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2779823 | prime ideals of [imath]\mathbb{R}[X,Y]/(X-Y^2,X+Y)[/imath]
I'm trying to understand an example in my lecture notes, it states that [imath]Spec(\mathbb{R}[X,Y]/(X-Y^2,X+Y))[/imath] has only two points. Can anyone develop more please ? and how would these points be like ? Thanks for the help. | 2775515 | [imath]\mathbb{R}[X,Y]/(Y-X^2,Y+X)\cong_\text{Ring} \mathbb{R}\times \mathbb{R}[/imath]
I have the following problem: Let [imath]R:=\mathbb{R}[X,Y]/(Y-X^2,Y+X)[/imath] i) Show that [imath]R\cong_\text{Ring} \mathbb{R}\times \mathbb{R}[/imath] ii) Conclude that [imath]|\operatorname{Spec}R|=2[/imath]. How do these two ideals look like? So for the first one I got the hint to eliminate one variable at a time and see what happens. Doing that I got: [imath]1.) \mathbb{R}[X]/(-X^2,X)\cong \mathbb{R}[X]/(-X^2)\times\mathbb{R}[X]/(X)\cong\mathbb{R}[X](-X^2)\times \mathbb{R}[/imath] [imath]2.) \mathbb{R}[Y]/(Y,Y)=\mathbb{R}[Y]/(Y)\cong\mathbb{R}[/imath] using the chinese remainder theorem in [imath]1.)[/imath]. But since [imath]\mathbb{R}[X]/(-X^2,X)\times\mathbb{R}[Y]/(Y)\ncong\mathbb{R}[X,Y]/(Y-X^2,Y+X)[/imath] and the fact that there is still [imath]\mathbb{R}[X]/(-X^2)[/imath] left in the first equation, I don't see how this leads me to the desired isomorphism. For the second one I don't see how an isomorphism can lead me to the number of ideals, because I'm pretty sure [imath]A\cong B[/imath] doesn't imply [imath]\operatorname{Spec}A \cong \operatorname{Spec}B[/imath] for rings [imath]A,B[/imath] (I've seen it in a book). Can someone help me with this problem? Thanks in advance. |
2780148 | Step in proof of [imath]\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/imath]
How was [imath]\pi^4/90[/imath] calculated here in the answer? Attempt: We have that [imath]c_0=\pi^2/3[/imath] and [imath]|c_{\pm k}|^2=4/k^4[/imath] thus [imath]|c_k|^2=|c_{\pm k}+c_0|^2\ ?[/imath] Or what should I put in the [imath]|c_k|^2[/imath] here [imath]\sum_{k=-\infty}^\infty |c_k|^2\ [/imath]? Thanks in advance for your help. Answer (By Christian Blatter user) Consider the function [imath]f(t):=t^2\ \ (-\pi\leq t\leq \pi)[/imath], extended to all of [imath]{\mathbb R}[/imath] periodically with period [imath]2\pi[/imath]. Developping [imath]f[/imath] into a Fourier series we get [imath]t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).[/imath] If we put [imath]t:=\pi[/imath] here we easily find [imath]\zeta(2)={\pi^2\over6}[/imath]. For [imath]\zeta(4)[/imath] we use Parseval's formula [imath]\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2\ .[/imath] Here [imath]\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}[/imath] and the [imath]c_k[/imath] are the complex Fourier coefficients of [imath]f[/imath]. Therefore [imath]c_0={\pi^2\over3}[/imath] and [imath]|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}[/imath] [imath]\ (k\geq1)[/imath]. Putting it all together gives [imath]\zeta(4)={\pi^4\over 90}[/imath]. | 2778906 | Understanding [imath]\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/imath]
In the answer, of this question Nice proofs of [imath]\zeta(4) = \frac{\pi^4}{90}[/imath]?, given by Christian Blatter, Why [imath]|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}[/imath] [imath]\ (k\geq1)[/imath] ? According to my calculations [imath]|c_{\pm k}|^2=\vert1/2\pi\int_{-\pi}^\pi t^2e^{-i(\pm k)x}dx\vert^2=\vert t^2\sin(\pi(\pm k))/\pi(\pm k)\vert^2=0[/imath] And how did he get [imath]\frac{\pi^4}{90}?[/imath] Consider the function [imath]f(t):=t^2\ \ (-\pi\leq t\leq \pi)[/imath], extended to all of [imath]{\mathbb R}[/imath] periodically with period [imath]2\pi[/imath]. Developping [imath]f[/imath] into a Fourier series we get [imath]t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).[/imath] If we put [imath]t:=\pi[/imath] here we easily find [imath]\zeta(2)={\pi^2\over6}[/imath]. For [imath]\zeta(4)[/imath] we use Parseval's formula [imath]\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2\ .[/imath] Here [imath]\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}[/imath] and the [imath]c_k[/imath] are the complex Fourier coefficients of [imath]f[/imath]. Therefore [imath]c_0={\pi^2\over3}[/imath] and [imath]|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}[/imath] [imath]\ (k\geq1)[/imath]. Putting it all together gives [imath]\zeta(4)={\pi^4\over 90}[/imath]. |
2780189 | Number of real roots of a polynomial
Find the number of real roots of the polynomial [imath]f(x)=x^5+x^3-2x+1[/imath] ( I think similar questions have already been asked. However, I am not sure whether the exact question has already been raised by some user. I urge the members of this mathematical community to mark my question as duplicate if it has already been asked. I tried an attempt below:-) What I attempted:- I know a little bit of theory of equations but not all. While I was in the examination, I did not ponder much more on any theorems relating to real roots. Rather I tried to do it with some intuition. We have [imath]f(1)=1[/imath],[imath]f(0)=1[/imath],[imath]f(2)=1[/imath], [imath]f(\frac{1}{2})=\frac{5}{32}[/imath],[imath]f(-\frac{1}{2})=\frac{59}{32}[/imath],[imath]f(-2)=-35[/imath] It just helped me to trace a root between [imath]-1[/imath] and [imath]-2[/imath]. Then I tried to study the nature of the graph. Here [imath]f'(x)=5x^4+3x^4-2[/imath] Let [imath]f'(x)=5x^4+3x^2-2=0[/imath] [imath]\Rightarrow 5y^2+3y-2=0[/imath] (where [imath]x^2=y[/imath]) [imath]\Rightarrow (5y-2)(y+1)=0[/imath] Therefore, [imath]y=\frac{2}{5}[/imath], or [imath]y=-1[/imath] (It is not of our interest, as we are dealing with real roots) Therefore [imath]x=\sqrt{\frac{2}{5}},-\sqrt{\frac{2}{5}}[/imath] These two values divides the real line into three disjoint subsets [imath](-\infty,-\sqrt{\frac{2}{5}})[/imath], [imath](-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})[/imath], [imath](\sqrt{\frac{2}{5}},\infty)[/imath] It s quite easy to notice that [imath]f[/imath] is decreasing in [imath](-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})[/imath] and increasing in [imath](-\infty,-\sqrt{\frac{2}{5}})[/imath], [imath](\sqrt{\frac{2}{5}},\infty)[/imath] Again, [imath]f(-\sqrt{\frac{2}{5}})=1.911[/imath] and [imath]f(\sqrt{\frac{2}{5}})=0.089[/imath] Looking at all these details it is clear that [imath]f(x)[/imath] comes close to zero in [imath](-\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}})[/imath] but is not equal to zero. There is no possibility for [imath]f(x)[/imath] to be zero in [imath](\sqrt{\frac{2}{5}},\infty)[/imath] as it is increasing here. Again, we have already discovered one root between [imath]-1[/imath] and [imath]-2[/imath]. Apart from that there is of course no root as [imath]f[/imath] is again increasing in [imath](-\infty,-\sqrt{\frac{2}{5}})[/imath]. So, te graph should look something like this Therefore there is only one real root of the polynomial. Is my approach correct? Is there any way to do the same thing using some more beautiful as well as advanced method ? (I have just made a rough plot of the graph in my answer sheet). I am not sure if the examiner would expect much better method than this one. | 1640561 | Determine the number of real roots
Determine number of real roots of [imath]f(x)=x^{5}+x^{3}-2x+1[/imath]. I used Descartes's rule of sign to determine the number of positive and negative roots; there is only 1 negative (hence minimum 1 real root) and two positive roots. These 2 positive roots can be anything- a complex pair, or two real roots- but what about the other 2 roots? Is there any flaw in my understanding of rule of signs? How can I find the exact number of real roots? Thanks in advance. |
2780430 | Set Theory - Elementary Question on Empty Sets
Ok so we are going to begin looking at Set theory next week in class so I thought I would try to learn a bit about it before we start. I am curious and perplexed by [imath](A \cap \emptyset)′[/imath] in regard to what it can be simplified to as when using the rules of set theory. For example if I have [imath]U[/imath] and [imath]A[/imath], [imath]B[/imath] are both subsets does that mean that [imath](A \cap \emptyset)′ = U[/imath] or does [imath](A \cap \emptyset)′ = (A)'[/imath]? If anyone can please provide some insight that would be much appreciated. | 2220993 | What is the complement of empty set?
We have learned that for universal set [imath]U[/imath] and for set [imath]A[/imath] belonging to [imath]U[/imath], [imath]A^c=U-A[/imath] but what about for the empty set? |
2780470 | To prove that triangles of side [imath]a^n, b^n, c^n[/imath] are isosceles.
Let [imath]a\ge b \ge c\gt 0[/imath] be real numbers such that for all [imath]n \in \mathbb N[/imath] there exist triangles of side [imath]a^n, b^n, c^n[/imath]. Prove that the triangles are isosceles. I tried proving it by writing [imath]c^n + b^n \gt a^n[/imath] and when I assumed some values for [imath]a, b \text{ and } c[/imath] I realized that it would be true for all [imath]n[/imath] only when [imath]c=b[/imath]. But I don't know how to generally prove this. | 2779345 | Condition on sides of triangle to proof it is isoceles
Let [imath] a\geq b \geq c > 0 [/imath] be real numbers such that for all [imath]n [/imath] element of natural number , there exists triangles of side length [imath] a^n , b^n , c^n. [/imath] Prove that the triangles are isosceles . I tried taking derivatives and proving that there will exist some [imath]n[/imath] at which inequality would stop holding. But it is not working out. Also I tried substituting triangle inequalities for [imath]n=1[/imath] ; but it is of no use I guess. I think binomial maybe of some use. I tried completing the binomial of [imath]a^n + b^n[/imath]. But I am not able to use it too. |
2780399 | For what values of ''[imath]a[/imath]'' , does this Improper Integral converges?
For which values of ''[imath]a[/imath]'', does the integral [imath]\int_{0}^{\infty} \frac{\sin x}{x^a}\;\mathrm dx[/imath] converges? I have shown that [imath]\left| \int_{0}^{\infty} \frac{\sin x}{x^a}\;\mathrm dx \right|\le \lim_{n\to\infty} \int_{0}^{n} \frac 1{x^a}\;\mathrm dx,[/imath] which converges if [imath]a > 1[/imath]. Are there other values of [imath]a[/imath] for which the integral converges? | 2779384 | For what values of [imath]\alpha[/imath], does this integral converge?
For what values of [imath]\alpha>0[/imath] this integral converges [imath]\int_0^{\infty} \frac {\sin x}{x^{\alpha}}dx [/imath] |
2758354 | Chromatic polynomial of non-tree graphs?
If x ≥ 3 and G is not a tree, show that [imath]χ(G, x) < x(x−1)^{n−1}[/imath] I think if G is connected, the result is trivial. How can I deal with the case that G is not connected? | 2758293 | Chromatic polynomial of connected graph [imath] \leq x(x-1)^{n-1}[/imath]
Number of proper [imath]x[/imath]-coloring, [imath]x \in {1,2,3,4,...}[/imath] of connected graph with [imath]n[/imath] vertices is less than or equal to [imath]x(x-1)^{n-1}[/imath] for [imath]n \geq 1 [/imath]. I think there are two cases that I need to consider [imath]Case[1][/imath] If G is connected and has no cycles, then it follows that G is a tree, where chromatic polynomial is just equal to [imath]x(x-1)^{n-1}[/imath] [imath]Case[2][/imath] If G is connecteded and has some cycles For second case, I am just not sure how to start proof that chromatic polynomial is less than [imath]x(x-1)^{n-1}[/imath] Also, would it suffice to consider above two cases to prove the claim? |
2781357 | Function Equation Question Involving [imath]f(xy)=f(x)f(y)-f(x+y)+1[/imath]
Let [imath]\mathbb{Q}[/imath] denote the set of rational numbers. Find all functions from [imath]\mathbb{Q}[/imath] to [imath]\mathbb{Q}[/imath] which satisfy [imath]f(1)=2[/imath]; and [imath]f(xy)=f(x)f(y)-f(x+y)+1[/imath]. I know that [imath]f(x)=x+1[/imath] satisfies the equation but I don't know how to prove if this is the only such function. This question was in the 'Induction' section of my textbook. Any hints? | 96316 | About finding the function such that [imath]f(xy)=f(x)f(y)-f(x+y)+1[/imath]
Define a function [imath]f\colon\mathbb{R}\to\mathbb{R}[/imath] which satisfies [imath]f(xy)=f(x)f(y)-f(x+y)+1[/imath] for all [imath]x,y\in\mathbb Q[/imath]. With a supp condition [imath]f(1)=2[/imath]. (I didn't notice that.) How to show that [imath]f(x)=x+1[/imath] for all [imath]x[/imath] that belong to [imath]\mathbb{Q}[/imath]? |
2782006 | Can we paint the plane in two colors such that there is no square of the same color?
We have our every day plane [imath]\mathbb{R}^2[/imath] which we want to paint using two colors in such a way that no four points form a square of the same color. This means that for any square on the plane its vertices have different colors. Is it possible to paint a plane in such a way? | 242724 | Monochromatic squares in a colored plane
Color every point in the real plane using the colors blue,yellow only. It can be shown that there exists a rectangle that has all vertices with the same color. Is it possible to show that there exists a square that has all vertices with the same color ? If it is not possible, please give me an example of a coloring of the real plane that does not have monochromatic squares. To give the viewers an idea about other similar results (which they might find useful), for any coloring (2 colors) of the real plane: 1) There exists three collinear points having the same color, such that one of the points is the midpoint of the line segment that joins the other two. 2)For any two angles [imath]\theta,\phi[/imath] there exists a monochromatic triangle that has angles [imath]\theta,\phi,180-(\theta+\phi)[/imath] 3)For any angle [imath]\theta[/imath], there exists a monochromatic parallelogram with angle [imath]\theta[/imath] Now its natural to ask if there are any monochromatic squares. Thank you |
2782384 | Prove that there exists a rational number raised to an irrational number that is an irrational number
Prove: There exists [imath]a \in \mathbb{Q}[/imath] and [imath]b \in \mathbb{R}\smallsetminus \mathbb{Q}[/imath] such that [imath]a^b \in \mathbb{R} \smallsetminus \mathbb{Q}[/imath]. I've tried using [imath]\log_23[/imath], [imath]\sqrt 2[/imath], and [imath]\frac{1}{\sqrt 2}[/imath] for the irrational number, but couldn't find a way to prove [imath]a^b[/imath] was irrational. Is there a way to prove this without using Gelfond–Schneider theorem? | 763190 | Is [imath]\text{rational}^{\text{irrational}}[/imath] rational or irrational?
Is the number [imath]\text{rational}^{\text{irrational}}[/imath] rational or irrational? For example [imath]2^{\sqrt{2}}[/imath]: is it rational or irrational? I tried using a logarithm but it didn't work. It seems by superficial studying that it will be irrational. But what is the proof? |
2781862 | Inequality [imath]\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1[/imath]
Show that [imath]\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1,\:\forall n\in\mathbb{N}[/imath] This is a 9th grade problem. I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only [imath]2n+1[/imath] terms. Right? After that I have no idea. Thx! | 1164493 | Mathematical induction for inequalities: [imath]\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1[/imath]
Prove by induction: [imath]\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1[/imath] adding [imath]1/(3m+4)[/imath] as the next [imath]m+1[/imath] value proves pretty fruitless. Can I make some simplifications in the inequality that because the [imath]m[/imath] step is true by the inductive hypothesis, the 1 is already less than all those values? |
2783255 | What is the sum [imath]\sum_{n=1}^\infty \frac 1{2^n} \tan \frac 1{2^{n+1}}[/imath]?
The problem is to determine the sum [imath]\sum_{n=1}^\infty \frac 1{2^n} \tan \frac 1{2^{n+1}}[/imath] I have been trying everything that comes to my mind (trigonometrical functions' formulas, transforming to integration, Taylor expansion, etc.) to no avail, so a slightest hint to a successful approach would be much appreciated. | 2764989 | Evaluate [imath]\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)[/imath]
Evaluate [imath]\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)[/imath] I tried to create a telescoping sum but I couldn't. The last step I could reach was turning the limit into [imath]\lim_{n\to \infty} \sum_{r=1}^n \left(\frac {1}{2^r(1-\tan (2^{-r+1})} -\frac {1}{2^r(1+\tan (2^{-r+1})}\right) [/imath] But couldn't proceed further. Also I thought about Riemann sums but it was a pure dead end. Any help would be greatly appreciated |
2782211 | Infinite differentiablility at [imath]0[/imath] of a piecewise function
I want to show that the function [imath] g = \begin{cases} 0, \quad \text{if } x \le 0, \\ e^{-\frac{1}{x^2}}, \quad \text{if } x > 0, \end{cases} [/imath] is infinitely differentiable at [imath]0[/imath] and that all derivative vanishes: [imath]g^{(n)}(0) = 0[/imath]. I was having trouble expanding the function into its Taylor expansion. Any help is appeciated! | 1258219 | formula for the [imath]n[/imath]th derivative of [imath]e^{-1/x^2}[/imath]
[imath]f(x) = \begin{cases} e^{-1/x^2} & \text{ if } x \ne 0 \\ 0 & \text{ if } x = 0 \end{cases}[/imath] so [imath]\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac {e^{-1/x^2}}x = \lim_{x \to 0} \frac {1/x}{e^{1/x^2}} = \lim_{x \to 0} \frac x {2e^{1/x^2}} = 0[/imath] (using l'Hospital's Rule and simplifying in the penultimate step). Similarly, we can use the definition of the derivative and l'Hospital's Rule to show that [imath]f''(0) = 0, f^{(3)}(0) = 0, \ldots, f^{(n)}(0) = 0[/imath], so that the Maclaurin series for [imath]f[/imath] consists entirely of zero terms. But since [imath]f(x) \ne 0[/imath] except for [imath]x = 0[/imath], we can see that [imath]f[/imath] cannot equal its Maclaurin series except at [imath]x = 0[/imath]. This is part of a proof question. I don't think the answer sufficiently proves that any [imath]n[/imath]th derivative of [imath]f(x)[/imath] is [imath]0[/imath]. Would anyone please expand on the answer? ps: I promise this is not my homework :) |
2242559 | A characterisation of planar complete [imath]k[/imath]-partite graphs
The complete [imath]k[/imath]-partite graph [imath]K_{n_1,n_2,\dots,n_k}[/imath] has vertex set [imath]V=V_1\dot\cup V_2\dot\cup\dots\dot\cup V_k[/imath], where [imath]V_1,\dots,V_k[/imath] are disjoint sets with [imath]|V_i|=n_i[/imath], and each vertex [imath]v\in V_i[/imath] is connected to all vertices of [imath]V\setminus V_i,i=1,2,\dots,k[/imath]. Describe all [imath]k[/imath]-tuples [imath](n_1,n_2,\dots,n_k)[/imath] of natural numbers, [imath]k=1,2,\dots[/imath], such that [imath]K_{n_1,n_2,\dots,n_k}[/imath] is a planar graph. | 256204 | Multipartite graphs which are not planar
Please take a look at these definitions: A multipartite graph is a graph of the form [imath]K_{r_1,\ldots, r_n}[/imath] where [imath]n > 1[/imath], [imath]r_1, \ldots, r_n\ge 1[/imath], such that The set of nodes of the graph is the disjoint union of n sets: [imath]V = V_1 \cup\cdots\cup V_n[/imath], and [imath]|V_i|=r_i[/imath] for all [imath]1\le i\le n[/imath]. The set of nodes of the graph is the set of all possible connections between nodes in that do no belong to the same set [imath]S_i[/imath]. Formally: [imath]E= \{(x,y)\mid x \in S_i, y \in S_j, i \ne j\}[/imath] A graph [imath]G[/imath] is planar if and only if every subdivision of [imath]G[/imath] is planar. A graph [imath]G[/imath] is planar if and only if it contains no subdivision of [imath]K_{3,3}[/imath] or [imath]K_5[/imath]. I need to determine all multipartite graphs that are not planar. |
2783941 | Solve the equation and Find the General Solution.
[imath]\sin 2x + \sin x = 0[/imath] This equation is getting on my nerves, partly because I think I have found the right answer and partly, my answer doesn't match in any one of the options given. | 2779555 | Solve [imath]\sin 7x+\sin 3x=0[/imath]
I tried it by graph plotting, but it is going so ugly. On solving it on paper it mixed up. Is there any other process to solve? |
2784088 | If [imath]f[/imath] is continuous and [imath]f(\frac{x+y}{2})\leq \frac{f(x)+f(y)}2[/imath] for all [imath]x , y \in (a,b)[/imath], prove that [imath]f[/imath] is convex.
I know a proof which starts with proving the convexity for "dyadic rational" numbers. I would like to know if someone has some other ideas. | 1002248 | If [imath]f[/imath] is continuous and [imath]\,f\big(\frac{1}2(x+y)\big) \le \frac{1}{2}\big(\,f(x)+f(y)\big)[/imath], then [imath]f[/imath] is convex
Let [imath]\,\,f :\mathbb R \to \mathbb R[/imath] be a continuous function such that [imath] f\Big(\dfrac{x+y}2\Big) \le \dfrac{1}{2}\big(\,f(x)+f(y)\big) ,\,\, \text{for all}\,\, x,y \in \mathbb R, [/imath] then how do we prove that [imath]f[/imath] is convex that is [imath] f\big(tx+(1-t)y\big)\le tf(x)+(1-t)f(y) , \forall x,y \in \mathbb R , t\in (0,1)? [/imath] I can prove it for dyadic rational's [imath]t=\dfrac k {2^n}[/imath] and then argue by continuity ; but I would like a more direct proof . Thanks in Advance |
2784298 | Proof of sum of cubes of odd numbers by induction
Prove by mathematical induction [imath]1^3 + 3^3 + \cdots + (2n - 1)^3 = n^2(2n^2-1),\forall n\geq1[/imath] Please help me solve this question using the mathematical induction method. | 959713 | Mathematical Induction: Sum of first n odd perfect cubes
The series is [imath]P_k: 1^3 + 3^3 + 5^3 + ... + (2k-1)^3 = k^2(2k^2-1)[/imath] and I have to replace [imath]P_k[/imath] with [imath]P_{k+1}[/imath] to prove the series. I have to show that [imath]k^2(2k^2-1) + (2k-1)^3 = (k+1)^2[2(k+1)^2-1][/imath] I'm sorry that I'm asking but their are just so many factors the algebra just passes over my head. Any help is appreciated. |
2784322 | Problem related to Fundamental Theorem of Integral Calculus
This is a problem from the book of Terence Tao, Analysis I and it looks like- Let [imath]a<b[/imath] be two real numbers, and let [imath]f:[a,b]\to\Bbb{R}[/imath] be a monotone increasing function. Let [imath]F:[a,b]\to\Bbb{R}[/imath] be the function [imath]F(x)=\int_{[a,x]} f[/imath]. Let [imath]x_0\in[a,b][/imath]. Show that [imath]F[/imath] is differentiable at [imath]x_0[/imath] if and only if [imath]f[/imath] is continuous at [imath]x_0[/imath]. Observe that, the "if" part can be easily proved by using Fundamental Theorem of Calculus. Just to recall ...Statement of Fundamental Theorem of Calculus: Let [imath]a<b[/imath] be two real numbers, and let [imath]f:[a,b]\to\Bbb{R}[/imath] be a Riemann Integrable function. Let [imath]F:[a,b]\to\Bbb{R}[/imath] be the function [imath]F(x)=\int_{[a,x]} f[/imath]. Then [imath]F[/imath] is continuous on [imath][a,b][/imath]. Furthermore, if [imath]x_0\in[a,b][/imath] and [imath]f[/imath] is continuous at [imath]x_0[/imath], then [imath]F[/imath] is differentiable at [imath]x_0[/imath], and [imath]F'(x_0)=f(x_0)[/imath]. Here, [imath]f[/imath] is monotone increasing on [imath][a,b][/imath], hence [imath]f[/imath] is Riemann Integrable on [imath][a,b][/imath] and [imath]f[/imath] is continuous at [imath]x_0\implies F[/imath] is differentiable at [imath]x_0[/imath] I can't understand how to tackle the "only if" part. There is a hint given in the book: Consider left and right limits of [imath]f[/imath] and argue by contradiction. Can anybody solve the part in a rigorous and analytical way? Thanks for your help in advance! | 2772636 | Differentiability of antiderivative and continuity of derivative
Let [imath]a,b\in \mathbb{R}: a<b[/imath]. Let [imath]f:[a,b]\to \mathbb{R}[/imath], [imath]f[/imath] is monotone increasing. Let [imath]F:[a,b]\to \mathbb{R}, x \mapsto \int_{[a,x]} f[/imath]. Show that if [imath]F[/imath] is differentiable at [imath]x_0 \in [a,b][/imath], then [imath]f[/imath] is continuous at [imath]x_0[/imath]. If f was not continuous, then the right and left limits would not match. Because the function jumps, its measure has to jump too, therefore integral would be discontinuous and therefore non differentiable. However I do not know how to formalize this intiuition. |
2783544 | A set of sentences true under an arbitrarily large finite models, then it is true under some infinite model
Here is what I want to prove [imath]\Sigma[/imath] a set of sentences such that there are arbitrarily large finite models in which [imath]\Sigma[/imath] is true. Prove that [imath]\Sigma[/imath] is true in some infinite model. Here is my attempt: Let [imath]A[/imath] be the domain of model [imath]\mathfrak{A}[/imath]. Let [imath]|A|=n[/imath] be the size of [imath]\mathfrak{A}[/imath]. If [imath]n=0[/imath], then [imath]A=\emptyset[/imath] and we have no variable assignment to talk about. Given [imath]n[/imath], consider sentence [imath]\mathcal{T}[/imath] such that \begin{align*} &\mathcal{T}:=\forall x_1\forall x_2\dots\forall x_n(\wedge_{j=1}^{n}\wedge_{i=1}^{n}(x_i\neq x_j))\text{ }\text{ when [imath]n>1[/imath]}\\ &\mathcal{T}:=\forall x_1 x_1=x_1\text{ }\text{ when [imath]n=1[/imath]} \end{align*} where [imath]\wedge_{j=1}^{n+1}\wedge_{i=1}^{n}(x_i\neq x_j)=(x_1\neq x_2)\wedge (x_1\neq x_3)\wedge\dots\wedge (x_1\neq x_{n})\wedge (x_2\neq x_1)\wedge\dots[/imath] If [imath]n=1[/imath], then any model with a smaller size is empty. And for any model with at least size [imath]n[/imath], since [imath]x_1=x_1[/imath] is true under any variable assignments, hence [imath]\mathcal{T}[/imath] is true in these models. For [imath]n>1[/imath], if size of [imath]\mathfrak{B}[/imath] is less than [imath]n[/imath], then there must be at least one pair of identical elements in [imath]\{b_1,\dots, b_{n}\}\subseteq B[/imath] by Pigeonhole principle. Hence for any variable assignment [imath]s[/imath], if [imath]s'\equiv s[/imath] except [imath]s'(x_i)=b_i[/imath] [imath]\forall 1\leq i\leq n[/imath], then [imath]s'(x_{i})=b_i= b_j=s'(x_{j})[/imath] for some [imath]1\leq i\neq j\leq n[/imath]. Hence [imath]\mathcal{T}[/imath] is never true in these model. If size of [imath]\mathfrak{B}[/imath] is at least [imath]n[/imath], then we can choose [imath]\{b_1,\dots, b_{n}\}\subseteq B[/imath] be pairwise distinct so that for any variable assignment [imath]s[/imath], if [imath]s'\equiv s[/imath] except [imath]s'(x_i)=b_i[/imath] [imath]\forall 1\leq i\leq n[/imath], then [imath]s'(x_{i})=b_i\neq b_j=s'(x_{j})[/imath] for all [imath]1\leq i\neq j\leq n[/imath], i.e. [imath]\mathcal{T}[/imath] is true under [imath]s[/imath] in [imath]\mathfrak{B}[/imath]. Now for each [imath]n[/imath], we get a sentence [imath]\mathcal{T}[/imath] that is true in all and only those models that have size [imath]\geq n[/imath]. Let [imath]\Sigma[/imath] be a set of [imath]\mathcal{T}[/imath] for all [imath]n[/imath]. Hence any finite subset of [imath]\Sigma[/imath] is true under some infinite model, so satisfiable. By Compactness, we know if every finite subset of a set of sentences is satisfiable then, the whole set is satisfiable, so we done. My questions: Since [imath]\Sigma[/imath] is true under some arbitrarily large finite models, so I assume that for each [imath]n[/imath], every sentence in it is true in all model with size greater than [imath]n[/imath]. However, I specifically choose my sentences to construct the [imath]\Sigma[/imath] instead of consider case in general. So am I wrong? And what will be a sentence look like in general such that given [imath]n[/imath], it is true in all and only those models that have size [imath]\geq n[/imath]. Thanks in advance. ==============update====================== Thanks @Asaf Karagila for pointing out the mistakes on the quantifier. [imath]\mathcal{T}[/imath] should be [imath]\exists x_1\exists x_2\dots\exists x_n(\wedge_{j=1}^{n}\wedge_{i=1}^{n}(x_i\neq x_j))\text{ }\text{ when [/imath]n>1[imath]}[/imath]. And the rest follow. And as mentioned by @Alex Kruckman, my question is a duplicate of a question asked by @MrTopology. I follow his idea to finish the problem in general case Now for each [imath]n[/imath], we get a sentence [imath]\mathcal{T}[/imath] that is true in all and only those models that have size [imath]\geq n[/imath]. Let [imath]\Sigma^*[/imath] be collection of [imath]\mathcal{T}[/imath] for all [imath]n[/imath]. Hence [imath]\Sigma^*[/imath] is true in infinite model. Let [imath]\overline{\Sigma}=\Sigma^*\bigcup\Sigma[/imath]. As [imath]\Sigma[/imath] is a set of sentences such that there are arbitrarily large finite models in which it is true, any finite subset of [imath]\overline{\Sigma}[/imath] is true under some finite model, so satisfiable. By Compactness, we know if every finite subset of a set of sentences is satisfiable then, the whole set is satisfiable, so [imath]\overline{\Sigma}[/imath] satisfiable. Since [imath]\Sigma^*[/imath] is true only in some infinite model under construction, hence [imath]\overline{\Sigma}[/imath] is true in some infinite model, then so is [imath]\Sigma[/imath] | 1804081 | Compactness theorem
Let [imath]\Sigma[/imath] be a set of theorems, such that for every [imath]\varphi\in\Sigma[/imath] exists an arbitrarily large (<--- edited) finite model [imath]\mathcal{M}[/imath], with [imath]\mathcal{M}\models\varphi[/imath]. Show: It exists an infinite modell [imath]\mathcal{M}[/imath] with [imath]\mathcal{M}\models\varphi[/imath] for every [imath]\varphi\in\Sigma[/imath]. Hello, I want to proof this statement, and might need some help. I think the compactness theorem is needed. My idea ist the following: I want to extend [imath]\Sigma[/imath], such that it has an infinite carrier set [imath]M[/imath] and conclude that for the resulting model [imath]\mathcal{M}=(M,\dotso )\quad\mathcal{M}\models\varphi[/imath] by using the compactness theorem. The actuall proof seems to be trivial: Let [imath]\Sigma':=\Sigma\cup\{v_i\neq v_j: i\neq j\}[/imath]. By assumption every finite [imath]\overline{\Sigma}\subset\Sigma[/imath] is satisfiable. Therefor [imath]\Sigma'[/imath] is satisfiable by the compactness theorem and has obviously an infinite carrier set, since we added an infinite amount of variables. My thoughts do not need that [imath]\Sigma[/imath] contains theorems (hence formulas without free variables) and not general formulas. Am I mistaken with my proof? Thanks in advance for your comments. |
2785055 | Ideal of a ring polynomial?
I'm doing a past exam paper and this is the question I am stuck on. Define the conditions on a subset [imath]I[/imath] of a ring [imath]R[/imath] to be an ideal of [imath]R[/imath]. Show that the set [imath]I = \{f ∈ \mathbb{R}[X]|f(1) = 0\}[/imath] is an ideal of [imath]\mathbb{R}[X][/imath]. I know the conditions for I to be an ideal of R, but I'm confused about what the set actually means? What is the result of [imath]f(a+b)[/imath] if [imath]a,b∈ \mathbb{R}[/imath] Thanks in advance for any help! | 550977 | Show that this set of polynomials is ideal in F[x]
In [imath]\mathbb{F}[x][/imath], where [imath]\mathbb{F}[/imath] is a field, let [imath]J[/imath] be the set of elements of polynomials that have coefficients that add to zero (so [imath]a_0 + a_1 + ... + a_n = 0[/imath]). Show that [imath]J[/imath] is an ideal of [imath]\mathbb{F}[x][/imath]. I know that the proof of this statement is meant to be very short, but I don't know how to go about it. |
741999 | How can I show that [imath]\lim\limits_{x \to \infty} \frac{\log(x+1)}{\log{x}}=1[/imath]?
Hello I have a silly question: How can I show that [imath]\lim\limits_{x \to \infty} \dfrac{\log(x+1)}{\log{x}}=1[/imath]. Thank you. | 196070 | Find [imath]\lim_{x\to \infty} \ln(x+1)/(\ln(x))[/imath]
I have to solve: [imath]\lim\limits_{x\to \infty} \frac{\ln(x+1)}{\ln(x)}[/imath]. Can you give me any hints to go? Thanks a lot! |
2785284 | Area between 2 rings
I would like to know if is possible calculate the area between 2 rings with opposite centers and the same radius, that is: [imath]r_1^2<(x-a)^2+(y-b)^2<r_2^2[/imath] [imath]r_1^2<(x+a)^2+(y+b)^2<r_2^2[/imath] I think that depend on the relation between center and radius, but I'm not sure. I'm working on... Edit: This is an example [imath]4<(x-1){}^2+(y-2){}^2<8[/imath] [imath]4<(x+1){}^2+(y+2){}^2<8[/imath] | 1337347 | Area of intersection of two Annulus
Given two separate annulus with centers [imath][C_1,C_2][/imath] and their corresponding radii being [imath][R_1,r_1][/imath] and [imath][R_2,r_2][/imath] respectively, larger radius being [imath]R[/imath]. There are methods to look at whether they are overlapping or not. If they overlap how to find the overlapping area. Is there any close form expression for it? equation of first annulus [imath]r_1^2<(x-x_{c1})^2 + (y-y_{c1})^2 <R_1^2[/imath] equation of second annulus [imath]r_2^2<(x-x_{c2})^2 + (y-y_{c2})^2 <R_2^2[/imath] If there are three circles overlapping, we get overlapping area as solution to set of 4 in-equations by least squares approach(Though we don't end up getting total area but, area within tangents of circles ). |
2489712 | [imath]5[/imath] points are randomly chosen on the plane, what is the probability that the resulting conic is parabola, ellipse, hyperbola.
[imath]5[/imath] points are randomly chosen on the plane, what is the probability that the resulting conic is a parabola, an ellipse, or a hyperbola? By randomly chosen on the plane I mean chosen uniformly. | 2395821 | Odds of ellipse from five random points
It's known that five points determine a conic section. Five random points can go right into the [imath]6\times6[/imath] matrix, and then the [imath]A x^2 + B xy + C y^2[/imath] part can be looked at. If [imath]B^2-4AC<0[/imath], it's an ellipse. Five random points will almost never produce circles or parabolas, so the results will be ellipses and hyperbolas. What are the odds of an ellipse? In a random run of 100000 trials, I got 27974 ellipses. "It's less than [imath]e/10[/imath]," seems like a solid answer. Anyone have anything more specific? EDIT: As Oscar points out, I should have said "It's more than [imath]e/10[/imath]." In my trial, real-values points were randomly picked from a unit square. Square Triangle Picking methods might be applicable. EDIT2: Aretino points out that odds of a convex pentagon are [imath]49/144≈0.34[/imath]. So how can points making a convex pentagon give a non-ellipse? Here's a picture. With the red points fixed, the black points are outside of the convex hull yet still yield a non-ellipse. EDIT3: That spray of points above goes back to Newton, Philosophiae naturalis principia mathematica, 1687, where he solved the 4 point parabola (another version). If a point is between one of the two parabolas and the degenerate lines, then it gives a hyperbola. |
2786019 | The Square Root of -i
I am having difficulty understanding [imath]\sqrt -i[/imath] from other sources (such as Wolfram Alpha) I have found it to equal [imath]-(-1)^\frac{3}{4}[/imath] but do not understand the steps in to reach this conclusion. I know it may be trivial but any assistance would be greatly appreciated. | 341252 | What does the square root of minus [imath]i[/imath] equal?
Can you enter the rabbit hole recursively? If the [imath] \sqrt{-1} = i [/imath] then, what does [imath] \sqrt{-i} [/imath] equal? |
2786789 | Find the closest number of the form [imath]a+b\sqrt2[/imath]
Is there a method that, for a given [imath]x\in \mathbb R[/imath] finds the closest number in the ring [imath]\mathbb Z[\sqrt2][/imath]? It is trivial for [imath]\mathbb Z[/imath] and also for comlex rings of integers like [imath]\mathbb Z[i][/imath] (if we [imath]\mathbb R[/imath] by [imath]\mathbb C[/imath]). The problem is I don't even see a reason why there should be a closest number of the required form. I actually think that you can get as close as you want to [imath]x[/imath] by the numbers [imath]a+b\sqrt2[/imath]. But I am unable to prove that. Do you have any ideas? | 889296 | [imath]x+y\sqrt{2}[/imath] infimum ([imath]x,y\in \mathbb{Z}[/imath])
I've looked for help with this question but I have not found anything, I hope this is not a duplicate. Define the set [imath]A=\{\mid x+y\sqrt{2}\mid \ : x,y\in \mathbb{Z}\ \mbox{and} \mid x+y\sqrt{2}\mid\gt0 \}[/imath], prove that the infimum of this set is zero. This is what I've thought: Proving that [imath][y\sqrt{2}][/imath] (where [imath][\cdot ][/imath] is the floor function) can be close to zero as we want would prove the proposition, so I suppose that this doesn't hold and I want to arrive to the contradiction that for some [imath]y_0[/imath]: [imath][y_0\sqrt{2}]=0[/imath], this would contradict the fact that [imath]\sqrt{2}[/imath] is irrational. Can someone give me a hand? Thanks! (This problem arised in the context of integer lattices) |
2786787 | Why [imath]O(3)[/imath] does not contain a normal subgroup [imath]\mathbb{Z}/2\mathbb{Z}[/imath]?
Why the orthogonal group [imath]G=O(3)[/imath] does not contain a normal subgroup, a cyclic group of order 2, [imath]N=\mathbb{Z}/2\mathbb{Z}[/imath]? It looks that any [imath]g \in G[/imath] satisfies [imath] g N g^{-1}=N [/imath] where [imath]N=\mathbb{Z}/2\mathbb{Z}[/imath] is generated by [imath]-diagonal(1,1,1)[/imath] of rank-3 matrix. So why the orthogonal group [imath]G=O(3)[/imath] does not contain a normal subgroup [imath]N=\mathbb{Z}/2\mathbb{Z}[/imath], but only an [imath]SO(3)[/imath]? | 881995 | What are the non-trivial normal subgroups of [imath]O(3)[/imath]?
What are the non-trivial normal subgroups of [imath]O(3)[/imath]? My guess is that the only one is [imath]SO(3)[/imath], but it's really only a guess, based on the fact that [imath]O(3)[/imath] is disconnected 3-manifold of [imath]SO(3)[/imath] and the set of reflective symmetries, which I'm not sure is even a subgroup (does it contain identity?)... I have no idea how to prove it. So first I want to prove that [imath](\forall o \in O(3)) \space oSO(3) = SO(3)o[/imath], then that there are no normal subgroups of [imath]SO(3)[/imath] and that the set of reflective symmetries is not a group. (Please, be easy on me. I obviously don't know much about abstract algebra yet.) Thanks! |
2784455 | Why not define a "Dolbeaut complex" with [imath]\partial[/imath] instead of [imath]\overline{\partial}[/imath]?
Yesterday I saw the definition of Dolbeaut cohomology: let [imath](M,J)[/imath] be a complex manifold, write the exterior derivative as [imath]{\rm d} = \partial +\overline{\partial}[/imath], where [imath]\partial: \Omega^{\ell,m}(M) \to \Omega^{\ell+1,m}(M)[/imath] and [imath]\overline{\partial}: \Omega^{\ell,m}(M) \to \Omega^{\ell,m+1}(M)[/imath], look at the complex [imath]\cdots \xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \Omega^{\ell,m-1}(M)\xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \Omega^{\ell,m}(M) \xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \Omega^{\ell,m+1}(M)\xrightarrow{\hspace{.4cm}\overline{\partial}\hspace{.4cm}} \cdots[/imath]and put [imath]H^{\ell,m}_{\rm Dolbeaut}(M) \doteq \frac{\ker(\overline{\partial}: \Omega^{\ell,m}(M) \to \Omega^{\ell,m+1}(M))}{{\rm Im}(\overline{\partial}: \Omega^{\ell,m-1}(M) \to \Omega^{\ell,m}(M))}.[/imath]Great. Question: Why did we took the above complex instead of [imath]\cdots \xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \Omega^{\ell-1,m}(M)\xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \Omega^{\ell,m}(M) \xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \Omega^{\ell+1,m}(M)\xrightarrow{\hspace{.4cm}\partial\hspace{.4cm}} \cdots?[/imath] I get that [imath]f: M \to \Bbb C[/imath] is [imath]J[/imath]-holomorphic if and only if [imath]\overline{\partial}f = 0[/imath] so that [imath]\overline{\partial}[/imath] sort of measures how stuff in general would be away from being holomorphic. But I'd like a more solid explanation. | 1754681 | Why only consider Dolbeault cohomology?
On a complex manifold we have the differential operators [imath]\partial:A^{p,q}\to A^{p+1,q}[/imath] [imath]\bar\partial:A^{p,q}\to A^{p,q+1}[/imath] which both square to zero. Hence one can define cohomology groups [imath]H^{p,q}_\partial=\frac{\ker\partial:A^{p,q}\to A^{p+1,q}}{\partial A^{p-1,q}}[/imath] [imath]H^{p,q}_{\bar\partial}=\frac{\ker\bar\partial:A^{p,q}\to A^{p,q+1}}{\bar\partial A^{p,q-1}}[/imath] But for some reason I only ever see people discussing the [imath]H_{\bar\partial}^{p,q}[/imath], i.e. Dolbeaut cohomology. What makes this more interesting the the [imath]H_\partial^{p,q}[/imath] groups? |
2787216 | Find the non negative continuous function defined over [imath][0,1][/imath] and satisfy given equations
Find the non negative continuous function defined over [imath][0,1][/imath] such that [imath]\int_0^1 x^kf(x){\rm d}x=a^k,[/imath] where [imath]k=0,1,2[/imath] and [imath]a\in[0,1][/imath] solution i try from [imath]\int^{1}_{0}f(x)dx=1\cdots (1)\times a^2[/imath] and [imath]\int^{1}_{0}x\cdot f(x)dx=a\cdots (2)\times 2a[/imath] and [imath]\int^{1}_{0}x^2\cdot f(x)dx=a^2\cdots (3)[/imath] getting [imath]\int^{1}_{0}(x-a)^2f(x)dx=0[/imath] how i get functions which satisfy given equation, help required | 2781926 | Find the continuous function defined over [imath][0,1][/imath] such that [imath]\int_0^1f(x){\rm d}x=1, \int_0^1 xf(x){\rm d}x=a[/imath] and [imath]\int_0^1x^2f(x){\rm d}x=a^2 .[/imath]
Find the continuous function defined over [imath][0,1][/imath] such that [imath]\int_0^1 x^kf(x){\rm d}x=a^k,[/imath] where [imath]k=0,1,2.[/imath] My Thought Indeed, the conditions imply that [imath]\int_0^1f(x)=1,~~~\int_0^1 xf(x){\rm d}x=a,~~~\int_0^1 x^2f(x){\rm d}x=a^2.[/imath] Thus, it's easy to obtain that [imath]\begin{align*}\int_0^1(1-x)^2f(x){\rm d}x&=\int_0^1(1-2x+x^2)f(x){\rm d}x\\&=\int_0^1f(x){\rm d}x-2\int_0^1xf(x){\rm d}x+\int_0^1 x^2f(x){\rm d}x\\&=1-2a+a^2\\&=(1-a)^2\end{align*}.[/imath] But how to go on with this? I'm stuck here. Please offer some help. Thanks. |
2787383 | Evaluating [imath]\sum_{n=1}^{N} 2^{-n}\sin(n\theta)[/imath]
Use de Moivre's theorem to find [imath]\sum_{n=1}^{N} 2^{-n}\sin(n\theta)[/imath] How to find sum of imaginary parts of geometric progression, or use exponential form of complex numbers to find the summation in terms of N, sine, and cosine?? | 1900142 | Sum from [imath]n=1[/imath] to [imath]N[/imath] of [imath]\sin(nx)/2^n[/imath]
The question states: I'm not sure what to do here. I used the geometric sum to do [imath]2^{-n}[/imath], but I can't think of a way to sum [imath]\sin(n\theta)[/imath] to get the required RHS. Further working Apologies for maybe being slow or missing something obvious here... I don't know how to get it into required form. |
2787541 | How to show that [imath]\sum_{i=1}^n|\lambda_i|\leq \sum_{i,j}|a_{ij}|[/imath]?
I was asked by a student that Given a matrix [imath](a_{ij})[/imath] with its eigenvalues (with their multiplicities) [imath]\{\lambda_i\}_{i=1}^n[/imath], show that [imath]\sum_{i=1}^n|\lambda_i|\leq \sum_{i,j}|a_{ij}|[/imath] Frankly speaking, I am not sure that it is true, but I cannot find any counterexample by try some easy example. | 1472420 | Upper bound for the sum of absolute values of the eigenvalues
Let [imath]\mathbf A = (a_{ij})[/imath] be an [imath]n\times n[/imath] real or complex matrix with eigenvalues [imath]\lambda_j[/imath], for [imath]j=1,...,n[/imath]. It is known that [imath]\max|\lambda_j|[/imath] is bounded above by the maximum row sum of [imath]A[/imath] (using entrywise absolute values). Does the following related bound also hold? Question: Is it true that [imath] \sum_{j=1}^n|\lambda_j| \le \sum_{i,j=1}^{n}|a_{ij}|, [/imath] where the right hand side is summed over all matrix elements[imath]\,[/imath]? For example, this inequality is obvious for triangular matrices as well as positive definite matrices and slightly less obvious for unitary matrices (using the fact that the rows are unit vectors). It can also be proved in other special cases but is it true in general? Thanks. |
2787879 | For all [imath]A, B \in \mathcal{M}_n(\mathbb{C})[/imath], if [imath][A, [A, B]] = 0[/imath], then [imath][A, B][/imath] is nilpotent
[imath]\newcommand{\tr}{\mathrm{Tr }}[/imath] Let be [imath]\mathcal{M}_n(\mathbb{C})[/imath] the set of complex-valued square matrices of order [imath]n[/imath] and [imath][A, B] = AB - BA[/imath] a commutator. If, for some [imath]A, B \in \mathcal{M}_n(\mathbb{C})[/imath], we have [imath][A, [A, B]] = 0[/imath], then [imath][A, B][/imath] is nilpotent. So: [imath][A, [A, B]] = 0[/imath] says that [imath]A[/imath] and [imath][A, B][/imath] commutes. Let be [imath]\lambda \in \mathbb{C}[/imath] and [imath]X \in \mathbb{C}^n[/imath] such that [imath][A, B]X = \lambda X[/imath] (an eigenvalue/eigenvector pair), such values always exist as [imath][A, B] \in \mathcal{M}_n(\mathbb{C})[/imath] and must have [imath]n[/imath] eigenvalues. Left-multiplication of the previous relation gives: [imath]A[A, B]X = \lambda AX[/imath]. By hypothesis, we have: [imath][A, B](AX) = \lambda (AX)[/imath]. If we denote [imath]E_{\lambda}[/imath] the set of eigenvectors of [imath][A, B][/imath] for the [imath]\lambda[/imath] value, we have: [imath]X \in E_{\lambda} \implies AX \in E_{\lambda}[/imath]. As a result, for all [imath]k \in \mathbb{N}, A^k X \in E_{\lambda}[/imath]. Now, we also have [imath]\tr [A, B] = 0[/imath] because of properties on the trace. If [imath]A[/imath] is not nilpotent or nilpotent for [imath]k \geq n[/imath], then: [imath](X, AX, \ldots, A^{n - 1} X)[/imath] is linearly independent, as [imath]\dim E_{\lambda} \leq n[/imath] and [imath]\dim E_{\lambda} \geq n[/imath]. We conclude that [imath]\dim E_{\lambda} = n[/imath], finally: [imath]n \lambda = 0[/imath], now: [imath]\lambda = 0[/imath]. We conclude that [imath][A, B][/imath] is nilpotent. If [imath]A[/imath] is nilpotent for [imath]k \leq n - 1[/imath], then [imath]\dim E_{\lambda} \geq k[/imath]. Here, I don't really know how to proceed, I'd like to show we cannot have any other value than [imath]\lambda[/imath] and it must be [imath]0[/imath] due to the trace. | 357566 | Show that a matrix is nilpotent.
Let [imath]A,B \in \mathbb{M}_n (\mathbb{C})[/imath]. If [imath]A^2B + BA^2 = 2ABA[/imath] then exist [imath]k \in \mathbb{N}[/imath] where [imath](AB-BA)^k = 0[/imath]. I tried solve with minimal polynomial, but I did not have much effect. |
1647250 | Showing a relation between binomial and negative binomial analytically
If [imath]X[/imath] is binomial random variable [imath]B(n,p)[/imath] and Y is negative binomial [imath](r,p)[/imath], How can I show that [imath]F_X(r-1) = 1- F_Y(n-r)[/imath]. While it is possible to show that using the definition of binomial and negative binomial, I am more interested in showing that directly using their p.m.fs. | 154236 | Relationship between binomial and negative binomial distributions (how to extend the probability space?)
I wonder a technique to extend the discrete probability space. Here's an example from Concrete Mathematics EXERCISE 8.17: Let [imath]X_{n,p}[/imath] and [imath]Y_{n,p}[/imath] have the binomial and negative binomial distributions, respectively, with parameters [imath](n,p)[/imath]. Prove that [imath]\Pr(Y_{n,p}\le m) = \Pr(X_{m+n,p}\ge n)[/imath]. The answer to the problem is also from Concrete Mathematics: \begin{align} \Pr(Y_{n,p}\le m) &= \Pr(Y_{n,p}+n \le m+n) \\ &= \hbox{probability that we need [imath]\le m+n[/imath]} \tag{1}\\ &= \hbox{probability that [imath]m+n[/imath] tosses yield [imath]\ge n[/imath] heads} \tag{2} \\ &= \Pr(X_{m+n,p}\ge n) \end{align} Well, (1) and (2) are describing the same thing, but they're in different probability spaces, so we should extend these two probability spaces into a unique probability space, ensuring that the probability of each event doesn't change. How can we do it? I haven't a clear idea. And the more general problem arises: How to extend a probability space? Is there any technique to do it, at least, treat part of problems? Thanks for your help! |
2787741 | What does it mean when one says that a point is lying on the circle?
[imath]{(x-h)^2 + (y-b)^2 = r^2}[/imath] What does a point lying on a circle mean? Does it mean that I can show this equation to anyone and they would be able to understand that it really is a circle? Edit: The equation of a circle is made out of coordinate axes 'X' and 'Y'. If I input numbers x,y that satisfy the equation then a Circle will emerge after joining those points? But it is also possible that the user might join points in random order and a circle might never emerge. I have a diagram below to show you what I understand in mathematics. construct circle by joining points? In previous post, I asked what is the meaning of equation of the circle and equation of a cube. The answer was, if I input x,y that satisfies the equation then the points lie on the circle. This post is not duplicate of my previous post. In this post I want to know what the meaning of points lying on the circle is. If points are lying somewhere then do I need to join them to make a circle? But there is a possibility that I can draw a line segment connecting those points in random order. I have illustrated my understanding in the diagram above. | 2786619 | What is the meaning of equation of a circle or equation of a cube?
We have an equation of a Circle [imath]x^2 + y^2 = 1[/imath] and equation of a Cube [imath]{||x-y|+|x+y|-2z|+||x-y|+|x+y|+2z|=1}[/imath] When I substitute x,y in first equation or x,y,z in the second equation then the equations are said to be satisfied when LHS and RHS are equal. I have a quick question. If the first equation or second equation get satisfied, because LHS becomes equal to RHS, then those values of x and y of first equation are said to be lying on the circumference of the circle and not inside or outside of the circle and x,y,z of the second equation are said to be lying on the surface of the cube and not on the inside and not on the circle of the cube. Correct? |
2781527 | I'm learning ring theory but just wanted some more information as my notes are limited
Let [imath]\theta : \mathbb{Z}_8 \to \mathbb{Z}_4[/imath] be defined by [imath]\theta([i]_8) = [i]_4[/imath] for all [imath]i \in \mathbb{Z}[/imath]. You may assume that this is a well-defined ring homomorphism. Find [imath]\mathrm{Ker}(\theta)[/imath] and exhibit the partition of [imath]\mathbb{Z}_8[/imath] into cosets of [imath]\mathrm{Ker}(\theta)[/imath]. Can anyone please explain to me what [imath]\mathrm{Image}(\theta)[/imath] and [imath]\mathrm{Kernel}(\theta)[/imath] are. I have the definition and I understand the mapping but how is it used? Why is it relevant? I don't even know if these are the correct questions to ask. | 2778020 | Partition of [imath]\mathbb{Z}_8[/imath] into cosets of Ker(θ)
Question: Let [imath]\theta : \mathbb{Z}_8 → \mathbb{Z}_4[/imath] be defined by [imath]\theta([i]_8) = [i]_4[/imath] for all [imath]i \in \mathbb{Z}[/imath]. You may assume that this is a well-defined ring homomorphism. Find Ker([imath]\theta[/imath]) Exhibit the partition of [imath]\mathbb{Z}_8[/imath] into cosets of Ker([imath]\theta[/imath]). My Answer: So I found the Ker([imath]\theta[/imath]) as Ker([imath]\theta[/imath]) [imath]=\{[0]_8, [4]_8\}[/imath] as [imath]\theta[0]_8=[0]_4[/imath] and [imath]\theta[4]_8=[4]_4=[0]_4[/imath]. But I'm stuck on part 2. I was thinking that I would need to show that the partition of [imath]\mathbb{Z}_8[/imath] are equal to [imath][0]_4[/imath] or [imath][4]_4[/imath]. Sorry about how I have typed it, haven't really asked questions on the site before. |
2787886 | Determine the convergence of [imath]\sum_{n=2}^{\infty} \frac{(-1)^n}{(-1)^n+n}[/imath]
I've been really stuck on this infinite series. [imath]\sum_{n=2}^{\infty} \frac{(-1)^n}{(-1)^n+n}[/imath] Trying to determine if it converges or diverges. As far as I've got is seperating it into two partial sums (odds and evens) but even this doesn't help me. Any ideas? Thanks | 1496386 | Determine if [imath]\sum_{n=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}[/imath] converges or diverges
I'm having a lot of trouble figuring this one out. Determine if [imath]\sum_{i=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}[/imath] converges or diverges Both ratio and root test are inconclusive and I'm at a loss. Can anyone help me? |
2788372 | What is the expansion for [imath]\frac{1}{1+x+x^{2}}[/imath]?
What is the expansion for [imath]\frac{1}{1+x+x^{2}}[/imath]? I know expansion for [imath]\begin{align}&(1+x)^{-1}=1-x+x^{2}-x^{3}+\dots\\ &(1+x)^{-2}=1+(-2)x+(-2)\frac{-3}{2!}+(-2)(-3)\frac{-4}{3!}+\dots\\ &(1+x^{2})^{-1}=1-x^{2}+x^{4}-x^{6}+\dots\end{align}.[/imath] But for [imath]\frac{1}{1+x+x^{2}}[/imath] I got problem. Can someone derive [imath]\frac{1}{1+x+x^{2}}[/imath] term expansion. | 1990704 | Find the power series of [imath]f(x)=\frac{1}{x^2+x+1}[/imath]
I want to find the power series of [imath]f(x)=\frac{1}{x^2+x+1}[/imath] How can I prove the following? [imath]f(x)=\frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \mathrm{sin}\frac{2\pi(n+1)}{3} x^n \,\,\,\, |x|<1[/imath] In particular I would like to know how to proceed in this case. The polinomial [imath]x^2+x+1[/imath] has no roots so here I cannot use partial fraction decomposition: what method should I use? |
2788320 | Determinant of [imath]\left(\begin{smallmatrix} -2a &a+b &a+c \\ b+a& -2b &b+c \\ c+a&c+b & -2c \end{smallmatrix}\right)[/imath]
Evaluate [imath]D=\begin{vmatrix} -2a &a+b &a+c \\ b+a& -2b &b+c \\ c+a&c+b & -2c \end{vmatrix}[/imath] My try: Applying [imath]R_1 \to R_1+R_2[/imath] we get [imath]D=\begin{vmatrix} b-a&a-b &a+b+2c \\ b+a& -2b &b+c \\ c+a&c+b & -2c \end{vmatrix}[/imath] Now apply [imath]C_1 \to C_1+C_2[/imath] [imath]D=\begin{vmatrix} 0&a-b &a+b+2c \\ a-b& -2b &b+c \\ 2c+a+b&c+b & -2c \end{vmatrix}[/imath] Now apply [imath]C_2 \to C_2 +C_3[/imath] [imath]D= \begin{vmatrix} 0&2a+2c &a+b+2c \\ a-b& c-b &b+c \\ 2c+a+b&b-c & -2c \end{vmatrix}[/imath] Now use [imath]R_3 \to R_3+R_2[/imath] [imath]D= \begin{vmatrix} 0&2a+2c &a+b+2c \\ a-b& c-b &b+c \\ 2c+2a&0 & b-c \end{vmatrix}[/imath] any way to proceed here using elementary operations? | 2782450 | Prove the equality: [imath]\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)[/imath]
i have to prove: [imath]\begin{vmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{vmatrix} = 4(a+b)(b+c)(c+a)[/imath] I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help? |
336082 | How to show that this set is not algebraic
How to show that the set: [imath] Z=\{ (\cos t ,\sin t, t) | t \in \mathbb{R}\}[/imath] is not algebraic? Thank you | 1592928 | Prove that a set in [imath]\mathbb R^3[/imath] is not an algebraic set
I want to prove that the set [imath]\{(\cos(t),\sin(t),t)\in A^3(\mathbb R); t\in \mathbb R \}[/imath] is not an algebraic set. I already proved that the set [imath]\{(\sin(t),t)\in A^2(\mathbb R);t\in \mathbb R \}[/imath] is not algebraic but the method that I used doesn't seems to be general. |
2789235 | Proving that [imath]f(x) = \ln(x)[/imath] is uniformly continuous on [imath][1, \infty)[/imath]
I was given the hint of using the Mean Value Theorem. I've gone over the definitions but am having a hard time seeing what to do. | 353480 | Uniform continuity of [imath]\ln(x)[/imath]
Is [imath]f(x)=\ln(x)[/imath] uniformly continuous on [imath](1,+\infty)[/imath]? If so, how to show it? I know how to show that it is not uniformly continuous on [imath](0,1)[/imath], by taking [imath]x=\frac{1}{\exp(n)}[/imath] and [imath]y = \frac{1}{\exp(n+1)}[/imath]. Also, on which interval does [imath]\ln(x)[/imath] satisfy the Lipschitz condition? |
2789264 | How to prove [imath]\det \begin{bmatrix}A & B \\ C & D \end{bmatrix} =\det(AD-BC)[/imath], where [imath]A, B, C,[/imath] and [imath]D[/imath] are [imath]n\times n[/imath] and commuting.
I can solve for the cases in which [imath]C=0[/imath] or [imath]B=0[/imath]. But I cannot find any idea for the case in which both [imath]B[/imath] and [imath]C[/imath] are nonzero matrices. | 1296257 | Determinant of block matrix with commuting blocks
I know that given a [imath]2N \times 2N[/imath] block matrix with [imath]N \times N[/imath] blocks like [imath]\mathbf{S} = \begin{pmatrix} A & B\\ C & D \end{pmatrix}[/imath] we can calculate [imath]\det(\mathbf{S})=\det(AD-BD^{-1}CD)[/imath] and so clearly if [imath]D[/imath] and [imath]C[/imath] commute this reduces to [imath]\det(AD-BC)[/imath], which is a very nice property. My question is, for a general [imath]nN\times nN[/imath] matrix with [imath]N\times N[/imath] blocks where all of the blocks commute with each other, can we find the determinant in a similar way? That is, by first finding the determinant treating the blocks like scalars and then taking the determinant of the resulting [imath]N\times N[/imath] matrix. Thanks in advance! |
2766543 | Discontinuity of Thomae Function on [imath]\mathbb{Q}[/imath]
I know there are many answers for this question. I had read almost every answer And try get arguments But something that goes missing in each time. I had been stuck on this problem since half day but not understating. Please help me in this regard. Let [imath]f[/imath] be defined on [imath][0,1][/imath]; [imath] \begin{align} f(x) = \begin{cases} 0 & \text{if $x$ is irrational}\\ \frac{1}{q} & \text{if $x = \frac{p}{q}$ where $(p,q) = 1$ and $q$ > 0}. \end{cases} \end{align} [/imath] For the rational numbers, I consider a sequence of irrational points [imath]x_n[/imath] converging to some rational number say [imath]m/n[/imath]. If this function is continuous for any rational say [imath]m/n[/imath] then this must hold: [imath]\forall \epsilon >0[/imath] ,[imath]\exists \delta>0[/imath] such that [imath]|1/n|<\epsilon [/imath] whenever [imath]|x_n-m/n|<\delta[/imath].But for any particular [imath]m/n,1/n[/imath] need not be small. Therefore function is discontinuous at the rational points. For the irrational points, consider rational sequence [imath]y_n[/imath] converging to [imath]a[/imath], irrational number. Consider [imath]y_n \to a[/imath] therefore [imath]\forall \epsilon >0 [/imath], [imath]\exists N[/imath] such that [imath]|y_n-a|<\epsilon [/imath] [imath]\forall n>N[/imath] In case the function is continuous at [imath]a[/imath] then it must satisfy following [imath]\forall \epsilon >0 [/imath] ,[imath]\exists \delta>0[/imath] such that|[imath]f(y_n)[/imath]|<[imath]\epsilon [/imath] whenever [imath]|y_n-a|<\delta[/imath]. I wanted to make that [imath]f(y_n)[/imath] small for that [imath]y_n[/imath] in neighborhood of [imath]a[/imath].From this I am not able convince myself from last answers. This is not a duplicate, since I specified my particular problem with this proof. Please help me. | 3020641 | Proof that Thomae's function doesn't have a limit as [imath]\lim_{x\rightarrow x_0}[/imath], [imath]x_0\in \mathbb{Q}[/imath]
I'm trying to proof that the Thomae's function doesn't have a limit as [imath]\lim_{x\rightarrow x_0}[/imath], [imath]x_0\in \mathbb{Q}[/imath]. I'm pretty sure, that I can solve this by using the [imath]\epsilon-\delta[/imath]-critereon, but I cant figure out how. If a limit would exist, that would mean: [imath]\forall\epsilon>0 \exists\delta : |f(x)-L|<\epsilon, \forall x\in D, 0<|x_i -x|<\delta[/imath] I'm pretty sure that you simply can't find a matching [imath]\delta[/imath] as in the environment of each rational number there are some irrational numbers where the Thomae's function is equal to [imath]1/q[/imath] and not [imath]0[/imath]. Could you help me to complete this proof by helping me arguing&writing this mathematicly correct? Edit: This isn't a duplicate because I'm looking for the limit and not discontinuity. I have to proof this with the [imath]\epsilon-\delta[/imath]-critereon for the limit and not for the discontinuity (even though they are kind of similar, they're different!) |
2788528 | Integrating [imath]\sin x\cos x[/imath]
Using substitution with [imath]u=\sin{x}[/imath], The integral of [imath]\sin{x}\cos{x}[/imath] is equal to [imath]\frac{1}{2}\sin^2{x}[/imath]. This is the solution I require. However, why is it that when I use the double angle formula [imath]\sin{2x} = 2\sin{x} \cos{x}[/imath], and integrate [imath]\frac{1}{2}\sin{2x}[/imath], I get a completely different answer of [imath]-\frac{1}{4}\cos{2x}[/imath] even though [imath]\sin{2x} = 2\sin{x} \cos{x}[/imath]. | 135723 | Integral of [imath]\sin x \cos x[/imath] using two methods differs by a constant?
[imath] \int \sin \theta\cos \theta~d \theta= \int \frac {1} 2 \sin 2\theta~ d \theta=-\frac {1} 4 \cos 2\theta[/imath] But, if I let [imath] u=\sin \theta , \text{ then }du=\cos \theta~d\theta [/imath] Then [imath] \int \sin \theta\cos \theta~d \theta= \int u ~ du =\frac { u^2 } 2 =\frac {1} 2 \sin^2 \theta [/imath] Since [imath] \sin^2 \theta =\frac {1} 2 - \frac {1} 2 \cos 2\theta[/imath] The above can be written as [imath] \int \sin \theta\cos \theta~d \theta= \frac {1} 2 \sin^2 \theta =\frac {1} 2 \left( \frac {1} 2 - \frac {1} 2 \cos 2\theta \right)=\frac {1} 4-\frac {1} 4 \cos 2\theta [/imath] Why are the two results differ by the constant [imath]1/4[/imath]? Thank you. |
2789725 | Evaluating [imath]\tan\left(\sum_{r=1}^{\infty} \arctan\left(\frac{4}{4r^2 +3}\right)\right)[/imath]
[imath]\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{4}{4r^2 +3}\right)\right)= ? [/imath] I wrote it in the form: [imath]\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{\dfrac43}{\dfrac{4r^2}{3} +1}\right)\right)[/imath] and tried to use: [imath]\arctan x- \arctan y = \arctan\left(\dfrac{x-y}{1+xy}\right)[/imath] but that trick doesn't help here. How to go about solving this problem then? | 1794175 | arccot limit: [imath]\sum_{r=1}^{\infty}\cot ^{-1}(r^2+\frac{3}{4})[/imath]
I have to find the limit of this sum: [imath]\sum_{r=1}^{\infty}\cot ^{-1}(r^2+\frac{3}{4})[/imath] I tried using sandwich theorem , observing: [imath]\cot ^{-1}(r^3)\leq\cot ^{-1}(r^2+\frac{3}{4})\leq\cot ^{-1}(r^2)[/imath] Now when I was calculating the limit of left hand expression, I convert it to [imath]\tan^{-1}[/imath], by using: [imath]\tan^{-1}\frac{1}{x} = \cot^{-1}x[/imath] but couldn't sum up the terms of arctan series. How can I proceed? Is there any better way ? |
2789702 | Graphs for complex numbers
The cartesian graphing system is based on using three mutually perpendicular lines to plot a point in [imath]3D[/imath] space. But that is for real numbers and no graphs for complex numbers exist. For eg. If I Have a function [imath]y=x^2-i,[/imath]where [imath]i=\sqrt{-1}[/imath] And i want to find its root graphically then how can i plot a graph which will help me? [imath]\because[/imath] real numbers require only one number to specify them while complex numbers need two it won't be very easy, Please help... | 398266 | Plotting in the Complex Plane
I just wonder how do you plot a function on the complex plane? For example,[imath]f(z)=\left|\dfrac{1}{z}\right|[/imath] What is the difference plotting this function in the complex plane or real plane? |
2790561 | Does [imath]|a_n-a_{n+1}|\to 0[/imath] imply [imath](a_n)[/imath] is Cauchy?
My textbook has this problem as a kind of "concept check", where one is supposed to find a counterexample to the following statement: A sequence of real numbers is cauchy iff. [imath] \forall \epsilon>0, \, \exists N \in \mathbb{N}, \, \forall n \geq N: |a_n-a_{n+1}|< \epsilon [/imath] I know that a sequence of real numbers is cauchy if [imath] \forall \epsilon>0, \, \exists N \in \mathbb{N}, \, \forall n,m \geq N: |a_n-a_m|< \epsilon [/imath] Finding such a counterexample is probably trivial, but I haven't been able to think of one. | 1237655 | Pseudo-Cauchy sequence
I have never seen this terminology before, so I will provide the given definition. A Pseudo-Cauchy sequence is : A sequence [imath](a_n)[/imath] if for any [imath]\epsilon > 0[/imath] there exists [imath]N \in \mathbb{N}[/imath] such that [imath]|a_{n+1} - a_n | \leq \epsilon \space \forall \space n \geq N[/imath] So then my question is that is a pseudo-cauchy sequence always converging? |
2790731 | Every prime divisor ([imath]p \neq 5[/imath]) of [imath]n^2+n-1[/imath] is of the form [imath]10k+9[/imath]
Now, what I have done so far is the following: Let [imath]p[/imath] be a prime such that [imath]p | n^2+n-1[/imath], then [imath]n^2+n-1 \equiv 0 \pmod p[/imath] This congruence has a solution if and only if [imath]x^2 \equiv \Delta \pmod p[/imath] has a solution, where [imath]\Delta = b^2 - 4ac = 1^2 - 4\cdot1\cdot(-1) = 5[/imath]. Since [imath]\Delta = 5[/imath], this has a solution iff [imath](\frac{5}{p}) = 1[/imath] (the Legendre symbol). And this Legendre symbol is [imath]1[/imath] iff [imath]p \equiv 1, 4 \pmod 5[/imath] Which, in turn, means [imath]p \equiv 1, 4, 6, 9 \pmod {10}[/imath]. But [imath]p[/imath] cannot be congruent to [imath]4[/imath] or [imath]6[/imath] modulo [imath]10[/imath], for otherwise it would not be a prime. Therefore, [imath]p \equiv 1, 9 \pmod {10}[/imath]. How do I get rid of the [imath]p \equiv 1 \pmod {10}[/imath] solution and prove that it can only be congruent to [imath]9[/imath], i.e. be of the form [imath]10k + 9[/imath], for some k? Is this proof valid for all choices of [imath]p\neq 5[/imath]? If so, mayhap my teacher forgot about the [imath]p[/imath] of the form [imath]10k+1[/imath]? | 573062 | Prime divisors of the integer [imath]n^2+n-1[/imath] (using the Legendre symbol)
What made me have a question is the following problem. The prime divisors [imath]p\not=5[/imath] of the integer [imath]n^2+n-1[/imath] are of the form [imath]10k+1[/imath] or [imath]10k+9[/imath]. I thought [imath]\left(\frac 5 p\right) = 1[/imath], because [imath]5 \equiv 1 \pmod 4[/imath]. So [imath]\left(\frac 5 p\right) = (-1)^{p-1}\left(\frac p 5\right)=1[/imath] by qudratic reciprocity. Then [imath]p[/imath] should be of the form [imath]2k+1[/imath] But I think there is something wrong, and I have lost. Is there someone to help me? |
2791075 | Infinite cartesian product from Herbert Enderton's Elements of set theory
I am reading Enderton's book Elements of set theory and I am having hard time understanding Infinite cartesian product section in which he define's product of [imath]H(i)[/imath]'s. Could you please explain in other simple words or by giving an example. PS : I am a self learner Thanks. | 195517 | Definition of the Infinite Cartesian Product
(1) If [imath]X[/imath] and [imath]Y[/imath] are two sets, we define the Cartesian product [imath]X \times Y[/imath] as the set of ordered pairs [imath](x,y)[/imath], such that [imath]x \in X[/imath] and [imath]y \in Y[/imath]. (2) On the other hand [Folland, Real Analysis, page 4], if [imath]\{X_\alpha\}_{\alpha \in A}[/imath] is infinite indexed family of sets, their Cartesian product [imath] \prod_{\alpha \in A}X_\alpha [/imath] is defined as the set of maps [imath]f: A \to \bigcup\limits_{\alpha \in A} X_\alpha[/imath] such that [imath]f(\alpha) \in X_\alpha[/imath] for every [imath]\alpha \in A[/imath]. After saying this, Folland remarks: it should be noted, and promptly forgotten, that when [imath]A = \{1,2\}[/imath], the previous definition of [imath]X_1 \times X_2[/imath] [that's (1) above] is set-theoretically different from the present definition of [imath]\prod_1^2 X_j[/imath] [that's (2) above]. Indeed, the latter concept depends on the mappings, which are defined in terms of the former one. I am not grasping this remark. Specifically, here are my questions. Question 1: How is (2) set-theoretically different from (1)? A simple illustrative example? Question 2: If (1) is extended to infinite families, which definition would be stronger? A simple illustrative example? Question 3: Why should this be "promptly forgotten"? I'll probably have more questions depending on the type of answers I'll get to these. Thanks! |
2789696 | How to draw the graph of [imath]F(x,y)=0[/imath] for given [imath]F[/imath].
Are there general method for drawing the graph of [imath]F(x,y)=0[/imath] for given [imath]F[/imath]? What if [imath]F[/imath] is a polynomial? | 1120582 | Contour graphing algorithm
I am not sure if this belongs on Mathematics Stack Exchange, but it is somewhat relevant here. The Problem If you've installed any graphing/plotting apps on your smartphone, you will notice that the app will only take the equation in form of a single variable for 2D (e.g [imath]y=f(x)[/imath]) or in 2 variables for a 3D plot (e.g [imath]z=f(x,y)[/imath]). This is problematic as many equations are tough to reduce to the form of [imath]y=f(x)[/imath] or [imath]z=f(x,y)[/imath]. The plotters like this because this is the only way they can plot it in form of lines connecting multiple points. These plotters compute the [imath]y[/imath] value at fixed intervals of [imath]x[/imath] and join the point to the previous one. The brute force approach to counter this is to check if the function satisfies it at each point on the image. The problem is that this not only takes more time, but gives discontinuous graph for the function as the image cannot have infinite resolution. The Question The only way to fix this is to identify which point comes after the next in the plot and join them together by a line. That is where I am stuck. How do you identify which point is sequentially the next one for a given function? Keep in mind that for a human, this might be very easy to do. But, for a computer, this requires a general algorithm. Any ideas? |
2791626 | Does there exists a finite index non-trivial subgroup of [imath]\mathbb{C}^*[/imath]
let [imath]\mathbb{C}^*[/imath] be a multiplicative group of non-zero complex numbers. Suppose that [imath]H[/imath] be a subgroup of finite index of [imath]\mathbb{C}^*[/imath].Prove that [imath]H=\mathbb{C}^*[/imath] Attempt If [imath][\mathbb{C}^*.H]=n[/imath] then if [imath]xH \in \frac{\mathbb{C}^*}{H}[/imath] then [imath]\left(xH\right)^n=H[/imath] implies [imath]x^n \in H[/imath]. Because [imath]x[/imath] is random we have [imath]\left(\mathbb{C}^*\right)^n=\mathbb{C}^* \in H[/imath] Note that [imath]\left(\mathbb{C}^*\right)^n=\mathbb{C}^*[/imath] | 1706207 | Subgroup of [imath]C^*[/imath] (nonzero complex) with finite index.
True or false: Let [imath]C^*[/imath] be the set of all nonzero complex numbers and [imath]H[/imath] be a subgroup of [imath]C^*[/imath](with respect to multiplication) be such that [imath][C^*:H][/imath] is finite then [imath]H=C^*[/imath]. I'm guessing it true as I am thinking that if suppose there is such a proper subgroup [imath]H[/imath] for which the number of coset will be finite then I'm guessing that there is a gap between [imath]C^*[/imath] and [imath]H[/imath] and that gap cannot be filled up by finite union. But I am unable to give a concrete prove. Thanks in advance. |
2788649 | An equation for Fibonacci numbers
Let [imath]F(n)[/imath] be [imath]n[/imath]-th Fibonacci-number. Then the following holds: [imath]F(n)^2+F(n+1)^2=F(2n+1).[/imath] I have proved, but I want a brief proof. Do you know anything about it? | 2248821 | Prove [imath]f_{n+1}^2+f_n^2=f_{2n+1}[/imath]
Prove that if [imath]f[/imath] is the Fibonacci sequence then [imath]f_{n+1}^2+f_n^2=f_{2n+1}[/imath] holds for all n. Instead of trying doing this by induction, I need to do it by trying to just replacing the explicit formula for [imath]f_n[/imath]. I am stuck at this step: [imath]f_{n+1}^2+f_n^2=\frac{1}{5}[\phi^{2n}(1+\phi^2)+(1-\phi)^{2n}(1+(1-\phi)^2)][/imath] Believe me that this is equivalent to [imath]\frac{1}{\sqrt{5}}[\phi^{2n+1}-(1-\phi)^{2n+1}][/imath] Maybe it's just a simple algebra problem, but I don't know how to get that expression. Thanks |
2791270 | [imath]\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{\infty}=?[/imath]
One of my friends has given me the following problem: [imath]\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{\infty} = ?[/imath] He said that the answer is [imath]1[/imath]. He has given his argument as that this is a part of the whole, and if we add together all the part, then we'd have a [imath]1[/imath]. But said, disagreed with him that it'd not be [imath]1[/imath], rather it'd be [imath]\frac{\infty - 1}{\infty}[/imath]. I'm not sure if this is correct. But I have given the argument with this view: If we add together only the first two fractions, that is [imath]\frac{1}{2}[/imath] and [imath]\frac{1}{4}[/imath], then the sum will be [imath]\frac{3}{4}[/imath], which is equal to [imath]\frac{4-1}{4}[/imath]. Likewise, if we add together the first three, four, five, then the sum will be [imath]\frac{x-1}{x}[/imath], where [imath]x[/imath] is the denominator of last fraction. These happen because every time divide ramainder(whole in the first case) in the equal halves, there remain one part of the equal halves and after we add together all the part(follows the pattern, never break the pattern) there remain [imath]\frac{1}{x}[/imath] of the whole part. So we can nonetheless say this that this only became 1(the whole), if we add add the remainder in the sum, that's mean there will be [imath]2[/imath] of the last fraction. But that will break the pattern. But he argued with me and wanted me to ask the question here. I also said that if we round it then it will be [imath]1[/imath], rather it never will be [imath]1[/imath]. So, what is your opinion about it? If included both, argument and mathematical calculation, then that will be great. | 2495413 | How to find the limit of series? (What should I know?)
There is a couple of limits that I failed to find: [imath]\lim_{n\to\infty}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}[/imath] and [imath]\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n-1}}{3^{n-1}}[/imath] There is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result. So my questions are: What should I do when I need a limit of infinite sum? (are there any rules of thumb?) What theorems or topics from calculus should I know to solve these problems better? I am new to math and will appreciate any help. Thank you! |
2791845 | Show that [imath]\{\frac{e^n}{n!}\}[/imath] diverge.
I know that the sequence [imath]\{\frac{e^n}{n!}\}[/imath] diverges and that for prove it i have to limit it, but i don't know how do it. In fact, i know that [imath]\{\frac{x^n}{n!}\}[/imath] diverges, but i don't know prove it. | 2791861 | Show that [imath]\{\frac{e^n}{n!}\}[/imath] converges.
I know that the sequence [imath]\{\frac{e^n}{n!}\}[/imath] converges and that for prove it i have to limit it, but i don't know how do it. In fact, i know that [imath]\{\frac{x^n}{n!}\}[/imath] converges, but i don't know prove it. i want to aclarate that is convergence of the sequence, isn't of te serie. |
2791165 | How can a ratio containing only real numbers have a complex value?
Solving this equation [imath]x = 1-\cfrac{2}{1-\cfrac{2}{1-\cfrac{2}{\ddots}}}[/imath] Sub in [imath]x[/imath] [imath]x=1-\frac{2}{x}\implies x^2=x-2 \implies x^2-x=-2[/imath] Solve through completing the square \begin{align} x^2-x+\frac{1}{4}&=-2+\frac{1}{4}\\ \left(x-\frac{1}{2}\right)^2&=-\frac74\\ x-\frac{1}{2}&=\pm\sqrt{\frac{7}{4}}i\\ x&=\frac{1}{2}\pm\sqrt{\frac{7}{4}}i\\ \end{align} When you substitute this value back in for [imath]x[/imath] it works. But i don't understand why a equation like this can equal a complex value Maybe I am missing something important here? EDIT: So this equation diverges right. Does that make my working invalid, or just explain the non-real part. Can anything useful be done by defining this recursively [imath]f(x)=1-\frac{2}{f(x)}[/imath] and then using a seed value I also wonder whether we would end up dividing by zero at some point. Can you prove we do or don't? | 1681993 | Why is [imath]1 - \frac{1}{1 - \frac{1}{1 - \ldots}}[/imath] not real?
So we all know that the continued fraction containing all [imath]1[/imath]s... [imath] x = 1 + \frac{1}{1 + \frac{1}{1 + \ldots}} [/imath] yields the golden ratio [imath]x = \phi[/imath], which can easily be proven by rewriting it as [imath]$x = 1 + \dfrac{1}{x}$[/imath], solving the resulting quadratic equation and assuming that a continued fraction that only contains additions will give a positive number. Now, a friend asked me what would happen if we replaced all additions with subtractions: [imath] x = 1 - \frac{1}{1 - \frac{1}{1 - \ldots}} [/imath] I thought "oh cool, I know how to solve this...": [imath]\begin{align} x &= 1 - \frac{1}{x} \\ x^2 - x + 1 &= 0 \end{align}[/imath] And voila, I get... [imath] x \in \{e^{i\pi/3}, e^{-i\pi/3} \} [/imath] Ummm... why does a continued fraction containing only [imath]1[/imath]s, subtraction and division result in one of two complex (as opposed to real) numbers? (I have a feeling this is something like the [imath]\sum_i (-1)^i[/imath] thing, that the infinite continued fraction isn't well-defined unless we can express it as the limit of a converging series, because the truncated fractions [imath]1 - \frac{1}{1-1}[/imath] etc. aren't well-defined, but I thought I'd ask for a well-founded answer. Even if this is the case, do the two complex numbers have any "meaning"?) |
846263 | Is there an analogue of Lebesgue’s Dominated Convergence Theorem for a net [imath] (f_{\alpha})_{\alpha \in A} [/imath] of measurable functions?
Is there an analogue of Lebesgue’s Dominated Convergence Theorem for a net [imath] (f_{\alpha})_{\alpha \in A} [/imath] of measurable functions defined on a measure space [imath] (\Omega,\Sigma,\mu) [/imath], where the index set [imath] A [/imath] (assumed to be directed) is not necessarily [imath] \mathbb{N} [/imath]? That is, if [imath] (f_{\alpha})_{\alpha \in A} [/imath] converges pointwise almost everywhere to a measurable [imath] f [/imath], there exists a measurable [imath] g [/imath] such that [imath] |f_{\alpha}| \leq |g| [/imath] almost everywhere for each [imath] \alpha \in A [/imath], and [imath] \displaystyle \int_{\Omega} |g| ~ \mathrm{d}{\mu} \leq \infty [/imath], then does [imath] \displaystyle \lim_{\alpha \in A} \int_{\Omega} f_{\alpha} ~ \mathrm{d}{\mu} = \int_{\Omega} f ~ \mathrm{d}{\mu} [/imath]? | 141198 | A net version of dominated convergence?
Let [imath]G[/imath] be a locally compact Hausdorff Abelian topological group. Let [imath]\mu[/imath] be a Haar measure on [imath]G[/imath], i.e. a regular translation invariant measure. Let [imath]f[/imath] be fixed in [imath]\mathcal{L}^1(G, \mu)[/imath]. Define the function from [imath]G[/imath] to [imath]\mathcal{L}^1(G)[/imath] that assigns to each [imath]y[/imath] in [imath]G[/imath] the function that takes [imath]x[/imath] in [imath]G[/imath] to [imath]f(xy^{-1})[/imath]. I.e. it assigns the [imath]y[/imath]-translate of [imath]f[/imath]. Is this a continuous function of [imath]y[/imath] into [imath]\mathcal{L}^1[/imath]? On p.94 3.10 of these notes it is asserted that it is. The usual proof of this sort of thing in cases where [imath]G=\mathbb{R}[/imath], say, is you use density of compactly supported continuous functions, and then dominated convergence. The former survives, but dominated convergence does not work on nets. In fact, a limit of a net of measurable functions need not be measurable. |
2793107 | How do I plot [imath]-\frac{\pi}{2} \leq \operatorname{Arg}\left(\frac{z}{\bar{z}}\right) \leq\frac{\pi}{2}[/imath] on the complex plane?
I'm new to Complex numbers, so I don't know a lot about them yet. How do I plot this condition on the complex plane? [imath]-\frac{\pi}{2} \leq \operatorname{Arg}\left(\frac{z}{\bar{z}}\right) \leq\frac{\pi}{2}[/imath] | 2793050 | What is the value of [imath]\operatorname{Arg}(\frac{z}{\bar{z}})[/imath]?
As part of an exercise, I was asked to plot this condition on the complex plane: [imath]-\frac{\pi}{2} \leq \operatorname{Arg}\left(\frac{z}{\bar{z}}\right) \leq\frac{\pi}{2}[/imath] My question is: What is the value of [imath]\displaystyle\operatorname{Arg}\left(\frac{z}{\bar{z}}\right)[/imath]? |
2792638 | Exponent in the prime factorization of [imath]3^{1024}-1[/imath] belonging to the prime [imath]p=2[/imath]
Find the largest positive integer [imath]n[/imath] such that [imath]3^{1024} - 1[/imath] is divisible by [imath]2^n[/imath]. I am trying all powers of [imath]2[/imath], beginning from [imath]2^0[/imath] onwards; I believe the answer will be [imath]n=\infty[/imath], since [imath]3^{1024} - 1[/imath] is an even number and it can be divided by [imath]2[/imath], so it is divisible by all powers of [imath]2[/imath]? | 2770741 | Maximum power of [imath]2[/imath] which divides [imath]3^{1024}-1[/imath]
What is the maximum power of [imath]2[/imath] which completely divides [imath]3^{1024}-1[/imath]? I proceeded thus: [imath]\phi(2^n)=2^{n-1}[/imath] for all [imath]n\ge1[/imath] [imath]3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}[/imath] [imath]3^{1024}-1\equiv0\pmod {2^{11}}[/imath] Since [imath]\phi(2^{11})=2^{10}[/imath]. So, maximum power of [imath]2[/imath] must be [imath]11[/imath]. But the answer says it is [imath]12[/imath]. Where am I wrong and how to solve it correctly? |
2793443 | PDF of [imath]Y=e^{2X}[/imath] given X~exp(3)
Consider the random variable X∼exp(3). And let [imath]Y=e^{2X}[/imath]. Find the PDF of Y. I know by properties of exp distribution that the PDF of [imath]f_X(x)[/imath] is given by [imath]f_X(x) = \begin{cases} 3e^{-3x}; & \text{ if, } x > 0 \\ 0; & \text{ otherwise } \end{cases}[/imath] The way I approched the problem is by first trying to find the CDF of Y and then taking the derivative to get the PDF of Y. This is a exercise on a previous exam so I know the result and I'm getting a wrong one. I think I'm having troubles finding the CDF. This is what I've tried [imath]F_Y(y) = P(Y\le y)=P(e^{2X}\le y)=P(X\le \frac{ln(y)}{2})[/imath] So I have the upper limit and the lower limit of the integral is 1, as [imath]e^{2\cdot 0}[/imath] is 1 and that is the lowest that [imath]x[/imath] gets. I then integrate the the whole thing to find the CDF. [imath]\int_{1}^{\frac{ln(y}{2}} 3e^{-3X} \ dx[/imath] But I'm doing something wrong, because it's suppose to give me the result of [imath]\frac{3}{2}y^\frac{-5}{2}[/imath], for [imath]y>1[/imath] | 2792899 | If [imath]X \sim \exp(3)[/imath] what is the PDF of [imath]Y=e^{2X}[/imath]?
If [imath]X \sim \exp(3)[/imath] and let [imath]Y = e^{2X}[/imath] what is the PDF of [imath]Y[/imath]? [imath] \begin{align} F_Y(y)& = P(Y\leq y) \\ & = P(e^{2X} \leq y) \\ &=P(X \leq \ln(\sqrt{y})) \end{align} [/imath] Then you need your CDF will look like [imath] \begin{align} P(X \leq \ln(\sqrt{y})) &= \int_0^{\ln(\sqrt{y})}3e^{-3x}dx \\ \end{align} [/imath] Using the fundamental theorem of calculus we get that the PDF of Y will be [imath] \begin{align} f_Y(y) &= 3(e^{-3ln(\sqrt{y})}-1) \\ & =3(y^{-3/2}-1) \quad y>1 \end{align} [/imath] The PDF doesnt integrate to 1 on it's support so there has to be a mistake however I don't where exactly please help me find it. |
2793854 | Show that the series converges
You have the sequence [imath](x_{n})_{n\in\mathbb{N}}[/imath], so that for every [imath]n \in \mathbb{N}[/imath], you have to [imath]x_ {n}:= \ln\frac{n + 1}{n}[/imath] Find an explicit formula for the partial sums of the series generated by [imath](x_{n})_{n \in\mathbb{N}}[/imath] and determine if the series converges. My development I'm supposed to find a formula that generalizes the sum, so what I did is \begin{align*} S_{n}&=a_{1}+a_{2}+a_{3}+\cdot+a_{n-1}+a_{n}\\&=S_{n-1}+a_{n} \end{align*} thus, \begin{align*} a_{n}&=S_{n}-S_{n-1}\\&=\ln\frac{n+1}{n}-\ln\frac{n}{n-1}\\&=\ln\frac{(n+1)(n-1)}{n^{2}} \end{align*} and you should determine if [imath] \sum\limits_{n=2}^{\infty}\ln\dfrac{(n-1)(n + 1)}{n ^ {2}} [/imath] is convergent, right? | 148075 | study the convergence of [imath]\sum_{n=1}^\infty \log \frac{n+1}{n}[/imath]
[imath]\sum_{n=1}^\infty \log \frac{n+1}{n}[/imath] [imath]\sum_{n=1}^\infty \log \frac{n+1}{n}[/imath] = [imath]\displaystyle\lim_{n \to{+}\infty}(\log2 - \log1)+(\log3-\log2)+...+(\log(n+1)-\log n)[/imath]=[imath]\displaystyle\lim_{n \to{+}\infty}\log(n+1)\to \infty[/imath]. So, the series diverges. Is my procedure correct? |
2793621 | How to solve the differential equation for power series [imath](1+x^2)y'' - 6xy=0[/imath] for [imath]x_{0}=0[/imath]
I have tried for the power series but when suppose the solution is [imath]y(x)=\sum _{n=0}^\infty a_nx^n[/imath] appeared sums dependents of [imath]a_n[/imath], [imath]a_{n+2}[/imath] and [imath]a_{n-1}[/imath] and this is not possible. assume [imath]y(x)=\sum _{n=0}^\infty a_nx^n[/imath] solution, so [imath] y'(x) = \sum_{n=1}^\infty na_nx^{n-1}[/imath] and [imath]y''(x)=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}[/imath] then the equation [imath](1+x^2)y'' - 6xy=0[/imath] is: [imath]\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=2}^\infty n(n-1)a_nx^{n} - \sum_{n=0}^\infty 6a_nx^{n+1}=0[/imath] We can shift the dummy indices, taking k=n−2 in the first sum, k=n in the second sum and k=n+1 in the last sum: [imath]\sum_{k=0}^\infty(k+2)(k+1)a_{k+2}x^k+\sum_{k=2}^\infty k(k-1)a_{k}x^k- \sum_{k=1}^\infty6a_{k-1}x^k=0[/imath] taking [imath]k=0[/imath] and [imath]k=1[/imath], we have [imath]2a_{2}=0[/imath] so [imath]a_2=0[/imath] and [imath] 6a_3x-6a_0x=0[/imath] so [imath]a_3=a_0[/imath]. For [imath]k>1[/imath] we have [imath](k+2)(k+1)a_{k+2} + k(k-1)a_k - 6a_{k-1}=0[/imath] so [imath]a_k=\frac{ 6a_{k-1} -(k+2)(k+1)a_{k+2} }{k(k-1)}[/imath] and [imath]a_2=0, a_3=a_0[/imath]. but this recorrence not determines the solution cause [imath]a_k[/imath] depends of [imath]a_{k-1}[/imath] and [imath]a_{k+2}[/imath]. | 2793155 | How to solve the diffencial equation for power series [imath](1+x²)y'' - 6xy=0[/imath] in [imath]x_{0}=0[/imath]
I have tried for the power series but when suppose the solution is [imath]y(x)=\sum _{n=0}^\infty a_nx^n[/imath] appeared sums dependents of [imath]a_n[/imath], [imath]a_{n+2}[/imath] and [imath]a_{n-1}[/imath] and this is not possible. |
2794096 | conditional expectation number of throws given that until throwing only even numbers
suppose you throw a fair die until [imath]6[/imath] appears for the first time. Compute the conditional expectation of the number of throws, given that until throwing only even numbers. This was an exercise in an old exam and I'm interested in a solution. | 2779681 | Expected number of die rolls to get 6 given that all rolls are even.
A fair 6-sided die is rolled repeatedly in till a 6 is obtained. Find the expected number of rolls conditioned on the event that none of the rolls yielded an odd number I had tried to figure out what will be the conditional distribution of [imath]\frac{X}{Y}[/imath] but I can't solved it yet Where [imath]X [/imath] is the face numbered 6 is obtained and [imath]Y[/imath] is the event only even number is occured |
2794329 | Geometric Proof for the shortest path cube sides problem
A spider in one edge of a cube (length of all side = [imath]l[/imath]) wants to get to an insect on the other edge of the cube. obviously the spider cannot fly and must walk on the sides of the cube to get to the insects. Find the shortest path possible. (See the image below) I know the differential approach to solve this question (optimization problem) and the path is drawn on the above picture. But I want a geometric proof without using derivative. Just using geometric theorems (like pythagoras) and using simple algebra. | 1683905 | How to find the shortest path between opposite vertices of a cube, traveling on its surface?
I am stuck with the following problem that says: Let [imath]A,B[/imath] be the ends of the longest diagonal of the unit cube . The length of the shortest path from [imath]A[/imath] to [imath]B[/imath] along the surface is : [imath]\sqrt{3}\,\,[/imath] 2.[imath]\,\,1+\sqrt{2}\,\,[/imath] 3.[imath]\,\,\sqrt{5}\,\,[/imath] 4.[imath]\,\,3[/imath] My Try: So, the length of the longest diagonal [imath]AB=\sqrt{3}[/imath]. If I reach from [imath]A[/imath] to [imath]B[/imath] along the surface line [imath]AC+CD+BD[/imath], then it gives [imath]3[/imath] units. But the answer is given to be option 3. Can someone explain? Thanks in advance for your time. |
2777865 | Let [imath]X, Y, Z[/imath] be independent and [imath]N(1,1).[/imath] Find the probability that [imath]X + Y ≥ 3Z[/imath] using convolution.
I did the following: Let [imath]W=X+Y-3Z,[/imath] then we have that [imath]W\sim N(1+1-3,1+1+9)=N(-1,11).[/imath] so [imath]P(X+Y\ge3Z)=P(W\ge 0)=1-P(W\le 0)=1-F_W(0)=1-\Phi\left(\frac{1}{\sqrt{11}}\right)\approx \\ \approx 1-0.618 = 0.382.[/imath] However, I'm in the chapter treating convolution. How can I solve this problem using this approach instead? | 2765119 | X,Y,Z independent and [imath]\sim N(1,1)[/imath]. Find [imath]P(X+Y \geqq 3Z)[/imath]
I'm stuck on this problem, and I dont know how to continue from here. It would be great if someone could just give me a hint of how to proceed from here. The problem is, "Let [imath]X,Y,Z[/imath] be independent and [imath]N(1,1)[/imath]. Find [imath]P(X+Y \geqq 3Z)[/imath]." Now, the way I see it there are two obvious ways to proceed. First off, one can simply condition on the different variables and because of their independence obtain [imath]P(X+Y \geqq 3Z) = \displaystyle \int_{-\infty}^{\infty}P(X+Y \geqq 3Z | Z=z)f_Z(z)dz = \int_{-\infty}^{\infty}P(X+Y \geqq 3z)f_Z(z)dz [/imath] [imath] (1)[/imath] and by doing the exact same reasoning we get [imath]P(X+Y \geqq 3z) = \int_{-\infty}^{\infty}F_X(3z-y)f_Y(y)dy[/imath] and thus (with (1)) [imath]P(X+Y \geqq 3Z) = \int_{-\infty}^{\infty}f_Z(z)\int_{-\infty}^{\infty}F_X(3z-y)f_Y(y)dydz[/imath]. This is problematic however, since [imath]X\sim N(1,1)[/imath] means that we cannot express [imath]F_X(3z-y)[/imath] explicitly. If the variables would have had some other distribution with known cdf, the calculation could continue in a straight forward way from here. This is not the case right now, however. I did, however, try to use the fact that [imath]F_X(3z-y) = \int_{0}^{3z-y}f_X(t)dt[/imath]. But then I couldnt make sense of the integration limits. So I doubt that this is the right way to go in solving this particular problem. I was also thinking about using the fact that [imath]X+Y =: W[/imath] then [imath]W\sim N(1+1,1+1)[/imath] and would simplify the calculation above but yield the same problem. Maybe there is a nice trick one can use in calculating the integral above. Either way, I would really appreciate if someone could give me some hints as to how to continue from here. Thanks! |
2794624 | Is the set inside a set subset of the bigger set?
I just started set theory (ncert maths 11th grade) and it had a question: Let [imath]A = \{1 , 2 , \{3,4\} , 5 , 6\}[/imath] Is [imath]\{3,4\}[/imath] a subset of A? The book said no. I think the answer should be yes, since [imath]A[/imath] contains a set which contains the elements [imath]3,4[/imath] hence [imath]A[/imath] contains [imath]3,4[/imath] and by the definition of subset [imath]\{3,4\}[/imath] should be a subset of [imath]A[/imath]. | 638560 | Why can a 1 element set be a member of another set but not a subset of it?
I have came across this in a textbook: [imath]\{2\}\nsubseteq\{\{2\}\}[/imath] but [imath]\{2\}\in\{\{2\}\}[/imath] I understand that [imath]\{2\}[/imath] is an element (member) of the other set but considering [imath]\{2\}[/imath] is a set itself, why is it not a subset? |
2794570 | Proving lemmas during construction of non-measurable sets.
Let [imath]I[/imath] be an interval: [imath]I=(0,1)[/imath] Define [imath]R[/imath] to be relation on [imath]I[/imath] such that: [imath](x,y)\in R\Leftrightarrow|x-y|\in\Bbb{Q}[/imath] It is straightforward to show that [imath]R[/imath] is an equivalence relation. Thus we can split [imath]I[/imath] into equivalence classes. The first class contains all rationals and then in each class there is some irrational number [imath]\xi[/imath] along with numbers which differ from [imath]\xi[/imath] by a rational. We can say that each equivalence class will correspond to one irrational number from [imath]I[/imath] (except from the first one). Let's invoke the axiom of choice to construct a set [imath]S[/imath], such that it contains an element from each class. [imath]s_1[/imath] from the first class, [imath]s_2[/imath] from the second ... and so on. We have: [imath]S=\{s_1,s_2,s_3\}[/imath] Now, let [imath]r_1,r_2,...[/imath] be a list of all rational numbers in [imath](-1,1)[/imath]. Now define family of sets [imath]S_k[/imath] for [imath]k\in \Bbb{N}[/imath]: [imath]S_k=\{s_1+r_k,s_2+r_k,s_3+r_k,\cdots\}[/imath] so [imath]S_k[/imath] is just [imath]S[/imath] shifted by [imath]r_k[/imath]. Now there are two big statements to consider: [imath]\forall i\neq j:S_i\cap S_j=\emptyset\tag{1}[/imath] [imath]\text{Every number from $I$ is in exactly one of $S_k$}\tag{2}[/imath] Let [imath]\mu[/imath] denote the Lebesgue measure. By (1) we have [imath]\mu\bigg(\bigcup_i S_i\bigg)=\sum_i\mu(S_i)[/imath] We also know that [imath]\bigcup S_i\subseteq (-1,2)[/imath] thus [imath]\mu(\bigcup S_i)\leq 3[/imath] and by (2) we have [imath]I\subseteq \bigcup S_i[/imath] thus [imath]\mu(\bigcup S_i)\geq 1[/imath]. Because [imath]S_k[/imath]'s are just shifted [imath]S[/imath]'s, we have [imath]\mu(S)=\mu(S_1)=\mu(S_2)\cdots[/imath] thus [imath]\mu\bigg(\bigcup S_i\bigg)=\sum \mu(S)\tag{a}[/imath] (The right hand side of (a) is an infinite sum) Which is impossible, because if [imath]\mu(S)=0[/imath] then [imath]\mu(\bigcup S_i)=0[/imath] and if [imath]\mu(S)>0[/imath] then [imath]\mu(\bigcup S_i)=\infty[/imath] which shows [imath]S[/imath] is non-measurable. Now, I'd be interested in proving the facts (1) and (2). (1) somehow makes sense to me, because there is no way how we can "jump" from rational number to irrational, because [imath]\Bbb{Q}[/imath] is closed under addition and subtraction. For (2), that would also make sense, because we know that each equivalence class except the first one corresponds to one irrational number in [imath]I[/imath] thus that will be some [imath]s_m+r_k[/imath] for [imath]m\neq 1[/imath] which we will find in the [imath]S_k[/imath]. For any rational, we will find it in some [imath]S_l[/imath] and it will correspond to [imath]s_1+r_l[/imath]. Is my reasoning correct? I'd like some rigorous proof for these two lemmas. | 238940 | Understanding equivalence class, equivalence relation, partition
I'm having difficulty grasping a couple of set theory concepts, specifically concepts dealing with relations. Here are the ones I'm having trouble with and their definitions. 1) The collection of equivalence classes w.r.t. [imath]R[/imath] Def: Let [imath]R[/imath] be an equivalence relation in a set [imath]X[/imath]. The collection of equivalence classes w.r.t. [imath]R[/imath] is the set: [imath]$$[X]/R =\{S|(\exists x)(x\in X\land S=[x]/R)\}=\{[x]/R|x\in X\}$$[/imath] 2) Partition Def: Let [imath]X[/imath] be a set. A collection of sets [imath]C[/imath] is a partition of [imath]X[/imath] if: (i) [imath]\bigcup_{S\in\ C} S=X.[/imath] (ii) [imath]\forall S \in C, S \neq \varnothing[/imath] (iii) [imath]\forall S, S' \in C, S' \neq S \Rightarrow S \cap S' = \varnothing[/imath] 3) Relation induced by Def: Let [imath]C[/imath] be a partition of [imath]X[/imath]. The relation induced by [imath]C[/imath], denoted by [imath]X/C[/imath], is a relation in [imath]X[/imath] such that [imath]$$X/C = \{(x,y) | (\exists S \in C)(\exists x \in S \land \exists y \in S)\}$$[/imath] |
2794569 | Surface and volumes of solids of revolution
The lemniscate [imath]\ r^2 =\ a^2 \cos (2\theta) [/imath] revolves about a tangent at the pole. What will be the volume generated? I know the answer will be [imath] \frac {\pi a^3}{4}[/imath] But my question is how? Any help possible is requested. | 941603 | Volume generated by lemniscate revolving about a tangent at the pole.
The lemniscates [imath]r^2 = a^2\cos2\theta[/imath] revolves about a tangent at the pole. What is the volume generated by it ? Please explain in detail. I found a couple of answers on finding surface areas, volumes but didn't understood it. Can someone solve this. I'm stuck in the middle with integrations. |
2786990 | What is parametric equations of a locus of a fixed point of a circle rolling along a ellipse in [imath]\mathbb{R}^2[/imath]?
I have learnt about cycloids and have a related question: What is parametric equation of a locus of a fixed point of a circle rolling along an ellipse in [imath]\mathbb{R}^2[/imath]? | 1308195 | Circle Rolling on Ellipse
I've gotten interested in describing a circle rolling on an ellipse; specifically, the curve traced out by a point on the circumference of the circle. I want a symbolic solution to the general case, radius [imath]r[/imath], axes [imath]a[/imath] and [imath]b[/imath]. I've written nine polynomial equations in terms of various angles and lengths. Exactly what "solution" means is subject to debate. Let [imath](u,v)[/imath] be the point on the circle. Similar to the cycloid, I would like an equation for [imath]u[/imath] in terms of a "natural" angle in the problem. Similarly, an equation for [imath]v[/imath]. Perhaps it is necessary to have a differential equation, so maybe [imath]du/dt[/imath], [imath]u[/imath], and [imath]t[/imath], where [imath]t[/imath] is an angle in the problem. I would have thought this was known, but I can't find it anywhere. |
2795963 | How to show that any subgroup of a cyclic group must be cyclic
If [imath]G[/imath] is a cyclic group that is generated by [imath]g\in G[/imath] so that [imath]G= <g>[/imath], how would I show that for any subgroup [imath]H \subseteq G[/imath] then [imath]H[/imath] must be cyclic? I've already attempted to start this proof: Let [imath]G[/imath] be a cyclic group and [imath]H[/imath] any subgroup of [imath]G[/imath]. I claim that for some [imath]m \in \mathbb{Z}[/imath] we have [imath]H=<g^m>[/imath]. Case 1: If [imath]H=\left \{ 1 \right \}[/imath] then [imath]H=<g^0>[/imath] and so H is cyclic. Case 2: If [imath]H=G[/imath] then [imath]H=G=<g^1>[/imath] and so H is cyclic, Case 3: [imath]H[/imath] is a proper, non trivial subgroup. Then [imath]\exists \ k\in H [/imath] such that [imath]k\neq 1 [/imath] and since [imath]k\in G \implies k=g^m[/imath] for some [imath]m \in Z[/imath] and so [imath]g^m \in H[/imath]. It then follows by closure under the binary operation on [imath]H[/imath] that [imath]\forall \ n\in \mathbb{Z}, (g^m)^n \in H \implies <g^m> \subseteq H[/imath] My question is how do I show that [imath]H \subseteq <g^m>[/imath] in order to show that they are equal? | 295564 | A subgroup of a cyclic group is cyclic - Understanding Proof
I'm having some trouble understanding the proof of the following theorem A subgroup of a cyclic group is cyclic I will list each step of the proof in my textbook and indicate the places that I'm confused and hopefully somewhere out there can clear some things up for me. Proof Let G be a cyclic group generated by "[imath]a[/imath]" and let H be a subgroup of G. If [imath]H = {\{e\}}[/imath], then [imath]H \text{ = <e>}[/imath] is cyclic. If [imath]H \neq \space {\{e\}}[/imath], then [imath]a^n \in H[/imath] for some [imath]n \in \mathbb{Z}^{+}[/imath]. Let [imath]m[/imath] be the smallest integer in [imath]\mathbb{Z}^{+}[/imath] such that [imath]a^m \in H[/imath]. We claim that [imath]c = a^m[/imath] generates H; that is, [imath]H \space = \space <a^m> \space = \space <c>.[/imath] We must show that every [imath]b \in H[/imath] is a power of c. Since [imath]b \in H[/imath] and [imath]H \leq G[/imath], we have [imath]b = a^n[/imath] for some [imath]n[/imath]. Find [imath]q[/imath] and [imath]r[/imath] such that [imath] n = mq + r \space \space \space \space for \space \space \space 0 \leq r < m,[/imath] Alright this is the first part in the proof where I start to get confused. Where does the division algorithm come from and why are we using it? The proof continues as follows: [imath]a^n = a^{mq + r} = (a^m)^q \cdot a^r,[/imath] so [imath] a^r = (a^m)^{-q} \cdot a^n.[/imath] Now since [imath]a^n \in H[/imath], [imath]a^m \in H[/imath], and [imath]H[/imath] is a group, both [imath](a^m)^{-q}[/imath] and [imath]a^n[/imath] are in [imath]H[/imath]. Thus [imath](a^m)^{-q} \cdot a^n \in H; \space \space \space \text{that is,} \space \space a^r \in H.[/imath] This is another point at which I’m a little confused. What exactly about [imath]a^n[/imath] and [imath]a^m[/imath] being elements of [imath]H[/imath] allows us to accept that [imath](a^m)^{-q}[/imath] and [imath]a^{n}[/imath] are in [imath]H[/imath]? The proof continues: Since [imath]m[/imath] was the smallest positive integer such that [imath]a^m \in H[/imath] and [imath]0[/imath] [imath]\leq r[/imath] [imath]< m[/imath], we must have [imath]r = 0[/imath]. Thus [imath]n = qm[/imath] and [imath]b \space = \space a^n \space = \space (a^m)^q \space = \space c^q,[/imath] so b is a power of c. This final step is confusing as well, but I think its just because of the previous parts I was confused about. Any help in understanding this proof would be greatly appreciated |
2796093 | Find the norm of the operator [imath]A: \ell^p \to \ell^p[/imath], [imath]A(x_{n})_{n\geq 1} = (\frac{1}{m}\sum_{n=1}^{m}\frac{x_{n}}{\sqrt{n}})_{m \geq 1}[/imath]
Note: this question has been asked before, here: How to find the norm of the operator [imath](Ax)_n = \frac{1}{n} \sum_{k=1}^n \frac{x_k}{\sqrt{k}}[/imath]?, but it wasn't definitively answered, so I'd like to post my thoughts on the problem, and ask for a solution if anyone has any ideas. So, the question is: For which [imath]p \geq 1[/imath] is [imath]A: \ell^p \to \ell^p[/imath], [imath] A(x_{n})_{n \geq 1} = \big( \frac{1}{m} \sum_{n=1}^{m}\frac{x_{n}}{\sqrt{n}} \big)_{m \geq 1} [/imath] a bounded linear operator? For those [imath]p[/imath] for which [imath]A[/imath] is bounded, find [imath]||A||[/imath], where [imath]||\cdot||[/imath] is the operator norm. Now, I had two ideas: use convexity of [imath]x \mapsto x^p[/imath], or use Hölder's inequality. However, each approach is far less sophisticated than Norbert's answer in the linked question, because I didn't know about Hardy's inequality before reading the question. The convexity approach got me nowhere, while the Hölder approach got me the following upper bound: [imath]||A(x_{n})_{n \geq 1}||_{p} \leq \sqrt[p]{\sum_{m=1}^{\infty} \frac{S_{m}}{m^p}} \cdot ||(x_{n})_{n \geq 1}||_{p},[/imath] where [imath]S_{m} = (\sum_{n=1}^{m} \frac{1}{n^{\frac{q}{2}}})^{\frac{p}{q}}[/imath]. The series converges because [imath]\sum_{n=1}^{m} \frac{1}{n^{q/2}} \sim m^{1-q/2}[/imath], and so the summands behave like [imath]m^{-p+\frac{p}{q}-\frac{p}{2}}[/imath], and [imath]\frac{p}{q}-\frac{p}{2}-p = p-1-\frac{p}{2}-p = -1-\frac{p}{2} <-1[/imath]. | 408017 | Prove that [imath]A[/imath] is bounded operator on [imath]\ell^p[/imath] and find [imath]\| A\|[/imath]
For which one [imath]p \ge 1[/imath] is with [imath]A(x_n)_{n=1}^{\infty}=\left(\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right)_{m=1}^{\infty}[/imath] defined bounded linear operator [imath]A:\ell^p \to \ell^p[/imath]? Find norm [imath]\| A\|.[/imath] First, I have to show that if [imath]x = (x_n)_{n=1}^{\infty} \in \ell^p[/imath] then [imath]\displaystyle y=(y_m)_{m=1}^{\infty} \stackrel{\text{def}}{=}\left(\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right)_{m=1}^{\infty} \in \ell^p[/imath], that is [imath]\displaystyle\sqrt[p]{\sum_{m=1}^{\infty}|y_m|^p} < +\infty.[/imath] All right, [imath]\begin{align*} |y_m|=\left|\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right|&\leq\frac{1}{m} \sum_{n=1}^{m} \frac{|x_n|}{\sqrt{n}}\\ &\stackrel{\star}{\leq}\frac{1}{m} \sqrt[p]{\sum_{n=1}^{m}|x_n|^p} \sqrt[q]{\sum_{n=1}^{m} n^{-q/2}} \\ &\leq \frac{\| x \|_p}{m}\sqrt[q]{\sum_{n=1}^{m} n^{-q/2}}.\end{align*}[/imath] So [imath]\begin{align*} \sum_{m=1}^{\infty} |y_m|^p &\leq \sum_{m=1}^{\infty} \left|\frac{\| x \|_p}{m}\sqrt[q]{\sum_{n=1}^{m} n^{-q/2}}\right|^p\\&=\sum_{m=1}^{\infty} \frac{1}{m^p} \| x\|^p_p \left(\sum_{n=1}^{m} n^{-q/2}\right)^{p/q}\\ &\leq \|x\|^p_p \sum_{m=1}^{\infty} \frac{1}{m^p} \left(\sum_{n=1}^{\infty} n^{-q/2}\right)^{p/q} \end{align*}[/imath] If I see well, last sum is convergent for [imath]q/2 >1 \iff q>2[/imath] and [imath]p>1[/imath] (but with condition [imath]1/p + 1/q =1[/imath]). I think my estimate is too sharp. And we find that [imath]\| A \| \leq \sqrt[p]{\sum_{m=1}^{\infty} \frac{1}{m^p} \left(\sum_{n=1}^{\infty} n^{-q/2}\right)^{p/q}}[/imath] But for part (if my approximation is good enough) with [imath]\geq[/imath] I don't have idea (to be honest, I used Hölder inequality, so we want [imath]x_n = c \frac{1}{\sqrt{n}}[/imath], that will give us something divergent I think so). [imath]\star[/imath] - one doubt also: I used here Hölder inequality, but that works only for [imath]p,q >1[/imath], but in my question we have [imath]p \geq 1[/imath], so basically I have to work case [imath]p=1[/imath] as separate? |
2795729 | A weighted geometric inequality
[imath]x,y,z[/imath] are positive reals. for a triangle of side lengths [imath] a, b, c[/imath] , and area [imath]A[/imath], the following inequality holds: [imath](xa^2+yb^2+zc^2)^2\ge 16(xy+yz+zx)A^2[/imath] This is like a weighted Weitzenböck's inequality. I hope to get some tips to prove this question | 2072087 | Oppenheim's Inequality for triangles, American Mathematical Monthly problems
I did a proof for the inequality below, and I would like know if anyone also has a trigonometric proof for this inequality.If you have a trigonometric demonstration, please post your solution. This problem appeared in the American Mathematical Monthly magazine in 1965, the inequality was proposed in that form by Sir Alexander Oppenheim: Let [imath]x,y,z[/imath] positive real numbers and [imath]\Delta ABC[/imath] a triangle. [imath]\displaystyle [ABC][/imath] denotes the triangle area and [imath]\displaystyle a,b,c[/imath] the sides of the triangle. The inequality below is true: [imath]a^2x+b^2y+c^2z\geq 4[ABC]\sqrt{xy+xz+yz}[/imath] Various inequalities can be deduced through this inequality, for example, Weitzenböck's inequality, Neuberg-Pedoe inequality, Hadwiger-Finsler inequality, and so on. I'll post my solution right below. [imath]Proof[/imath] Let [imath]\alpha,\beta,\gamma[/imath] denote the opposite angles to the sides [imath]a, b, c[/imath], respectively. [imath]R[/imath] is the circumradius of [imath]\Delta ABC[/imath]. Observe that: [imath]a^2x+b^2y+c^2z\geq 4[ABC]\sqrt{xy+xz+yz}[/imath] [imath]a^2x+b^2y+c^2z\geq \frac{abc}{R}\sqrt{xy+xz+yz}[/imath] [imath]\frac{aRx}{bc}+\frac{bRy}{ac}+\frac{cRz}{ab}\geq \sqrt{xy+xz+yz}[/imath] [imath]\frac{1}{2}\left(\frac{4aR^2x}{2Rbc}+\frac{4bR^2y}{2Rac}+\frac{4cR^2z}{2Rab}\right)\geq \sqrt{xy+xz+yz}[/imath] [imath]x\frac{\sin\alpha }{\sin\beta \sin \gamma}+y\frac{\sin\beta }{\sin\alpha \sin \gamma}+z\frac{\sin\gamma }{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}[/imath] [imath]x\frac{\sin(\pi-\alpha) }{\sin\beta \sin \gamma}+y\frac{\sin(\pi-\beta )}{\sin\alpha \sin \gamma}+z\frac{\sin(\pi-\gamma )}{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}[/imath] [imath]x\frac{\sin(\alpha+\beta+\gamma-\alpha) }{\sin\beta \sin \gamma}+y\frac{\sin(\alpha+\beta+\gamma-\beta )}{\sin\alpha \sin \gamma}+z\frac{\sin(\alpha+\beta+\gamma-\gamma )}{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}[/imath] [imath]x\frac{\sin(\beta+\gamma) }{\sin\beta \sin \gamma}+y\frac{\sin(\alpha+\gamma )}{\sin\alpha \sin \gamma}+z\frac{\sin(\alpha+\beta )}{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}[/imath] [imath]x\frac{(\sin\beta \cos\gamma+\sin\gamma \cos \beta) }{\sin\beta \sin \gamma}+y\frac{(\sin\alpha \cos\gamma+\sin\gamma \cos \alpha) }{\sin\alpha \sin \gamma}+z\frac{(\sin\alpha \cos\beta+\sin\beta \cos \alpha) }{\sin\alpha \sin \beta}\geq 2\sqrt{xy+xz+yz}[/imath] \begin{equation} (\cot\beta+\cot\gamma)x+(\cot\alpha+\cot\gamma)y+(\cot\alpha+\cot\beta)z\geq 2\sqrt{xy+xz+yz} \tag{1} \end{equation} Since inequality is homogeneous in the variables [imath]x,y,z[/imath], do it [imath]\displaystyle xy+xz+yz=1[/imath] and take the substitution [imath]\displaystyle x=\cot\alpha',y=\cot\beta',z=\cot\gamma'[/imath], we have que [imath]\displaystyle \alpha',\beta',\gamma'[/imath] are angles of a triangle, and our inequality will be equivalent to the inequality below: \begin{equation} (\cot\beta+\cot\gamma)\cot\alpha'+(\cot\alpha+\cot\gamma)\cot\beta'+(\cot\alpha+\cot\beta)\cot\gamma'\geq 2 \tag{2} \end{equation} Suppose without loss of generality that (the reverse case is analogous) : \begin{equation} \cot\alpha \geq \cot \alpha' \tag{3} \end{equation} \begin{equation} \cot\beta \geq \cot \beta' \tag{4} \end{equation} \begin{equation} \cot\gamma'\geq \cot \gamma \tag{5} \end{equation} Because these variables are angles of a triangle, we can not have [imath]\cot \alpha \geq \cot \alpha' , \cot\beta \geq \cot \beta', \cot \gamma\geq \cot\gamma'[/imath].In fact, this can not occur, since it supposes without loss of generality that [imath]\displaystyle \alpha'\geq\alpha[/imath] and [imath]\displaystyle \beta'\geq\beta[/imath](as the cotangent is decreasing, this implies that [imath]\displaystyle \cot\alpha \geq \cot \alpha' [/imath] and [imath]\displaystyle \cot\beta \geq \cot \beta'[/imath]), summing up these first two inequalities we have: [imath]\\ \\ \displaystyle \alpha'+\beta'\geq \alpha+\beta \Rightarrow \cot(\alpha+\beta)\geq \cot(\alpha'+\beta')\Rightarrow -\cot(\pi-\alpha+\beta)\geq- \cot(\pi-\alpha'+\beta') \Rightarrow -\cot(\alpha+\beta+\gamma-(\alpha+\beta))\geq- \cot(\alpha'+\beta'+\gamma'-(\alpha'+\beta')) \Rightarrow -\cot(\gamma)\geq- \cot(\gamma') \Rightarrow \cot(\gamma')\geq \cot(\gamma)\\ \\[/imath] Now set the [imath]\displaystyle f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma'):\mathbb{R}^6\rightarrow \mathbb{R}[/imath] and [imath]\displaystyle f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma'):\mathbb{R}^6\rightarrow \mathbb{R}[/imath] such that: \begin{equation*} f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')= \end{equation*} \begin{equation} (\cot\beta+\cot\gamma)(\cot\alpha'-\cot\alpha)+(\cot\alpha+\cot\gamma)(\cot\beta'-\cot\beta)+(\cot\alpha+\cot\beta)(\cot\gamma'-\cot\gamma) \tag{6} \end{equation} \begin{equation*} f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma')= \end{equation*} \begin{equation} (\cot\beta'+\cot\gamma')(\cot\alpha-\cot\alpha')+(\cot\alpha'+\cot\gamma')(\cot\beta-\cot\beta')+(\cot\alpha'+\cot\beta')(\cot\gamma-\cot\gamma') \tag{7} \end{equation} Note now that by inequalities (3), (4) and (5) it follows that: \begin{equation} 0 \geq \cot\alpha'-\cot\alpha \tag{8} \end{equation} \begin{equation} 0 \geq \cot \beta' -\cot\beta \tag{9} \end{equation} \begin{equation} \cot\gamma'-\cot\gamma \geq 0 \tag{10} \end{equation} We know that [imath]\displaystyle \alpha',\beta',\gamma'[/imath] are angles of a triangle, so there exists [imath]\displaystyle a',b',c'[/imath] such that [imath]\displaystyle a'^2=b'^2+c'^2-2b'c'\cos\alpha',b'^2=a'^2+c'^2-2a'c'\cos\beta',c'^2=a'^2+b'^2-2a'b'\cos\gamma'[/imath].Let [imath]\displaystyle R'[/imath] the circumradius of the triangle of sides [imath]\displaystyle a',b',c'[/imath]. Let [imath]\displaystyle k_{\alpha',\beta',\gamma'}:=\frac{a'}{b'c'}+\frac{b'}{a'c'}+\frac{c'}{a'b'}[/imath], therefore: \begin{equation} \frac{a'}{b'c'}+\frac{b'}{a'c'}+\frac{c'}{a'b'}=k_{\alpha',\beta',\gamma'} \tag{11} \end{equation} Where [imath]\displaystyle k_{\alpha',\beta',\gamma'}[/imath] is a real variable of any kind.And since our original inequality is homogeneous in the variables a, b, c, suppose without loss of generality that the equality below occurs: \begin{equation} \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=k_{\alpha',\beta',\gamma'} \tag{12} \end{equation} For each real value of fixed [imath]\displaystyle k_{\alpha',\beta',\gamma'}[/imath].Since x, y, z do not depend of the circumradius R ', suppose that [imath]\displaystyle R'\geq R[/imath].Take the inequality (3) and consider the development (applying the law of cosines and law of sines): [imath]\\ \displaystyle \cot\alpha \geq \cot\alpha' \Rightarrow \frac{(b^2+c^2-a^2)R}{abc} \geq \frac{(b'^2+c'^2-a'^2)R'}{a'b'c'} \Rightarrow \left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}-2\frac{a}{bc}\right)R\geq \left(\frac{a'}{b'c'}+\frac{b'}{a'c'}+\frac{c'}{a'b'}-2\frac{a'}{b'c'}\right)R' \Rightarrow Rk_{\alpha',\beta',\gamma'}-2\frac{aR}{bc}\geq Rk_{\alpha',\beta',\gamma'}-2\frac{a'R'}{b'c'} \Rightarrow \frac{a'R'}{b'c'}\geq \frac{aR}{bc} \Rightarrow [/imath] \begin{equation} \frac{a'R'}{b'c'}\geq \frac{aR}{bc} \tag{13} \end{equation} Applying the same rationale for inequality (4), we conclude: \begin{equation} \frac{b'R'}{a'c'}\geq \frac{bR}{ac} \tag{14} \end{equation} Suppose by contradiction that it occurs: \begin{equation} \cot\alpha+\cot\gamma> \cot\alpha'+\cot\gamma' \tag{15} \end{equation} See that: [imath]\\ \displaystyle \cot\alpha+\cot\gamma> \cot\alpha'+\cot\gamma' \Rightarrow \frac{(b^2+c^2-a^2)R}{abc}+\frac{(a^2+b^2-c^2)R}{abc}> \frac{(b'^2+c'^2-a'^2)R'}{a'b'c'}+\frac{(a'^2+b'^2-c'^2)R'}{a'b'c'} \Rightarrow \frac{bR}{ac}>\frac{b'R'}{a'c'} \\[/imath] This contradicts the inequality (14). On the other hand, suppose by contradiction that it occurs: \begin{equation} \cot\beta+\cot\gamma> \cot\beta'+\cot\gamma' \tag{16} \end{equation} See that: [imath]\\ \displaystyle \cot\beta+\cot\gamma> \cot\beta'+\cot\gamma' \Rightarrow \frac{(a^2+c^2-b^2)R}{abc}+\frac{(a^2+b^2-c^2)R}{abc}> \frac{(a'^2+c'^2-b'^2)R'}{a'b'c'}+\frac{(a'^2+b'^2-c'^2)R'}{a'b'c'} \Rightarrow \frac{aR}{bc}>\frac{a'R'}{b'c'} \\[/imath] This contradicts the inequality (13).Therefore: \begin{equation} \cot\alpha+\cot\gamma \leq \cot\alpha'+\cot\gamma' \tag{17} \end{equation} \begin{equation} \cot\beta+\cot\gamma \leq \cot\beta'+\cot\gamma' \tag{18} \end{equation} Multiplying (17) by [imath]\displaystyle \cot\beta'-\cot\beta[/imath] and multiplying (18) by [imath]\displaystyle \cot\alpha'-\cot\alpha[/imath], note that these inequalities will reverse, since we are multiplying by non-positive quantities, we will have, respectively: \begin{equation} (\cot\alpha+\cot\gamma) (\cot\beta'-\cot\beta)\geq (\cot\alpha'+\cot\gamma')(\cot\beta'-\cot\beta) \tag{19} \end{equation} \begin{equation} (\cot\beta+\cot\gamma) (\cot\alpha'-\cot\alpha)\geq (\cot\beta'+\cot\gamma')(\cot\alpha'-\cot\alpha) \tag{20} \end{equation} On the other hand of inequalities (3) and (4) we know that: \begin{equation} \cot\alpha+\cot\beta \geq \cot\alpha'+\cot\beta' \tag{21} \end{equation} Multiplying the above inequality by [imath]\displaystyle \cot\gamma'-\cot\gamma[/imath], that by the inequality (10) we know to be greater than or equal to zero, we will have: \begin{equation} (\cot\alpha+\cot\beta) (\cot\gamma'-\cot\gamma)\geq (\cot\alpha'+\cot\beta')(\cot\gamma'-\cot\gamma) \tag{22} \end{equation} Adding (19), (20) and (22), we will have: \begin{equation*} f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')\geq \end{equation*} \begin{equation} (\cot\alpha'+\cot\gamma')(\cot\beta'-\cot\beta)+(\cot\beta'+\cot\gamma')(\cot\alpha'-\cot\alpha)+(\cot\alpha'+\cot\beta')(\cot\gamma'-\cot\gamma) \tag{23} \end{equation} Adding the LHS of (23) with the LHS of (7) and the RHS of (23) with the RHS of (7), the terms will cancel and we will have: \begin{equation} f_1(\alpha,\beta,\gamma,\alpha',\beta',\gamma')+f_2(\alpha,\beta,\gamma,\alpha',\beta',\gamma')\geq 0 \end{equation} And this implies, finally, that: \begin{equation} (\cot\beta+\cot\gamma)\cot\alpha'+(\cot\alpha+\cot\gamma)\cot\beta'+(\cot\alpha+\cot\beta)\cot\gamma'\geq 2 \end{equation} That is precisely the inequality (2), which is equivalent to the desired inequality.Thus, the inequality yields. |
2796695 | Can we say that [imath]\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} [/imath]
If [imath]a,b,c,d[/imath] are positive real numbers, can we say that [imath]\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} [/imath] is always true? If no, can you please give insignts on under which conditions this might be true. Any references to a similar type of inequalities are also welcome Thank you, | 506152 | Is [imath]\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}?[/imath], for [imath]a,b,c,d>0[/imath]?
Is [imath]\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}[/imath] for [imath]a,b,c,d>0[/imath] If it is true, then can we generalize? EDIT:typing mistake corrected. EDIT, WILL JAGY. Apparently the real question is Is [imath]\color{magenta}{\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}}[/imath] for [imath]a,b,c,d>0,[/imath] where letters on the left hand side and in the numerator stay in the numerator on the right-hand side, and letters on the left hand side and in the denominator stay in the denominator on the right-hand side. |
2796946 | [imath]g[/imath] is monotonic strictly increasing, continuous in [imath] [0,\infty)[/imath] and positive function.
[imath]\lim\limits_{n \to \infty} g(t)=L [/imath] [imath]g[/imath] is monotonic strictly increasing, continuous in [imath] [0,\infty)[/imath] and positive function. I need to find : [imath]\lim\limits_{n \to \infty} \int^2_1 g (nx) \,\mathrm{d}x [/imath] I have started with: [imath]|g(t)-L|<\epsilon\>[/imath] $L-\epsilon\>[imath]=<[/imath]g(t)[imath]<=[/imath]L+\epsilon\>$ $\int^2_1 (L-\epsilon\>) \,\mathrm{d}x[imath]=<[/imath]\int^2_1 g (nx) \,\mathrm{d}x[imath]<=[/imath]\int^2_1 (L+\epsilon\>) \,\mathrm{d}x$ I'm not sure if I'm in the right direction ... | 2796739 | Given function [imath]g[/imath] continuous in line [imath][0,\infty)[/imath]...
Given function [imath]g[/imath] continuous in line [imath][0,\infty)[/imath], positive and strongly increasing that upholds [imath]\lim_{t\to\infty} g(t) = L[/imath]. Calculate the limit: [imath] \lim_{n\to\infty} \int\limits_1^2 g(nx) \, dx. [/imath] At the moment what I think that I should do is swap the [imath]g(nx)[/imath] with [imath]L[/imath] as it approaches infinity but from there on I am just not sure whats my next step is. |
2154958 | Using the Fitch system, how do I prove [imath]((p \implies q) \implies p)\implies p[/imath]?
Using the Fitch system, how do I prove [imath]((p \implies q) \implies p)\implies p[/imath]? I started with the hypotheses [imath](p \implies q) \implies p[/imath] and [imath]\sim p[/imath]. However, from these hypotheses I did not get the desired contradiction to solve the question. Thanks sb45 | 2942822 | Fitch proof of Peirce's Law
This was a question on my exam today that I couldn't figure out. Show [imath]((P \supset Q) \supset P) \supset P[/imath] with no premises using Fitch. The first attempt is I assumed [imath](P \supset Q) \supset P[/imath]. Then I assume [imath]\neg P[/imath] and attempted to derive a contradiction. However, I couldn't get anywhere productive. Any hints or direction? |
2769292 | If [imath]X[/imath] and [imath]Y[/imath] are both independent from [imath]\mathscr {F}[/imath], is [imath]f(X,Y)[/imath] also independent from [imath]\mathscr{F}[/imath]?
Let [imath]X[/imath] and [imath]Y[/imath] are random variables and they are both independent from the [imath]\sigma[/imath]-algebra [imath]\mathscr F[/imath]. Given a Borel measurable function [imath]f(x,y)[/imath], is [imath]f(X,Y)[/imath] also independent from [imath]\mathscr F[/imath]? I guess it is right, but actually not quite sure. Here is what I do: Let [imath]A[/imath] be a Borel set, then [imath]B = \{ f(X,Y) \in A \}[/imath] is also a Borel set. I consider the case where [imath]B = B_1 \times B_2[/imath] is a rectangular area. Then given [imath]E \in \mathscr{F}[/imath], [imath]P(\{f(X,Y)\in A \}\cap E) = P((\{ X \in B_1\} \cap E) \cap (\{Y \in B_2\} \cap E))[/imath]. It seems I cannot proceed any more. And I even get skeptical whether it is right. Any help is appreciated! | 2792286 | [imath]X_1,...,X_n[/imath] be independent RVs and [imath]X_i \perp \mathcal F [/imath] for [imath]1\ \leq \forall i \leq n[/imath]. Show that [imath]\sigma (X_1,...,X_n) \perp \mathcal F[/imath]
Let [imath]X_1,...,X_n[/imath] be mutually independent RVs. Suppose [imath]X_i \perp \mathcal F [/imath] for [imath]1\ \leq \forall i \leq n[/imath]. How can I show that: [imath]\sigma (X_1,...,X_n) \perp \mathcal F[/imath] ? What I have tried: [imath]\sigma (X_1...X_n) = \sigma(\cup_{i=1} ^n \sigma(X_i))[/imath] holds. I think I need some Dynkin's lemma-ish argument but can't see how. |
2789737 | Independent Events Conceptual Meaning - probability theory
When I check for definition of independent events I get the following definitions Type 1 (Reference ) When two events are said to be independent of each other, what this means is that the probability that one event occurs in no way affects the probability of the other event occurring. Type 2 [imath]P(A \cap B) = P(A)P(B)[/imath] Then it follows some useful information such as ( Reference page 14) a) Do not say that two events are independent if one has no influence on the other b) No circumstances say that A and B are independent if [imath] A \cap B = \phi [/imath] (this is the statement that A and B are disjoint, which is quite a different thing!) b) Also, do not ever say that P(A ∩ B) = P(A) · P(B) unless you have some good reason for assuming that A and B are independent (either because this is given in the question, or as in the next-but-one paragraph). Example Consider example where events are not independent My conclusion is that [imath]P(A/B) \ne P(A)[/imath] and [imath]P(B/A) \ne P(B)[/imath] . It says given precondition of event B has an effect on P(A) ([imath]P(A/B) \ne P(A)[/imath]) and given precondition of event A has an effect on P(B) ([imath]P(B/A) \ne P(B)[/imath]) . So these are dependent events My conclusion is that [imath]P(A/C) = P(A)[/imath] and [imath]P(C/A) = P(C)[/imath] . It says given the precondition of an event C has no effect on P(A) ([imath]P(A/C) = P(A)[/imath]) and an event A has no effect on P(C) ([imath]P(C/A) = P(C)[/imath]). So these independent events Doubt 1) Mathematically we can say whether two events are independent by formulas such as [imath]P(A/C) = P(A)[/imath] and [imath]P(C/A) = P(C)[/imath] or [imath]P(A\cap C) = P(A)P(C)[/imath] . But is it possible to show meaning using Venn diagrams as we show [imath]P(A \cup B )[/imath] etc.? Can we make some statements about the participating event sets if know prior two events are independent. I am looking for getting a conceptual meaning derived from the formula with visual identification from Venn diagram if possible 2) But if you check the definitions I got from two sources as mentioned above as type 1 and type 2, type 1 says When two events are said to be independent of each other, what this means is that the probability that one event occurs in no way affects the probability of the other event occurring. and type 2 says Do not say that two events are independent if one has no influence on the other.No circumstances say that A and B are independent if [imath] A \cap B = \phi [/imath] (this is the statement that A and B are disjoint, which is quite a different thing!) What is the difference between influence of events on other and affecting other event as mentioned above? What is meaning of two independent events which never affects each other but influence each other? I am looking for a precise conceptual definition or identification method for independent events | 1898596 | Can one infer independence by simple reasoning/intuition?
From my recent experience in probability, it feels as though independence is something we "discover" from the system via the equation: [imath]P(A)*P(B)=P(A\cap B)[/imath] Could one ever conclude independence from the "system" by intuition? Is it wise to conclude independence for events that are "seemingly" independent? What would be some interesting examples where this would fail. |
2797592 | Prove that [imath]^nC_r = {^{n - 1}C_r} + {^{n - 1}C_{r - 1}}[/imath]
I see that there are a few ways to prove this, but I don't know why my method isn't working. I set [imath]{^{n - 1}C_r} = \dfrac{(n - 1)\cdot(n-2)\cdot\cdots\cdot(n - 1 - r + 1)}{r!}[/imath] so the last term in the numerator is [imath]n-r[/imath]. Call the numerator [imath]N_1[/imath]. Then [imath]{^{n - 1}C_{r - 1}} = \dfrac{(n - 1)\cdot(n - 2)\cdot\cdots\cdot(n - 1 - (r - 1) + 1)}{(r - 1)!}[/imath] so the last term in the numerator is [imath]n-r+1[/imath]. Call this numerator [imath]N_2[/imath]. Then the sum is [imath]\dfrac{N_1 + r\cdot N_2}{r!}[/imath]. I don't know if I made a mistake or I'm failing to see how this equals [imath]^nC_r[/imath]. | 807763 | Binomial coefficient proof
Prove that for any [imath]0\lt r\lt n[/imath] we have [imath]\binom nr=\binom{n-1}{r-1}+\binom{n-1}r.[/imath] How do prove this and what step do i take in order for it to be true? |
2797711 | Problem understanding a question concerning the following function: [imath]z=a+bi \mapsto f(z)=a+i(a+b)[/imath]
I'm preparing for an exam, and I have come over the following problem: I have the function f, where C is a complex number [imath]f: C \to C,[/imath] [imath]z=a+bi \mapsto f(z)=a+i(a+b)[/imath] Is the function of the form [imath]f(z) = z·w[/imath], where [imath]w[/imath] does not depend on [imath]z[/imath]? I have written [imath]w=c+di[/imath], which gives [imath]z·w=(ac-bd)+ (bc+ad)i[/imath]. This gives us two equations, namely: [imath]ac-bd=a[/imath] and [imath]bc+ad=a+b[/imath] I tried solving this system of linear equations for [imath]c[/imath] and [imath]d[/imath], and ended up with [imath]c=d=1[/imath]. Seeing as this does not hold for any z, I concluded with the answer 'no', to the question asked in this task. My teacher, however, concludes with the opposite, using the following logic: In his solution, my teacher ends up with the same system of linear equations as I did in my post. From there he says that the system has a unique solution if and only if the determinant of the matrix [imath] \begin{matrix}a&-b\\b&a\end{matrix}[/imath] is non-zero. That is [imath]a^2+b^2 \neq 0[/imath].Then he says: But if [imath]a^2+b^2=0 \Rightarrow a=b=0 \Rightarrow f(z) = 0 = 0 · w[/imath] Hence the answer to the question must be 'yes'. If anyone could help me understand what is meant, I would greatly appreciate it. Please let me know if something was unclear, or if I should explain more clearly what I have done. Thank you in advance! | 2796883 | Question regarding the function [imath]z=a+bi \mapsto f(z)=a+i(a+b)[/imath].
I'm preparing for an exam, and I have come over the following problem: I have the function [imath]f[/imath], where C is a set of complex numbers [imath]f: C \to C,[/imath] [imath]z=a+bi \mapsto f(z)=a+i(a+b)[/imath] Is the function of the form [imath]f(z) = z + w[/imath], where [imath]w[/imath] does not depend on [imath]z[/imath]? Is it of the form [imath]z\cdot w[/imath] (where [imath]w[/imath] does not depend on [imath]z[/imath])? For the first part, my answer is 'no', seeing as [imath]w=ai[/imath] for the identity to hold, and it is then clear that [imath]w[/imath] depends on [imath]z[/imath]. My teacher's solution says the same thing, but that the answer to the question is 'yes'. Am i wrong, or is my teacher wrong? For the second part of the exercise: I have written [imath]w=c+di[/imath], which gives [imath]z·w=(ac-bd)+ (bc+ad)i[/imath]. This gives us two equations, namely: [imath]ac-bd=a[/imath] and [imath]bc+ad=a+b[/imath] I tried solving this system of linear equations for [imath]c[/imath] and [imath]d[/imath], and ended up with [imath]c=d=1[/imath]. However this only holds if [imath]b=0[/imath]. I am a bit stuck, and can't really see clearly how to proceed, or how to come to a general conclusion. Please let me know if something was unclear, or if I should explain more clearly what I have done. Thank you in advance! EDIT (included my teacher's answer, which I do not understand): In his solution, my teacher ends up with the same system of linear equations as I did in my post. From there he says that the system has a unique solution if and only if the determinant of the matrix [imath] \begin{matrix}a&-b\\a&b\end{matrix}[/imath] is non-zero. That is [imath]a^2+b^2 \neq 0[/imath].Then he says: But if [imath]a^2+b^2=0 \Rightarrow a=b=0 \Rightarrow f(z) = 0 = 0 · w[/imath] Hence the answer to the question must be 'yes'. If anyone could help me understand what is meant, I would greatly appreciate it. NB: There is a possibility that my teacher has read the task wrong, and therefore answers 'yes' when the answer to both questions is 'no'. |
2797046 | In [imath]\Bbb R^3[/imath], If tangent map of F preserves inner product, then F is an isometry?
I am differential geometry beginner. I have one question. Following statement is well known. If [imath]F[/imath] : [imath]\Bbb R^m\rightarrow\Bbb R^n[/imath] is a map & diffeomorphism. And if [imath]F_*[/imath](tangent map of [imath]F[/imath]) preserves inner product, then [imath]F[/imath] is an isometry. But In [imath]m=n=3[/imath], [imath]F[/imath] is still Isometry even if we subtract the condition of '[imath]F[/imath] is a diffeomorphism' I tried several ways to prove this, but it was useless. I really want to know this elementary proof. Thanks. | 1238376 | A diffeomorphism whose tangent map preserves dot products is an isometry.
I'm having trouble solving the following problem. If [imath]F:\mathbb{R^3} \to \mathbb{R^3}[/imath] is a diffeomorphism such that [imath]F\ast[/imath](the tangent map of [imath]F[/imath]) preserves dot products, show that [imath]F[/imath] is an isometry. (Hint: Show that [imath]F[/imath] preserves lengths of curve segments and deduce that [imath]F^{-1}[/imath] does also.) I'm trying to proceed as in the hint, but I'm stuck on the first step, showing the [imath]F[/imath] preserves lengths of curve segments. Moreover, how does showing that both [imath]F[/imath] and its inverse preserves length of curve segments show that it is an isometry. And how do I characterize the curve segments. I would appreciate any solutions, hints or suggestions. |
2797819 | Given [imath]\sum a_n [/imath] convergent, are [imath]\sum n^{1/n} a_n[/imath] and [imath]\sum a_n/({1+ |a_n|})[/imath] convergent?
Let [imath]\sum a_n [/imath] be a convergent series. Is [imath]\sum n^{1/n} a_n[/imath] convergent? [imath]\sum a_n/({1+ |a_n|})[/imath] convergent? If not, provide examples in the contrary. I feel like 1 is maybe convergent, however I cannot seem to prove it or come up with an example to contradict. Please help. | 1025896 | Convergent or divergent series ? Given that [imath]\sum a_n [/imath] is already convergent.
Could anyone please give me some hints ? Let [imath]\sum a_n[/imath] be a convergent serie of real number. Prove or disprove that [imath]\sum a_n \sin n[/imath] and [imath]\sum n^{\frac{1}{n}} a_n[/imath] are also convergent. |
2797948 | Taylor series of [imath]\sqrt{1+x}[/imath]
I have to give the Taylor series of [imath]\sqrt{1+x}[/imath] with development point [imath]0[/imath]. But I am not quite sure what to do. Normally you have to give the Taylor series up to a certain order. Well [imath]T_f(x,0)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n[/imath] I calculated the derivative of [imath]f(x)=\sqrt{1+x}[/imath] up to order 5 and locked for a regularity. [imath]f^{(n)}(0)=(-1)^n\cdot\frac{\prod_{k\leq n\,\text{with}\, k\,\text{odd}}k}{2^{n}}[/imath] Then the Taylor series is given by: [imath]T_{f}(x,0)=\sum_{n=0}^\infty \frac{(-1)^n}{\prod_{1\leq k\leq n\,\text{with}\, k\,\text{even}}k\cdot 2^n}x^n[/imath] Which is pretty ugly to work with. Can you confirm this solution? Thanks in advance. | 732540 | Taylor series of [imath]\sqrt{1+x}[/imath] using sigma notation
I want help in writing Taylor series of [imath]\sqrt{1+x}[/imath] using sigma notation I got till [imath]1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots[/imath] and so on. But I don't know what will come in sigma notation. |
2798132 | Minimum value of [imath]\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}[/imath]
If [imath]x,y,z>0[/imath] and [imath]x+y+z=60[/imath]. Then minimum value of [imath]\displaystyle \frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}[/imath] Try: [imath]\frac{(x+y)^2-2xy}{x+y}+\frac{(y+z)^2-2yz}{y+z}+\frac{(z+x)^2-2zx}{z+x}[/imath] So [imath]2(x+y+z)-2\bigg[\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x}\bigg][/imath] Now Using arithmetic geometric inequality [imath]x+y\geq 2\sqrt{xy}[/imath] and [imath]y+z\geq 2\sqrt{yz}[/imath] and [imath]z+x\geq 2\sqrt{zx}[/imath] Could some help me how to solve further, Thanks | 424150 | Prove that [imath]\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c[/imath]
If [imath]a,b,c[/imath] are positive , show that [imath]\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c[/imath] Trial: Here I proceed in this way [imath]\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\dfrac{2ca}{c+a}+\dfrac{2ab}{a+b}[/imath] then how I proceed. Please help. |
2798036 | How to understand the function [imath]o(h) + o(h) = o(h)[/imath]?
On page 150 of the book Introduction to Mathematical Statistics, there's the following passage: Let the symbol [imath]o(h)[/imath] represent any function such that [imath]lim_{h→0}[o(h)/h] = 0;[/imath] for example, [imath]h^2 = o(h)[/imath] and [imath]o(h) + o(h) = o(h)[/imath]. I can understand how [imath]h^2 = o(h)[/imath] fits the description. But I couldn't understand the function [imath]o(h) + o(h) = o(h)[/imath]. | 86076 | What are the rules for equals signs with big-O and little-o?
This question is about asymptotic notation in general. For simplicity I will use examples about big-O notation for function growth as [imath]n\to\infty[/imath] (seen in algorithmic complexity), but the issues that arise are the same for things like [imath]\Omega[/imath] and [imath]\Theta[/imath] growth as [imath]n\to\infty[/imath], or for little-o as [imath]x\to 0[/imath] (often seen in analysis), or any other combination. The interaction between big-O notation and equals signs can be confusing. We write things like [imath]\tag{1} 3n^2+4 = O(n^2)[/imath] [imath]\tag{2} 5n^2+7n = O(n^2)[/imath] But we're not allowed to conclude from these two statements that [imath]3n^2+4=5n^2+7n[/imath]. Thus it seems that transitivity of equality fails when big-O is involved. Also, we never write things such as [imath]\tag{3} O(n^2)=3n^2+4,[/imath] so apparently commutativity is also at risk. Many textbooks point this out and declare, with varying degrees of indignation, that notations such as [imath](1)[/imath] and [imath](2)[/imath] constitute an "abuse of notation" that students are just going to have to get used to. Very well, but then what are the rules that govern this abuse? Mathematicians seem to be able to communicate using it, so it can't be completely random. One simple way out is to define that the notation [imath]O(n^2)[/imath] properly denotes the set of functions that grow at most quadratically, and that equations like [imath](1)[/imath] are just conventional abbreviations for [imath]\tag{4} (n\mapsto 3n^2+4)\in O(n^2)[/imath] Some authors even insist that writing [imath](1)[/imath] is plain wrong, and that [imath](4)[/imath] is the only correct way to express the estimate. However, other authors blithely write things like [imath]\tag{5} 5 + O(n) + O(n^2)\log(O(n^3)) = O(n^2\log n)[/imath] which does not seem to be easily interpretable in terms of sets of functions. The question: How can we assign meaning to such statements in a principled way such that [imath](1)[/imath], [imath](2)[/imath] and [imath](4)[/imath] are true but [imath](3)[/imath] is not? |
2798105 | Proving stirling approximation result.
Finding value of [imath]\displaystyle \frac{e^n\cdot n!}{n^n\sqrt{n}}[/imath] Try: Using Stirling Approximation result [imath]n!\approx\bigg(\frac{n}{e}\bigg)^n\sqrt{2\pi n}[/imath] So [imath]\lim_{n\rightarrow \infty }\frac{e^n\cdot n!}{n^n\sqrt{n}}=\sqrt{2\pi}[/imath] Could anyone explain me How can i prove that result, Thanks | 1984702 | Stirling's Approximation Proof
I was searching for the reason why Stirling's Approximation holds true. I found the website Stirling's Approximation which apparently shows why this is the case. Does this part of the equation make sense in the proof? If so why? [imath]\int_{1}^{n}\ln(x)dx\approx\sum_{k=1}^{n}\ln(k)[/imath] |
2798718 | Convergence of [imath]1-\frac12-\frac13+\frac14+\frac15+\frac16-....[/imath]
How do i establish the convergence of the series [imath]1-\frac12-\frac13+\frac14+\frac15+\frac16-\frac17-\frac18-\frac19-\frac1 {10}+...[/imath] where the number of signs increases by 1 in each "block"? I cannot apply the Dirichlet test because the sequence of partial sums of [imath]1,-1,-1,1,1,1,...[/imath] is not bounded. Please help. | 336035 | Establish convergence of the series: [imath]1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...[/imath]
Establish convergence of the series: [imath]1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...[/imath] The number of signs increases by one in each "block". I have an idea. Group the series like this: [imath]1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...[/imath] We can show that [imath]1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...[/imath] converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing. Any idea? Or any other method to establish the convergence? |
2798541 | What is the base of [imath]\log x[/imath]?
I've seen "[imath]\log x[/imath]" being used in some papers (and by Wolfram|Alpha), and I was confused because so far I have only ever seen the [imath]\log[/imath] used with a base ( so e.g. [imath]\log_y x[/imath]). Am I correct that [imath]\log x = \log_e x = \ln x[/imath]? If so, why is [imath]\log x[/imath] used over [imath]\ln x[/imath]? Isn't the letter more expressive and less confusing? If not, what is the base of [imath]\log x[/imath]? | 293783 | When log is written without a base, is the equation normally referring to log base 10 or natural log?
For example, this question presents the equation [imath]\omega(n) < \frac{\log n}{\log \log n} + 1.4573 \frac{\log n}{(\log \log n)^{2}},[/imath] but I'm not entirely sure if this is referring to log base [imath]10[/imath] or the natural logarithm. |
2799123 | Prove [imath]1+2+2^2+\dots+2^{n-1}=2^n-1[/imath]
Prove the following equation by counting the non-empty subsets of [imath]\{1,2,\ldots,n\}[/imath] in [imath]2[/imath] different ways: [imath]1+2+2^2+2^3\ldots+2^{n-1}=2^n-1[/imath]. Let [imath]A=\{1,2\ldots,n\}[/imath]. I know from theory that it has [imath]2^n-1[/imath] non-empty subsets, which is the right-hand side of the equation but, how do count the left one? I've proven it using induction but how can i get to the first part of the equation by counting subsets differently? | 1703282 | How to prove with double counting technique that [imath]1+2+\dots+2^n=2^{n+1} -1[/imath]?
How to prove with double counting technique that [imath]1+2+\dots+2^n=2^{n+1} -1[/imath]? I can see, for example, that the right-hand side of the equation counts the cardinality of the powerset of a set with n+1 elements (excluding the empty set): could this be a good idea? Because I don't seem to be able to make the left-hand side fit this idea... |
2799903 | In [imath]\mathbb{R}^2[/imath], let [imath]V^2=\overline{B(0,1)}[/imath]. Prove [imath]V^2[/imath] is homeomorphic to [imath]I\times I[/imath] where [imath]I=[0,1][/imath]
In [imath]\mathbb{R}^2[/imath], let [imath]V^2=\overline{B(0,1)}[/imath]. Prove [imath]V^2[/imath] is homeomorphic to [imath]I\times I[/imath] where [imath]I=[0,1][/imath] My work: I know two spaces [imath]A,B[/imath] are homeomorphic if exists a function [imath]f:A\rightarrow B[/imath] homeomorphic. Definition: Let [imath]f:A\rightarrow B[/imath] a function. We say f is homeomorphic is [imath]f[/imath] is bijective, continuous and [imath]f^{-1}[/imath] is continuous. Then, We need find a function [imath]f[/imath] such that [imath]f:\overline{B(0,1)}\rightarrow[0,1]\times[0,1][/imath] bijective, continuous such that [imath]f^{-1}[/imath] is continuous. I'm stuck here, can someone help me? | 996453 | Homeomorphism from [imath][0,1]\times[0,1][/imath] to [imath]\overline{D}(0,1)[/imath]?
I'm trying to construct a homeomorphism from [imath][0,1]\times[0,1][/imath] to [imath]\overline{D}(0,1)[/imath]. I'm pretty sure there is one. I've been trying to work geometrically : mapping [imath][0,1]\times[0,1][/imath] to [imath][-1/2,1/2]\times[-1/2,1/2][/imath] (which is realatively easy) and then mapping the boundary of the latter to the boundary of [imath]\overline{D}(0,1)[/imath] and generalizing to the inside of both... I think it could work but I can't construct the map, can you help me? Thanks a lot! |
2799937 | In the connected component of [imath]\{z \in \mathbb{C} : |p(z)| \le 1\}[/imath], [imath]p[/imath] must vanish at least once.
If [imath]p[/imath] is a non-constant polynomial, and. [imath]G[/imath] is an open connected component of [imath]\{z \in \mathbb{C} : |p(z)| \le 1\}[/imath], then [imath]p[/imath] has at least one zero in [imath]G[/imath] My thoughts so far. Suppose that [imath]p[/imath] is nonzero in [imath]G[/imath]. By the minimum modulus principle, [imath]p(z)[/imath] then achieves its minimum (say [imath]m[/imath]) on the boundary of [imath]G[/imath]. From here, I either want to show that [imath]G[/imath] cannot be connected, or maybe that [imath]p[/imath] must be constant, and hence we have a contradiction? Any thoughts? | 426162 | Does every connected component of [imath]\{z : |P(z)|<1 \}[/imath] contain a zero?
Let [imath]P(z)[/imath] be a complex non-constant polynomial. Let [imath]G[/imath] be a connected component of open set [imath]\{z : |p(z)|<1 \}[/imath]. How to prove that [imath]G[/imath] contains a zero of [imath]P[/imath]? I have no idea how to even start; maybe I don't know the required theorem? Can anybody help? |
2800029 | Showing that [imath] \int_{10}^\infty \frac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx[/imath] diverge
Is it true that [imath] \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx[/imath] diverge since [imath] \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx > \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}}e^x }dx=\displaystyle \int_{10}^\infty \cfrac{1}{x^{1 + \frac{1}{x}} }dx[/imath] and [imath]\displaystyle \int_{10}^\infty \cfrac{1}{x^{1 + \frac{1}{x}} }dx[/imath] diverges by [imath]p[/imath]-test? | 2799545 | Convergence or divergence of the integral [imath]\int\limits_{10}^\infty \frac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)} \,dx[/imath]
I want to calculate [imath] \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx = \lim_{R \to \infty}\int_{10}^R \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)} \, dx [/imath]. I think that it will diverge by bounding it below with [imath]\frac{1}{x}[/imath]. So how do we calculate this integral? |
2800133 | Solution of a Riccati ODE
I have the following ODE: [imath]\frac{d y}{d x}=y^2-(A \csc^2 x+B^2 \sin^2 x-C)[/imath] where [imath]A,B,C[/imath] are constants. Can one come up with an appropriate integrating factor to make it exact? | 2800409 | Solution of a Riccati ODE
I have the following ODE: [imath]\frac{d y}{d x}=y^2-(A \csc^2 x+B^2 \sin^2 x-C)[/imath] where [imath]A,B,C[/imath] are constants. Can one come up with an appropriate integrating factor to make it exact? I have been trying various ways such as change of variable: [imath]y=z \sqrt{A \csc^2 x + B^2 \sin ^2 x-C}[/imath], but none of them seem to work. Edit: As pointed out in the comment, I am sorry to not have linked my duplicate question in MSE in the first place. I had already asked the same question in MSE, and as I was not getting any response I thought of asking here. Duplicate question in MSE: https://math.stackexchange.com/questions/2800133/solution-of-a-riccati-ode |
2799761 | Smith transformation
How to transfer the following matrix into Smith normal form? [imath]\left[\begin{matrix} 2 & -2b & 0 \\ 0 & 2 & -2c \\ -2a & 0 & 2 \end{matrix}\right][/imath] The final answer is [imath]\left[\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2|abc-1| \end{matrix}\right][/imath] But how we get it? | 133076 | Computing the Smith Normal Form
Let [imath]A_R[/imath] be the finitely generated abelian group, determined by the relation-matrix [imath]R := \begin{bmatrix} -6 & 111 & -36 & 6\\ 5 & -672 & 210 & 74\\ 0 & -255 & 81 & 24\\ -7 & 255 &-81 & -10 \end{bmatrix}[/imath] Reduce this matrix using Smith Normal Form and determine the isomorphism type of [imath]A_R[/imath]. I know that the Smith Normal Form of this matrix is: [imath]\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 21 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} [/imath] However, this was computed using Maple and I need to understand the method of computing this manually which I am struggling to grasp. Can anyone help? |
2800715 | Proving [imath](\cos x +i\sin x)^n= \cos nx+i\sin nx[/imath] by mathematical induction
Prove [imath](\cos x +i\sin x)^n= \cos nx+i\sin nx[/imath] by mathematical induction. Hence, express [imath]1-i[/imath] in the form of [imath]r(\cos x+i\sin x )[/imath]. | 2420291 | Review of a proof for De Moivre's theorem using mathematical induction
The following is a rigorous proof of De Moivre's theorem by means of mathematical induction. The theorem put simply is that: Any complex number, [imath]z = a+bi[/imath], on a cartesian plane can be expressed in polar form, where [imath]a=r\cos\theta[/imath] and [imath]b=r\sin\theta[/imath] and [imath]r[/imath] is the absolute distance from the origin to the point [imath]z[/imath]. In light of this, the expression for the complex number [imath]z[/imath] in polar form is [imath]z=r(\cos\theta+i\sin\theta)[/imath] In order to find the [imath]n^{th}[/imath] power of [imath]z[/imath], the following rule applies: [imath]\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)[/imath] where [imath]n\in\mathbb N[/imath] (I have confined [imath]n[/imath] to the natural numbers given that I intend on using mathematical induction to prove the theorem. Base step: We must prove that what is stated is true for [imath]n=1[/imath], hence [imath]z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)[/imath] [imath]=r(\cos\theta+i\sin\theta)[/imath] which is true. Next we assume it is true for [imath]n=k[/imath], and therefore proceed to the inductive step. Inductive step: When calculating [imath]z^{n+1}[/imath], it's the same as calculating [imath]z^nz[/imath], hence [imath]z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))[/imath] [imath]=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)[/imath] [imath]=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)[/imath] from here, [imath]i^2=-1[/imath], and therefore [imath]+i^2\sin n\theta\cdot\sin\theta[/imath] becomes [imath]-\sin n\theta\cdot\sin\theta[/imath]. In addition to this, the following trigonometric identities will be used: [imath]\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta[/imath] [imath]or[/imath] [imath]k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta[/imath] [imath]and[/imath] [imath]\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta[/imath] by using these, the original equation has now become [imath]=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))[/imath] then, after factoring out [imath]\theta[/imath], the theorem becomes [imath]=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))[/imath] which completes the proof. Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for [imath]\{n|n\in\mathbb R,n\geq1\}[/imath] |
2800474 | [imath]F=\mathbb{Q}(\alpha_1,\alpha_2, ... , \alpha_n)[/imath], where [imath]\alpha_i^2 \in \mathbb{Q}[/imath] implies [imath]\sqrt[3]{2} \notin F[/imath]
Let [imath]F=\mathbb{Q}(\alpha_1,\alpha_2, ... , \alpha_n)[/imath], where [imath]\alpha_i^2 \in \mathbb{Q}[/imath] for [imath]i=1,2,...,n.[/imath] I want to show that [imath]\sqrt[3]{2} \notin F[/imath]. I am trying to prove by contradiction assuming [imath]\sqrt[3]{2} \in F[/imath]. However, I am not able to complete the proof. How should I obtain a contradiction from the assumption? | 2768158 | Showing that [imath]\sqrt[3]{2}\notin\Bbb Q(\alpha_1,...,\alpha_k)[/imath] where [imath]\alpha_i^2\in\Bbb Q\ \forall i[/imath]
Let [imath]F[/imath] be the field [imath]\Bbb Q(\alpha_1,...,\alpha_k)[/imath] where [imath]\alpha_i^2\in\Bbb Q\ \forall i[/imath] I want to show that [imath]\sqrt[3]{2}\notin F[/imath] I thought if [imath]\sqrt[3]{2}\in F[/imath] then [imath]\sqrt[3]{2}\in\Bbb Q\cup(\alpha_1,...,\alpha_k)[/imath] but this is not a disjoint union so I'm not sure what to do... taking the square of [imath]\beta:=\sqrt[3]{2}[/imath] gives [imath]2^{2/3}=\sqrt[3]{4}\notin\Bbb Q[/imath] but that doesn't really advance things So I'm looking for a hint |
2801611 | When should one stop playing this game?
You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game. When should one stop playing this game? Let [imath]X[/imath] represents the number that comes up on the die. Therefore the game continues until [imath]X<6[/imath], So, [imath]P(X=6)=nCr p^r q^{n-r}[/imath] where [imath]r=1[/imath] [imath] \dfrac{1}{6}=nC_1 \times \dfrac{1}{6} \times \biggr (\dfrac{5}{6}\biggr )^{n-1}[/imath] [imath]1=n \times \biggr(\dfrac{5}{6}\biggr)^{n-1}[/imath] [imath]\boxed {n = 1}[/imath] Am I wrong? | 1977081 | When to stop rolling a die in a game where 6 loses everything
You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game. When should one stop playing this game? Obviously, one wants to maximize total score. As I was asked to show my preliminary results on this one, here they are: If we simplify the game to getting 0 on 6 and 3 otherwise, we get the following: [imath]\begin{align} EV &= \frac{5}{6}3+\frac{25}{36}6+\frac{125}{216}9+\ldots\\[5pt] &= \sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n3n \end{align}[/imath] which is divergent, so it would make sense to play forever, which makes this similar to the St. Petersburg paradox. Yet I can sense that I'm wrong somewhere! |
2801917 | Give an example of a series [imath]\sum a_n[/imath] such that [imath]\sum a_n[/imath] is convergent but [imath]\sum a_{3n}[/imath] is divergent.
Give an example of a series [imath]\sum a_n[/imath] such that [imath]\sum a_n[/imath] is convergent but [imath]\sum a_{3n}[/imath] is divergent. Here an example by zhw. is given as [imath]\frac 1 2 + \frac 1 2 - \frac 2 2 + \frac 1 3 + \frac 1 3 − \frac 2 3 + \frac 1 4 + \frac 1 4 − \frac 2 4 + \frac 1 5 + \frac 1 5 − \frac 2 5+\cdots.[/imath] Give an example of a series such that [imath]\sum a_n[/imath] is convergent but [imath]\sum a_{3n}[/imath] is divergent But I need a better result with functional expression. | 2783399 | Example of a series!
Give an example of a convergent series [imath]\sum_{n=1}^{\infty}a_n[/imath] such that the series [imath]\sum_{n=1}^{\infty}a_{3n}[/imath] is divergent.. I cannot find find any kind of series... I am also be very thankful if you find a divergent series (changing the term convergent)... |
2801728 | ODE IVP : [imath]u_t + uu_x = 0, \; \; u(x,0) = \cos \pi x[/imath]
Exercise : Show that a smooth solution of the initial value problem [imath]\begin{cases} u_t + uu_x = 0 \\ u(x,0) = \cos \pi x \end{cases}[/imath] must satisfy [imath]u = \cos[\pi(x-ut)][/imath]. Also, show that when [imath]t=1/u[/imath], the function [imath]u[/imath] stops to exist (as a single-value function). Attempt : [imath]\frac{\mathrm{d}x}{u} = \frac{\mathrm{d}t}{1} = \frac{\mathrm{d}u}{0} [/imath] [imath]\implies u_1 = u, \; \; u_2 = x - ut[/imath] Which means that the solutions are given by [imath]u_1 = F(u_2) \Rightarrow u = F(x-ut)[/imath]. Thus, taking into account the initial value : [imath]F(x) = \cos \pi x.[/imath] Now, for [imath]x:= x-ut[/imath] we get [imath]F(x-ut) = \cos[\pi(x-ut)][/imath] which [imath]\in C^\infty[/imath] and thus a smooth solution. Question : How would one proceed to show the second part of the exercise ? It gives as a hint to check the graphs of [imath]\arccos(u)[/imath] and [imath]\pi(x-ut)[/imath] as functions of [imath]u[/imath], which makes sense, since : [imath]\arccos(u)=\pi(x-ut)[/imath] But I can't seem how to grasp out a graph or figure out a solution to it. | 2799257 | Why is the solution single-valued?
I have shown that a smooth solution of the problem [imath]u_t+uu_x=0[/imath] with [imath]u(x,0)=\cos{(\pi x)}[/imath] must satisfy the equation [imath]u=\cos{[\pi (x-ut)]}[/imath]. Now I want to show that [imath]u[/imath] ceases to exist (as a single-valued continuous function) when [imath]t=\frac{1}{\pi}[/imath]. When [imath]t=\frac{1}{\pi}[/imath], then we get that [imath]u=\cos{(\pi x-u)}[/imath]. With single-valued function is it meant that the function is 1-1 ? If so, then we have that [imath]\cos{(2 \pi-u)}=\cos{(4 \pi -u)}[/imath], i.e. for two different values of [imath]x[/imath], we get the same [imath]u[/imath], and so for [imath]t=\frac{1}{\pi}[/imath], [imath]u[/imath] is not 1-1. But if this is meant, how are we sure that for [imath]t \neq \frac{1}{\pi}[/imath] the function is single-valued? |
2802372 | Isolating two variables from multiple trig ratios
I have 3 functions of the same form, [imath]x(p, q)[/imath], [imath]y(p, q)[/imath], and [imath]z(p, q)[/imath] for which I am trying to evaluate [imath]p[/imath] and [imath]q[/imath]. I know that [imath]x+y+z=0[/imath]. I know I need to express p and q in terms of each other, but as you can see below, the three functions together are quite complex. I've been futzing around with trig identities to try and isolate either variable, but am stuck. Perhaps fresh eyes will help; can anyone here see a way forward? All values are known, except [imath]p[/imath] and [imath]q[/imath]. [imath]x(p,q) = [v_{1x} \cos(p)-u_{1x}\sin(p)][(C_{2x}+u_{2x}\cos(q)+v_{2x}\sin(q))-(C_{1x}+u_{1x}\cos(p)+v_{1x}\sin(p))][/imath] [imath]y(p,q) = [v_{1y}\cos(p)-u_{1y}\sin(p)][(C_{2y}+u_{2y}\cos(q)+v_{2y}\sin(q))-(C_{1y}+u_{1y}\cos(p)+v_{1y}\sin(p))][/imath] [imath]z(p,q) = [v_{1z} \cos(p)-u_{1z}\sin(p)][(C_{2z}+u_{2z}\cos(q)+v_{2z}\sin(q))-(C_{1z}+u_{1z}\cos(p)+v_{1z}\sin(p))][/imath] | 2768145 | Solving a multi-variable trig sum to find the max separation btwn 2 ellipses
For a hobby project gone too far, I'm trying to find the maximum separation between two ellipses. I have only a high-school-graduate-level grasp of algebra and calculus, and am happy to dig deeper than that... but I need some guidance. I have 3 functions of the same form, [imath]X(p, q)[/imath], [imath]Y(p, q)[/imath], and [imath]Z(p, q)[/imath] for which I am trying to evaluate [imath]p[/imath] and [imath]q[/imath]. I know that [imath]X+Y+Z=0[/imath]. I know I need to express p and q in terms of each other, but as you can see below, the three functions together are quite complex. I've been futzing around with trig identities to try and isolate either variable, but am stuck. Perhaps fresh eyes will help; can anyone here see a way forward? All values are known, except [imath]p[/imath] and [imath]q[/imath]. [imath]X(p,q) = [b_1V_{x1}\cos(p)-a_1U_{x1}\sin(p)][(C_{x2}+a_2U_{x2}\cos(q)+b_2V_{x2}\sin(q))-(C_{x1}+a_1U_{x1}\cos(p)+b_1V_{x1}\sin(p))][/imath] [imath]Y(p,q) = [b_1V_{y1}\cos(p)-a_1U_{y1}\sin(p)][(C_{y2}+a_2U_{y2}\cos(q)+b_2V_{y2}\sin(q))-(C_{y1}+a_1U_{y1}\cos(p)+b_1V_{y1}\sin(p))][/imath] [imath]Z(p,q) = [b_1V_{z1}\cos(p)-a_1U_{z1}\sin(p)][(C_{z2}+a_2U_{z2}\cos(q)+b_2V_{z2}\sin(q))-(C_{z1}+a_1U_{z1}\cos(p)+b_1V_{z1}\sin(p))][/imath] Edit - A couple threads I've been trying to springboard off of: Furthest distance between two circles in 3D How to calculate minimum distance between two arbitrary ellipses in 2D a \sin x + b \cos x = c">Solving trigonometric equations of the form [imath]a \sin x + b \cos x = c[/imath] I arrived here by taking the partial derivatives of the distance function for two points, one on each ellipse. Once I can find the critical points from that, I should have my solution(s). The distance function is from the ellipses' parametric equations: [imath]x(θ)=C_x+aU_x\cos(θ)+bV_x\sin(θ)[/imath] [imath]y(θ)=C_y+aU_y\cos(θ)+bV_y\sin(θ)[/imath] [imath]z(θ)=C_z+aU_z\cos(θ)+bV_z\sin(θ)[/imath] (Where [imath]C[/imath] is the centrepoint of the ellipse, [imath]a[/imath] and [imath]b[/imath] are its semimajor and semiminor axes, respectively, and [imath]U[/imath] and [imath]V[/imath] are unit vectors defining the plane of the ellipse. Angle [imath]θ[/imath] describes position on the ellipse) So the distance function is: [imath]D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}[/imath] As the derivative of [imath]\sqrt{n}[/imath] is [imath]{n'\over2\sqrt{n}}[/imath], I only need concern myself with where the numerator ([imath]n'[/imath]) is equal to [imath]0[/imath]: [imath]n'=-2f'(p)[-f(p)+j(q)]-2g'(p)[-g(p)+k(q)]-2h'(p)[-h(p)+l(q)][/imath] (Where functions [imath]f[/imath], [imath]g[/imath], and [imath]h[/imath] refer to the parametrics for the 1st ellipse, and functions [imath]j[/imath], [imath]k[/imath], and [imath]l[/imath] to those for the 2nd. [imath]p[/imath] and [imath]q[/imath] are used instead of [imath]θ[/imath]) [imath]n'=0[/imath] is essentially of the form [imath]0=X+Y+Z[/imath] as outlined above. |
2795103 | [imath]F[/imath] closed, [imath]K[/imath] compact [imath]\implies FK[/imath] is closed in group [imath]G[/imath]
Let [imath]G[/imath] be a Hausdorff topological group, let [imath]F[/imath] be a closed and let [imath]K[/imath] be compact, both subsets of [imath]G[/imath]. Then [imath]FK[/imath] is closed in [imath]G[/imath]. Attempt: [imath]aF[/imath] is closed in [imath]G[/imath] for each [imath]a \in G[/imath]. All we have to do is show that [imath]FK[/imath] is compact since a compact subset of a Hausdorff space is closed. [imath]FK \subset \bigcup\limits_i U_i[/imath] | 71983 | Product of compact and closed in topological group is closed
This could be classified as "homework", but I tried to solve this, made research online, and still failed, so I'll be glad to get some hints. Let [imath]G[/imath] be a topological group, let [imath]A[/imath] be a compact subset of [imath]G[/imath], and let [imath]B[/imath] be a closed subset of [imath]G[/imath]. Prove that [imath]AB[/imath] is closed. If both [imath]A[/imath] and [imath]B[/imath] are not compact, but closed, this can fail, for example, if we let [imath]A[/imath] be the set of integers and [imath]B[/imath] the set of integer multiples of [imath]\pi[/imath], then both are closed, but [imath]A+B[/imath] is a proper dense subset of [imath]\mathbb R[/imath], so can't be closed. Also if [imath]A[/imath] is compact but [imath]B[/imath] is not closed, this easily fails. Thanks |
2800013 | In a proof by contradiction, what if both the proposition and its negation lead to contradictions?
I'm learning math. I've recently thought more about the proof by contradiction technique, and I have a question that I would like cleared up. Let me set the stage. Suppose I am trying to prove a theorem. Theorem: If A and [imath]\neg[/imath]B, then [imath]\neg[/imath]C. Proof (contradiction): Let us suppose that A is true and [imath]\neg[/imath]B is true. Let us assume that C is true ([imath]\neg[/imath]C is false). [blah blah blah] From this, we arrive at a contradiction because we see that B is true ([imath]\neg[/imath]B is false), but we know that [imath]\neg[/imath]B is true (because we assumed it to be true). Thus, since assuming that C is true lead us to a contradiction, it must be the case that C is false ([imath]\neg[/imath]C is true). QED. My issue with this: why is it that C leading to a contradiction must mean that [imath]\neg[/imath]C is true? What if [imath]\neg[/imath]C also leads to a contradiction? In that case, doesn't a proof by contradiction not prove anything? Why can we be sure that C leading to a contradiction must mean that [imath]\neg[/imath]C doesn't lead to a contradiction? I'm sorry if this question has already been asked. I searched for a bit before asking to see if anyone had this same specific question, but most results just asked why a proof by contradiction works in general without any clear question. | 1667884 | Are proofs by contradiction really logical?
Let's say that I prove statement [imath]A[/imath] by showing that the negation of [imath]A[/imath] leads to a contradiction. My question is this: How does one go from "so there's a contradiction if we don't have [imath]A[/imath]" to concluding that "we have [imath]A[/imath]"? That, to me, seems the exact opposite of logical. It sounds like we say "so, I'll have a really big problem if this thing isn't true, so out of convenience, I am just going to act like it's true". |
717961 | Countable axiom of choice: why you can't prove it from just ZF
This is a follow-up question to the discussion about the finite axiom of choice here. Suppose we have a countable collection of non-empty sets [imath]\{A_1, A_2, A_3,\cdots\}[/imath] Reasoning as indicated in that discussion, we can prove the existence of a choice function [imath]c_n[/imath] such that [imath]c_n(i)[/imath] belongs to [imath]A_i[/imath] for [imath]i = 1,\cdots,n[/imath] Again reasoning as suggested and using induction, we can prove the existence of a function [imath]C[/imath] defined over [imath]\{1, 2, \cdots\}[/imath] such that [imath]C(n)[/imath] is a choice function defined for [imath]i=1,\cdots,n[/imath] with [imath]C(n)(i)[/imath] an element of A_i. We can require in addition that [imath]C(n)(i)[/imath] and [imath]C(n-1)(i)[/imath] are equal for [imath]i=1,\cdots,n-1[/imath]. Now it would it seem that we can prove the countable axiom of choice by taking the union of [imath]C(1)[/imath], [imath]C(2)[/imath], [imath]\cdots[/imath] The question is as follows: why is that proof wrong? I suspect the answer may have something to do with the axiom of union. | 2820246 | Can I use induction to prove this?
Let [imath]A_1, A_2, A_3,\ldots[/imath] be pairwise disjoint denumerable sets. Prove that [imath]\bigcup_{i=1}^\infty A_i\text{ is denumerable.}[/imath] This question comes from exercise [imath]10.11[/imath] of Mathematical Proofs: A Transition to Advanced Mathematics (Second Edition) by chartrand, polimeni, and zhang. I'm using this book for self study. I used induction in this proof attempt. Sorry for the messy handwriting. I figured that the natural numbers were equivalent to integers from 1 to infinity, but a buddy pointed out that infinity was not an element of the natural numbers, so induction might not be possible. So, was it possible to use induction on this problem? Or should I just redo the proof by constructing a table of the denumerable sets and coming up with some diagonal arrows to order them like my book suggests. |
1872716 | Induction - Countable Union of Countable Sets
Stephen Abbott has a an exercise in Chapter 1 (1.2.12) that suggests that one cannot use induction to prove that a countable union of countable sets is countably infinite. One answer is that n=infinity cannot be demonstrated via induction, as inifinity is not a natural number. This seems sketchy. Rudin in chapter 2 clearly distinguishes the use of inifinity symbol for a union of sets to indicate a countably infinite union of sets and distinguishes it from the infinity used to extend the reals. All of this also appears to ignore the fact [imath]N[/imath] is countably infinite by definition. Therefore any bijection with [imath]N[/imath] is also proved for countably infinite cases. So why cannot induction be used to argue countable union of countable sets is countable? Here is an example where induction is being used in the context of countably infinite sets. Using induction to prove that the infinite set of polynomials is countably infinite | 2896439 | Does 'A countable infinite Cartesian product of countable sets is not necessarily countable' violate induction principle?
We have two theorems about Cartesian product: A finite Cartesian product of countable sets is countable; A countable infinite Cartesian product of countable sets is not necessarily countable. To prove the first theorem, we can show Firstly, we can prove [imath]A\times B[/imath] is countable, Secondly, suppose it is true for [imath]n-1[/imath], i.e. [imath]X=A_1\times A_2\times \ldots \times A_{n-1}[/imath] is countable, we prove the case for [imath]n[/imath] by the fact [imath]A_1\times A_2\times \ldots \times A_{n-1}\times A_{n}=X\times A_{n}[/imath] is in fact the product of two countable sets, in above we know it is still countable. The above method to prove the first theorem is exactly the mathematical induction, which should show the theorem is true for all [imath]n\in\mathbb{Z}_{+}[/imath], so it should works for countable infinite unions. However, we may use Cantor diagonal method to prove at least [imath]\{0,1\}^w[/imath] is not countable, thus the second theorem is also right. Does this means that the mathematical induction is wrong? EDIT: After the following discussions, I have the answer: The key point is that any element [imath]n\in\mathbb{Z}_+[/imath] is always a finite number, thus induction principle only guarantee the theorem is true for any [imath]n\in\mathbb{Z}_+[/imath], i.e. for any finite number. When discuss a countable infinite union, it is infinite number of unions, this infinite number no more belong to [imath]\mathbb{Z}_+[/imath], thus can not be guaranteed by induction principle. |
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