lim_{h\rightarrow \infty}\int_{-\infty}^\infty\left|f(x+h)-f(x)\right|^p \, dx[/imath] | 1018716 | Translation operator and continuity
I came across a text that proves that translation operator [imath]T_a(f):=f(x-a)[/imath] where [imath]a\in\mathbb{R}^n[/imath] and [imath]f\in L^p(\mathbb{R}^n)[/imath] is continuous. The proof follows: [imath]||f(x-a)-f(x)||_p=||f(x-a)-g(x-a)+g(x-a)-g(x)+g(x)-f(x)||_p\leq ||f(x-a)-g(x-a)||_p+||g(x-a)+g(x)||_p+||g(x)-f(x)||_p<3\varepsilon[/imath] Where [imath]g[/imath] is some continuous function and hence [imath]C^\infty[/imath] is dense in [imath]L^p[/imath] the inequality holds. Wouldn't that meant [imath]f[/imath] is also continuous which it doesn't have to be? Why does the proof hold, respectively what is the catch with the proof, why isn't it the proof of continuity of [imath]f[/imath]? |
1378691 | Disequality in Type Theory
Is it possible to prove [imath]0 \neq 1[/imath] in (non-univalent) Martin-Löf type theory, where [imath]0[/imath] and [imath]1[/imath] are natural numbers (defined using the usual inductive type [imath]0 : \mathbb{N}[/imath], [imath]S : \mathbb{N} \to \mathbb{N}[/imath])? | 1376680 | Constructing a HoTT proof term of 1≠0
As an exercise in HoTT basics, I am trying to construct a term that has the type [imath]Id_{Nat}(S(O),O)\to\bot[/imath]. If this were a Coq proof, I'd be done after a single inversion on the premise, as the impossible identity would leave zero cases to consider. I guess I could do something similar here, but I'm not sure how. What would be an HoTT term that has the type? |
645287 | Integral [imath]\int_{0}^{+\infty}\frac{t \sin(t)}{t^{2}+b^{2}}dt[/imath]
I want to solve the integral [imath]\int_{0}^{+\infty}\frac{t \sin(t)}{t^{2}+b^{2}}dt[/imath] Which function and contour should I consider ? | 1377706 | Improper rational/trig integral comes out to [imath]\pi/e[/imath]
During my studying to integration I find this integration. So I tried to prove but I got stuk. So I need help in this integration. [imath]\displaystyle\int_{-\infty}^{\infty} \frac{x \sin (x)}{1+x^2} dx = \frac {\pi}{e} [/imath] |
1100957 | Is it true that the Fibonacci sequence has the remainders when divided by 3 repeating?
About this Fibonacci sequence, is it true that the remainders when divided by three repeat along with the sequence like this: Fibonacci sequence: [imath]1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946\cdots[/imath] Remainder when divided by [imath]3[/imath] sequence: [imath]1, 1, 2, 0, 2, 2, 01, 00, 01, 01, 02, 000, 002, 002, 001, 000, 0001, 0001, 0002, 0000, 00002\cdots[/imath] I made each one the same number of digits to align them to avoid confusion. Is it true that it happens every eight numbers? If so, how and why does this happen? | 631171 | Proof that Fibonacci Sequence modulo m is periodic?
It's well known that the Fibonacci sequence [imath]\pmod m[/imath] (where [imath]m \in \mathbb N[/imath]) is periodic. I have figured out a proof for this, but upon googling, I found proofs online that were far more complicated. This leads me to suspect that my proof may be fallacious - that's why I am posting here. Proof: Let us list out the Fibonacci sequence modulo m, where m is some integer. It will look something like this at first (for [imath]10[/imath] at least): [imath] 1,2, 3,5,8, 3,1, 4, 5, 9, 4, 3, 7 {\dots}[/imath] Obviously, any number in the sequence is the sum of the last two numbers modulo [imath]m[/imath]. Therefore, if at any point in the series modulo m a pair of numbers repeat, the numbers following that pair must repeat as well. eg. if at some point later we see the pair [imath]1, 1[/imath], then [imath]2, 3, 5, \dots[/imath] must follow that pair. Now there are [imath]m^2[/imath] possible pairs in the series [imath]\text{mod} m[/imath]. By the pigeonhole principle, after [imath]m^2 + 1[/imath] terms, a pair must repeat. If a pair repeats once, it must repeat again the same number of terms later. Therefore, the Fibonacci sequence [imath]\pmod m[/imath] is periodic [imath]\forall m[/imath] . Is my proof correct? |
1207547 | Find the sum of the series [imath]\sum_{n=1}^{\infty} \frac{1}{n \cdot (n+1) \cdot (n+2)}[/imath]
I am trying to find the sum of the following series [imath]\sum_{n=1}^{\infty} \frac{1}{n \cdot (n+1) \cdot (n+2)}[/imath] I've already figured it out that it converges and, as it looks like a telescoping series, I also tried decomposing it into partial fractions, which would be [imath]\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}[/imath] But that doesn't seem to be helping me at all... Any ideas? | 560816 | Find the sum of the series [imath]\sum \frac{1}{n(n+1)(n+2)}[/imath]
I got this question in my maths paper Test the condition for convergence of [imath]\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}[/imath] and find the sum if it exists. I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you. |
60641 | There exists [imath]C\neq0[/imath] with [imath]CA=BC[/imath] iff [imath]A[/imath] and [imath]B[/imath] have a common eigenvalue
Question: Suppose [imath]V[/imath] and [imath]W[/imath] are finite dimensional vector spaces over [imath]\mathbb{C}[/imath]. [imath]A[/imath] is a linear transformation on [imath]V[/imath], [imath]B[/imath] is a linear transformation on [imath]W[/imath]. Then there exists a non-zero linear map [imath]C:V\to W[/imath] s.t. [imath]CA=BC[/imath] iff [imath]A[/imath] and [imath]B[/imath] have a same eigenvalue. ===========This is incorrect========== Clearly, if [imath]CA=BC[/imath], suppose [imath]Ax=\lambda x[/imath], then [imath]B(Cx)=CAx=C(\lambda x)=\lambda (Cx)[/imath], so [imath]A[/imath] and [imath]B[/imath] have same eigenvalue [imath]\lambda[/imath]. On the other hand, if [imath]A[/imath] and [imath]B[/imath] have a same eigenvalue [imath]\lambda[/imath], suppose [imath]Ax=\lambda x, By=\lambda y[/imath]. Define [imath]C:V\to W[/imath] s.t. [imath]Cx=y[/imath], then [imath]BCx=By=\lambda y=C\lambda x=CAx[/imath]. But I don't kow how to make [imath]BC=CA[/imath] for all [imath]x\in V[/imath]. ====================================== | 2333851 | Let [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] be complex matrices such that [imath]C\neq 0, [/imath] [imath]AC=CB[/imath]. Prove that [imath]A[/imath] and [imath]B[/imath] have a common eigenvalue.
Let [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] be complex matrices such that [imath]C\neq 0, [/imath] [imath]AC=CB[/imath]. Prove that [imath]A[/imath] and [imath]B[/imath] have a common eigenvalue. There is a hint in the question, these facts can be used for the prove: For a complex matrices [imath]A, B[/imath], If [imath]AB = 0[/imath], and [imath]B[/imath] is invertible, [imath]A = 0[/imath]. For a complex matrices [imath]A, B[/imath] and [imath]C[/imath], if [imath]AB = BC[/imath] than for each natural number [imath]k[/imath], [imath]\\\\A^kB = BC^k [/imath]. Any ideas? |
1379206 | Evaluate [imath]\iint dydx[/imath] on the domain [imath]0\leq r\leq1[/imath], [imath]{\pi}/{3}\leq\theta \leq{2\pi}/{3}[/imath]
I need to evaluate [imath]\displaystyle\iint \color{red}{dydx}\;\;\;,\bigg\{\frac{\pi}{3}\leq\theta \leq\frac{2\pi}{3}\bigg\}\;\;\;\;,0\leq r\leq1[/imath] [imath]\color{blue}{\text{without using polar coordinates}}[/imath]. My attempt: The equation of the two lines will be [imath]y=\pm\tan(60)x=\pm\sqrt3x[/imath] The intersection points of the lines with the circle will be [imath]\pm\frac{1}{2}[/imath] Therefore: [imath]\int_{x=-1/2}^{x=0}\int_{y=-\sqrt3 x}^{\sqrt{1-x^2}}dydx+\int_{x=0}^{x=1/2}\int_{y=\sqrt3x}^{y=\sqrt{1-x^2}}dydx=\dots\approx \boxed{2\times0.478..}[/imath] But area of circle is [imath]\pi r^2[/imath] and here the area should be [imath]\boxed{\frac{\pi}{6}}[/imath] I can't find my mistake It looks similar to Evaluate [imath]\iint dy\,dx;\frac{\pi}{4}\leq\theta \leq\frac{3\pi}{4};0\leq r\leq2[/imath] but here the radius is diffrente and also the angles | 1377261 | Evaluate [imath]\iint dy\,dx;\frac{\pi}{4}\leq\theta \leq\frac{3\pi}{4};0\leq r\leq2[/imath]
I need to evaluate [imath]\displaystyle\iint \color{red}{dydx}\;\;\;,\frac{\pi}{4}\leq\theta \leq\frac{3\pi}{4}\;\;\;\;,0\leq r\leq2[/imath] [imath]\color{blue}{\text{without using polar coordinates}}[/imath]. My attempt: [imath]\int_{x=-2}^{x=0}\int_{y=-x}^{\sqrt{4-x^2}}dydx+\int_{x=0}^{x=2}\int_{y=x}^{y=\sqrt{4-x^2}}dydx[/imath] Is the correct? |
1378051 | What is the next prime with this form?
The following are primes, [imath]P_1 = 2^2 + 3^3[/imath] [imath]P_2 = 2^2 + 3^3 +5^5 + 7^7[/imath] After these two, the only prime of such form I've found is, [imath]P_3 = 2^2 + 3^3 + 5^5 + 7^7 + 11^{11} + 13^{13} +\dots+ 83^{83} + 89^{89}[/imath] Is there a prime number with this form after these three? (I've checked prime numbers up to 5500, but I couldn't find any others.) | 1148222 | Prime number expressible as [imath]2^2+3^3+5^5+7^7+11^{11}+\cdots[/imath]
I found that [imath]2^2+3^3=31[/imath] and [imath]31[/imath] is a prime, also [imath]2^2+3^3+5^5+7^7[/imath] is a prime. After these the only prime I found is [imath]2^2+3^3+5^5+7^7+11^{11}+\cdots+83^{83}+89^{89}[/imath]. And I've checked [imath]p(n)[/imath] up to [imath]1009[/imath], but I couldn't find another prime of the form [imath] 2^2+3^3+5^5+7^7+11^{11}+\cdots+p(n-1)^{p(n-1)}+p(n)^{p(n)}. [/imath] Is there any other prime with that form? |
1380236 | For what [imath]p[/imath] is the condition number of a given matrix [imath]A[/imath], using the [imath]p[/imath] norm on matrices, minimal?
For what p is the condition number of a given matrix [imath]A[/imath], using the p norm on matrices, minimal? The condition number on [imath]A[/imath] is given as: [imath]K(A,p) = \|A\|_p \times \|A^{-1}\|_p[/imath] I tried simplifying this further but got nowhere, I feel like I'm missing something. Any help would be appreciated! | 1377074 | Choose [imath]\rho[/imath] such that [imath]\rho[/imath]-norm minimizes the matrix condition number
I'm solving questions from am exam that I failed miserably, so I would love it if someone can take a look at my proof and make sure I'm not making any gross mistakes. Question Let [imath]A[/imath] a symmetric matrix. Which [imath]\rho[/imath]-norm minimizes [imath]A[/imath]'s condition number: [imath]\kappa(A,\rho)[/imath]? Edit Well, my solution was clearly wrong. I would love any suggestion or direction for solution you might have. Thanks! :) |
743383 | In how many steps is it necessary to construct the Borel [imath]\sigma[/imath]-algebra?
We only consider Borel sets on [imath]\mathbb R[/imath]. As we know, the Borel [imath]\sigma[/imath]-algebra is constructed transfinitely as follows: Let [imath]B_0=\mathcal T[/imath] be the set of open sets on [imath]\mathbb R[/imath]; If [imath]\alpha[/imath] is a limit ordinal, [imath]B_\alpha=\bigcup_{\beta<\alpha}B_\beta[/imath]; Otherwise, let [imath]\beta=\alpha-1[/imath], and [imath]B_\alpha=(B_\beta)_{\delta\sigma}[/imath], where [imath]A_\delta[/imath] denotes the collection of countable intersections of sets in [imath]A[/imath], and [imath]A_{\delta\sigma}[/imath] denotes the collection of countable unions of sets in [imath]A_\delta[/imath]; Let [imath]\mathcal B=B_{\omega_1}[/imath] where [imath]\omega_1[/imath] is the smallest uncountable ordinal, then [imath]\mathcal B[/imath] is the Borel [imath]\sigma[/imath]-algebra. However, I wonder whether [imath]\omega_1[/imath] is the smallest ordinal [imath]\alpha[/imath] such that [imath]B_\alpha=\mathcal B[/imath], i.e. the minimal step to finish the process. As we know, [imath]((0,1)\cap\mathbb Q)\cup([2,3]\setminus\mathbb Q)[/imath] is a Borel set which is neither [imath]F_\sigma[/imath] nor [imath]G_\delta[/imath]. Any idea? Thanks! | 509326 | Borel hierarchy doesn't "collapse" before [imath]\omega_1[/imath]
I know that [imath]\bigcup_{\alpha < \omega_1} \Sigma_\alpha^0 = \mathcal{B}(X)[/imath] for any Polish space [imath]X[/imath], but I want to know if it's really necessary to take the union for every ordinal less than [imath]\omega_1[/imath]. Is there an example of a Polish space where [imath]\bigcup_{\alpha < \beta} \Sigma_\alpha^0 \varsubsetneq \mathcal{B}(X)[/imath] for every countable ordinal [imath]\beta[/imath]? What's the case of [imath]\mathbb{R}[/imath] or [imath][0, 1][/imath]? Many thanks in advance. |
1373612 | [imath]\int\dfrac{dx}{x^2(x^4+1)^{3/4}}[/imath]
Evaluate [imath]\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}[/imath] I thought of rewriting this as [imath]\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}[/imath] and substituting [imath]u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}[/imath] and subsequently I got [imath]du=\dfrac{1}{4}\left(1+\frac{1}{x^4}\right)^{-3/4}\times (-4x^{-5})dx[/imath] However, I can not think of how to proceed further. Any help would be truly appreciated. Many thanks in advance! | 1250935 | Evaluation of [imath]\int\frac{1}{x^2.(x^4+1)^{\frac{3}{4}}}dx[/imath]
Evaluate the integral [imath]\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx[/imath] My Attempt: Let [imath]x = \frac{1}{t}[/imath]. Then [imath]dx = -\frac{1}{t^2}\,dt[/imath]. Then the integral converts to [imath] -\int \frac{t^3}{(1+t^4)^{3/4}}\,dt [/imath] Now Let [imath](1+t^4) = u[/imath]. Then [imath]t^3\,dt = \frac{1}{4}du[/imath]. This changes the integral to [imath] \begin{align} -\frac{1}{4}\int t^{-3/4}\,dt &= -u^{1/4}+\mathcal{C}\\ &= -\left(1+t^4\right)^{1/4}+\mathcal{C} \end{align} [/imath] So we arrive at the solution [imath]\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx = - \left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C.}[/imath] Question: Is there any other method for solving this problem? |
1380810 | Proving [imath]\sum_{i=1}^n 2^i = 2^{n+1} - 2[/imath] using strong induction
I just started learning proof by induction in class, but got a problem requiring proof by strong induction. Here is the problem. Prove by strong induction: [imath]\sum_{i=1}^n 2^i = 2^{n+1} - 2[/imath] I've done the base, showing that the statement holds for [imath]n=1[/imath], [imath]n=2[/imath], and [imath]n=3[/imath]. (I won't show the simple math here). For [imath]n=k[/imath], the statement would be [imath]2^{k+1}-2[/imath]. But that's where I get stuck, as I'm still trying to grasp the concept of strong induction. For [imath]n=k+1[/imath], do I do the following and simplify? [imath]\sum_{i=1}^{k+1} 2^i = \sum_{i=1}^k 2^i + 2^{(k+1)+1} - 2[/imath] [imath]=[2^{k+1}-2]+[2^{k+2}-2][/imath] [imath]=\text{etc}\ldots?[/imath] | 1193942 | Proving [imath]\sum_{i=0}^n 2^i=2^{n+1}-1[/imath] by induction.
Firstly, this is a homework problem so please do not just give an answer away. Hints and suggestions are really all I'm looking for. I must prove the following using mathematical induction: For all [imath]n\in\mathbb{Z^+}[/imath], [imath]1+2+2^2+2^3+\cdots+2^n=2^{n+1}-1[/imath]. This is what I have in my proof so far: Proof: Let [imath]p\left(n\right)[/imath] be [imath]\sum_{j=0}^n 2^j=1+2+2^2+\cdots+2^n=2^{n+1}-1[/imath]. Then [imath]P\left(0\right)[/imath] is \begin{align} \sum\limits_{j=0}^0 2^j=1=2^1-1.\tag{1} \end{align} Hence [imath]P\left(0\right)[/imath] is true. Next, for some [imath]k\in\mathbb{Z^+}[/imath], assume [imath]P\left(k\right)[/imath] is true. Then \begin{align} \sum\limits_{j=0}^{k+1}2^j&=\left(\sum\limits_{j=0}^k 2^j\right)+2^{k+1}\tag{2}\\ &=2^{k+1}-1+2^{k+1}\tag{3} \\ &=2\cdot 2^{k+1}-1\tag{4}\\ &=2^{k+2}-1\tag{5}\\ &=2^{\left(k+1\right)+1}-1.\tag{6} \end{align} Therefore, [imath]P\left(n\right)[/imath] holds for all [imath]n[/imath].[imath]\;\;\;\;\blacksquare[/imath] Is everything right? My professor went over the methodology in class quite fast (and I had a bit too much coffee that night) and I want to make sure I have it down for tomorrow night's exam. |
1381115 | Greatest of the numbers given
To find out the greatest among the number given below: [imath]3^{1/3}, 2^{1/2}, 6^{1/6}, 1, 7^{1/7}[/imath] I have plotted the following graph using graph plotter which is shown below: It can be concluded that [imath]3^{1/3}[/imath] is the greatest. I want to know that is there any other method to find greatest among the such numbers. | 1331015 | Which of the numbers [imath]1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}[/imath] is largest, and how to find out without calculator?
[imath]1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}[/imath]. I got this question in an Application of Derivatives test. I think log might be used here to compare the values, but even then the values are very close to each other and differ by less than 0.02, which makes it difficult to get some specific answer to this question. How to solve this by a definite method? |
1381056 | How many 3 letter words can you form from 'EEAAP'
How many 3 letter words can you form from 'EEAAP' I think the answer is [imath]{3\choose 3} * 3! + {2\choose 1} * 3 + {2\choose 1} *3=18[/imath]. Is this right? [imath]{3\choose 3} * 3![/imath] = You pick all distinct [imath]{2\choose 1} * 3[/imath] = You pick 2 e's and either a or p [imath]{2\choose 1} * 3[/imath]= You pick 2 a's and either e or p | 1380641 | Permutation count of AABBC
Given a string: [imath]AABBC=A^2B^2C^1[/imath] I am trying to find the Total Permutations (this may be incorrect): [imath]\dfrac{5!}{2!\cdot2!}=30[/imath] My question is how would I find the partial sums (perhaps the wrong choice in words) when added together would give an index such that: [imath]AABBC=1[/imath] [imath]ABABC=2[/imath] [imath]CBABA=29[/imath] [imath]CBBAA=30[/imath] (30 might be the wrong number here but is suppose to be the last) I have tried to draw out a tree structure to get an idea and thought with as the minimum: [imath]CAABB[/imath] [imath]4(\dfrac{4!}{2!\cdot2!})[/imath] + 1 And: [imath]CABAB[/imath] I would get something like: [imath]4(\dfrac{4!}{2!\cdot2!}) + 0 + 1(\dfrac{2!}{2!}) + 0 + 0[/imath] But I am missing something here. How would I find the sums needed for [imath]BACAB[/imath] ? More specifically I am trying to find the index of a word that is arbitrarily long in alphanumeric. So the string may also be something like: [imath]AAAAABBBBDDEEFFFGGGGGHIIIIJJJJKLMOOOPZZZZZ[/imath] |
1381183 | Geometric series and complex numbers
I'm new to this site, english is not my mother tongue, and I'm just learning LaTeX. I'm basically a noob, so please be indulgent if I break any rule or habits. I'm stuck at proving the following equation. I suppose I should use the formula for geometric series ([imath]\sum\limits_{k=0}^{n}q^k=\frac{1-q^{n+1}}{1-q}[/imath]), and also use somewhere that [imath]e^{î\theta}=cos(\theta)+i\sin(\theta)[/imath] So here is the equation I have to prove : [imath]\frac{1}{2}+cos(\theta)+cos(2\theta)+...+cos(n\theta)=\frac{sin(n+\frac{1}{2}\theta)}{2sin(\frac{\theta}{2})}[/imath] Thanks for your help | 300114 | Sum of a complex, finite geometric series and its identity
I have the formula for summing a finite geometric series as [imath]1+z+z^2\cdots +z^n = \frac{1-z^{n+1}}{1-z},[/imath] where [imath]z\in\mathbb{C}[/imath] and [imath]n=0,1,...[/imath]. I am asked to infer the identity [imath]1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta = \frac{1}{2}+\frac{\sin(n+1/2)\theta}{2\sin\theta /2}.[/imath] Now, I understand that on the left hand side I'm going to get [imath]1+\cos\theta +\cdots + \cos n\theta + i[\sin\theta + \sin 2\theta +\cdots + \sin n\theta][/imath] using [imath]z=e^{i\theta}[/imath] for any complex [imath]z[/imath]. However, when I make that substitution on the right hand side, a monstrous expression occurs and I cannot simplify it down to the desired result. For instance, I get [imath]\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}[/imath] and using identities I get [imath]\frac{1-\cos [(n+1)\theta]+i(\sin[(n+1)\theta])}{1-\cos n\theta -i\sin n\theta}.[/imath] From here I did a whole lot of manipulating, but never getting any closer to the identity asked. If anyone could shed some light it would be greatly appreciated! ~Dom |
1380983 | Variant of CMVT
If [imath]\Phi(x)[/imath] and [imath]\Psi (x)[/imath] are continuous for [imath]a \leq x \leq b[/imath] , differentiable for [imath]a \lt x \lt b, [/imath] and [imath]\Psi'(x)[/imath] never vanishes , then for some [imath]\xi[/imath] in [imath](a,b)[/imath] [imath]\frac{\Phi(\xi)-\Phi(a)}{\Psi(b)-\Psi(\xi)}=\frac{\Phi'(\xi)}{\Psi'(\xi)}[/imath] Prove or derive the following expression. I am having quite a hard time on this, well the problem is, this one resembles a lot to Cauchy mean value theorem except for the fact, there the [imath]\xi[/imath] is also introduced into the LHS which i have no idea how. I tried to have some guess function like [imath]h(x)=f(x)+Ag(x)| h(a)=h(b)[/imath] and apply rolle's theorem and etc but all failed. Most of them always lead to cauchy mean value theorem and not this. | 860423 | Variation on the Cauchy mean value theorem
From Spivak's Calculus, 4th edition, problem 11-50: Prove that if [imath]f[/imath] and [imath]g[/imath] are continuous on [imath][a,b][/imath] and differentiable on [imath](a,b)[/imath], and [imath]g'(x)\neq 0[/imath] for [imath]x[/imath] in [imath](a,b)[/imath], then there is some [imath]x[/imath] in [imath](a,b)[/imath] with [imath]\frac{f'(x)} {g'(x)} = \frac {f(x) - f(a)} {g(b)-g(x)}.[/imath] Hint: Multiply out first, to see what this really says. I've struggled with this one for a while, to no avail. I'm sure it's just something small I'm missing. |
1381372 | Proof of x = 0 modulo 3 only if the sum of its digits 0 modulo 3
Okey, lets beggin from a helpfull proposition I've already proved: [imath]$$ if $a_i\equiv b_i\:\forall 0\le i\le m$ then to any $m$ numbers: $p_1,p_2,...,p_m\in \mathbb{Z}$ [/imath]\sum _{i=0}^m\left(k_ia_i\right)=\sum _{i=0}^m\left(k_ib_i\right)[imath][/imath] Now, what I want to prove, is: for number $x$ who's digits are: $x_m...x_0$ x\equiv 0\left(mod\:3\right)\:\:<==>\:\:\sum \:_{i=0}^m\left(x_i\right)\equiv 0\left(mod\:3\right)[imath] [/imath][imath] I guess the same exmaple with modulo 11 is the same proof.[/imath] So, in the first direction, we'll treat $\sum \:_{i=0}^m\left(x_i\right)\equiv 0\left(mod\:3\right)$ as given data and we'll prove $x\equiv \:0\left(mod\:3\right)\:$ That direction of proof is simple: As to the first proposition, we'll take $k_0=10^0,\:k_1=10^1,...,\:k_m=10^m$ and accroding to it: \sum \:\:_{i=0}^m\left(x_i10^i\right)\equiv \:\sum \:\:_{i=0}^m\left(0\cdot \:10^i\right)\left(mod\:3\right)[imath] then also [/imath]\sum \:_{i=0}^m\left(x_i10^i\right)\equiv 0\left(mod\:3\right)[imath] As we want.[/imath] Now, in second direction, i have a little trouble: Now we are given: $x\equiv \:0\left(mod\:3\right)\:$ and we want to prove: \sum \:_{i=0}^m\left(x_i\right)\equiv 0\left(mod\:3\right)[imath][/imath] So, from the data that is given we can conclude that: x=\sum \:\:_{i=0}^m\left(x_i\cdot 10^i\right)\equiv \:0\left(mod\:3\right)[imath][/imath] And from here I dont sure how to get rid of the $10^i$, because the first proposition as i understand doesnt work on both directions (am I wrong?), because if n is a devider of a sum, we dont know for sure if it devides every element or only the sum of it, it can be the second case either. So how do I show that \sum \:\:_{i=0}^m\left(x_i\cdot 10^i\right)\equiv \:0\left(mod\:3\right)\:\:\:===>\:\:\sum \:\:\:_{i=0}^m\left(x_i\right)\equiv \:\:0\left(mod\:3\right)$$ | 1108450 | Prove that if the sum of digits of a number is divisible by 3, so is the number itself.
Here is the proof of the converse: Iff a number [imath]n[/imath] is divisible by [imath]3[/imath], then the sum of its digits is also divisible by [imath]3[/imath]. Proof: We know [imath]n \mod 3 = 0[/imath]. By the basis representation theorem, [imath]n[/imath] can be re-written as [imath]n_k10^k + n_{k-1}10^{k-1} + \cdots + 10n_1 + n_0 \equiv 0 \mod 3[/imath]. By the modular arithmetic addition rule, we can take [imath]n_k10^k \mod 3 + n_{k-1}10^{k-1} \mod 3 + \cdots + 10n_1 \mod 3 + n_0 \mod 3[/imath]. But then we just get [imath]n_k + n_{k-1} + \cdots + n_1 + n_0[/imath], since [imath]10 \mod 3 \equiv 1[/imath]. Now suppose that the sum of the digits was divisible by 3. How can I prove that the representation in base [imath]10[/imath] is divisible by [imath]3[/imath]? |
1381413 | Spaces whose all their metrizations are complete
Which metrizable topological spaces [imath](X,\tau)[/imath] posses the following property: Every compatible metric (i.e one which induces the same topology [imath]\tau[/imath]) is complete. Compact metrizable spaces satisfy this. Are there any non-compact examples? Note: It turns out that if you impose additional structure the answer is yes. In particular, if [imath]M[/imath] is a smooth manifold, and every Riemannian metric on [imath]M[/imath] is complete, then [imath]M[/imath] must be compact. Reference: Nomizu, Katsumi, and Hideki Ozeki. "The existence of complete Riemannian metrics." Proceedings of the American Mathematical Society 12.6 (1961): 889-891. http://www.jstor.org/discover/10.2307/2034383?uid=2&uid=4&sid=21105114015163is | 944416 | Is there always an equivalent metric which is not complete?
I have seen that completeness is not a topological property like compactness or connectedness. I have seen some examples also showing that there are two equivalent metrics one of which is complete and the other one is incomplete. I want to know some general result. Consider any metric space [imath](X,d)[/imath]. Let [imath]d[/imath] be complete. Does there exist an equivalent metric [imath]d'[/imath] in [imath]X[/imath] which is incomplete? When [imath]d[/imath] is compact, such [imath]d'[/imath] does not exist (Since compactness implies completeness). So answer me when [imath]d[/imath] is noncompact. |
1381684 | Does [imath]\displaystyle\sum^{\infty}_{n=1}\left(\frac{n!}{n^n}\right)[/imath] converge or diverge?
Does [imath]\displaystyle\sum^{\infty}_{n=1}\left(\frac{n!}{n^n}\right)[/imath] converge or diverge? I've tried the ratio test, but i'm unsure if I can continue this way. [imath]\displaystyle\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\lim_{n\to\infty}\frac{(n+1)n!n^n}{(n+1)^n(n+1)n!}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}[/imath] If this approach works, how do we proceed? If not, what test might be worth trying? | 1196708 | Convergence of [imath]\sum_n \frac{n!}{n^n}[/imath]
I'm working on a problem sheet and it ask to discuss the convergence of [imath]\sum \frac{n!}{{n}^{n}}[/imath] By D'Lembert's ratio test, [imath]\lim_{n->\infty}\frac{{a}_{n+1}}{{a}_{n}} = 1[/imath] and so, is inconclusive. Using Cauchy's root test, [imath]\lim_{n->\infty}({\frac{n!}{{n}^{n}}})^\frac{1}{n}=1[/imath] What are my alternatives? Should I take the integral of the term of the series above? Would integrating factorial works? |
1381765 | For nonnegative continuous [imath]f[/imath], if [imath]f'(x)-f(x)\leq 0, \forall x\geq 0[/imath] and [imath]f(0)=0[/imath], find the value of [imath]f(1)[/imath].
Let [imath]f(x)[/imath] be a non-negative continuous function such that [imath]f'(x)-f(x)\leq 0, \forall x\geq 0[/imath] and [imath]f(0)=0,[/imath]find the value of [imath]f(1)[/imath]. $f'(x)-f(x)\leq 0[imath]\Rightarrow f'(x)\leq f(x)[/imath]\Rightarrow \frac{f'(x)}{f(x)}\leq 1[imath]\Rightarrow \int\frac{f'(x)}{f(x)}dx\leq \int1 dx[/imath]\Rightarrow \log f(x) \leq x$.Then could not solve further.Can someone assist me find [imath]f(1)[/imath]? | 671932 | A non-trivial, non-negative, function bounded below by its derivative with [imath]f(0)=0[/imath]?
I did not know what to search to see if this existed elsewhere. But, I could not find it. Here's the question: Does there exist a continuously differentiable function, [imath]f: [0,1] \rightarrow [0, \infty)[/imath], such that the following hold? (i) [imath]f(0) = 0[/imath] (ii) [imath]f'(x) \le c f(x)[/imath] for all [imath]x[/imath] ([imath]c[/imath] is a fixed constant) (iii) [imath]f \not\equiv 0[/imath] This was a fun problem someone asked me a long time ago. I have always been convinced there is no such function, but I cannot prove it. I have mainly tried using the mean value theorem (in multiple forms), and re-wording the problem in terms of the integral of a continuous function to no avail. I spent some time trying to use (ii), the definition of derivative, and squeeze theorem, but no luck. After much time trying to prove non-existence, I spent a little time trying to find such a function... also with no luck. This problems ability to avoid being solved has since taken away the original fun and replaced it with a twinge of annoyance. |
1302650 | find Jordan form
Determine the jordan form of [imath]A = \begin{pmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 4 \end{pmatrix} [/imath] First, I find the characteristic polynomial. [imath]C_A(x)=(x-1)(x-4)^2[/imath]. Therefore, the minimal polynomial will be [imath]m_A(x)=(x-1)(x-4)^n, n\leq2[/imath]. Next, I find the eigenspace of 4 ([imath]E_4[/imath]): [imath]Null(A-4I) = \begin{pmatrix} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 0 \end{pmatrix}[/imath]. hence, [imath]dim(E_4)=1 \implies m_A(x)=(x-1)(x-4)[/imath]. therefore the jordan form will be : [imath]\begin{pmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4 \end{pmatrix}[/imath] the problem my solution does not agree with the key. | 1379998 | Jordan canonical form of an upper triangular matrix
Find the Jordan canonical form of the matrix. Justify your answer. [imath]A=\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 4 \end{bmatrix} [/imath] My Try: The eigenvalues are [imath]1[/imath] and [imath]4[/imath]. We have to find [imath]P[/imath] such that [imath]P^{-1}AP=J[/imath]. But it is difficult to find two linearly independent vectors for the eigenvalue [imath]4[/imath], since [imath]A-4I[/imath] is of rank [imath]2[/imath]. How would I find [imath]J[/imath] in such case? Can somebody please help me? |
1382799 | [imath]\lim x_n^{x_n}=4[/imath] prove that [imath]\lim x_n=2[/imath]
Let [imath](x_n)[/imath] be a sequence of real numbers, such that: [imath]\lim x_n^{x_n}=4[/imath], prove that [imath]\lim x_n=2[/imath] I'm not sure if my proof is right. I assumed that [imath]\lim x_n [/imath] isn't 2 and using Cauchy's criterion: [imath]|x_n-2|>\epsilon[/imath] so [imath] x_n>\epsilon+2[/imath] or [imath]x_n<-\epsilon+2[/imath] [imath]|x_n^{x_n}-4|<\epsilon [/imath] so [imath]x_n<\sqrt[x_n]{\epsilon+4}[/imath] and then we combine what we've found and get: [imath]\epsilon+2<\sqrt[x_n]{\epsilon+4}[/imath] [imath]\epsilon+4<(\epsilon+2)^2<(\epsilon+2)^{\epsilon+2}<(\epsilon+2)^{x_n}<\epsilon+4[/imath] and it's not true so [imath]\lim x_n=2[/imath]. Is that okay? Edit: I just wanted to know if my solution was right but the other post helped as well, thanks. | 1178164 | Prove [imath]\lim_{n\to\infty}x_n=2[/imath] Given [imath]\lim_{n \to \infty} x_n^{x_n} = 4[/imath]
Given a sequence [imath]\{x_n\}_{n=1}^\infty[/imath] of positive numbers, satisfying [imath]\lim_{n \to \infty} x_n^{x_n} = 4[/imath], prove [imath]\lim_{n\to\infty}x_n=2[/imath]. My attampt: from [imath]\lim_{n \to \infty} x_n^{x_n} = 4[/imath] we know [imath]\exists n_0\in\mathbb{N}:\forall n\ \geq n_0: |x_n^{x_n} -4| < \epsilon[/imath] so [imath]|x_n-2| = \frac{|x_n^2-4|}{|x_n+2|} \leq |x_n^2-4| \leq |x_n^{x_n} -4| <\epsilon[/imath] Which completes the proof if correct. Is it? |
1382889 | Do the initial value problem [imath]dy/dx=2y^{1/2}[/imath] , y(0)=a has infinitely many solution for a=0.
please solve the initial value problem [imath]dy/dx=2y^{1/2}[/imath], y(0)=a. I wanted to know that this problem admits infinitely many solutions for a=0 or admits infinitely many solutions for [imath]a\geq 0[/imath] | 199086 | How can I show that [imath]y'=\sqrt{|y|}[/imath] has infinitely many solutions?
Show that the first order differential equation [imath]y'(x)=\sqrt{|y(x)|}[/imath] with intial value [imath]y(1/2)= 1/16[/imath] has infinitely many solutions on the interval [−1, 1]. My thought were to show that this equation has two solutions (just by ansatz, or just looking at the interval [imath]]0, 1][/imath] to get [imath]y'(x) = \sqrt{|y(x)|}[/imath]) and deduce from that that there must be infinitely many solutions since the solution set forms a vector space. I've got a nagging feeling this isn't the way to go though. Can anybody give me a hint where to look? |
1383377 | What is Laurent series expansion of [imath]\frac{1}{e^z-1}[/imath] around [imath]z=0[/imath]?
Consider [imath]f(z)=\frac{1}{e^z-1}[/imath] I want to expand it over [imath]z_0=0[/imath] in a Laurent series. We know that [imath]e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots[/imath] And I know that [imath]\frac{1}{x-1}=-1-x-x^2-x^3-x^4-x^5+\cdots[/imath] But I can't get the correct answer for my desired series which is, [imath]\frac{1}{x}-\frac{1}{2}+\frac{x}{12}-\frac{x^3}{720}+\frac{x^5}{30240}+O(x^6)[/imath] Any help? | 893708 | Laurent series for [imath]1/(e^z-1)[/imath]
Trying to compute the first five coefficients of the Laurent series for [imath]\frac{1}{e^z-1}[/imath] centered at the point [imath]0[/imath]. I'm not seeing a way to use the geometric series due to the exponential. Any ideas? |
1382366 | [imath]\zeta(2n)[/imath] proof
Can anybody pass me on a good source to see the steps in proving, \begin{equation} \zeta(2n) = \frac{(-1)^{k-1}B_2k (2 \pi)^{2k}}{2(2k)!} \end{equation} I know how we start by looking at the product of sine and use the generatinf function for the Bernoulli numbers to connect them. I am finding it hard to find a source that doesn't just assume the result or say that it is fairly trivial. Any help would be appreciated, Thanks | 1322604 | Ways to prove Eulers formula for [imath]\zeta(2n)[/imath]
I recently, out of interest, tried to prove Euler's formula [imath]\zeta{(2n)}=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}[/imath] for all [imath]n\in\mathbb{N}[/imath]. I adapted Euler's original proof for [imath]\zeta(2)=\frac{\pi^2}{6}[/imath]: We have the well known formulas [imath]\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}[/imath] and [imath]\sin(x)=x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)[/imath]. The product formula can be transformed into a series expansion as well but in order to keep it neat, some definitions first: For the real sequence [imath]\left(a_i\right)_{i\in\mathbb N}[/imath] with [imath]\sum a_i[/imath] convergent, define [imath] A_n:=\sum_{i=1}^{\infty} a_i^n [/imath] [imath] \alpha_n:=\sum_{(x_1,\dots,x_n)\in M_n}\prod_{r=1}^{n} a_{x_r} [/imath] Where [imath]M_n:=\{(x_1,\dots,x_n)\in\mathbb N\space|\space x_1<\dots<x_n\}[/imath]. Now we write [imath]c_i=-\frac{1}{i^2\pi^2}[/imath]. When we expand the product, we get only odd exponents of [imath]x[/imath] and the coefficient of [imath]x^{2k+1}[/imath] for [imath]k≥1[/imath] is [imath]\gamma_k[/imath]. By comparison of the coefficients, we obtain [imath]\gamma_k=\frac{(-1)^k}{(2k+1)!}[/imath]. Now, through telescoping, it can be shown algebraically that for every sequence [imath]\left(a_i\right)_{i\in\mathbb N}[/imath] with [imath]\sum a_i[/imath] convergent we have: [imath]\sum_{i=1}^{n}{(-1)^{n-i}\alpha_{n+1-i}\space A_{i}}=A_{n+1}+(-1)^{n+1}(n+1)\alpha_{n+1}[/imath] For all [imath]n\in\mathbb N_0[/imath]. Since [imath]C_n=\sum_{i=1}^{\infty} \left(-\frac{1}{i^2\pi^2}\right)^n=\frac{(-1)^n}{\pi^{2n}}\zeta{(2n)}[/imath] we have: [imath] \sum_{i=1}^{n}{(-1)^{n-i}\gamma_{n+1-i}\space C_{i}}=\sum_{i=1}^{n}{(-1)^{n-i}\frac{(-1)^{n+1-i}}{(2n-2i+3)!}\frac{(-1)^i}{\pi^{2i}}\zeta{(2i)}}=\sum_{i=1}^{n}{(-1)^{i-1}\frac{\zeta{(2i)}}{(2n-2i+3)!\pi^{2i}}}=\frac{(-1)^{n+1}}{\pi^{2n+2}}\zeta{(2n+2)}+(-1)^{n+1}(n+1)\frac{(-1)^{n+1}}{(2n+3)!}=\frac{(-1)^{n+1}}{\pi^{2n+2}}\zeta{(2n+2)}+\frac{(n+1)}{(2n+3)!} \iff \sum_{i=1}^{n+1}{(-1)^{i-1}\frac{\zeta{(2i)}}{(2n-2i+3)!\pi^{2i}}}=\frac{(n+1)}{(2n+3)!} [/imath] For all [imath]n\in\mathbb N_0[/imath]. Now we are ready to prove the formula [imath]\zeta{(2n)}=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}[/imath] with induction. With the equation above we can easily verify that it is true for [imath]n=1[/imath]; therefore we assume that it holds for all [imath]1≤i≤n[/imath]. The assertion [imath]\zeta{(2n+2)}=(-1)^{n}\frac{(2\pi)^{2n+2}}{2(2n+2)!}B_{2n+2}[/imath] is then equivalent to: [imath] \sum_{i=1}^{n+1}{(-1)^{i-1}\frac{(-1)^{i-1}\frac{(2\pi)^{2i}}{2(2i)!}B_{2i}}{(2n-2i+3)!\pi^{2i}}}=\frac{(n+1)}{(2n+3)!}\iff \sum_{i=1}^{n+1}{\binom{2n+3}{2i}2^{2i-1}B_{2i}}=n+1\iff \sum_{i=0}^{2n+2} \binom{2n+3}{i}2^iB_i=0 [/imath] This identity can be proven by comparing the coefficients in the equation [imath]\frac{e^x-1}{x}\frac{2x}{e^{2x}-1}=\frac{2}{e^x+1}[/imath]. Is this proof valid? Nonetheless, if it is, it isn't very elegant. Are there faster ways to prove it? |
203383 | Proof of [imath]\gcd(a,b)=ax+by[/imath]
Here is my proof of [imath]\gcd(a,b)=ax+by[/imath] for [imath]a, b, x, y \in \mathbb{Z}[/imath]. Am I doing something wrong? Are there easier proofs? [imath]a,b \in \mathbb{Z}, g=\gcd(a,b)[/imath] and suppose [imath]g \neq ax + by[/imath]. Let [imath]c[/imath] be a common divisor of [imath]a[/imath] and [imath]b[/imath]. Then [imath]\forall x', y' \in \mathbb{Z}: c | ax' + by'\Longrightarrow\exists q_1, q_2 \in \mathbb{Z}\,\,\, s.t.\,\,\, c q_1 = ax' + by'\,\,,\,\,cq_2 = g[/imath] So [imath]gcd(a, b)=g=c q_2 = c q_1 \frac{q_2}{q_1} = \left(\frac{q_2}{q_1}x'\right)a + \left(\frac{q_2}{q_1}y'\right)b[/imath] So if we have [imath]q_1|q_2[/imath] then we found [imath]\gcd(a,b)=ax'+by'[/imath] for all [imath]x', y' \in \mathbb{Z}[/imath]. Now [imath]\frac{q_1}{q_2}=\frac{c q_1}{c q_2} = \frac{ax'+by'}{g}[/imath] but [imath]g|a[/imath] and [imath]g|b[/imath] so [imath]\exists q_3 \in \mathbb{Z}: \frac{q_1}{q_2}=q_3 \Rightarrow q_1=q_2 q_3 \Rightarrow q_1 | q_2\,[/imath] . QED. | 2964551 | Greatest Common Divisor question (involving Bezout's theorem)
Assume [imath]a[/imath] and [imath]b[/imath] are positive integers. Consider the following set of positive integers [imath]X =\{xa + yb > 0\,|\,x,y ∈ \mathbb{Z}\}[/imath] (where [imath]x[/imath] or [imath]y[/imath] can be negative or [imath]0[/imath]). [imath]X[/imath] is non-empty since it contains both [imath]a[/imath] and [imath]b[/imath]. Therefore it contains a smallest element [imath]d[/imath]. Prove that [imath]d = \gcd(a, b)[/imath] I have to solve the question above, but I'm not sure how to do it. I think it has to do with Bézout’s theorem, but I have have no idea how to apply it in this particular question. Any help? |
455480 | How were hyperbolic functions derived/discovered?
Trig functions are simple ratios, but what do [imath]\cosh[/imath], [imath]\sinh[/imath] and [imath]\tanh[/imath] compute? How are they related to Euler’s number anyway? | 1711141 | How was [imath]\cosh (x) = \frac{e^x + e^{-x}}{2} [/imath] computed?
I recently came across this fact in my differential equations book, and haven't been able to figure out how this was computed. |
1384245 | looks like Vandermonde determinant
Calculate the determinant of [imath]M = \left( {\begin{array}{*{20}c} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{array}} \right)\;[/imath]. How can one calculate this? Is there a general method like in the case of Vandermonde determinants? | 735883 | Calculate determinant of Vandermonde using specified steps.
[imath]V_n(a_1,a_2\dots, a_n)[/imath] is a [imath]n\times n[/imath] Vandermonde matrix = [imath]\left|\begin{array}[cccc] 11&a_1&\cdots&a^{n-1}_1\\ 1&a_2&\cdots&a^{n-1}_2\\ 1&a_3&\cdots&a^{n-1}_3\\ \vdots&\vdots&\ddots&\vdots\\ 1&a_n&\cdots&a_n^{n-1} \end{array}\right|[/imath] Replace [imath]a_1[/imath] by [imath]x[/imath] so that the first row is [imath]1,x, \dots ,x^{n-1}[/imath] [imath]V_n(x,a_2\dots, a_n) = \left|\begin{array}[cccc] 11&x&\cdots&x^{n-1}\\ 1&a_2&\cdots&a^{n-1}_2\\ 1&a_3&\cdots&a^{n-1}_3\\ \vdots&\vdots&\ddots&\vdots\\ 1&a_n&\cdots&a_n^{n-1} \end{array}\right|[/imath] Let [imath]P(x) = V_n(x,a_2\dots, a_n)[/imath]. (a) Show that [imath]P(x)[/imath] is a polynomial in [imath]x[/imath] of degree [imath]\leq n-1[/imath]. [imath]z = 1\det() -x\det() + x^2\det() -\cdots+ (-1)^{n-1}x^{n-1} \cdot \det()[/imath], where each [imath]\det()[/imath] stands for the determinant of a smaller matrix after removing the appropriate columns, rows. Is this right? But isn't [imath]z[/imath] a polynomial of degree [imath]n[/imath]? (b) Show that [imath]P(x)[/imath] has [imath]n-1[/imath] distinct roots [imath]a_2, \dots a_n[/imath] and therefore has factorization [imath]P(x) = k \prod_{i=2}^n(x-a_i)[/imath] where the constant factor [imath]k[/imath] is the coefficient of [imath]x^{n-1}.[/imath] I'll be honest, I have no clue how to do this. Not even sure where to start. Nor do I know where to start on the rest. (c) Show that [imath]k = (-1)^{n-1}V_{n-1}(a_2,\dots a_n)[/imath]. (d) Use parts (b) and (c) to deduce the recursion formula [imath]V_n(a_1, \dots a_n) = \left(\prod_{i=2}^n(a_i-a_1)\right)V_{n-1}(a_2,\dots a_n)[/imath] (e) Use part (d) to deduce that [imath]V_n(a_1,a_2,\dots a_n) = \prod^n_{1\leq i< j\leq n} (a_j - a_i)[/imath] |
1384207 | Exercise about differential forms and pull-back: [imath]\omega_p=0[/imath] if and only if [imath](\omega|_U)_p=0[/imath]
Let [imath]M[/imath] be a manifold, [imath]\omega\in\Omega^q(M)[/imath]. Let [imath]U\subset M[/imath] be an open subset embedded as manifold in [imath]M[/imath]. Let [imath]p\in U[/imath]. Show that [imath]\omega_p=0[/imath] if and only if [imath](\omega|_U)_p=0[/imath]. Notation: Let [imath]F:U\to M[/imath] be the inclusion map which is an embedding. [imath]F^*:\Omega^*(M)\to\Omega^*(U)[/imath] be its pull-back. We denote [imath]F^*\omega[/imath] by [imath]\omega|_U[/imath]. Suppose [imath]\omega_p =0[/imath]. Then [imath](\omega|_U)_p=(F^*\omega)_p=dF^*_p(\omega_{F(p)})=dF^*_p(\omega_{p})=0[/imath], where the last equality follows by linearity of [imath]dF^*_p[/imath]. I can't prove the reverse implication. Any suggestion? | 654975 | If [imath]\omega \in \Omega^q(M)[/imath], and [imath](\omega|_{U})_p = 0[/imath], is [imath]\omega_p = 0[/imath]?
If I have a manifold [imath]M[/imath] and [imath]\omega \in \Omega^q(M)[/imath], considered an open subset [imath]U[/imath] of [imath]M[/imath], and choose a point [imath]p \in M[/imath], is it true that [imath](\omega|_{U})_p=0 \rightarrow \omega_p=0[/imath]? With [imath]\omega|_{U}[/imath] I mean the pull-back of [imath]\omega[/imath] by the embedding [imath]F:U\to M[/imath]. The converse is obviously true, but this implication? |
1384730 | Are there expansions of the expression [imath](a+b)^{1/n}[/imath]?
Is there an expansion of the expression in the bracket such as [imath] \sqrt{a + b} = (a + b)^{1/2}[/imath] If not do you know of a method that lets us solve such expression and ones with higher roots? | 640895 | Binomial theorem for non integers ? O_o ??
Could we use the binomial theorem for non integers? This comes from: [imath]\sqrt{(a+b)}[/imath] which I can write as [imath](a+b)^{1/2}[/imath] Could I then use the binomial theorem to figure out the value of this expression? |
1384917 | Relation between number of digits of a number and its logarithm?
I found a couple of questions where, for example, they ask you to calculate the number of digits in [imath]18^{200}[/imath] and only the value of [imath]\log 18[/imath] is given. Can anyone tell me a way? | 231742 | Proof: How many digits does a number have? [imath]\lfloor \log_{10} n \rfloor +1[/imath]
I read recently that you can find the number of digits in a number through the formula [imath]\lfloor \log_{10} n \rfloor +1[/imath] What's the logic behind this rather what's the proof? |
1382504 | Prove that for a group with even order [imath]2k[/imath], [imath]k[/imath] odd, there is a subgroup [imath]K[/imath] with order [imath]k[/imath]
I'm trying to understand the proof my teacher did: Consider a subgroup [imath]H[/imath] of [imath]G[/imath]. If [imath]H[/imath] is not contained in [imath]A_n[/imath], then we can say that there exists at least one permutation in [imath]H[/imath] that is odd (remember that [imath]A_n[/imath] is the group of even permutations). So, if we assign the homomorphism [imath]\phi: H\to \{1,-1\}[/imath] such that: [imath]\phi(\sigma) = \ \ \ \ 1, \sigma \mbox{ is even}[/imath] [imath]\phi(\sigma) = -1, \sigma \mbox{ is odd}[/imath] then, the kernel of this homomorphism is [imath]H\cap A_n[/imath], because the identity of [imath]\{1,-1\}[/imath] is [imath]1[/imath], and the ones that are sent to [imath]1[/imath] are the even permutations. But since they're not entirely contained in [imath]A_n[/imath], then we have to pick the intersection of both sets. So, by the isomorphism theorem, if we have an homomorphism [imath]\phi: H \to \{1,-1\}[/imath], the following is true: [imath]\frac{H}{\ker(\phi)}\cong \{1,-1\}[/imath] so we have: [imath]\frac{H}{H\cap A_n}\cong \{1,-1\}\implies\\\frac{|H|}{|H\cap A_n|} = |\{1,-1\}| = 2 \implies \\ \frac{|H|}{|H\cap A_n|} = 2[/imath] This is all from my head, as I understand, so please correct me if something's wrong. The part that I didn't understand of the proof follows now: Ok, so, I think that we need to find an homomorphism from [imath]G[/imath] to [imath]\{1,-1\}[/imath] such that [imath]K[/imath] is the kernel, and also [imath]K[/imath] is not contained in [imath]A_n[/imath]. If we can find this, we have that: [imath]\frac{G}{K}\cong \{1,-1\} \implies \frac{|G|}{|K|} = \frac{2K}{|K|} = 2 \implies |K| = K \implies K \mbox{ is a group of odd order}[/imath] I understand all above. What I don't understand is how the teacher picked up the homomorphism. It tells me to pick an isomorphism from [imath]G[/imath] to [imath]S_{2k}[/imath] and consider the subgroup [imath]H = \phi(G) < S_{2k}[/imath]. Then it proves that [imath]H[/imath] is not contained in [imath]A_n[/imath] in this way: Picks an [imath]a[/imath] such that [imath]o(a) = 2[/imath]. Then consider a [imath]\phi(a)\in S_{2k}[/imath]. Now [imath]\phi(a)[/imath] is the permutation [imath]\pi_a: G\to G[/imath] defined in this way: [imath]\pi_a(x) = ax[/imath] Then it decomposes [imath]\pi_a[/imath] in cycles so we can know its parity: [imath]\pi_a = (x_1 \ \ \pi_a(x_1))\cdots (x_2 \ \ \pi_a(x_k))[/imath] and since there are [imath]k[/imath] transpositions, [imath]\pi[/imath] is odd. So [imath]\pi[/imath] is not contained in [imath]A_n[/imath] It then concludes that [imath][G:K] = 2[/imath] Could somebody explain why it didn't choose an homomorphism from [imath]H[/imath] to [imath]\{1,-1\}[/imath] and what did happen in the proof in general? UPDATE What I think should be done: If I mimic the result [imath]1[/imath] to the second problem, we should think about an homomorphism from [imath]G[/imath] to [imath]\{1,-1\}[/imath] such that [imath]G\cap A_{2k}[/imath] is the kernel. Then, we could have: [imath]\frac{G}{G\cap A_{2k}}\cong \{1,-1\}\implies \frac{|G|}{|K|} = |\{1,-1\}| \implies \\ \frac{2k}{|K|} = 2 \implies |K| = k[/imath] where the group [imath]K[/imath] is [imath]K = G\cap A_{2k}[/imath]. - UPDATE 2 - 20/08/2015: This question, besides having been asked in the past, did not get answered using the theorem I proved here. I would like to know how to proof works using this theorem, because my book says this theorem is useful to proving my question. | 55964 | Let [imath]G[/imath] be a group of order [imath]2m[/imath] where [imath]m[/imath] is odd. Prove that [imath]G[/imath] contains a normal subgroup of order [imath]m[/imath]
I searched in the existing post and didn't find this problem. I am sorry if someone else have already posted. Let [imath]G[/imath] be a group of order [imath]2m[/imath] where [imath]m[/imath] is odd. Prove that [imath]G[/imath] contains a normal subgroup of order [imath]m[/imath]. There is a hint: Denote by [imath]\rho[/imath] the regular represetation of [imath]G[/imath]: find an odd permutation in [imath]{\rho}(G)[/imath]. I don't know how to find an odd permutation in the regular representation. I am wondering whether all the elements of [imath]G[/imath] of odd order form this subgroup in this case. Thanks. |
1384479 | Solve the trig equation [imath]\cos\theta − \sin\theta = 1[/imath]
Solve the given equation. Let k be any integer. [imath]\cos θ − \sin θ = 1[/imath] What do I do? Am I allowed to square everything? I was thinking about squaring everything and then substituting in [imath]1-\sin^2θ[/imath] for [imath]\cos^2θ[/imath] and then factoring out a [imath]\sinθ[/imath], then setting both equations [imath]=0[/imath] but I'm not sure if I am allowed to square everything. | 618192 | How to solve [imath]\sin x +\cos x = 1[/imath]?
No matter how I do it, I always end up with [imath]x = 0, 90, 270[/imath] and [imath]360[/imath]. All of those except [imath]270[/imath] is right, but I can't quite figure out how to get the [imath]270[/imath] degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with [imath]2\sin x\cos x[/imath] which then leads me to [imath]x = 0, 90, 270, 360[/imath]. But [imath]\sin (270) + \cos (270) = -1[/imath] so I'm doing something wrong. |
1384830 | [imath]F = \langle yz-2xy^2, axz-2x^2y+z, xy+y \rangle[/imath]
Given that: [imath]F = \langle yz-2xy^2, axz-2x^2y+z, xy+y \rangle[/imath] in which [imath]a[/imath] is some constant. Now, for what [imath]a[/imath] would make the vector field of [imath]F[/imath] conservative? How can we find an [imath]f[/imath] with [imath]\nabla f=F[/imath]? Also for what [imath]a[/imath] would [imath]F[/imath] be the curl of another vector field? | 1383229 | Vector Field Conceptual Question
Given that: [imath]F = \langle yz-2xy^2, axz-2x^2y+z, xy+y \rangle[/imath] in which [imath]a[/imath] is some constant. Now, for what [imath]a[/imath] would make the vector field of [imath]F[/imath] conservative? Why is there only one, or are there many? How can we find an [imath]f[/imath] with [imath]\nabla f=F[/imath]? Also for what [imath]a[/imath] would [imath]F[/imath] be the curl of another vector field? For to find for which [imath]a[/imath] the vector field is conservative, do we have to go through the process of finding partial derivative, or is there a much shorter approach. I don't know how to approach my other questions either. |
355365 | Infinitely many positive integers [imath]n[/imath] such that [imath]\phi(n) = \frac{n}{4}[/imath]?
Do there exist infinitely many positive integers [imath]n[/imath] such that [imath]\phi(n) = \dfrac{n}{4}[/imath]? | 2524136 | Are there any positive integers n for which [imath]\phi(n) = \frac{n}{4}[/imath]?
For starters, I established that n cannot be prime, as [imath]\phi(n) = n - 1[/imath] for prime numbers. So, if [imath]n[/imath] is composite, I wrote [imath]n[/imath] as its prime factorization: [imath]n = p_1^{r_1}...p_k^{r_k}[/imath] and thus the phi-totient of n is [imath]\phi(n) = n (\frac{p_1 - 1}{p_1})...(\frac{p_k - 1}{p_k})[/imath]. However, now I'm stuck. I know since [imath]\phi(n) = \frac{n}{4}[/imath] then [imath](\frac{p_1 - 1}{p_1})...(\frac{p_k - 1}{p_k}) = \frac{1}{4}[/imath] but I'm not really sure how to prove this, or if I can. Where do I go from here? |
1385587 | Does the dihedral group [imath]D_{12}[/imath] have a subgroup of order [imath]4[/imath]?
Let the dihedral group [imath]D_{12}=\{xy\mid x^6=y^2=1,xy=yx^{-1}=yx^5\}[/imath] Order 2 subgroups are: [imath]\{1, x^3\}[/imath], [imath]\{1, y\}[/imath], [imath]\{1, xy\}[/imath], [imath]\{1, x^2 y\}[/imath],[imath]\{1, x^3 y\}[/imath], [imath]\{1, x^4 y\}[/imath],[imath]\{1, x^5y\}[/imath]. Order 3 subgroups are: [imath]\{1, x, x^5\}[/imath], [imath]\{1, x^2, x^4\}[/imath]. Order 6 subgroups are: [imath]\{1, x, x^2, x^3, x^4, x^5\}[/imath]. Does [imath]D_{12}[/imath] have a subgroup of order 4? | 1327191 | Is there a general formula for finding all subgroups of dihedral groups?
It seems that [imath]\{e\}, \{e,s\}, \{e,rs\}, \{e,r^2s\},...,\{e,r^{n-1}s\}, \{e,r,r^2,...,r^{n-1}\}, D_n[/imath] are always subgroups of [imath]D_n[/imath]. Especially when [imath]n[/imath] is odd, these seem to be the only subgroups. But when n is even, say [imath]n=4[/imath], then there are also [imath]\{e,s,r^2,r^r2s\}[/imath] and [imath]\{e,rs,r^2,r^3s\}[/imath]. It makes me wonder, is there a general formula/algorithm for finding all subgroups of [imath]D_n[/imath] when [imath]n[/imath] is even? |
532229 | Subgroups of [imath]D_4[/imath]
I need to determine the subgroups of the dihedral group of order 4, [imath]D_4[/imath]. I know that the elements of [imath]D_4[/imath] are [imath]\{1,r,r^2,r^3, s,rs,r^2s,r^3s\}[/imath] But I don't understand how to get the subgroups.. | 2990947 | Find subgroups of [imath]D_4[/imath].
If r stands for counter-clockwise 90 degree rotation, s stands for horizontal flip. [imath]D_4= \{1, r, r^2, r^3, s, rs, r^2s, r^3s\}[/imath]. What rule should I apply to find the subgroups of [imath]D_4[/imath]? Should I just put elements with same order in the same subgroup? |
1385917 | Polynomial products
This problem [imath] \large \displaystyle\prod \limits^{14}_{k=1}\cos \left( \frac{k \pi }{15} \right) =\ ? [/imath] I solved it in this way [imath] x = \displaystyle \prod \limits^{14}_{k=1}\cos \left( \frac{k\pi }{15} \right) y \\ = \displaystyle\prod \limits^{14}_{k=1}\sin\left( \frac{k\pi }{15} \right) x. y\\ =\displaystyle\prod \limits^{14}_{k=1}\sin\left( \frac{k\pi }{15} \right) \cos \left( \frac{k\pi }{15} \right) x. y \\ = \displaystyle\prod_{k=1}^{14} \frac{1}{2} \sin\left( \frac{2k\pi}{15} \right) \\ = \bigg(\frac{1}{2}\bigg)^{14}\sin \left( \frac{2\pi }{15} \right) \sin \left( \frac{4\pi }{15} \right) \cdots \sin \left( \frac{14\pi }{15} \right) \sin \left( \frac{16\pi }{15} \right) \sin \left( \frac{18\pi }{15} \right) \cdots \sin \left( \frac{28\pi }{15} \right) \\ \text {Now} :\sin \left( \frac{16\pi }{15} \right) = \left( -1\right) \sin \left( \frac{\pi }{15} \right) \& \sin \left( \frac{18\pi }{15} \right) \\ = \left( -1\right) \sin \left( \frac{3\pi }{15} \right)\& \cdots \&\sin \left( \frac{28\pi }{15} \right)\\ = \left( -1\right) \sin \left( \frac{13\pi }{15} \right) x. y \\ = \left( \frac{1}{2} \right) ^{14}\left( -1\right) ^{7} \bigg[ \sin \left( \frac{2\pi }{15} \right) \sin \left( \frac{4\pi }{15} \right) \cdots \sin \left( \frac{14\pi }{15} \right) \sin \left( \frac{\pi }{15} \right) \sin \left( \frac{3\pi }{15} \right) \cdots \sin \left( \frac{13\pi }{15} \right) \bigg] x. y\\ =\left( \frac{1}{2} \right) ^{14}\left( -1\right) ^{7}. y \text{ so} : x =-\left( \frac{1}{2} \right) ^{14} [/imath] the question is how I can solve it by Chebyshev Polynomials??? . any help will appreciate | 173238 | Evaluating the product [imath]\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)[/imath]
Recently, I ran across a product that seems interesting. Does anyone know how to get to the closed form: [imath]\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}[/imath] I tried using the identity [imath]\cos(x)=\frac{\sin(2x)}{2\sin(x)}[/imath] in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something. This gives the correct solution if [imath]n[/imath] is odd, but of course evaluates to [imath]0[/imath] if [imath]n[/imath] is even. So, I tried taking that into account, but must have approached it wrong. How can this be shown? Thanks everyone. |
1385736 | Almost surely tends to problem
[imath]\dfrac{\max|X_i|}{n}[/imath] tends almost surely to [imath]0[/imath] where '[imath]X_i[/imath]' has a distribution with finite [imath]2[/imath]nd order moments. Can anyone here provide me with the proof? | 351081 | Asymptotics of [imath]\max\limits_{1\leqslant k\leqslant n}X_k/n[/imath]
I found an assertion in this paper at the beginning of page 6, but i can't see how to justify it: Let [imath]X_n \geq 0[/imath] i.i.d. with finite expectation then: [imath] \frac1n\max\limits_{k \leq n}X_k \to 0 \quad\text{almost surely as } \,n \to \infty[/imath] As Davide posted there is convergence in probability to zero. I was wondering if it is evident that we have almost sure convergence also. |
1385919 | Find the value of [imath](x^{2}+y^{2})[/imath]
If [imath]x^{3}-3xy^{2}=10[/imath], [imath]y^{3}-3x^{2}y=30[/imath] Find [imath](x^{2}+y^{2})[/imath] I tried but I got nothing, any help please? | 979252 | If [imath](a + ib)^3 = 8[/imath], prove that [imath]a^2 + b^2 = 4[/imath]
If [imath](a + ib)^3 = 8[/imath], prove that [imath]a^2 + b^2 = 4[/imath] Hint: solve for [imath]b^2[/imath] in terms of [imath]a^2[/imath] and then solve for [imath]a[/imath] I've attempted the question but I don't think I've done it correctly: [imath] \begin{align*} b^2 &= 4 - a^2\\ b &= \sqrt{4-a^2} \end{align*} [/imath] Therefore, [imath] \begin{align*} (a + ib)^3 &= 8\\ a + \sqrt{4-a^2} &= 2\\ \sqrt{4-a^2} &= 2 - a\\ 2 - a &= 2 - a \end{align*} [/imath] Therefore if [imath](a + ib)^3 = 8[/imath], then [imath]a^2 + b^2 = 4[/imath]. |
1386335 | Prove : The polynomial has no integral roots.
Q. Prove that a polynomial [imath]f(x)[/imath],with integer coefficients has no integral roots if [imath]f(0)[/imath] and [imath]f(1)[/imath] are both odd integers. My attempt: Let [imath]f(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n[/imath] now [imath]f(0)=a_0[/imath] which is an odd integer. and [imath]f(1)=(a_0+a_1+a_2+\dots+a_n[/imath]) an odd integer. Now, what is the strategy I need to imply to prove that [imath]f(x)[/imath] can't have integral roots. | 83592 | Prove that [imath]f(x)\in \mathbb{Z}[x][/imath] such that [imath]f(0)[/imath] and [imath]f(1)[/imath] are odd has no integer roots
Suppose [imath]f(x) \in \mathbb{Z}[x][/imath] is such that [imath]f(0)[/imath] and [imath]f(1)[/imath] are odd. How do I show that [imath]f(x)[/imath] has no integer roots? |
1387044 | calculate [imath] \lim \limits_{x \to 0} x^a \int_x^1 \frac{f(t)}{t^{a+1}} \, dt[/imath]
Let [imath] f[/imath] be a continious function and [imath]a>0[/imath] calculate the limit [imath] \lim \limits_{x \to 0} x^a \int_x^1 \frac{f(t)}{t^{a+1}} \, dt[/imath] i tried to divied to 2 cases, if [imath] \lim \limits_{x \to 0} \int_x^1 \frac{f(t)}{t^{a+1}} \, dt = L < \infty[/imath] then the answer is clearly [imath]0[/imath], else we can use l'Hopital rule, but I'm not sure how to apply it on the integral, I know that if [imath]F(x) = \int_0^x \frac{f(t)}{t^{a+1}} \, dt[/imath] then the derivative will be [imath]\frac{f(x)}{x^{a+1}} [/imath] but what do i do with [imath]\int_x^1[/imath] thx | 107675 | Correct way to prove the limit: [imath]\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha + 1}}}}dt} [/imath]?
Some days ago I answered a question that asked to find [imath]\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha + 1}}}}dt} [/imath] given that [imath]f[/imath] is continuous in [imath][0,1][/imath] I proceeded as follows: [imath]\eqalign{ & t = x\cdot u \cr & dt = x\cdot du \cr} [/imath] So this is produces: [imath]\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} [/imath] I then thought: "Well, if [imath]f[/imath] is continuous in the closed interval, then it is also uniformly continuous, so I can assume [imath]\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} = f\left( 0 \right)\int\limits_1^\infty {\frac{{du}}{{{u^{\alpha + 1}}}}} = \frac{1}{\alpha }f\left( 0 \right)[/imath] This turned out to be true. However, I wasn't very comfortable with such "move". So now I'm thinking, one can put [imath]\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} [/imath] And then [imath]\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} } \right| < \int\limits_1^{\frac{1}{x}} {\frac{{\left| {f\left( {xu} \right) - f\left( 0 \right)} \right|}}{{{u^{\alpha + 1}}}}du} < \epsilon \frac{{1 - {x^\alpha }}}{\alpha }[/imath] However, this is still insufficient since I need to adress the behaviour of the upper limit too. Can someone show me how to adress both behaviours simultaneously? Would this work? Let [imath]P[/imath] be the statement that [imath]\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} = \frac{{f\left( 0 \right)}}{\alpha }[/imath] Then [imath]P[/imath] is true if and only if [imath] \forall \epsilon > 0\exists \delta > 0[/imath] Such that if [imath]\left| x \right| < \delta [/imath] then [imath] \left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} - \frac{{f\left( 0 \right)}}{\alpha }} \right| < \epsilon [/imath] But then [imath]\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| < [/imath] [imath]\left| {\int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| \leqslant [/imath] [imath]\varepsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } < [/imath] And since [imath]\frac{{1 - {\delta ^\alpha }}}{\alpha } < \frac{1}{\alpha }[/imath] [imath]\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } < \frac{\epsilon }{\alpha } < \epsilon [/imath] |
556322 | Strange combinatorial identity [imath]\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\binom{2n-2k}{n-1}=0[/imath]
I need to find a combinatorial proof of this identity [imath]\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\binom{2n-2k}{n-1}=0.[/imath] I think inclusion exclusion is the best method here. But I"m having a really hard time coming up with a set to count. Permutations don't work here. A hint would be really nice. Thanks. | 260666 | Sum of binomial coefficients [imath]\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n - 2k}{n - 1} = 0[/imath]
How do I prove the following identity: [imath]\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n - 2k}{n - 1} = 0[/imath] I am trying to use inclusion-exclusion, and this will boil down to a sum like inclusion-exclusion, and the [imath]\binom{2n-2k}{n-1}[/imath] term wouldn't matter (it will be equivalent to set sizes). Is this a correct way to go? |
1387239 | Prove that [imath]2730[/imath] divides [imath]n^{13} - n[/imath] for all integers [imath]n[/imath].
Prove that [imath]2730[/imath] divides [imath]n^{13} - n[/imath] for all integers [imath]n[/imath]. What I attempted is breaking [imath]2730[/imath] into [imath]2, 3, 5[/imath], and [imath]7, 13[/imath]. Thus if I prove each prime factor divides by [imath]n^{13} - n[/imath] for all integers [imath]n[/imath], it will prove the original problem. I already proved [imath]2, 3[/imath], and [imath]5[/imath] work. Struggling with how to do [imath]91[/imath]? Other solutions are welcome, but they should be associated with mod and solutions of this kind, because my answer should pertain to what was learned in class. (Edit: Fixed the 91 problem, being slow today...In any case, I've got this problem now, so no answers are necessary. Posting my own soon.) | 596074 | How to show that [imath]2730\mid n^{13}-n\;\;\forall n\in\mathbb{N}[/imath]
Show that [imath]2730\mid n^{13}-n,\;\;\forall n\in\mathbb{N}[/imath] I tried, [imath]2730=13\cdot5\cdot7\cdot3\cdot2[/imath] We have [imath]13\mid n^{13}-n[/imath], by Fermat's Little Theorem. We have [imath]2\mid n^{13}-n[/imath], by if [imath]n[/imath] even then [imath]n^{13}-n[/imath] too is even; if [imath]n[/imath] is odd [imath]n^{13}-n[/imath] is odd. And the numbers [imath]5[/imath] and [imath]7[/imath], how to proceed? |
1387784 | How do we identify if a number can be represented as the sum of squares of[imath] 3[/imath] integers?
How do we identify if a number can be represented as the sum of squares of [imath]3[/imath] integers? For eg: [imath]434 = 11^2 + 12^2 + 13^2[/imath] but [imath]432[/imath] is not? | 793643 | Expressing an integer as the sum of three squares
I'm trying to determine if 1317 and 116 can be written as the sum of three squares? I have the condition that if it is not of the form [imath]4^{\alpha}(8k+7)[/imath] then it can be written as a sum of three squares, but how do I use this condition to work it out? Thank you |
1061927 | when does a partial differential equation have unique solution?
The differential equation [imath] xu_x + yu_y = 2u[/imath] satisfying the initial conditions [imath]y = xg(x), u=f(x)[/imath] with [imath]f(x) = 2x, g(x) = 1[/imath], has no solution [imath]f(x) = 2x^2, g(x) =1[/imath], has infinite number of solutions [imath]f(x) = x^3, g(x) = x[/imath], has a unique solution [imath]f(x) = x^4, g(x) = x[/imath], has a unique solution I don't know when a partial differential equation has unique solution, infinite solutions or no solution. can I solve this by Cauchy's method of characteristics? I have no idea. please help. | 259865 | The differential equation [imath]xu_{x}+yu_{y}=2u[/imath] satisfying the initial condition [imath]y=xg(x),u=f(x)[/imath]
I was thinking about the following problem : Find out which of the following option(s) is/are correct? The differential equation [imath]xu_{x}+yu_{y}=2u[/imath] satisfying the initial condition [imath]y=xg(x),u=f(x)[/imath], with (a)[imath]f(x)=2x,g(x)=1[/imath] has no solution, (b)[imath]f(x)=2x^2,g(x)=1[/imath] has infinite number of solutions, (c)[imath]f(x)=x^3,g(x)=x[/imath] has a unique solution, (d)[imath]f(x)=x^4,g(x)=x[/imath] has a unique solution. My Attempt: Using lagrange's method, we see that [imath]\frac{dx}{x}=\frac{dy}{y}=\frac{du}{2u} [/imath] gives [imath]y/x=c_1,[/imath] and [imath]y=\sqrt uc_2[/imath] ,where [imath]c_1,c_2[/imath] being constants. Now we have to apply to the given initial conditions and check the given options.Here ,I observe that [imath]y=xg(x)=x.1=x[/imath],[as [imath]g(x)=1[/imath]] and [imath]u=f(x)=2x^2[/imath]But,this relation does not satisfy [imath]xu_{x}+yu_{y}=2u[/imath] and so the choice"(a)[imath]f(x)=2x,g(x)=1[/imath] has no solution," is right. Now I am stuck here and could not progress further.Am i going in the right direction? Please help.Thanks in advance for your time. |
270566 | How to calculate the Fourier transform of a Gaussian function?
I would like to work out the Fourier transform of the Gaussian function [imath]f(x) = \exp \left(-n^2(x-m)^2 \right)[/imath] It seems likely that I will need to use differentiation and the shift rule at some point, but I can't seem to get the calculation to work. Does anyone have any advice? By the way, I am using [imath]\mbox{FT}(f)(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,\mathrm d x[/imath] as my definition of a Fourier transform. Many thanks. | 1521744 | How can I make some progress on this Gaussian-looking integral?
An old complex analysis exam question: Evaluate [imath]\large I(a) = \int_{-\infty}^{\infty} e^{-\frac{1}{2}x^2+iax}dx[/imath] So far, I have completed the square in exponent, and now I have the integral [imath]\large I(a) = \int_{-\infty}^{\infty} e^{(x+\frac{ia}{2})^2-\frac{3}{2}x^2+\frac{a^2}{4}}dx[/imath] Perhaps completing the square above gives me an integral where I now don't have to worry about whether a is positive or negative -- and so I don't need to break the integration out into two cases. But I'm not sure what to do from here. Most integration that I come across in old exam questions is an application of the Residue Theorem, but currently the integrand looks pole-free / entire, so there'd be no residues to compute. One last thing I thought of so far is that the integral looks a bit like the Gaussian integrals. Would this be the better / correct path to follow? If so, what should I start with? The [imath]iax[/imath] term makes it tricky to know what sort of substitution I could go with to perhaps get something like [imath]e^{-ax^2}[/imath] Any hints or suggestions are welcome. Thanks, |
1388335 | If [imath]\dim X=n[/imath] then for any norm in [imath]X[/imath], [imath]X[/imath] is complete.
I know there are standard proofs for this theorem, but I need to prove it by contradiction or proving that [imath]\dim X=\infty[/imath]. I thought maybe using Hahn-Banach? Thanks. | 168275 | Proof that every finite dimensional normed vector space is complete
Can you read my proof and tell me if it's correct? Thanks. Let [imath]V[/imath] be a vector space over a complete topological field say [imath]\mathbb R[/imath] (or [imath]\mathbb C[/imath]) with [imath]\dim(V) = n[/imath], base [imath]e_i[/imath] and norm [imath]\|\cdot\|[/imath]. Let [imath]v_k[/imath] be a Cauchy sequence w.r.t. [imath]\|\cdot\|[/imath]. Since any two norms on a finite dimensional space are equivalent, [imath]\|\cdot\|[/imath] is equivalent to the [imath]l^1[/imath]-norm [imath]\|\cdot\|_1[/imath] which means that for some constant [imath]C[/imath], [imath]\varepsilon > 0[/imath], [imath]k,j[/imath] large enough, [imath] \varepsilon > \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|[/imath] for each [imath]1 \leq i \leq n[/imath]. Hence [imath]v_{ki}[/imath] is a Cauchy sequence in [imath]\mathbb R[/imath] (or [imath]\mathbb C[/imath]) for each [imath]i[/imath]. [imath]\mathbb R[/imath] (or [imath]\mathbb C[/imath]) is complete hence [imath]v_i = \lim_{k \to \infty} v_{ki} [/imath] is in [imath]\mathbb R[/imath] (or [imath]\Bbb C[/imath]) for each [imath]i[/imath]. Let [imath]v = (v_1, \dots , v_n) = \sum_i v_i e_i[/imath]. Then [imath]v[/imath] is in [imath]V[/imath] and [imath]\|v_k - v\| \to 0[/imath]: Let [imath]\varepsilon > 0[/imath]. Then [imath] \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon[/imath] for [imath]k[/imath] large enough. |
1388400 | Showing [imath]5^{1/3}[/imath] is not in the field [imath]\mathbb{Q}(2^{1/3})[/imath]
I want to show [imath]5^{1/3}[/imath] is not in the field [imath]\mathbb{Q}(2^{1/3})[/imath]. My reasoning was [imath][\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3[/imath] and [imath][\mathbb{Q}(5^{1/3}):\mathbb{Q}]=3[/imath]. Similarly we know that [imath][\mathbb{Q}(2^{1/3}, 5^{1/3}):\mathbb{Q}]=9[/imath] so by multiplicativity of dimensions, [imath][\mathbb{Q}(5^{1/3}):\mathbb{Q(2^{1/3})}]=3[/imath] which implies that [imath]5^{1/3} \not\in \mathbb{Q}(2^{1/3})[/imath]. Is this the right approach? | 1106670 | Is [imath]\sqrt[3]{5}[/imath] in [imath]\mathbb Q(\sqrt[3]{2})[/imath]?
I am trying to solve question 3.7 (b) from Chapter 15 in Artin's book "Algebra". The problem is: Is it true that [imath]\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})[/imath]? It is clear by Eisenstein's criterion that both [imath]\sqrt[3]{2}[/imath] and [imath]\sqrt[3]{5}[/imath] have their minimal polynomails of egree [imath]3[/imath] over [imath]\mathbb Q[/imath]. Thus if we assume that [imath]\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})[/imath], then we must have [imath]\mathbb Q(\sqrt[3]{5})=\mathbb Q(\sqrt[3]{2})[/imath]. This didn't get me anywhere. Then I tried the most simple minded approach. Say [imath]\sqrt[3]{5}=a+b\sqrt[3]{2}+c\sqrt[3]{4}[/imath], for some [imath]a,b,c\in \mathbb Q[/imath]. By cubing and rearranging, it is likely that we will arrive at some absurdity. But clearly this shouldn't be what the author must have intended. Can somebody point me towards a slick solution? |
608538 | Godel's pairing function and proving c = c*c for aleph cardinals
I have a few questions about Godel's pairing function and proving that c = c * c for aleph cardinals. Mostly, though, I'm concerned that most of the proofs I've seen are erroneous, and this concerns me. (And I've searched many times for a good proof.) So we order the class On x On lexicographically, and this is a proper well-ordering, so that On x On is order-isomorphic to On by a unique mapping P: On x On -> On. This makes perfect sense. But in every "proof" of c = c * c that I've seen, there's been some glaring mistake. For example, one "proof" relies on the "proposition" that "If [imath]\gamma[/imath] = max([imath]\alpha[/imath],[imath]\beta[/imath]) and at least one of [imath]\alpha[/imath],[imath]\beta[/imath] is infinite, then P([imath]\alpha[/imath],[imath]\beta[/imath]) <= card([imath]\gamma[/imath]) x card([imath]\gamma[/imath])." But it seems to me this is false, and here's why: so it seems clear that (0,[imath]\omega[/imath]) is the first limit ordinal in On x On under the lexicographical ordering, and of course [imath]\omega[/imath] is the first limit ordinal in On--consequently, we must have P(0,[imath]\omega[/imath]) = [imath]\omega[/imath]; but then P(1,[imath]\omega[/imath]) = [imath]\omega[/imath] + 1, which contradicts the "proposition" (since P(1,ω) = ω + 1 is not <= [imath]\omega[/imath], which is card([imath]\gamma[/imath]) x card([imath]\gamma[/imath]), as [imath]\gamma[/imath] = [imath]\omega[/imath]). Am I wrong here? I guess what I'm asking for is some kind of real explanation as to how Godel's pairing function is used to prove that c = c * c for all aleph cardinals c. Even if this just involves pointing me toward a decent source, that would be awesome. | 2832882 | Proof: [imath]\kappa \cdot \kappa = \kappa[/imath] for infinite cardinals
Im looking for a detailed proof for [imath]\kappa \cdot \kappa = \kappa [/imath] with [imath]\kappa [/imath] beeing a infinite cardinal number. The problem is described in a book (Frank R. Drake, Set Theory: An Introduction to Large Cardinals) as a exercise but without any solution. As far as i know i can proof that [imath]\kappa \cdot \kappa = \kappa [/imath] holds for the cardinal [imath]\aleph_0[/imath]([imath]\mathbb{N}[/imath]) wich can be proven with "Cantor's Diagonal Proof". But i need it for any infinite cardinal so it should also be valid for [imath]\aleph_1,\aleph_2,...[/imath] So im missing the induction part. I know that it should be possible cause very roughly explained to show something like [imath]| \mathbb{N} \cdot \mathbb{R}| = |\mathbb{R}|[/imath]. We can make a bijection if we put all the [imath]\mathbb{N}[/imath] numbers in between 0,1 of the [imath]\mathbb{R}[/imath] and still have enough space to adress all [imath]\mathbb{R}[/imath]. TL;DR im looking for the proof of [imath]\kappa \cdot \kappa = \kappa [/imath] and [imath]\kappa \cdot \lambda = max(\kappa, \lambda) [/imath] |
1389994 | General formula of a sequence [imath]a_{n+1} = 2a_n + 1/a_n[/imath]
What is the exact formula for [imath]a_n[/imath] in the sequence [imath]a_{n+1} = 2a_n + 1/a_n, a_1=1[/imath]? I discovered that there are no elementary answers, but I don't know how to solve it. | 1385197 | explicit formula for recurrence relation [imath]a_{n+1}=2a_n+\frac{1}{a_n}[/imath]
For [imath]n\in\mathbb N[/imath], [imath]a_{n+1}=2a_n+\frac{1}{a_n},\quad a_1=1. [/imath] Can any one give an explicit formula for all [imath]a_n[/imath]? If such an explicit general formula doesn't exist, please explain it. I've tried to figure out the [imath]n[/imath]-iterated function [imath]f^{(n)}[/imath] where [imath]f(x)=2x+1/x[/imath] or even [imath]f(\tan(t))[/imath]. But in either cases, I failed.Since the recurrence isn't linear nor homogeneous,the generating function method doesn't apply here. |
873628 | How to solve [imath]2\cdot (2x-1)^\frac13=x^3+1[/imath]?
This was one of the questions in my entry exam and I couldn't manage to solve it. I struggled with this equation for more than an hour, so I'd like a direct solution instead of hints please. [imath]2\cdot (2x-1)^\frac13=x^3+1[/imath] | 929053 | Solving an equation over the reals: [imath] x^3 + 1 = 2\sqrt[3]{{2x - 1}}[/imath]
Solve the following equation over the reals:[imath] x^3 + 1 = 2\sqrt[3]{{2x - 1}} [/imath] I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by [imath](x-1)[/imath].. But I can't see the solution, how do I go from here? |
1391056 | Show that polynomial is primitive in GF(5)
How can I show that [imath]x^2 + 2x + 3[/imath] is primitive in [imath]GF(5)[/imath]? My idea: [imath] x^1 = x\\ x^2 = -2x - 3 = 3x + 2\\ x^3 = (3x + 2)x = 3x^2 + 2x = 3(3x + 2) + 2x = x + 1\\ ...\\ x^a = 1\\ [/imath] This would take quite long. Is there a better way? Please only basic Algebra! | 358182 | Primitive polynomials
I am revising for a discrete mathematics exam and as quite stuck on this question. Show that the polynomial [imath]f = x^2 + 2 x + 3 \in \mathbb{Z}_5[x][/imath] is primitive. How many monic primitive quadratic polynomials are there in [imath]\mathbb{Z}_5[x][/imath]? Any help would be greatly appreciated. |
1390990 | Prove that the set of eigenvalues of block matrix with blocks [imath]A[/imath] and [imath]B[/imath] is the union of eigenvalues of [imath]A[/imath] and [imath]B[/imath].
Let [imath]A,B,U,O[/imath] four matrix of real entries. [imath]A[/imath] is a square matrix of size [imath]m[/imath], [imath]B[/imath] is a square matrix of size [imath]n[/imath], while [imath]U[/imath] is a [imath]n\times m[/imath] matrix of all entries [imath]=-1[/imath] and [imath]O[/imath] is a [imath]m\times n[/imath] matrix of all entries [imath]=0[/imath]. Let us consider the square (block) matrix of size [imath]n+m[/imath]: [imath]M=\left( \begin{array}{cc} A & U \\ O & B \\ \end{array} \right) [/imath] Prove that the set of eigenvalues of [imath]M[/imath] is the union of eigenvalues of [imath]A[/imath] and [imath]B[/imath]. This property of determinants should follow immediately if we can show [imath]\det M=\det A \det B[/imath] but I do not know to prove this statement. Any suggestions please? | 21454 | Prove that the eigenvalues of a block matrix are the combined eigenvalues of its blocks
Let [imath]A[/imath] be a block upper triangular matrix: [imath]A = \begin{pmatrix} A_{1,1}&A_{1,2}\\ 0&A_{2,2} \end{pmatrix}[/imath] where [imath]A_{1,1} ∈ C^{p \times p}[/imath], [imath]A_{2,2} ∈ C^{(n-p) \times (n-p)}[/imath]. Show that the eigenvalues of [imath]A[/imath] are the combined eigenvalues of [imath]A_{1,1}[/imath] and [imath]A_{2,2}[/imath] I've been pretty much stuck looking at this for a good hour and a half, so any help would be much appreciated. Thanks. |
1387820 | A proper definition of [imath]i[/imath], the imaginary unit
Back when I was in high school, which was a long time ago, I recall my math teacher telling me that the definition of [imath]i[/imath], the imaginary unit, is [imath]\sqrt{-1}[/imath]. Knowing little, at the time, I accepted it without thinking twice. Several years later when I was in my Complex Analysis class, one of my classmates asked for the distinction between defining [imath]i[/imath] in the normal way, i.e., [imath]i=\sqrt{-1}[/imath], and more ambiguous way, [imath]i^2=-1[/imath]. Now that I thought deeply about it, I have some suspicions about what my high school teacher taught me at the time. Firstly, at least at the level of high school, one normally defines a square root of [imath]x[/imath], [imath]\sqrt{x}[/imath] as the positive quantity of either number that satisfies the property [imath]\sqrt{x}\sqrt{x}=x[/imath]. Clearly, when [imath]x<0[/imath], such notion of sign makes no sense, so one cannot honestly talk about [imath]\sqrt{-1}[/imath] with the naive definition of the square root. Fine, we are better than that, and we may say that [imath]i[/imath] is the principal root of the equation [imath]x^2=-1[/imath]. We then just denote it by [imath]\sqrt{-1}[/imath]. But this way of denoting [imath]i[/imath] brings with its convenience a litany of disasters, including the famous [imath]1=-1[/imath] fallacy. Namely, one can show that [imath]1=-1[/imath] by [imath]1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i^2=-1[/imath]. Many a people have pointed out the haphazardness of assuming that the familiar law [imath]\sqrt{x}\sqrt{y}=\sqrt{xy}[/imath] holds when [imath]x,y<0[/imath]. But at the same time we have no shame in writing [imath]\sqrt{-5}[/imath] as [imath]\sqrt{5}i[/imath] (in fact, I think this is the very reason of inventing the imaginary unit). Is it not terribly unnatural that the law holds for odd number of negative factors, and not so for the even ones? In fact, is there an example where this native rule (that when you have a negative radicand, you can pretty much apply the familiar laws of exponents)? (I guess it makes the first question.) Secondly (so this officially marks the second, and the last question), which ought to be the definition of [imath]i[/imath], in your opinion? I believe that many people choose to write [imath]i=\sqrt{-1}[/imath] as it gives some illusion of determinancy, whereas [imath]i^2=-1[/imath] does not. But I still prefer the latter definition, and it seems to be the consensus of every complex analysis textbook that I've ever laid my hands on. Better yet, I believe that the complex numbers shold be defined as the algebraic completion of reals or an isomorphic field to [imath]\mathbb{R}\times\mathbb{R}[/imath], with some special addition and multiplication rules, but I guess it is a little bit out of high school students' league (at least for most of them). EDIT: Thank you all for your insightful responds, but there still one thing none of you has yet answered... Is there a conunter example to the law where we have [imath]\sqrt{-A}[/imath] for [imath]A>0[/imath], we have [imath]\sqrt{A}i[/imath]? | 887724 | Refining my knowledge of the imaginary number
So I am about halfway through complex analysis (using Churchill amd Brown's book) right now. I began thinking some more about the nature and behavior of [imath]i[/imath] and ran into some confusion. I have seen the definition of [imath]i[/imath] in two different forms; [imath]i = \sqrt{-1} [/imath] and [imath]i^2 = -1[/imath]. Now I know that these two statements are not equivalent, so I am confused as to which is the 'correct' definition. I see quite frequently that the first form is a common mistake, but then again Wolfram Math World says otherwise. So my questions are: What is the 'correct' definition of [imath]i[/imath] and why? Or are both definitions correct and you can view the first one as a principal branch? It seems that if we are treating [imath]i[/imath] as the number with the property [imath]i^2 = -1[/imath], it is implied that we are treating [imath]i[/imath] as a concept and not necessarily as a "quantity"? If we are indeed treating [imath]i[/imath] as a concept rather than a "quantity", how would things such as [imath]i^i[/imath] and other equations/expressions involving [imath]i[/imath] be viewed? How would such an equation have value if we treat [imath]i[/imath] like a concept? I've checked around on the various imaginary number posts on this site, so please don't mark this as a duplicate. My questions are different than those that have already been asked. |
1388369 | Show that [imath]\sqrt{6+4\sqrt{2}}-\sqrt{2}[/imath] is rational using the rational zeros theorem
Let [imath]r=\sqrt{6+4\sqrt{2}}-\sqrt{2}[/imath], then [imath]r+\sqrt{2}=\sqrt{6+4\sqrt{2}}[/imath]. Squaring both sides, we get [imath]r^2+2r\sqrt{2}+2=6+4\sqrt{2}[/imath] which is the same as [imath]r^2-4=2\sqrt{2}[/imath]. Squaring both sides again, we get [imath]r^4-8r^2+16=8[/imath] or [imath]r^4-8r^2+8=0.\tag{$\star$}[/imath] The rational zeros theorem tells us that the only possible rational solutions to ([imath]\star[/imath]) are [imath]±1[/imath], [imath]±2[/imath], [imath]±4[/imath], [imath]±8[/imath]. I do not know where to go from here. Please help me complete this proof. | 1388206 | Show that [imath]\sqrt{4+2\sqrt{3}}-\sqrt{3}[/imath] is rational using the rational zeros theorem
What I've done so far: Let [imath]r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.[/imath] Thus, [imath]r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7[/imath] and [imath]r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.[/imath] I did this because in a similar example in class, we related [imath]r^2[/imath] and [imath]r^4[/imath] to find a polynomial such that [imath]mr^4+nr^2 = 0[/imath] for some integers [imath]m,n[/imath]. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems. |
1388282 | Irreducible polynomial [imath]X^q-2[/imath]
Prove that the polynomial [imath]X^q-2[/imath] is irreducible in the ring [imath]\Bbb Q(\sqrt[p]{2})[X][/imath] What method i can use for proving, that this polynomial is irreducible in this specific ring? | 1369148 | Problem on Galois theory and irreducible polynomial
Let [imath]p,q[/imath] be primes, estimate the degree [imath][\Bbb Q(\sqrt[p]{2}\cdot\sqrt[q]{2}):\Bbb Q][/imath] and prove that the polynomial [imath]X^q-2[/imath] is irreducible in the ring [imath]\Bbb Q(\sqrt[p]{2})[X][/imath] I found this problem on mathematical competition without any solution and it would be very interesting if we can find a proof for this. |
666662 | When does the quotient metric reduce to the infimum of the distances of only two points?
Given a metric space [imath]X[/imath] and an equivalence relation [imath]\sim[/imath], the quotient (pseudo-)metric on [imath]X/\sim[/imath] is defined as follows: [imath]d'([x],[y]) = \inf \left \{ d(p_1,q_1) + d(p_2,q_2) + ... + d(p_n,q_n)\right \}[/imath] where [imath][p_1] = [x], [q_n] = [y],[/imath] and [imath][q_i] = [p_{i+1}][/imath]. Under what circumstances does this reduce to the much simpler equation [imath]d'([x],[y]) = \inf \left \{ d(p,q): [p] = [x], [q] = [y] \right \}[/imath] ? It appears to work for [imath]\mathbb{R}^n/\mathbb{Z}^m[/imath] for [imath]m \leq n[/imath], but I'm curious about the general case. | 613347 | Why does the natural quotient metric works in this case?
There have been at least two discussions in this forum about why a (pseudo) metric is defined in the quotient of a metric space by: [imath] d([x],[y]) = \inf \{ d(p_{1}, q_{1}) + \ldots, d(p_{n}, q_{n}) \} [/imath] where [imath][p_{1}] = [x][/imath], [imath][q_{n}] = [y][/imath] and [imath][q_{i}] = [p_{i+1}][/imath] for [imath]i = 1, \ldots, n-1[/imath]. I understood that this is defined in order to guarantee the triangle inequality. I wasn't able, myself, to verify it, so some help here would be appreciated. Nevertheless, the main point of my question is the following: in Anatole Katok's book A First Course in Dynamics, proposition 2.6.7 asserts that the far more natural function [imath] d([x],[y]) = \min \{ |b-a| : a \in [x], b \in [y] \} [/imath] is actually a metric in [imath]\mathbb{R} / \mathbb{Z}[/imath]. That left me wondering: is there some general extra condition ensuring that the more natural function [imath]\inf[/imath] of distances over all representatives is a metric in the quotient? What's special about this example that doesn't work in the general case? Reading Katok's proof, I had a guess that it lies in the fact that there is a minimum-attaining representative in this case, which can be fixed. |
1392684 | Does [imath]F[/imath] is an isomorphism [imath]\implies F^{-1}[/imath] is an isomorphism?
Does [imath]F[/imath] is an isomorphism [imath]\implies F^{-1}[/imath] is an isomorphism? Let [imath]\mathfrak{D, E} [/imath] be some algebraic structures of the same 'kind' (say, groups, graphs, fields, vector spaces, etc.), and [imath]F:\mathfrak D \to \mathfrak E[/imath] is an isomorphism, is there any way to prove that [imath]F^{-1}:\mathfrak E \to \mathfrak D[/imath] is an isomorphism? By isomorphism I mean a bijection that preserves whatever structure [imath]\mathfrak D[/imath] and [imath]\mathfrak E[/imath] have, clearly [imath]F^{-1}[/imath] is a bijection, so the important part is if it is a structure preserving map. | 552867 | [imath]F:G\to B[/imath] is an isomorphism between groups implies [imath]F^{-1}[/imath] is an isomorphism
[imath]F[/imath] is an isomorphism from group [imath]G[/imath] onto group [imath]B[/imath]. Prove [imath]F^{-1}[/imath] is an isomorphism from [imath]B[/imath] onto [imath]G[/imath]. I do not know which direction to go in. |
1392476 | Prove a matrix is invertible
The [imath]2 \times 2[/imath] matrix [imath]{A}[/imath] satisfies [imath]A^2 - 4 {A} - 7{I} = {0},[/imath] where [imath]{I}[/imath] is the [imath]2 \times 2[/imath] identity matrix. Prove that [imath]{A}[/imath] is invertible. What is the best way to do this? | 1369372 | If [imath]\mathbf{A}[/imath] is a [imath]2\times 2[/imath] matrix that satisfies [imath]\mathbf{A}^2 - 4\mathbf{A} - 7\mathbf{I} = \mathbf{0}[/imath], then [imath]\mathbf{A}[/imath] is invertible
[imath]\mathbf{A}[/imath] is a [imath]2\times 2[/imath] matrix which satisfies [imath]\mathbf{A}^2 - 4\mathbf{A} - 7\mathbf{I} = \mathbf{0}[/imath], where [imath]\mathbf{I}[/imath] is the [imath]2\times 2[/imath] identity matrix. Prove that [imath]\mathbf{A}[/imath] is invertible. Hint: Find a matrix [imath]\mathbf{B}[/imath] such that [imath]\mathbf{A}\mathbf{B}=\mathbf{I}[/imath]. I tried substituting a variable matrix for [imath]\mathbf{A}[/imath] and substituting [imath]\mathbf{I}[/imath]'s value into both the original equation and the hint, but the result was full of equations and didn't seem to help much at all. I would appreciate advice for this problem. Thanks. --Grace |
1392851 | Is [imath]2^n-1[/imath] always a prime for odd values of [imath]n[/imath]?
Is [imath]2^n-1[/imath] always a prime for odd values of [imath]n[/imath]? [imath]n\not=1[/imath] Taking some odd values of [imath]n[/imath], I observed outcome is coming as a prime number. How to verify it? Or at-least, is [imath]2^n-1[/imath] always coprime to [imath]n[/imath]? | 1349878 | Proof that [imath]2^n-1[/imath] does not always generate primes when primes are plugged in for [imath]n[/imath]?
Exactly what the name entails. The function [imath]2^n-1[/imath] I see largely tends to generate primes when [imath]n[/imath] is prime. However, a week ago I heard that this was horribly false. Please show me a disproof. |
1392971 | Prove that [imath]4[/imath] divides [imath]n[/imath]
Let [imath]a_1[/imath],[imath]a_2[/imath],[imath]a_3[/imath],.......,[imath]a_n[/imath] be [imath]n[/imath] such that each [imath]a_i[/imath] either [imath]1[/imath] or [imath]-1[/imath].If [imath]a_1 a_2 a_3 a_4+a_2 a_3 a_4 a_5+......+a_n a_1 a_2 a_3=0[/imath], then prove that [imath]4[/imath] divides [imath]n[/imath]. I tried this for small n and also observed different specific cases such as [imath]a_1=1[/imath] [imath]a_2=-1[/imath] [imath]a_3=1[/imath] [imath]a_4=-1[/imath].... etc. But I am unable to circumvent anyway. Any kind of hints and full answers will be appreciated. Thank you all in advance. | 445946 | Prove that n is divisible by 4 in a cylic sum with variables which have only two possible values
It is known that [imath]a_1, a_2, a_3, ... , a_n \in \left\{-1, 1 \right\}[/imath] and [imath]S = a_1a_2a_3a_4 + a_2a_3a_4a_5 + ... + a_na_1a_2a_3 = 0[/imath] Prove that [imath]n \equiv 0\space(mod\space 4)[/imath] I know this problem can be solved using number theory, but I am looking for a solution utilizing the concept of invariance. I was able to identify an invariant of sorts: even if you change the sign of any [imath]a_i[/imath], the sum will still remain congruent to same number [imath]mod \space 4[/imath]. It can be proved using this invariant that no matter what values you choose for each of the [imath]a_i[/imath], [imath]S \equiv 0 \space (mod \space 4)[/imath]. It is not yet clear how I can prove the required statement using this result. |
1392892 | Relation between eigen values of [imath]AB[/imath] and [imath]BA[/imath]
Let say [imath]A \in \mathbb{F}^{n\times m}[/imath] and [imath]B \in \mathbb{F}^{m\times n}[/imath]. I am wondering weather there is any standard result stating the relationship between eigen values of [imath]AB[/imath] and [imath]BA[/imath] matrices. I know that eigen values are same for [imath]AB[/imath] and [imath]BA[/imath] when both [imath]A[/imath] and [imath]B[/imath] are square matrices. But in a general setup, we can not use the result that- [imath]\det(AB) = \det(BA)[/imath]. | 378868 | The relationship between the eigenvalues of matrices [imath]XY[/imath] and [imath]YX[/imath]
If [imath]X \in \mathbb{C}^{m \times n}[/imath] and [imath]Y \in \mathbb{C}^{n \times m}[/imath] ([imath]m \geq n[/imath]), how to prove the following? [imath]\lambda (XY) = \lambda (YX) \cup \underbrace{\left \{ 0, ..., 0 \right \}}_{m-n}[/imath] Here, [imath]\lambda \left ( \cdot \right )[/imath] denotes the set of eigenvalues. |
1392555 | schauder basis for [imath]\ell_\infty[/imath]
I know that [imath]\ell_\infty[/imath] is not separable, therefore has no Schauder basis. However I cannot understand why the set [imath]\{e_1, e_2, e_3, \dotsc \}[/imath] where [imath]e_1=(1,0,0,\dotsc), e_2=(0,1,0,0,\dotsc), \dotsc[/imath] can not work as a Schauder basis. Every sequence [imath]x=(x_1,x_2,x_3,\dotsc)[/imath] that belongs to [imath]\ell_\infty[/imath] can be written in one and only one way as [imath]\sum_{i=1}^\infty x_i e_i[/imath]. Could you explain why this is wrong? Thank you! | 1271935 | Why is [imath](e_n)[/imath] not a basis for [imath]\ell_\infty[/imath]?
Let [imath](e_n)[/imath] (where [imath] e_n [/imath] has a 1 in the [imath]n[/imath]-th place and zeros otherwise) be unit standard vectors of [imath]\ell_\infty[/imath]. Why is [imath](e_n)[/imath] not a basis for [imath]\ell_\infty[/imath]? Thanks. |
1393031 | A Residue problem
Preparing myself for qualifying exam, I found this problem in residues in some previous qualifying exam, and I am stuck and don't know how to solve it, any helps? [imath]f(z)=(ze^{2-z}+(2-z)e^z)e^{(z-1)^{2}+(z-1)^{-2}} [/imath] at z = 1 | 1391435 | A Tough Problem about Residue
I tried my best to solve this problem from what I learned in residues, but the solution seems very far from what I was doing!! Is there any way other than using Laurent series expansion? Here is the problem: [imath]f(z)=(ze^{2-z}+(2-z)e^z)e^{(z-1)^{2}+(z-1)^{-2}} [/imath] at z = 1 Thank you |
1393160 | How to integrate [imath]sec^{3}x[/imath]
How do I integrate [imath]sec^{3}x[/imath]. I have done this by converting [imath]sec[/imath] to [imath]cos[/imath] and then using [imath]cos^{3}x= (1-sin^{2}x)cosx[/imath] and then putting sinx=t and then using partial fractions. But this is very long method. I want shorter method for this Thanks | 1392749 | Finding [imath]\int \sec^3x\,dx[/imath]
I've tried substitution, replacing [imath]\sec^2x[/imath] with [imath]\tan^2x + 1[/imath], and parts and I just hit dead ends every time... Do you need knowledge of higher-level calculus to solve this? |
1393265 | Limit of [imath](n!)^{1/n}[/imath] as [imath]n\rightarrow \infty[/imath].
How to prove that[imath](n!)^{1/n}[/imath] tends to infinity as limit tends to infinity? I tried to do this by expanding [imath]n![/imath] as [imath]n\times (n-1)\times (n-2)\cdots 4\times3\times2\times 1[/imath] and taking out n common from each factor so that I can have [imath]n[/imath] outside the radical sign, But then the last terms would be [imath](4/n)\times(3/n)\times(2/n)\times (1/n)[/imath], which would tend to zero and would present indeterminate form of [imath]0\cdot \infty[/imath], but how should I further solve it. I would appreciate a little help. | 293473 | Prove that [imath]n! ≥ (⌈n/2⌉)^{⌈n/2⌉}[/imath]
Prove that : [imath]n! ≥ (⌈n/2⌉)^{⌈n/2⌉}[/imath] |
1393352 | Intuitive explanation of div(curlF)=0
If we consider [imath]\mathbf{F}[/imath] as a vector field, then we say that [imath]\mathrm{div}(\mathrm{curl}(\mathbf{F}))=0[/imath]. We can prove this in mathematics easily. But I' am not getting an intuitive explanation due to which it is zero. Can someone explain intuitively why it is zero? Thanks | 26854 | What is an intuitive explanation for [imath]\operatorname{div} \operatorname{curl} F = 0[/imath]?
I am aware of an intuitive explanation for [imath]\operatorname{curl} \operatorname{grad} F = 0[/imath] (a block placed on a mountainous frictionless surface will slide to lower ground without spinning), and was wondering if there were a similar explanation for [imath]\operatorname{div} \operatorname{curl} F = 0[/imath]. |
1393985 | Show that [imath]\int_0^\infty \!\frac{e^{-x}-e^{-xt}}{x}\,\mathrm{d}x=\log(t)[/imath]
I have tried a lot of different integration methods and I can't solve it. [imath]\int_0^\infty\! \frac{e^{-x}-e^{-xt}}{x}\,\mathrm{d}x=\log(t)[/imath] | 164400 | Show [imath]\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),[/imath] for [imath]t \gt 0[/imath]
The problem is to show [imath]\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),[/imath] for [imath]t \gt 0[/imath]. I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution [imath]u=e^{-x}[/imath] and came to a new integral (hopefully after no mistakes) [imath] \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, [/imath] but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself. Thanks a lot! |
1394059 | What is the least possible value of [imath](999)?[/imath]
Let [imath] be a one-to-one function from the set of natural numbers to itself such that $() = ()()$ for all natural numbers [/imath] and [imath].[/imath] What is the least possible value of [imath](999)?[/imath] What I think is that I have a hunch for the answer and follow the intuitions, any help? | 971342 | Functional Equation [imath]f(mn)=f(m)f(n)[/imath].
If [imath]f: \mathbb N \mapsto \mathbb N[/imath] is one-to-one and [imath]f(mn) = f(m)f(n)[/imath], what is the smallest possible value of [imath]f(999)[/imath]? Easily [imath]f(1)=1[/imath], and I think [imath]f(n)=n[/imath] must be the only map, but not able to prove it. The other alternative is if [imath]f(999) = f(3)^3f(37) \ge 24[/imath], which looks somehow odd. |
1394283 | Does [imath]1+x+x^2 ... x^{p -1}[/imath] being a prime number imply p is prime?
Let [imath]p[/imath] and [imath]x[/imath] be two positive integers greater than 2. If it is given that the sum : [imath]1+x+x^2+x^3... x^{p -1}[/imath] is a prime, is it possible to prove or disprove that [imath]p[/imath] is prime? If so, what would this proof be? | 132633 | Prime number in a polynomial expression
Will be glad for a little hint: let x and n be positive integer such that [imath]1+x+x^2+\dots+x^{n-1}[/imath] is a prime number then show that n is prime |
1394225 | Finding the n-th arrangement of items with repetitions
I'm new to Stackexchange and maybe I do not have the correct mathematical terms for the question I'm about to ask. I'm given a multiset of given size [imath]N[/imath] which consists of zeros and ones. Example: Multiset size [imath]5[/imath] with number of ones: [imath]2[/imath], number of zeros: [imath]3[/imath] How many distinct arrangements (sequences) of all these items are possible? First arrangement is: [imath]0,0,0,1,1[/imath] Listing all arrangements in ascending order: 1: [imath]0,0,0,1,1[/imath] 2: [imath]0,0,1,0,1[/imath] 3: [imath]0,0,1,1,0[/imath] 4: [imath]0,1,0,0,1[/imath] 5: [imath]0,1,0,1,0[/imath] 6: [imath]0,1,1,0,0[/imath] 7: [imath]1,0,0,0,1[/imath] 8: [imath]1,0,0,1,0[/imath] 9: [imath]1,0,1,0,0[/imath] 10: [imath]1,1,0,0,0[/imath] Because in a real task the set size could be much larger than this example, is it possible without iterating through all arrangements: To find the [imath]n^{\text{th}}[/imath] arrangment in the lexicographically ordered list (example [imath]7^{\text{th}}[/imath] permutation = [imath]1,0,0,0,1[/imath]) ? To calculate the index if I'm given the arrangement (example [imath]0,1,0,1,0 = 5[/imath]) ? I'm looking for an algorithm suitable for implementation in some programming language. | 1363239 | Fast way to get a position of combination (without repetitions)
This question has an inverse: (Fast way to) Get a combination given its position in (reverse-)lexicographic order What would be the most efficient way to translate a combination of [imath]k[/imath]-tuple into its positions within the [imath]\left(\!\!\binom{n}{k}\!\!\right)[/imath] combinations? I need this to be fast for combinations of [imath]\left(\!\!\binom{70}{7}\!\!\right)[/imath] order of magnitude - very large, but not exceeding 2 billion (fits into int32 maximum value). Below is an example of [imath]\left(\!\!\binom{6}{3}\!\!\right)[/imath], where the aim is to quickly translate (a, d, f) tuple to value 9, while in the real problem [imath]k[/imath] ranges between 5 and 8. [imath]\begin{array} {cccccc|c} a&b&c&d&e&f&^{combination}/_{sort\_order}& \\\hline x&x&x& & & &1\\ x&x& &x& & &2\\ x&x& & &x& &3\\ x&x& & & &x&4\\ x& &x&x& & &5\\ x& &x& &x& &6\\ x& &x& & &x&7\\ x& & &x&x& &8\\ x& & &x& &x&9\\ x& & & &x&x&10\\ .&.&.&.&.&.&.\\ & & &x&x&x&20\\ \end{array}[/imath] I know that I could pre-calculate all the combinations and reverse the lookup dictionary. However, such dictionary would not be efficient in terms of memory usage. Therefore I am looking for either calculation-based approach, or a more efficient data structure to perform this mapping. |
1394732 | Prove that [imath]\int_{-\infty}^{\infty} (F(x+a)-F(x))dx=a[/imath] for all [imath]a>0[/imath].
This is a qualifying exam problem. Suppose [imath]F[/imath] is a distribution function of a Borel measure [imath]\mu[/imath] with [imath]\mu (\Bbb R)=1[/imath]. Prove that [imath]\int_{-\infty}^{\infty} (F(x+a)-F(x))dx=a[/imath] for all [imath]a>0[/imath]. | 958265 | For a distribution function [imath]F(x)[/imath] and constant [imath]a[/imath], integral of [imath]F(x + a) - F(x)[/imath] is [imath]a[/imath].
For any distribution function and any [imath]a \geq 0[/imath], [imath]\int_{-\infty}^{\infty} (F(x+a)-F(x))dx = a[/imath]. In this case, "distribution function" means a right continuous function F with [imath]F(-\infty) = 0[/imath], [imath]F(\infty)= 1[/imath]. This problem comes after a section in the book on mathematical expectation, so I suspect I should use some property of expectation to prove it. Expectation appears to be something that applies to random variables, so I think I should be somehow transforming the problem into something in terms of expectations of random variables, but I'm not sure how. I know that given a distribution function, I can get a probability measure, but I'm not sure how to go from there to a random variable and actually get a relation between the expectation of the random variable and this integral. I'm sure whether this is the right way to go about this problem, but if it is, I'm missing a connection somewhere. |
1394846 | Is there a formula for [imath]1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{N}[/imath]?
Is there a known formula to the sum [imath]1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{N}[/imath] where [imath]N[/imath] is some natural number? Thanks | 544222 | sum of square roots
I was wondering what the estimate for the value [imath] S= \sum_{j=1}^N \sqrt{j} [/imath] is? Is there a way (or a formula) to estimate it well? I am sure it is close to [imath]\int_1^N \sqrt{t} dt[/imath], but I guess I was interested in estimating it better. Thanks! |
1394815 | Question on finitely generated algebras
While reading Reid's "Introduction to Algebraic Geometry", I came across the following passage: "A finitely generated [imath]k[/imath]-algebra is a ring of the form [imath]A = k[a_1,\cdots,a_n][/imath], so that [imath]A[/imath] is generated as a ring by [imath]k[/imath] and [imath]a_1, \cdots , a_n[/imath]; clearly, every such ring is isomorphic to a quotient of the polynomial ring, [imath]A = k[X_1, \cdots ,X_n]/I[/imath]" My question is: why must such ring be isomorphic to a ring of the form [imath]k[X_1, \cdots ,X_n]/I[/imath]? | 35403 | finitely generated k-algebra and polynomial ring
Let [imath]k[/imath] be a field and let [imath]A \neq 0[/imath] be a finitely generated [imath]k[/imath]-algebra, and [imath]x_1, \cdots, x_n[/imath] generate [imath]A[/imath] as a [imath]k[/imath]-algebra. Is there any relationship(inclusion, homomorphism, etc.) between [imath]A[/imath] and [imath]k[x_1,\cdots,x_n][/imath]? How about the case if [imath]x_1,\cdots,x_n[/imath] are algebraically independent over [imath]k[/imath]? |
1394993 | Proof exercide of [imath]\sigma[/imath]-algebra
If [imath]M(\epsilon)[/imath] is the [imath]\sigma[/imath]-algebra generated by [imath]\epsilon[/imath], then [imath]M(\epsilon)[/imath] is the union of the [imath]\sigma[/imath]-algebras generated by [imath]F[/imath] as [imath]F[/imath] ranges over all countable subsets of [imath]\epsilon[/imath] Proof: Let [imath]M = \bigcup_{F\subset \epsilon}M(F)[/imath], we need to show that [imath]M = M(\epsilon)[/imath]. First, we will show that [imath]M\subset M(\epsilon)[/imath], observe that for each countable subset of [imath]F[/imath] of [imath]\epsilon[/imath] that [imath]M(F)\subset M(\epsilon)[/imath]. Since, [imath]F\subset \epsilon \subset M(\epsilon) \Rightarrow M(\epsilon)[/imath] is a [imath]\sigma[/imath]-algebra containing [imath]F[/imath] therefore [imath]M(F)\subset M(\epsilon)[/imath]. Now we need to show [imath]M\subset \epsilon[/imath] but this is where I am sort of lost, any suggestions is greatly appreciated. I know this is similar to another question that was asked but I didn't really like the notation and I wish to follow through with what I have set up already | 1370946 | Showing a sigma algebra is the union of certain collection of sigma algebras
Suppose [imath]\mathcal{A}[/imath] is a [imath]\sigma[/imath]-algebra generated by [imath]\mathcal{E}[/imath]. Then [imath]\mathcal{A}[/imath] is the union of the [imath]\sigma[/imath]-algebras generated by [imath]\mathcal{F}[/imath] where [imath]\mathcal{F}[/imath] ranges over all countable subsets of [imath]\mathcal{E}[/imath] I am kind of confused when the propostion says "ranges over all countable ... " Am I asked to show that [imath] \mathcal{A} = \bigcup \{ \mathcal{M} : \mathcal{M} = \sigma( \mathcal{F}), \mathcal{F} \; \; \text{countable subset of $\mathcal{E}$} \} \text{ ??}[/imath] |
1394302 | Can We Always Realize the Value of the Quotient Norm.
Let [imath](V, \|\cdot\|)[/imath] be a Banach space over [imath]\mathbf R[/imath] and [imath]W[/imath] be a closed subspace of [imath]V[/imath]. We know that [imath]V/W[/imath] becomes a normed linear space under the quotient norm [imath]\|\cdot\|_q[/imath] defined as [imath]\|v+W\|_q=\inf_{w\in W} \|v+w\|[/imath] for all [imath]v\in V[/imath]. Suppose [imath]v\in V[/imath] and we have [imath]\|{v+W}\|_q=\lambda[/imath]. Does there necessarily exist a [imath]w\in W[/imath] such that [imath]\|{v+w}\|=\lambda[/imath]. Clearly, the above is true if [imath]\lambda=0[/imath] or if [imath]V[/imath] is finite dimensional (because we can use the compactness of [imath]S^n[/imath]). But I am getting nowhere with the statement in block quotes. We get a sequence [imath](w_n)[/imath] in [imath]W[/imath] such that [imath]\|v+w_n\|>\lambda+1/n[/imath] for all [imath]n[/imath]. But this is weaker than saying [imath]\|v+W\|_q=\lambda[/imath]. | 299735 | On the norm of a quotient of a Banach space.
Let [imath]E[/imath] be a Banach space and [imath]F[/imath] a closed subspace. It is well known that the quotient space [imath]E/F[/imath] is also a Banach space with respect to the norm [imath] \left\Vert x+F\right\Vert_{E/F}=\inf\{\left\Vert y\right\Vert_E\mid y\in x+F\}. [/imath] Unfortunately in a set of lecture notes on (Lie) group representations (material for our study group) the author accidentally used here [imath]\min[/imath] instead of [imath]\inf[/imath]. Probably a mostly harmless booboo, because at that point it was only needed to get a Banach space structure on the quotient, and we will probably be concentrating on Hilbert spaces anyway, where the problem does not arise. Namely from Rudin's Functional Analysis I could not find a proof that the minimum should always be attained. Except in the case of a Hilbert space, where an application of parallelogram law (the sum of the squared norms of the two diagonals of a parallelogram equals that of the four sides) allows us to find a Cauchy sequence among a sequence of vectors [imath](y_n)\subset x+F[/imath] such that [imath]\lim_{n\to\infty}\left\Vert y_n\right\Vert_E=\left\Vert x+F\right\Vert_{E/F}.[/imath] But anyway, the suspicion was left that the infimum is there for a reason (other than conveniently allowing us to sweep this detail under the rug at that point of the development of theory), so in the interest of serving our study group I had to come up with a specific example, where the minimum is not achieved. It's been 25 years since I really had to exercise the Banach space gland in my brain, so it has shrunk to size of a raisin. Searching this site did help, because I found this question. There we have [imath]E=C([0,1])[/imath], the space of continuous real functions on [imath][0,1][/imath] equipped with the sup-norm. If we denote by [imath]\Lambda[/imath] the continuous functional [imath] \Lambda: E\to\mathbb{R},f\mapsto\int_0^{1/2}f-\int_{1/2}^1f [/imath] and let [imath]F=\ker\Lambda[/imath], then the answer to the linked question proves that there is no minimum sup-norm function in the coset [imath]\Lambda^{-1}(1)[/imath]. So I have a (counter)example, and the main question has evolved to: When can we use minimum in place of infimum in the definition of the quotient space norm? My thinking: It seems to me that the answer is affirmative, if [imath]F[/imath] has a complement, i.e. we can write [imath]E=F\oplus F'[/imath] as a direct sum of two closed subspaces such that the norm on [imath]E[/imath] is equivalent to the sum of the norms on [imath]F[/imath] and [imath]F'[/imath]-components. But the first point also raises the suspicion that the question may be a bit ill-defined (and uninteresting) in the sense that the answer might depend on the choice of the norm [imath]\left\Vert\cdot\right\Vert_E[/imath] among the set of equivalent norms. However, if we, for example, perturb the sup-norm of [imath]C([0,1])[/imath] in the above example by multiplying the functions with a fixed positive definite function before taking the sup-norm, the argument seems to survive, so may be replacing the norm with an equivalent one is irrelevant? So to satisfy my curiosity I also welcome "your favorite example" (one with a finite-dimensional [imath]F[/imath] would be nice to see), where we absolutely need the infimum here. Bits about any sufficient or necessary conditions for the minimum to be sufficient or (as a last resort :-) pointers to relevant literature are, of course, also appreciated. |
1302651 | Theorem of Lagrange multipliers - Extremas of [imath]f[/imath]
I have to find the extremas of [imath]f(x, y, z)=x+y+z[/imath] subject to [imath]x^2-y^2=1[/imath], [imath]2x+z=1[/imath]. I have done the following: We will use the theorem of Lagrange multipliers. The constraints are [imath]g_1(x,y,z)=x^2-y^2-1=0 \ \ , \ \ g_2(x, y, z)=2x+z-1=0[/imath] We are looking for [imath]x[/imath], [imath]y[/imath], [imath]z[/imath], [imath]\lambda_1[/imath] and [imath]\lambda_2[/imath] such that [imath]\nabla f(x, y, z)=\lambda_1 \nabla g_1(x, y, z)+\lambda_2 \nabla g_2(x, y, z) \tag 1[/imath] and [imath]g_1 (x, y, z)=0 \tag 2[/imath] [imath]g_2(x, y, z)=0\tag 3[/imath] [imath](1) \Rightarrow (1, 1, 1)=\lambda_1(2x, -2y, 0)+\lambda_2(2, 0, 1)[/imath] So, we have that [imath]1=2\lambda_1 x+2\lambda_2 \\ 1=-2\lambda_1 y \\ 1=\lambda_2[/imath] So, we get [imath]x=y[/imath], right?? When we apply this at the first constraint [imath]g_1(x, y, z)=0[/imath] we get [imath]-1=0[/imath], or not?? What have I done wrong?? | 1004983 | Global maxima/minima of [imath]f(x,y,z) = x+y+z[/imath] in [imath]A[/imath]
Find the global maxima/minima of [imath]f(x,y,z) = x+y+z[/imath] for points inside of [imath]A = \{ (x,y,z) \in \mathbb{R}^3: x^2-y^2 = 1 \wedge 2x+z = 1 \}[/imath] I renamed the conditions of [imath]A[/imath] to a function [imath]g(x,y,z) = x^2-y^2-2x-z = 0[/imath] in order to be able to use Lagrange multipliers. Deriving [imath]f[/imath]: [imath]\nabla(x,y,z) = (1,1,1)[/imath] Deriving [imath]g[/imath]: [imath]g_x = 2x-2[/imath] [imath]g_y = -2y[/imath] [imath]g_z = -1[/imath] By solving the Lagrange system: \begin{cases} 1 = \lambda(2x-2) \\ 1 = \lambda (-2y) \\ 1 = \lambda (-1) \\ g(x,y) = 0 \end{cases} I get that [imath](x,y,z) = (\frac{1}{2}, \frac{1}{2}, -1)[/imath] for [imath]\lambda = -1[/imath]. But that point is not inside A. Is this enough to guarantee that [imath]f|_A[/imath] does not have either maxima nor minima? (Since the second derivatives are all zero, the second derivative test doesn't give any information.) Thanks! EDIT: By parametrizing [imath]y = \pm \sqrt{x^2-1}[/imath] and [imath]z=1-2x[/imath] and doing the composition with [imath]f[/imath] I get that [imath]f(x, \sqrt{x^2-1}, 1-2x)[/imath] is monotonically increasing and [imath]f(x, -\sqrt{x^2-1}, 1-2x)[/imath] monotonically decreasing. Would that prove that [imath]f[/imath] never reaches either maxima nor minima? |
1395761 | Why must [imath]A(V)[/imath] be an open subset of [imath]A[/imath]'s range?
Suppose [imath]A[/imath] is a linear transformation from [imath]\Bbb R^n[/imath] to [imath]\Bbb R^m[/imath] with [imath]\text{rank}A=r>0[/imath], [imath]V[/imath] is an open subset of [imath]\Bbb R^n[/imath], then is [imath]A(V)[/imath] an open (relative to [imath]\text{Im}(A)[/imath]) subset of [imath]\text{Im} (A)[/imath]? I don't know how to prove this. I cannot construct a diffeomorphism from [imath]V[/imath] to [imath]A(V)[/imath] because there is a dimension shrinkage. Can you help me with this? Thanks a lot! | 273195 | A surjective linear map into a finite dimensional space is open
I'm in search of different proofs of the following proposition: [imath]\bf{Proposition}[/imath]: Suppose [imath]X[/imath] and [imath]Y[/imath] be topological vector spaces, [imath]\text{dim }Y<\infty[/imath], and [imath]\Lambda:X\to Y[/imath] is a surjective linear map. Then [imath]\Lambda[/imath] is open. Any and all proofs are welcomed. |
1396227 | A bounded Lebesgue measurable function and integral.
I honestly have no idea where to even begin with this question: Let [imath]f: [0,1]\rightarrow \mathbb{R}[/imath] be bounded and Lebesgue measurable. Suppose that for every [imath]0\leq a<b\leq 1[/imath], we have [imath] \int_a^b f(x)\ \mathsf dx=0. [/imath] Prove that [imath]f=0[/imath] almost everywhere. | 274702 | Integral vanishes on all intervals implies the function is a.e. zero
I am having trouble with the following problem: [imath]f:\mathbb{R}\to \mathbb{R}[/imath] is a measurable function such that for all [imath]a[/imath]: [imath]\int_{[0,a]}f\,dm=0.[/imath] Prove that [imath]f=0[/imath] for [imath]m[/imath] almost every [imath]x[/imath] (here [imath]m[/imath] is the Lebesgue measure). I have no problem proving this for [imath]f[/imath] non-negative, or under the assumption that [imath]f[/imath] is integrable. But the question only assumes that [imath]f[/imath] is measurable and no more. My idea was the usual thing; we look at the set of points where [imath]f[/imath] is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where [imath]f[/imath] is integrable. Alternatively, if it were integrable one could simply estimate [imath]f[/imath] by a continuous function, where the result is quite obvious. Ultimately we are going to show that [imath]f[/imath] is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions? |
1394140 | Please help me this hard circle geometry question
Let [imath]T_1[/imath] be a circle with centre at the point [imath]O[/imath] and radius [imath]R[/imath]. Two other circles [imath]T_2[/imath] and [imath]T_3[/imath] with centres at [imath]O_2[/imath] and [imath]O_3[/imath] respectively are tangent internally to [imath]T_1[/imath] and meet each other ([imath]T_2[/imath] and [imath]T_3[/imath]) at the points [imath]A[/imath] and [imath]B[/imath]. Find the sum of the radii of [imath]T_2[/imath] and [imath]T_3[/imath]. [imath]R_2[/imath] + [imath]R_3[/imath], if [imath]\angle OAB = \pi/2[/imath]. [OP's thoughts, added by Brian Tung] I got [imath]O_2AO_3B[/imath] is a Kite. Also if extend [imath]OA[/imath] and intersects [imath]T_3[/imath] to point [imath]W[/imath] in [imath]T_2[/imath] for example, and extend [imath]AO[/imath] to [imath]T_2[/imath] then intersect [imath]T_2[/imath] to [imath]V[/imath], can get [imath]O_2O_3[/imath] equal to half of [imath]WV[/imath]. If I can approve [imath]WV[/imath] is paralleling to [imath]O_2O_3[/imath] or [imath]OO_3[/imath] is paralleling to [imath]WB[/imath], may get the result of [imath]R_2 + R_3 = R[/imath]. Not sure if this is the right direction? | 1396302 | Sum of radii of intersecting circles internally tangent to another circle
Let [imath]T[/imath] be a circle with centre at [imath]O[/imath] and radius [imath]R[/imath]. Two other circles [imath]T_1[/imath] and [imath]T_2[/imath] with centres at [imath]O_1[/imath] and [imath]O_2[/imath], respectively, are tangent internally to [imath]T[/imath]. [imath]T_1[/imath] and [imath]T_2[/imath] intersect one another at the points [imath]A[/imath] and [imath]B[/imath]. Find the sum of the radii of [imath]T_1[/imath] and [imath]T_2[/imath], [imath]R_1 + R_2[/imath], if [imath]\angle OAB = \pi/2[/imath]. The answer should be expressed in terms of [imath]R[/imath] and the coordinates of the points [imath]O_1[/imath] and [imath]O_2[/imath] (relative to [imath]O[/imath]). (This was a question posted by someone who subsequently deleted his/her account. The question originally generated a lot of comments - and was eventually put on hold - because the OP hadn't made it clear enough. Some of us thought the question is sufficiently interesting to be re-opened so I've cleaned it up and re-posted it anew. The original question is found here but may have been deleted by the time you read this) |
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