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2365146 | Get angle of point in circle with center [imath](0, 0)[/imath].
I have a circle with a centre at position [imath](0, 0)[/imath] and I have a point in the circle at [imath](-5, 2)[/imath]. How can I get the angle of the point towards the centre? I thought to use the theorem Al Kashi (law of cosines) but we need an another point to form a triangle but I don't have it. Thanks, your help will be appreciated. | 94379 | calculating angle in circle
How to calculate angle in a circle. Please see the diagram to get the idea what I want to calculate? I have origin of circle that is [imath](x_1,x_2)[/imath]. I have a point on circumstance of circle that is [imath](x_2,y_2)[/imath]. Also I know the radius of circle that is R. How to calculate the angle between these two lines -- line starting from origin and ending on circumstance at 0 degree -- line starting from origin and ending on [imath](x_2,y_2)[/imath]. |
2363687 | What is the concept (or idea) behind "induced subgraph" definition?
Let [imath]G, G'[/imath] be graphs. If [imath]G' \subseteq G[/imath] and [imath]G'[/imath] contains all the edges [imath]xy \in E[/imath] with [imath]x,y \in V'[/imath], then [imath]G'[/imath] is an induced subgraph of [imath]G[/imath]. We say that [imath]V'[/imath] induces or spans [imath]G'[/imath] in [imath]G[/imath], and write [imath]G'=:G[V'][/imath]. Thus if [imath]U \subseteq V[/imath] is any set of vertices, then [imath]G[U][/imath] denotes the graph on [imath]U[/imath] whose edges are precisely the edges of [imath]G[/imath] with both ends in [imath]U[/imath]. Can I say that the term "induced" brings some sort of equivalence meaning? So that, if some equivalence holds, I can use smaller structures for my convenience. In other words, does the graph [imath]G'[/imath] preserve some sort of structure or properties belonging to graph [imath]G[/imath]? | 1013143 | Difference between a sub graph and induced sub graph.
I have the following paragraph in my notes: If [imath]G=(V,E)[/imath] is a general graph . Let [imath]U\subseteq V[/imath] and let [imath]F[/imath] be a subset of [imath]E[/imath] such that the vertices of each edge in [imath]F[/imath] are in [imath]U[/imath] , then [imath]H=(U,F)[/imath] is also a general graph and [imath]H[/imath] is a subgraph of [imath]G[/imath] . If [imath]F[/imath] consists of all edges of [imath]G[/imath] which have endpoints in [imath]U[/imath] ,then [imath]H[/imath] is called induced subgraph of [imath]G[/imath] and is denoted by [imath]G_U. [/imath] From here both the definition of a subgraph and a induced subgraph seem same to me.. I can't understand what is the difference between them... Please help with this.. |
2365371 | Prove [imath] \frac{1}{x^{2n}}+x^{2n}[/imath] is an integer for all [imath]n[/imath].
Question: Suppose [imath] \frac{1}{x^2}+x^2[/imath] is an integer. Prove that [imath] \frac{1}{x^{2n}}+x^{2n}[/imath] is an integer for all natural [imath]n[/imath]. Hint: Use Strong Induction My attempt: Base Case is trivial. I.H: Assume the result is true for [imath]n = 1,2, ...., k.[/imath] Consider [imath]n = k+1[/imath]. [imath] \frac{1}{x^{2\left(k+1\right)}}+x^{2\left(k+1\right)}\ =\ \frac{1}{x^{\left(2k+2\right)}}+x^{2k+2}[/imath]. I am not sure what to do from here and how to use the induction hypothesis. | 558027 | Prove by induction that a^n+a^-n is an integer.
I am to prove by induction that given that [imath]a+1/a[/imath] is an integer (i.e. belongs to Z ) then [imath]a^n+1/a^n[/imath] is an integer too. I'm pretty much clueless here. Thanks in advance. |
2365686 | Let S be an infinite set and A a finite subset. Prove that [imath]|S| = |S -A|[/imath].
I don't know if my solution is correct. This is what I have so far: Let [imath]S = \{ s_1, s_2,.....\}[/imath] and [imath]S-A = \{a_1, a_2,.....\}[/imath] where all the elements are arranged in an fixed order. let [imath]f(x): S \to A , f(s_i) = a_i[/imath] if I prove [imath]f[/imath] is bijective will my solution be correct? | 71199 | Show that A has the same cardinality as [imath]A\setminus\{x,y,z\}[/imath]
This is a homework problem for class credit and I would appreciate a hint. Let [imath]A[/imath] be an infinite set and suppose [imath]x, y[/imath] and [imath]z[/imath] are points in [imath]A[/imath]. Show that [imath]A[/imath] has the same cardinality as [imath]A\setminus\{x,y,z\}[/imath]. So far I am trying to solve this for the more general case, let [imath]A[/imath] be an infinite set and [imath]B[/imath] be a finite subset of [imath]A[/imath], show that [imath]A[/imath] has the same cardinality as [imath]A\setminus B[/imath]. If [imath]B[/imath] is finite and [imath]A[/imath] is infinite, then [imath]A\setminus B[/imath] is infinite because if [imath]A\setminus B[/imath] were finite then [imath]A = A\setminus B \cup B[/imath], the union of two finite sets and hence [imath]A[/imath] would be finite. Using Cantor-Bernstein we know that if there is a one-to-one function from [imath]A[/imath] to [imath]A\setminus B[/imath] and a one-to-one function from [imath]A\setminus B[/imath] to [imath]A[/imath], then [imath]A[/imath] and [imath]A\setminus B[/imath] are equivalent. It is easy to define a function [imath]f:A\setminus B \to A[/imath] that is one-to-one, [imath]f(x) = x[/imath]. Now I need to define a function [imath]g:A \to A\setminus B[/imath] that is one-to-one. I have a feeling that this involves using the axiom of choice. Is this the right strategy? Have I made any errors? And can anyone give me a hint on how to define a one to one function from [imath]A[/imath] to [imath]A\setminus B[/imath]? |
2365781 | how many isomorphism are there from [imath]Z_{12}[/imath] to [imath]Z_4\oplus Z_3[/imath]??
Isommorphism that maps identity elements of each group. But how to find the number of such isomorphism. | 1883438 | How many isomorphisms are there from [imath]\Bbb Z_{12}[/imath] to [imath]\Bbb Z_{4} \oplus \Bbb Z_{3}[/imath]?
How many isomorphisms are there from [imath]\Bbb Z_{12}[/imath] to [imath]\Bbb Z_{4} \oplus \Bbb Z_{3}[/imath]? I can see that if [imath]f[/imath] is the isomorphism, the isomorphism is determined by the value of [imath]f(1)[/imath] because [imath]f(1) = (1,1)[/imath] and [imath]f(k) = f(k) + f(1) = f(k) + (1,1)[/imath]. But since [imath]\Bbb Z_{12}[/imath] has [imath]4[/imath] generators, there are [imath]4[/imath] different ways to set [imath]f(1)[/imath] equal to a generator. Therefore there are [imath]4[/imath] isomorphisms. Is this correct? |
2365867 | A question about derivative
[imath]f(x)=\int_{0}^{x}\cos(1/t)dt[/imath] Then[imath]f'(0)=?[/imath] I tried [imath]f'(x)=\cos\left(1/x\right)[/imath], but then I get into trouble. Hope for your help | 1551332 | Prove that [imath]\lim_{h \to 0}\frac{1}{h}\int_0^h{\cos{\frac{1}{t}}dt} = 0[/imath]
I'm trying to prove that [imath]\lim_{h \to 0}\frac{1}{h}\int_0^h{f(t)dt} = 0[/imath] where [imath]f(t) = \begin{cases}\cos{\frac{1}{t}} &\text{ if } t \neq 0\\ 0&\text{otherwise}\end{cases}.[/imath] Can someone give me a hint where to start? Darboux sums somehow seem to lead me nowhere. NOTE: I cannot assume that [imath]f[/imath] has an antiderivative [imath]F[/imath]. |
2366381 | Need help with proof of [imath]\frac{1}{2}n^{3/2} \leq \sqrt{1} + \sqrt{2} + ... + \sqrt{n}[/imath]
Can someone please help me with this proof? I tried approaching it using a proof by induction. The base case with [imath]n = 0[/imath] is intuitively true. For the inductive step, I let [imath]k \in \mathbb{N}[/imath], let [imath]P(n) = \frac{1}{2}n^{3/2} \leq \sqrt{1} + \sqrt{2} + ... + \sqrt{n}[/imath], assumed [imath]P(k)[/imath] and tried to prove [imath]P(k+1)[/imath]. For the inductive step, I got [imath]\frac{1}{2}k^{3/2} + \sqrt{k+1} \leq \sqrt{1} + \sqrt{2} + ... + \sqrt{k} + \sqrt{k+1}[/imath] by adding [imath]\sqrt{k+1}[/imath] to both sides of the inductive hypothesis. Some basic algebra allows us to rewrite the statement: [imath]\frac{1}{2}k \cdot k^{1/2} + \sqrt{k+1} \leq \sqrt{1} + \sqrt{2} + ... + \sqrt{k} + \sqrt{k+1}[/imath]. I'm not exactly sure how to move forward from here. I've tried manipulating the statement in various ways, but I can never get back to [imath]P(k+1)[/imath]. Can someone please help to solve this proof, and understand the proof behind it? Thank you!! | 191204 | Induction proof of lower bound for [imath]\sum \sqrt n[/imath]
I'm having some trouble proving the following statement using mathematical induction: [imath]\frac{1}{2}n^{\frac{3}{2}} \leq \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} + ... + \sqrt{n} ,\text{ (for all sufficiently large n)}[/imath] I'm sort of confused because [imath]n^{\frac{3}{2}}[/imath] = [imath]n n^{\frac{1}{2}}[/imath] which means that there are n of the largest possible root values on the left hand side while the right hand side has values like [imath]\sqrt{1}[/imath], [imath]\sqrt{2}[/imath], etc. As n grows sufficiently large wouldn't the left hand side outgrow the right hand side no matter what the constant multiple is? If I'm thinking about this correctly just let me know, if not, any help would be appreciated. |
2365746 | Prove the series [imath]\sum_{n=1}^{\infty}\frac{1}{{5^n}^!}[/imath] converges to an irrational number
This is the sum [imath]\sum_{n=1}^{\infty}\frac{1}{{5^n}^!}[/imath] My first attempt was to assume that the series does converge to a rational number [imath]a/b[/imath]. But the [imath]n![/imath] bothered me and I failed in my proof. How would you try to prove this series? | 2373896 | Prove an infinite sum is irrational
I'm trying to prove that [imath] \sum_{k=1}^{\infty} 7^{-k!} [/imath] is irrational but I'm so lost. Any tips for where to begin, thanks in advance. |
2363769 | If [imath]P_1[/imath] and [imath]P_2[/imath] are coprime, how do I show that [imath]\ker ( (P_1 \times P_2 )(f) )=\ker(P_1 (f)) \oplus \ker(P_2(f))[/imath]?
Suppose that [imath]P_1[/imath] and [imath]P_2[/imath] are coprime between them in [imath]\mathbb{K}[X][/imath]. Show that [imath] \ker ( (P_1 \times P_2 )(f) )=\ker(P_1 (f)) \oplus \ker(P_2(f)).[/imath] I tried by starting off by showing that the right hand equality is true, that is [imath]\mathbb{K}[X][/imath] is indeed equal to [imath]\ker(P_1 (f)) \oplus \ker(P_2(f))[/imath]. My first attempt was to show that [imath]\ker(P_1 (f)) \cap \ker(P_2(f))= \{ 0 \} .[/imath] But I got nowhere. I think I could have used the Bezout theoreme, that is: If [imath]P_1[/imath] and [imath]P_2[/imath] are coprime, then [imath]\exists U,V \in \mathbb{K}[X][/imath] so that [imath]P_1U(X) + P_2V(X)=1[/imath] But nothing comes to my mind. | 462748 | Primary Decomposition of [imath]\mathbb{R^n}[/imath] under T
I have a question related to the Primary Decomposition Theorem for a vector space under a linear transformation. The following question is a direct corollary of this theorem. However, I wish to prove it without calling upon theorem. I am currently studying for a comprehensive exam and ran across this question. I never really felt comfortable with the proof I saw of the Primary Decomposition Theorem so I hoping with your help with this problem I will understand this particular problem and better understand that proof as I believe their proofs will be essentially the same techniques. Question: Suppose we have a linear transformation [imath]T: \mathbb{R^n}\to \mathbb{R^n}[/imath] with minimal polynomial [imath]m(x)=f(x)g(x)[/imath] and [imath]f(x),g(x)[/imath] are relatively prime. Show that we can write [imath]\mathbb{R^n}=M\oplus N[/imath] where [imath]M[/imath] and [imath]N[/imath] are [imath]T[/imath]-invariant subspaces s.t [imath]f(x)[/imath] is the minimal polynomial for [imath]T|_M[/imath] and [imath]g(x)[/imath] is the minimal polynomial for [imath]T|_N[/imath]. So I know that I want to use [imath]M=ker(f(T))[/imath] and [imath]N=ker(g(T))[/imath] but after that I really don't have much. Since this isn't a homework problem for a class I am essentially looking for someone to present me a proof of this simpler question and walk me through their logic so that I can hopefully understand the general theorem and its proof. Thanks in advance for the help! |
2260482 | Hoffman and Kunze, linear algebra chapter 1 exercise
Suppose [imath]R_1[/imath] and [imath]R_2[/imath] are [imath]2\times 3[/imath] row-reduced echelon matrices and that the systems [imath]R_1X=0[/imath] and [imath]R_2X=0[/imath] have the same solutions. Prove that [imath]R_1=R_2[/imath] I wrote down the [imath]5[/imath] different types of row reduced echelon [imath]2\times 3[/imath] matrices but it got me nowhere. It was tedious. I tried contradiction; suppose [imath]R_1\ne R_2,[/imath] then there exists [imath]i,j[/imath] such that [imath] 1\le i\le 2,1\le j\le 3[/imath] and [imath]R_{1_{ij}}\ne R_{2_{ij}}.[/imath] But I am not able to make use of this information to lead to a contradiction. EDIT: I think I might be able to use this lemma : If two homogeneous systems of linear equations have the same solutions, then they are equivalent Coming back to the solution - So now, I can conclude that since [imath]R_1[/imath] and [imath]R_2[/imath] are row-equivalent and they must be the same since row-reduced echelon matrices are unique. Please help me verify. | 577838 | Same solution implying row equivalence?
Suppose [imath]R[/imath] and [imath]R '[/imath] are [imath] 2 \times 3[/imath] row-reduced echelon matrices and that the system [imath]Rx=0[/imath] and [imath]R'x=0[/imath] have exactly the same solutions. Prove that [imath]R=R'[/imath]. In general, is it true that any 2 [imath]m \times n[/imath] matrices that have the same solution must be row-equivalent? |
2368174 | Show that [imath]HK\cong H\times K[/imath]
Let [imath]H\lhd G[/imath] and [imath]K\lhd G[/imath] s.t. [imath]H\cap K=\{1\}[/imath]. Show that [imath]HK\cong H\times K.[/imath] Attempts Using second isomorphism theorem, we have that [imath]HK/K\cong H[/imath] and [imath]HK/H\cong K[/imath], so I need to prove that [imath]HK\cong HK/K\times HK/H.[/imath] To simplify notatino let [imath]W=HK[/imath]. I consider the group morphism [imath]W\longrightarrow W/H\times W/K[/imath] defined by [imath]w\longmapsto (wH,wK).[/imath] The injectivity is clear, but I have problem to show surjectivity. Could someone help ? | 361396 | Let [imath]G[/imath] a group with normal subgroups [imath]M,N[/imath] such that [imath]M\cap N=\{e\}[/imath]. Show that if [imath]G[/imath] is generated by [imath]M\cup N[/imath] then [imath]G\cong M \times N[/imath].
Let [imath]G[/imath] a group with normal subgroups [imath]M,N[/imath] such that [imath]M\cap N=\{e\}[/imath]. i) Show that for every [imath]m\in M[/imath] and for every [imath]n\in N[/imath], [imath]mn=nm[/imath]. ii) If [imath]G[/imath] is generated by [imath]M\cup N[/imath] then [imath]G\cong M \times N[/imath]. This is an homework question for abstract algebra. I think I can prove [imath]i)[/imath]. But I have no idea how to prove [imath]ii)[/imath]. Can anybody give me a hint for this one ? Here my proof for i): Let [imath]n \in N[/imath] and let [imath]m\in M[/imath]. Let [imath]x=mnm^{-1}n^{-1}[/imath]. As [imath]N[/imath] is normal subgroup, [imath]mnm^{-1} \in N[/imath]. Therefore [imath]x \in N[/imath]. As [imath]M[/imath] is normal subgroup , [imath]nmn^{-1} \in M[/imath]. Since [imath]m\in M[/imath], then [imath]x=m(nm^{-1}n^{-1})\in M[/imath]. This implies that [imath]c\in M \cap N[/imath], then [imath]c\in \{e\}[/imath], then [imath]c=e[/imath], and we get [imath]mn=nm[/imath]. |
2367923 | Is there a systematic way of finding the matrix of a quadratic form?
For example i have this quadratic form [imath]q(x_1,x_2)=8{x_1}^2-4x_1x_2+5{x_2}^2[/imath] , here it's a simple factoring: [imath]q\begin{bmatrix}x_1 \\x_2 \\x_3\\\end{bmatrix}=\begin{bmatrix}x_1 \\x_2 \\x_3\\\end{bmatrix} \cdot \begin{bmatrix}8x_1 &-2x_2\\-2x_1&5x_2\end{bmatrix}=\vec{x}^{T}A\vec{x} ,A=\begin{bmatrix}8 &-2\\-2&5\end{bmatrix}[/imath]. But this is not always the case where one can simply see how the matrix is going to be ,so is there a certain method of finding this matrix? | 1856016 | How to find the matrix of a quadratic form?
I was wondering. If I have a bilinear symmetric form, it is easy to find its matrix. But, when I have a quadratic form, which is the procedure to do that? I heard that one possibility is: If [imath]q[/imath] is my quadratic form, then [imath]f(x,y) = \frac{1}{4}q(x+y) - \frac{1}{4}q(x-y)[/imath] is the bilinear symmetric form associated, so the method reduces to find the matrix of [imath]f(x,y).[/imath] The point is that seems a little noising... How to find, in practice, the matrix of quadratic form? Thanks in advance. |
2368402 | Bump function derivative control
Does there exist a smooth bump function [imath]\Phi: \mathbb{R} \rightarrow \mathbb{R} [/imath] such that the sequence [imath]M_k = \max_{x \in \mathbb{R}}|\Phi^{(k)}(x)|[/imath] does not grow faster than exponentially? In general, are there examples of [imath]\Phi[/imath] where the [imath]M_k[/imath] grow (much) slower than in the case of [imath]e^{\frac{1}{x^2-1}}[/imath] (or the like)? By smooth bump function, I mean a [imath]C^\infty[/imath] function that is [imath]1[/imath] in a neighbourhood of [imath]0[/imath] with compact support. Pointers to other resources on this are appreciated as well. | 101480 | Are there other kinds of bump functions than [imath]e^\frac1{x^2-1}[/imath]?
I've only seen the bump function [imath]e^\frac1{x^2-1}[/imath] so far. Where could I find examples of functions [imath]C^∞[/imath] on [imath]\mathbb{R}[/imath] that are zero everywhere except on [imath](-1,1)[/imath]? Are there others that do not involve the exponential function? Are there any with a closed form integral? Is there a preferred function? |
2364350 | Alternate definition for sup norm
My definition for the essential supremum norm as follows. For measurable [imath]f[/imath] defined on a measurable set [imath]E[/imath], the essential supremum norm of [imath]f[/imath] over [imath]E[/imath] is [imath] ||f||_{L^{\infty}(E)} = ||f||_{\infty} = \inf \left\{\sup_{x \in A} |f(x)|: A \text{ measurable}, m(E \setminus A) = 0\right\}. [/imath] I was reading a proof showing that [imath]||f||_{\infty} = 0 \implies f = 0[/imath] almost everywhere, which used the fact that [imath] ||f||_{\infty} = \inf \left\{ \alpha \in \mathbb{R}: m\left\{x: \left|f(x)\right| > \alpha\right\} = 0\right\}. [/imath] I am struggling to prove this alternate form of [imath]||f||_{\infty}[/imath] from my definitions of measurable functions, sets, etc. though, and would appreciate some help. | 2272732 | Alternative definition of [imath]\|f\|_{\infty}[/imath] as the smallest of all numbers of the form [imath]\sup\{|g(x)| : x \in X \}[/imath], where [imath]f = g[/imath] almost everywhere
Recall that [imath]\|f\|_\infty = \inf\{M\ |\ |f(x)|\leq M \text{ for almost every } x \in X\}[/imath], for every function [imath]f[/imath] defined on [imath]X[/imath]. Show that, if [imath]\|f\|_\infty[/imath] is finite, then [imath]\|f\|_\infty[/imath] is also the smallest of all numbers of the form [imath]\sup\{|g(x)| : x \in X \}[/imath], where [imath]f = g[/imath] almost everywhere with respect to the Lesbegue measure. I am trying to understand how [imath]L^{\infty}[/imath] is defined. Could someone help me see how to prove the above fact? |
2368560 | Prove that the square is not a variety?
I am looking for a proof that the standard unit square in [imath]\mathbb{R}^2[/imath] is not a variety. To be a variety I only require that it be the solution to a system of polynomial equations. | 1007299 | Square is not an algebraic set.
I am trying to show that square with vertices at [imath](\pm1,0)[/imath] and [imath](0,\pm1)[/imath] is not a zero set of a polynomial in [imath]\mathbb{R}[x,y][/imath]. Clearly, square is the zero set of a function [imath] f: \mathbb{R}^2\to \mathbb{R}; [/imath] [imath] (x,y)\mapsto |x|+|y|-1, [/imath] which is not a polynomial. But how to show that there is no polynomial which has square as the zero set? |
2368250 | Is this approximation for a factorial a known formula?
I saw this approximation in some notes: [imath]n! \approx e^{-n}*n^n*\sqrt{2\pi n}[/imath] What is this approximation? How is it derived? Is it a known formula? | 1867985 | Proof of Stirling's Formula using Trapezoid rule and Wallis Product
I need a proof of stirling's formula which uses the riemann's sum and trapezoid approximation to come up with [imath] \frac {n!}{(n/e)^n \sqrt n}[/imath] [imath] \rightarrow C[/imath] where [imath]C[/imath] is derived from Wallis product. I tried searching the internet but was not able to come up with anything. |
2368661 | Avg Number of random numbers between 0 and 1 required to add up to 1
A reddit comment claimed that you would require [imath]e[/imath] random numbers between [imath]0[/imath] and [imath]1[/imath] to add up to [imath]1[/imath]. So I wrote a python program to verify this. Conducted the experiment with 100 thousand times 200 times. The Standard deviation of the results was [imath]0.00271225[/imath] ( from [imath]e[/imath], not from their mean ). I Have two questions, first, while conducting the experiment, the result was undoubtedly over 1 most of the times. What should I do to the number of random number required in these cases. The above [imath]SD[/imath] does NOT take in account those excess over 1. It treats them as if the total was [imath] = 1[/imath] Second, how to prove this? Here's the program if you can understand Python Program (Sorry, cant post image directly) | 2336978 | Compute [imath]P(X_1+\cdots+X_k\lt 1)[/imath] for [imath](X_i)[/imath] i.i.d. uniform on [imath](0,1)[/imath]
Consider [imath]N=\min\{n: S_n>1\}[/imath], where [imath]S_n=X_1+\cdots+X_n[/imath] and [imath](X_i)_{i=1}^\infty[/imath] is i.i.d. uniform on [imath](0,1)[/imath]. So, [imath]N[/imath] is the first time that [imath](S_n)_{n=1}^\infty[/imath] crosses [imath]1[/imath]. I'd like to calculate [imath]E(N)[/imath]. To this end I'd like to calculate [imath]\Pr(N\gt k)=\Pr(S_k\lt 1)[/imath]. Is there a way to actually do this for any arbitrary [imath]k[/imath]? I mean, certainly there is as I know about the distribution of [imath]S_n[/imath] but the formula is just formidable when you move with [imath]k[/imath] toward [imath]\infty[/imath] and even though tractable with a computer, I don't think it's tractable without it. Is there maybe a better way to calculate the expectation? Thanks for any help. |
2368681 | Finding [imath]\int \frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}[/imath]
Find [imath]\displaystyle\int \frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}[/imath], where [imath]a,b[/imath] are constants. I tried it some specific values of [imath]a,b[/imath] and tried many substitutions but it doesn't seem to work. Just throw me a hint. Thanks. | 1221897 | integrate [imath]dx/(a^2 \cos^2x+b^2 \sin^2x)^2[/imath]
Integrate [imath]\dfrac{dx}{(a^2 \cos^2x+b^2 \sin^2x)^2}[/imath]. I can go up to the residue formula like in this example here but then I just can't end up with the result for when [imath]n=2[/imath]. I keep messing up my math. Can someone please show me how to get to the answer when [imath]n=2[/imath]? I think I've spent 3+ hours trying to get to it and no luck unfortunately. Thank you! |
2368691 | How can I prove submultiplicative of [imath]p[/imath]-norm of matrix?
How can I prove [imath]\|AB\|_p ≤ \|A\|_p \|B\|_p[/imath] where [imath]\|A\|_p = \max_{x≠0}\frac{\|Ax\|_p}{\|x\|_p}, \qquad p = 1, 2, 3, \ldots[/imath] and known information below: (i) [imath]\|A\| ≥ 0[/imath], with equality iff [imath]A = 0.[/imath] (ii) [imath]\|cA\| = |c| \|A\|[/imath], for any [imath]c ∈ \mathbb R.[/imath] (iii) [imath]\|A + B\| ≤ \|A\| + \|B\|[/imath]. *Not duplicate of Proof of matrix norm property: submultiplicativity, I don't know whats the connection between that answer an my question. | 487855 | Proof of matrix norm property: submultiplicativity
I've been searching for the definition of the submultiplicative (I think it has multiple names from what I've seen) property in proof form. Some books define it as part of the properties that define matrix norms, and some include it as an additional property. I still haven't been able to work it out for myself or find it anywhere. Let [imath]A[/imath] and [imath]B[/imath] be [imath]n\times m[/imath] and [imath]m\times l[/imath] matrices respectively, prove that: [imath]\begin{align} \|AB\| \le \|A\|\|B\| \end{align}[/imath] |
2075396 | Power of Series
In taking the power of a series [imath]\left(\sum_{k=0}^{\infty} a_k x^k \right)^n = \sum_{k=0}^{\infty} c_k x^k[/imath] do you know an expression for [imath]c_k[/imath] solely in terms of the coefficients [imath]a_k[/imath]? | 1782316 | Series expansion of infinite series raised to the [imath]n[/imath]th power
So I know there is a well-known straightforward way to expand something like [imath](a+b)^n[/imath] and that there are formulas which allow us to expand trinomials and multinomials in general. My question is, Is there any known way to expand something like [imath]\left[\sum_{k=0}^{\infty} a_k\right]^n[/imath] or at least to determine the first few terms? |
2367945 | Rational integral
[imath]\int\dfrac{x^2}{1+x^4}dx[/imath] I tried many standard approaches, but I didn't get too far! Here's the most promising of them: [imath]\int\dfrac{dx}{\frac{1}{x^2}+x^2}[/imath] knowing that [imath]\left(1/x+x\right)^2=\frac{1}{x^2}+x^2+2[/imath] we can change variables [imath]1/x+x=t[/imath]. Unluckily this doesn't work either. | 349424 | Integrate rational function [imath]\frac{x^2}{1+x^4}[/imath]
Integrate [imath]\int\frac{x^2dx}{1+x^4}[/imath] I've factored the denominator to [imath](x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)[/imath] and got stuck. |
2368962 | Abstract algebra ring of polynomials
Let [imath]K[/imath] be a field and [imath]K[x][/imath] be the ring of polynomials over [imath]K[/imath] in a single variable [imath]x[/imath] for a polynomial [imath]f[/imath] belong to [imath]K[x][/imath]. Let [imath](f)[/imath] denote the ideal in [imath]K[x][/imath] generated by [imath]f[/imath]. show that [imath](f)[/imath] is a maximal ideal in [imath]K[x][/imath] if and only iff [imath]f[/imath] is an irreducible polynomial over [imath]K[/imath]. | 1888679 | [imath]\langle r\rangle[/imath] maximal [imath]\iff r[/imath] irreducible
Let [imath]R[/imath] be a PID, and [imath]r\in R- \{0\}[/imath]. Prove that [imath]\langle r\rangle[/imath] maximal [imath]\iff r[/imath] irreducible. "[imath]\Leftarrow[/imath]"Easy. "[imath]\Rightarrow[/imath]"If [imath]J=\langle r \rangle[/imath] then we will prove that [imath]r[/imath] is irreducible. If [imath]r=ab[/imath], we want to prove that [imath]a\in U(R)[/imath] or [imath]b \in U(R)[/imath]. If we take the ideal which is generated by [imath]\langle a\rangle[/imath] then (because [imath]J[/imath] is maximal)[imath]\langle a\rangle \subseteq\langle r \rangle \iff r\mid a \iff a=kr, k\in R \Longrightarrow r=krb\iff r(1-kb)=0_R \iff kb=1_R[/imath] so [imath]b\in U(R)[/imath]. Same way if we work with [imath]\langle b \rangle[/imath]. Is this proof right? |
2368473 | If [imath]n[/imath] elements generate a free module of rank [imath]n[/imath], are they necessarily linearly independent?
Let [imath]A[/imath] be a nonzero commutative ring with [imath]1[/imath], and [imath]n[/imath] a positive integer. If [imath]b_1,\ldots,b_n\in A^n[/imath] generate [imath]A^n[/imath] as an [imath]A[/imath]-module, is it true that [imath]b_1,\ldots,b_n[/imath] are linearly independent over [imath]A[/imath] (hence a basis)? I suspect this is true (It is true if [imath]n=1[/imath]), but I can't seem to prove it. I guess an equivalent way of asking this question is: if a linear map [imath]A^n\to A^n[/imath] is surjective, is it necessarily injective? Any help is welcome! | 241379 | Generators of a finitely generated free module over a commutative ring
Let [imath]L[/imath] be a finitely generated free module over a commutative ring [imath]A[/imath]. Set [imath]n=\operatorname{rank} L[/imath]. Let [imath]x_1,\dots,x_m[/imath] be generators of [imath]L[/imath]. Then [imath]m \ge n[/imath]? If [imath]m = n[/imath], then is [imath]x_1,\dots,x_m[/imath] a basis of [imath]L[/imath]? |
2369197 | Finding the matrix of a specific scalar product
Let [imath]V[/imath] be the vector space over [imath]\mathbb{R}[/imath] whose basis is [imath](\sin(t),\cos(t))[/imath]. Let the scalar product be defined by [imath]\langle f,g \rangle =\int_{-\pi}^{\pi} f(t)g(t)\,\mathrm dt[/imath] What is the matrix of this scalar product with respect to the given basis? [imath]\langle f,g \rangle = \int_\limits{-\pi}^{\pi} f(t)g(t)\,\mathrm dt = \int_\limits{-\pi}^{\pi}(\sin (t)+\cos(t))(\sin (t)+\cos(t))\,\mathrm dt = 2\pi[/imath] From a previous exercise with polynomials, I acquired this method. However I do not get a matrix but a number. Am I doing anything wrong? I do not know why people consider it a duplicate, if I am not getting the same result using the same method. Thanks in advance. | 2368876 | Find the matrix of [imath]\langle f, g \rangle = \displaystyle\int_0^1 f(t) g(t) \, \mathrm dt[/imath] with respect to the basis [imath]\{1,t,\dots,t^n\}[/imath]
Let [imath]V[/imath] be the vector space over [imath]\mathbb{R}[/imath] consisting of all polynomials of degree [imath]\leqslant n[/imath]. If [imath]f,g\in V[/imath], let [imath]\langle f,g \rangle = \int_\limits{0}^{1}f(t)g(t)dt[/imath] Find the matrix of this scalar product with respect to the basis [imath]\{1,t,\dots,t^n\}[/imath]. What does it mean to find a matrix of the scalar product? Thanks in advance! |
2369671 | [imath](1+x/n)^{1/n}[/imath] monotone increasing?
Given the sequence [imath]\{x_n\}=(1+1/n)^n[/imath], by using Bernoulli's inequality, [imath](1+a)^n>1+{na}[/imath], and a little creative, clever algebra, it is a straightforward proof to show that [imath]x_{n+1}/x_n>1[/imath] and hence [imath]\{x_n\}[/imath] is a monotone increasing sequence. I thought that the same approach (the ratio of consecutive elements) and the same basic set of algebraic constructions might be used to show that [imath]\{x_n\}=(1+x/n)^n[/imath] for [imath]x>0[/imath] is also a monotone increasing sequence but it's escaping me, I'm afraid. I've seen other solutions that refer to the arithmetic-geometric mean inequality but I'd like to attempt a more basic solution that doesn't refer to that. Any hints to push me in the right direction? | 2052328 | Show that [imath]\left(1+\frac xk\right)^k[/imath] is monotonically increasing
Note: I don't want to get the full solution, but only a hint. I have to show that for [imath]x \in [0, \infty)[/imath], the sequence [imath]\left(1+\frac xk\right)^k[/imath] is monotonically increasing. We were already given the hint that we could use the inequality of arithmetic and geometric means, but I don't see how to apply it yet. I tried to do it by induction. While the case [imath]k = 1[/imath] works easily, I don't know how to start further from here. I already tried to use the definition of the binomial theorem, but it didn't lead me anywhere. |
2370127 | Continuity of the linear operator [imath]A:V\to V'[/imath] induced by the bilinear continuous form [imath]a:V\times V\to{\bf R}[/imath]?
Let [imath]V[/imath] be a Hilbert space with inner product [imath](\ ,)[/imath] and norm [imath]\|\cdot\|[/imath]. Suppose [imath]a:V\times V\to{\bf R}[/imath] is a bilinear continuous form. One can then define [imath]A:V\to V'[/imath] with [imath]A(u)=\xi_u[/imath] where [imath]\xi_u(v)=a(u,v)[/imath] for all [imath]v\in V[/imath]. One can show by checking the definition that [imath]A[/imath] is linear. Question: how to show that [imath]A[/imath] is continuous? Suppose [imath]u_n\to u[/imath] in [imath]V[/imath]. I end up with showing the estimate [imath] \sup_{v\in V, \|v\|=1}|a(u-u_n,v)|<\epsilon. [/imath] For each [imath]v[/imath], by the continuity of [imath]a[/imath], one has for some constant [imath]M_v[/imath], [imath] |a(u-u_m,v)|\leq M_v\cdot \|u-u_m\|\tag{*} [/imath] But I don't see how to get a bound for [imath]\{M_v:\|v\|=1\}[/imath] so that (*) can be used. | 808983 | Condition for continuity of bilinear form
In my numeric script there is a unproved theorem, saying that a bilinear form [imath]a \colon V\times V \to \mathbb{R}[/imath] on a normed vector space [imath]V[/imath] is continuous if and only if [imath]|a(v,w)| \leq c \, \|v\| \, \|w\|[/imath] holds for all [imath]v,w\in V[/imath] for some [imath]c > 0[/imath]. My first question is: what is ment by a continuous bilinearform? Is it according to the norm [imath]\| (v,w) \| := \max \{ \|v\| , \|w\| \}[/imath] (which is equivalent to [imath] \|(v,w)\| = \|v\| + \|w\|[/imath]) ? If so, then I agree that such a bilinear form is continuous but I don't see that a continuous bilinear form is bounded as above. Can anyone explain this to me? |
2370269 | A question about the free resolution of an abelian group.
On pg 195 of the book "Algebraic Topology", Hatcher says the following: Every abelian group [imath]H[/imath] has a free resolution of the form [imath]0\to F_1\to F_0\to H\to 0[/imath], with [imath]F_i=0[/imath] for [imath]i>1[/imath]. He justifies this by saying that the kernel of the map [imath]F_0\to H[/imath] is a subgroup of the free group [imath]F_0[/imath], and hence also free. This would imply that [imath]F_1[/imath] is free, and hence the map [imath]F_1\to F_0[/imath] would have a trivial kernel. Why isn't this true for non-abelian groups? Wouldn't the same argument be applicable even there? | 1555753 | Two term free resolution of an abelian group.
This is probably a very easy question but I think I am missing some background regarding free abelian groups to answer it for myself. In Hatcher's Algebraic Topology, the idea of a free resolution is introduced in the section on cohomology. A [imath]\textbf{free resolution}[/imath] of an abelian group is an exact sequence [imath]F_2 \to F_1 \to F_0 \to H \to 0[/imath] such that each [imath]F_i[/imath] is free. Let [imath]f_0:F_0 \to H[/imath] and choose a set of generators of [imath]H[/imath]. Let [imath]F_0[/imath] be the free abelian group with basis in one-to-one correspondence with this set of generators. Then we can easily form the two term free resolution [imath]\to 0 \to Ker(f_0) \to F_0 \to H \to 0[/imath] Why is [imath]H[/imath] an abelian group a necessary condition so that there necessarily exists resolution of the form [imath]0 \to F_1 \to F_0 \to H \to 0[/imath]? For what non-abelian group does such a free resolution not exist? |
2370410 | How to prove equality of the cardinality two sets?
The question states: Prove that [imath]|(0,1)| = |(-e, \pi)|[/imath]. I am assuming this has something to do with the Cantor-Bernstein-Schröeder Theorem, but I don't really have a good understanding as to how to apply it. How would I show this proof? | 1659870 | The intervals [imath](2,4)[/imath] and [imath](-1,17)[/imath] have the same cardinality
I have to prove that [imath](2,4)[/imath] and [imath](-1,17)[/imath] have the same cardinality. I have the definition of cardinality but my prof words things in the most confusing way possible. Help! |
2370669 | In probability theory what does the notation [imath]\int_{\Omega} X(\omega) P(d\omega)[/imath] mean?
Where [imath]P[/imath] is a probability measure on the space [imath]\Omega[/imath] and [imath]X[/imath] is a real-valued random variable? Specifically what does [imath]P(d\omega)[/imath] mean and how is it different from [imath]dP[/imath] together with the integral for [imath]\Bbb{E}(X)[/imath] involving supremum of approximative simple functions? I'm going through the book A Basic Course in Probability Theory and some things they don't explain. That's fine though, other than that it's a great book, and I already know some measure theory. So I'm at the part about simple functions (which I've read about / worked with before) and the change of variables formula, where the above notation is used: [imath]\Bbb{E}(h(X)) \equiv \int_{\Omega} h(X(\omega))P(d\omega) = \int_{S} h(x) Q(dx)[/imath], provided one of the two integrals exists, and [imath]h: \Bbb{S} \to \Bbb{R}[/imath] is Borel measurable. | 1217971 | Abstract Integration in Elementary Probability Theory
In measure theoretic probability I often see these two notations for the expectation of a random variable expressed as an abstract integral. [imath] \int_\Omega X(\omega) \mathbb{dP(\omega)} = \int_\Omega X(\omega) \mathbb{P(d\omega)} [/imath] Is each case trying to express a different view of the abstract integral? What is an intuitive way to interpret [imath]\mathbb{dP(\omega))}[/imath] and [imath]\mathbb{P(d\omega)}[/imath]? |
2370277 | Differentiating a matrix function
In the book "Elements of Statistical Learning", early on the author is discussing linear regression, and naturally discusses the residual sum of squares (RSS) based on the parameter space [imath]\boldsymbol{\beta}[/imath]. In the general formulation, [imath]\text{RSS}(\beta) = (\boldsymbol{y} - \boldsymbol{X}\beta)^T(\boldsymbol{y} - \boldsymbol{X}\beta)[/imath] where [imath]\boldsymbol{X}[/imath] is an [imath]N \times p[/imath] matrix and [imath]\beta[/imath] is a [imath]p \times K[/imath] matrix. The author then says to minimize RSS, you differentiate with respect to [imath]\beta[/imath] and get [imath]\boldsymbol{X}^T(\boldsymbol{y} - \boldsymbol{X}\beta) = 0[/imath] My question is, what are the mechanics of differentiating with respect to the matrix [imath]\beta[/imath]? I have a B.S. in physics, so I have a reasonably sophisticated math background, but I never covered this in my undergraduate education. I tried looking a bit into "Matrix calculus", but it wasn't much help. Is that the correct term? If this is the language used in the remainder of the textbook, what are some good resources somewhat familiar with vector calc and linear algebra to learn "matrix calc"? | 2232764 | Taking derivatives with respect to a matrix
I am reading Elements of Statistical Learning, and they derive the equation for linear regression using differentiation on a function which takes a matrix as input. This is making my head spin a bit and I am looking for some resources to practice this type of differentiation, and become more comfortable with it. This is the problem. Let [imath]X \in \mathbb{R}^{N \times p}[/imath], [imath]b \in \mathbb{R}^{p \times k}[/imath], and [imath]y \in \mathbb{R}^{N \times k}[/imath]. Define [imath]RSS(X, y) = \operatorname{tr}[(Xb - y)^t(Xb - y)][/imath]. I want to differentiate [imath]RSS[/imath] with respect to [imath]b[/imath] and set the derivative to zero since I want to minimize the function, [imath]RSS[/imath]. The book claims this is [imath]X^t(Xb - y) = 0[/imath], which gets us, [imath]b = (X^tX)^{-1}Xy[/imath] if [imath](X^tX)[/imath] is singular. It does not show how it applied the derivatives (which is expected since thats not what this book is about). I can sort of reason that we have some sort of product rule happening. It seems likely that [imath]d/db(\operatorname{tr}(A)) = \operatorname{tr}(d/db(A))[/imath]. But right now I am just writing down symbols that somehow represent some ideas in my head. In my Principles of Mathematical Analysis book (by Rudin) he covers when the input is a vector, and the output can be a vector. However, this seems to be input is a matrix, and the output in some of the subproblems is a matrix. Any recommendations for books I can look at? |
2370738 | Efficient way to find [imath][\mathbb{Q}(\sqrt{5}, \sqrt{3}, \sqrt{2}): \mathbb{Q}(\sqrt{3}, \sqrt{2})][/imath]
I want to show rigorously that this is 2. I'm sure there's a faster way than by trying to see if [imath]\sqrt{5} = a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}[/imath] (linear combination of the basis elements for [imath]\mathbb{Q}(\sqrt{3}, \sqrt{2})[/imath]). I'm just not sure what. | 1713529 | Finding basis of [imath]\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})[/imath] over [imath]\mathbb{Q}[/imath]
I'm having trouble finding a basis for [imath]\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})[/imath]. So far I know that [imath][K=\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2[/imath] and [imath][L=\mathbb{Q}(\sqrt{2}, \sqrt{3}):K] = 2[/imath], but it's harder to extend [imath]L[/imath] to [imath]L(\sqrt{5})[/imath]. Namely I'm trying to prove that [imath]x^2-5[/imath] is irreducible over [imath]L[/imath] by showing that [imath]\sqrt{5} \notin L[/imath] as follows: Assume by contradiction that [imath]\sqrt{5} \in L[/imath]. Then we have [imath]\sqrt{5} = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}[/imath] with [imath]a,b,c \in \mathbb{Q}[/imath]. Squaring both sides and using independence of basis gives the following nasty system of equations: [imath]-5+a^2+2b^2+3c^2+6d^2 = 0[/imath] [imath]ab + 3cd = 0[/imath] [imath]ac + 2bd = 0[/imath] [imath]ad + bc = 0[/imath] I can't see how these equations contradict the assumption that [imath]\sqrt{5} \in L[/imath]. |
2370832 | How to solve the integral [imath]\int \sqrt {\sin x+1}[/imath]
How to solve the integral [imath] \int{\sqrt{\sin x+1} }\hspace{.1cm} dx[/imath] with steps. | 1901857 | How to evaluate the integral [imath]\int \sqrt{1+\sin(x)} dx[/imath]
To find: [imath]\int \sqrt{1+\sin(x)} dx[/imath] What I tried: I put [imath]\tan(\frac{x}{2}) = t[/imath], using which I got it to: [imath]I = 2\int \dfrac{1+t}{(1+t^2)^{\frac{3}{2}}}dt[/imath] Now I am badly stuck. There seems no way to approach this one. Please give a hint. Also, can we initially to some manipulations on the original integral to make it easy? Thank you. |
2340695 | Solution of nonlinear ODE: [imath]x= yy'-(y')^2[/imath]
How to solve [imath]x= yy'-(y')^2.[/imath] Can somebody please hint at some substitution or refer any text related to these type of ode. | 2375942 | Solving ODE of first order
Solve the differential equation [imath]x = py - p^2[/imath] where [imath]p = \frac{dy}{dx}.[/imath] I have differentiated the equation with [imath]y[/imath] and replaced LHS by [imath]\frac1p[/imath] but after that I couldn't solve the equation. |
2365015 | Minimum value of [imath] f(x,y,z)=\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2. [/imath]
If [imath]x>0[/imath], [imath]y>0[/imath], [imath]z>0[/imath] and [imath]x+y+z=6[/imath] then find the minimum value of [imath] f(x,y,z)=\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2. [/imath] Thanks in advance. | 2287621 | Minimum value of [imath]\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2[/imath]
Given [imath]a,b,c \in \mathbb{R^+}[/imath] such that [imath]a+b+c=12[/imath] Find Minimum value of [imath]S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2[/imath] My Try: By Cauchy Schwarz Inequality we have [imath]\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)\le \sqrt{3}\sqrt{S}[/imath] [imath]\implies[/imath] [imath]\sqrt{3S} \ge 12+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/imath] Now by [imath]AM \ge HM [/imath] inequality we have [imath]\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}[/imath] [imath]\implies[/imath] [imath]\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{3}{4}[/imath] hence [imath]\sqrt{3S} \ge 12+\frac{3}{4}=\frac{51}{4}[/imath] hence [imath]3S \ge \frac{2601}{16}[/imath] so [imath]S \ge \frac{867}{16}[/imath] is this approach correct and any better approach please share. |
2371100 | Finite Abelian group with the given property
Let [imath]G[/imath] be a finite Abelian group such that it contains a subgroup [imath]H_0 \neq (e)[/imath] which lies in every subgroup [imath]H \neq (e)[/imath]. Prove that [imath]G[/imath] must be cyclic. What can you say about order of [imath]G[/imath] ? I have examples to illustrate this result, but i can no idea about the general proof. Here's my examples: 1) Take [imath]G=\Bbb{Z}_8[/imath] and [imath]H_0=\{0,4\}[/imath]. Clearly [imath]H_0[/imath] lies every other subgroups [imath](\neq(e))[/imath], namely lies in [imath]\{0,2,4,6\}[/imath] and lies in [imath]\Bbb{Z}_8[/imath] Here, [imath]o(G)=8=2^3[/imath] Edit: 2) Similarly, take [imath]G=\Bbb{Z}_{16}[/imath] and [imath]H_0=\{0,8\}[/imath]. Clearly [imath]H_0[/imath] lies every other subgroups [imath](\neq(e))[/imath], namely lies in [imath]\{0,4,8,12\}[/imath], [imath]\{0,2,4,6,8,10,12,14\}[/imath] and lies in [imath]\Bbb{Z}_{16}[/imath] Here, [imath]o(G)=16=2^4[/imath] 3) Likewise, [imath]G=\Bbb{Z}_{4}[/imath] and [imath]H_0=\{0,2\}[/imath]. Clearly [imath]H_0[/imath] lies every other subgroups [imath](\neq(e))[/imath], namely lies in [imath]\Bbb{Z}_{4}[/imath] Here, [imath]o(G)=4=2^2[/imath] From this examples, can i conclude [imath]o(G)=p^n[/imath], for a prime [imath]p[/imath] ? Because, if [imath]o(G)=p^n[/imath], then there are unique subgroups of order [imath]1,p,p^2,p^3,..,p^{n-1}[/imath] and in this case we take [imath]H_0[/imath] as a subgroup of order [imath]p[/imath], then it is immediately lies in other subgroups. Is this correct ? What is the general proof? | 1863141 | A finite abelian group containing a non-trivial subgroup which lies in every non-trivial subgroup is cyclic
Let [imath]G[/imath] be a finite abelian group s.t. it contains a subgroup [imath]H_{0} \neq (e)[/imath] which lies in every subgroup [imath]H \neq (e) [/imath]. Prove that [imath]G[/imath] must be cyclic. Also what can be said about [imath]o(G)[/imath] ? I'm clueless about this problem from Herstein. I tried to think [imath]G[/imath] might be a solvable group and its subgroups are all normal (as [imath]G[/imath] is abelian) . But I'm unable to get further. |
2371668 | Converse of Lagranges theorem question:
For a finite group [imath]G[/imath] and a subgroup [imath]H,[/imath] Lagrange's theorem says that [imath]|G|=|G:H||H|,[/imath] where [imath]|G:H|[/imath] is the number of cosets of [imath]H[/imath] in [imath]G.[/imath] My question is for any subgroup [imath]H[/imath] can we find another subgroup with order [imath]|G:H|[/imath]? | 2039378 | Why is the Converse to Lagrange's Theorem False?
Can somebody explain me how is the converse of the Lagrange's theorem false by saying that [imath]A_4[/imath] (alternating group) doesn't contain any subgroup of order [imath]6[/imath]? |
2371499 | Limit of a periodic function existing
If [imath]\lim_{n \rightarrow \infty}f{\left(n\right)}=k[/imath] and [imath]f{\left(x\right)}[/imath] is periodic and continuous, is that enough to imply that [imath]f{\left(x\right)}[/imath] is a constant function? | 1582055 | Limit of a periodic function
I stumbled upon this question in my course, and I am out of ideas. Let [imath]f[/imath] be a periodic function [imath]f(x)=f(x+l), \qquad l>0[/imath] Prove that if it is not constant, then [imath]\lim_{x\to 0}f\left(\frac1x\right)[/imath] does not exist. I don't understand why it's true, let alone how to prove it. |
2370686 | if [imath]f:[0,1] \to \mathbb{R}[/imath] is increasing, show that [imath]f[/imath] is the pointwise limit of a sequence of continuous functions over [imath][0,1][/imath]
if [imath]f:[0,1] \to \mathbb{R}[/imath] is increasing, show that [imath]f[/imath] is the pointwise limit of a sequence of continuous functions over [imath][0,1][/imath] Intuitively this makes sense but I am having trouble with showing why there would be a sequence of continuous functions converging pointwise to [imath]f[/imath]. Clearly there is a sequence converging pointwise to [imath]f[/imath], I can set: [imath]\forall n \in \mathbb{N}, f_n = f[/imath]. How to prove there is at least one which is made up of continuous functions [imath]f_n, \forall n \in \mathbb{N}[/imath] over [imath][0,1][/imath] I can't quite figure out the argument. | 540241 | Increasing functions are Baire one
[imath]\newcommand{\R}{\mathbb R}[/imath] Let [imath]f: \R \to \R[/imath] be an increasing function. [imath](a \leq b \Rightarrow f(a) \leq f(b))[/imath]. I want to prove that [imath]f \in \mathcal B[/imath], the Baire one functions. I first tried to prove that [imath]g(x) := \lfloor f(x)+x \rfloor[/imath] is in [imath]\mathcal B[/imath]. The points of discontinuity of [imath]g[/imath] are countable since [imath]g[/imath] is increasing. Now, let [imath]\mathcal D[/imath] be the points in [imath]\R[/imath] where [imath]g[/imath] is not continuous. If [imath]z \in \mathcal D[/imath] then I want to prove that [imath]\exists \delta > 0: (z-\delta,z+\delta) \cap \mathcal D = \{z\}[/imath], i.e. [imath]z[/imath] is the only discontinuity in a certain neighborhood. Then it is not hard to prove that [imath]g \in \mathcal B[/imath] and that [imath]f[/imath] is the uniform limit of [imath]\mathcal B[/imath] functions and thus is also [imath]\mathcal B[/imath]. I already tried to prove this (existence of [imath]\delta > 0[/imath]) by contradiction. However, I could not see a nice way to get here a contradiction. |
2371875 | Let [imath]G = \langle a\rangle[/imath] a cyclic group of order m. Prove that G is isomorphic with [imath]Z_m[/imath].
Let [imath]f[/imath]: [imath]Z_m \rightarrow G[/imath], [imath]f(n)[/imath] = [imath]a^n[/imath]. Show that f is a function and an homomorphism. Doing a review on the whole semester, and a little refresh on this topic would be great. By definition I know [imath]G[/imath] has the same order of [imath]Z_m[/imath], but not sure how to attack this problem with ease. I also know that cyclic groups of the same order are isomorphic. (I cannot use the fact that cyclic groups of the same order are isomorphic, that's why I ask for another approach, because I can't use theorems or lemmas that have not been discussed in class.) | 2182419 | Proving that any finite cyclic group of order [imath]n[/imath] is isomorphic to [imath][/imath]
The proof goes on like this : [imath]Z[/imath] denotes integers. [imath]a^m=e[/imath] for some positive integer [imath]m[/imath]. Let [imath]n[/imath] be the smallest positive integer such that [imath]a^n=e[/imath] if [imath]s \epsilon Z[/imath] and [imath]s=nq+r[/imath] for [imath]0\le r < m[/imath] then [imath]a^s=a^{nq+r}=a^r \ \ \ (*)[/imath]. Then after this statement, the book proves that all elements of the finite cyclic group is distinct and creates a homorphism between this group and [imath]<Z_{n},+_{n}>[/imath] What I couldn't get in the proof is what does [imath](*)[/imath] tell us? I couldn't understand it and why is it needed in the proof? Also why do we work with positive integers, why do we let [imath]n[/imath] be the smallest integer instead of minimizing [imath]|n|?[/imath] |
2372231 | Evaluate expression in terms of Beta function
Evaluate [imath]\int_0^1\frac{x^{(m-1)} + x^{(n-1)}}{(1+x)^{(m+n)}}dx[/imath] in terms of Beta function. I am able to obtain the answer if the limits were 0 to infinity, but not with 0 to 1. Please help. | 1843621 | Evaluate[imath]\int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}}dx[/imath] in terms of Beta function
I have no idea of Beta functions. Based on some properties of beta functions like, [imath]B(m,n) = \int_{0}^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx \;\; ;B(m,n) = B(n,m)[/imath] I arrived at [imath]2B(m,n) = I+ \int_{1}^\infty \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}[/imath] where I is my required integral. How to express the 2nd integral in terms of Beta function or is there any other way of solving. Need help. Thank You. |
2372583 | An equivalent condition to normality of topological spaces
I wonder if this claim is true: Let [imath]X[/imath] be a topological space. Then [imath]X[/imath] is normal iff for every two open subsets [imath]U,V[/imath] such that [imath]U\cup V=X[/imath], there exist two closed subsets [imath]A\subset U[/imath], [imath]B\subset V[/imath] such that [imath]A\cup B=X[/imath]. Thanks in advance. | 486434 | Normal space- equivalent condition: [imath]\forall U, V[/imath] - open [imath]: U \cup V = X \ \ \ \exists F \subset U \ , G \subset V[/imath] - closed [imath]: F \cup G = X[/imath]
Could you help me prove that [imath]X[/imath] is normal [imath]\iff[/imath] for all open [imath]U, V[/imath]such that [imath]U \cup V = X[/imath] there are closed [imath]F \subset U,\ G \subset V,\ F \cup G = X?[/imath] I've been trying to prove it using other equivalent condition for [imath]X[/imath] to be normal, namely that [imath]\forall F \subset U,\ \ \ [/imath] [imath]F[/imath]-closed, [imath]U[/imath]-open [imath]\exists V[/imath]-open [imath]F \subset V \subset \overline{V} \subset U[/imath], but I keep failing. Could you help me with that? |
2372762 | Prove that a function [imath]f:\mathbb R \to\mathbb R[/imath] given by [imath]f(x) = x\left|x\right|[/imath] is a bijection
So I know in order to prove a function is bijective, you need to prove that it is both injective and surjective. I know that to prove it is an injection, I need to make [imath]f(x) = f(y)[/imath], and try to get [imath]x=y[/imath] from that, but I can't seem to manipulate the equations to do so. Also, how would I prove that this is surjective? | 2366409 | Proving [imath]f(x)=x\cdot |x|[/imath] is a bijection
Prove that the function [imath]f: R\to R[/imath] given [imath]f(x)=x\cdot |x|[/imath] is a bijection Proof 1: f is injective Suppose [imath]f(a) = f(b) \implies a\cdot |a| = b\cdot |b|[/imath] [imath]\implies (a\cdot |a|)^2 = (b\cdot |b|)^2[/imath] [imath]\implies a^4 =b^4[/imath] [imath]a=b[/imath] Thus, f is injective Proof 2: f is surjective Pick [imath]n\in R[/imath], we want to show [imath]n=f(x)[/imath] for some [imath]x\in R[/imath] [imath]\implies n= a\cdot |a|[/imath] for some [imath]a\in R[/imath] I dont know how to complete the surjective proof |
2371700 | Showing that the event: [imath]X[/imath] is continuous on [imath][0,t_0)[/imath] belongs to the given filtration.
Let [imath]( \Omega, F, P)[/imath] be a probability space. Let [imath]X[/imath] be a stochastic process defined on such probability space whose sample paths are RCLL (right continuous on [imath][0, \infty)[/imath] with finite left hand limits on [imath](0, \infty)[/imath]. Let [imath]A[/imath] be the event that [imath]X[/imath] is continuous on [imath][0,t_0)[/imath]. I am tasked in showing that [imath]A \in F^X_{t_0}[/imath], where [imath]F^X_{t_0}:= \sigma(X_s ; 0 \le s \le t) = \{ \{ X_s \in B \}\subset \Omega : B \in \mathcal{B}(\Bbb{R}), 0 \le s \le t \}[/imath] Some may recognize this exercise as one of the first from Brownian Motion and Stochastic Calculus by Karatzas and Shreve, I am in fact trying to self study this book through august. My problem here is that I can't figure out exactly what [imath]A[/imath] represents, I know that for a fixed [imath]w \in \Omega[/imath] I have that [imath]X_s(w)[/imath] is RCLL but I can't figure out how to start. | 112477 | Show that a certain set is measurable
I just started to working through Karatzas and Shreve "Brownian Motion and Stochastic Calculus". Solving an exercise there's some questions around. I want to prove. Let [imath]X[/imath] be a stochastic process, every sample path of which is RCLL. Let [imath]A[/imath] be the event that [imath]X[/imath] is continuous on [imath][0,t_0)[/imath]. Show that [imath]A\in \mathcal{F}_{t_0}^X[/imath], where the latter is defined as, [imath]\mathcal{F}_{t_0}^X:=\sigma{(\{X_s;0\le s\le t_0\})}[/imath] What I did (sketch): Write [imath]A[/imath] as, [imath] A=\{\omega \in \Omega; \lim_{s\uparrow t}X_s(\omega)=\lim_{s\downarrow t}X_s(\omega)\}[/imath] for [imath]t\in [0,t_0)[/imath]. Since the Limit of a sequence of measurable functions is measurable, this is in [imath]\mathcal{F}_{t_0}^X[/imath]. For more details see the answer of Byron Schmuland below. Now here's the part which is unclear. There's a second claim, very similar to the first one: Let [imath]X[/imath] be a stochastic process whose sample paths are RCLL a.s. Let [imath]A[/imath] be the event that [imath]X[/imath] is continuous on [imath][0,t_0)[/imath]. Show that [imath]A[/imath] can fail to be in [imath]\mathcal{F}_{t_0}^X[/imath]. But if [imath]\{\mathcal{F_t;t\ge 0}\}[/imath] is a filtration satisfying [imath]\mathcal{F}^X_t\subset\mathcal{F}_t,t\ge 0[/imath] and [imath]\mathcal{F}_{t_0}[/imath] is complete under P, then [imath]A \in \mathcal{F}_{t_0}[/imath]. There's a sample solution to this second question. But what I don't get is the difference between the two questions. What exactly is the difference? As mentioned in my comment after Byron Schmulands answer I think this is more a linguistic problem. cheers math |
2373358 | Miller-Rabin special case
I have question about a homework. Why can we be [imath]100\%[/imath] sure after [imath]4[/imath] tests for the special case of [imath]p=13[/imath]? I don't get it. I have a formula that states, that every test increases certainty by [imath]\left(\frac14\right)^k[/imath], where [imath]k[/imath] is the number of tests. So after [imath]4[/imath] tests I get around [imath]99.6\%[/imath] certainty, not [imath]100\%.[/imath] | 2370079 | Miller-Rabin-Test : Why can we be [imath]100\%[/imath] certain after [imath]4[/imath] tests for [imath]p=13[/imath]?
I have question about a homework. The question is: Why can we be [imath]100\%[/imath] sure after [imath]4[/imath] tests for the special case of [imath]p=13[/imath]? I don't get it. I have a formula that states, that every test increases certainty by [imath]\left(\frac14\right)^k[/imath], where [imath]k[/imath] is the number of tests. So after [imath]4[/imath] tests I get around [imath]99.6\%[/imath] certainty, not [imath]100\%.[/imath] |
2373510 | Differentiable function which sends rationals to rationals and irrationals to irrationals
Function [imath]f : \mathbb{R} \to \mathbb{R}[/imath] defined for all [imath]x\in\mathbb{R}[/imath]. It is also continuous and has derivative for all [imath]x \in \mathbb{R}[/imath]. Does the condition that there are no points [imath]x\in \mathbb{Q}[/imath] such that [imath]f(x) \in \mathbb{R}\setminus \mathbb{Q}[/imath] and no points [imath]x \in \mathbb{R}\setminus\mathbb{Q}[/imath] such that [imath]f(x) \in \mathbb{Q}[/imath] imply that [imath]f[/imath] is a linear function? I tried to construct a counterexample, however I did not succeed. Are there any ideas, please? | 654344 | [imath]f[/imath] differentiable, [imath]f(x)[/imath] rational if [imath]x[/imath] rational; [imath]f(x)[/imath] irrational if [imath]x[/imath] irrational. Is [imath]f[/imath] a linear function?
Let [imath]f[/imath] be an everywhere differentiable function whose domain consists of all real numbers. Assume that [imath]f(x)[/imath] is rational for rational [imath]x[/imath] and irrational for irrational [imath]x[/imath]. Can we conclude that [imath]f[/imath] is a linear function? |
2373783 | Trigonometry and Complex Numbers with Series
The number [imath]\text{cis}75^\circ + \text{cis}83^\circ + \text{cis}91^\circ + \dots + \text{cis}147^\circ[/imath] is expressed in the form [imath]r \, \text{cis } \theta,[/imath] where [imath]0 \le \theta < 360^\circ[/imath]. Find a value of [imath]\theta[/imath] in degrees. I have no idea on how to deal with sequences in any way shape or form, so I am totally lost on this problem! Please help! | 907262 | Polar form of the sum of complex numbers [imath]\operatorname{cis} 75 + \operatorname{cis} 83 + \ldots+ \operatorname{cis} 147[/imath]
The number [imath]\operatorname{cis} 75 + \operatorname{cis} 83 + \operatorname{cis} 91 +\dots+ \operatorname{cis} 147[/imath] is expressed in the form [imath]r\operatorname{cis}(\theta)[/imath], where [imath]0\leq \theta< 360[/imath]. Find [imath]\theta[/imath] in degrees I'm having major trouble with this problem. |
2373897 | If [imath]f[/imath] is continuous on all of [imath]\mathbb{C}[/imath] and analytic on [imath]\mathbb{C}[/imath] minus [imath] [-1,1][/imath] then [imath]f[/imath] is entire
I'm not sure what to do here. It seems like Morera's Theorem, but even then I'm not sure how to choose the triangular path. | 2373844 | If [imath]f: \mathbb{C} \to \mathbb{C}[/imath] is continuous and analytic off [imath][-1,1][/imath] then is entire.
This is a problem from Complex Variable (Conway's book) 2nd ed. (Section 4.4) 9. Show that if [imath]f: \mathbb{C}\to\mathbb{C}[/imath] is a continuous function such that [imath]f[/imath] is analytic off [imath][-1,1][/imath] then [imath]f[/imath] is an entire function. I already have a solution by Morera's theorem that split this problem in 5 cases. I think this solution is too long and I'm trying to solve this using a different approach. Any ideas ? |
397280 | Semi-simple Lie algebra [imath]L[/imath] coincides with its derived algebra [imath]L'[/imath]
If [imath]L[/imath] is a semi-simple Lie algebra, then [imath]L=L'[/imath]. Since [imath]L[/imath] is semi-simple we can write it as a direct sum of simple ideals [imath]L_i[/imath], i.e. [imath]L=\oplus_{i=1}^r L_i[/imath]. Then [imath]L'=\oplus_{i=1}^r L_i'[/imath] and since every [imath]L_i[/imath] is simple we have [imath]L_i'=L_i[/imath] and consequently [imath]L'=\oplus_{i=1}^r L_i'=\oplus_{i=1}^r L_i=L[/imath] This proof seems surprisingly short. So my question is, did I make any mistakes or did I forget any crucial assumptions? Is there perhaps a more elementary proof? | 658275 | Semisimple Lie algebras are perfect.
Can anyone explain why a semi-simple finite dimensional Lie algebra [imath]\mathfrak{g}[/imath] has to be perfect ? The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when [imath]\mathfrak{g}[/imath] is not perfect would be expected to be solvable. But it doesn't seem to work. |
1514073 | Sum of series: [imath]1^1 + 2^2 + 3^3 + … + n^n[/imath]
Searched every where on web but I couldn't find out the formula for this series. That's why I am asking here. I tried following following formula: [imath]((n*(n+1))/2)^n[/imath] | 1487637 | Formula : (Exact) Sum of [imath]1^1+2^2+3^3+..+n^n[/imath] (modulo [imath]10^m[/imath]) with relatively small [imath]m[/imath]
I am trying to programmatically solve mathematical problem - get sum of all powers from [imath]1^1[/imath] to [imath]1000^{1000}[/imath]. So far I have found solution by simply summing powers, but it takes way to long time and not getting any near to final 1000 integer. Question - is there a formula for this case : get sum of all number powers [imath]1^1 + 2^2 + 3^3 + .. 1000^{1000}[/imath] ? |
2374336 | Self-complementary graphs having constant vertex degree
A self-complementary graph [imath]G[/imath] on [imath]n[/imath] vertices is one which is isomorphic to its complementary graph [imath]\bar{G}.[/imath] Some such graphs, for example the cyclic graph on [imath]5[/imath] vertices, have the same degree for each vertex. I'm interested in examples of these. It's fairly easy to show in this situation one needs [imath]n=4k+1.[/imath] If such [imath]n[/imath] are also prime, I have a construction of an example based on a primitive root for [imath]n.[/imath] I'm wondering about composite [imath]n=4k+1.[/imath] Are there any of these for which a constant degree self-complementary graph exists? The first composite case is [imath]n=9,[/imath] next is [imath]n=21,[/imath] etc. Of interest would be an edample in one of these composite cases, or an impossibility proof, even for a few low values of [imath]n.[/imath] I tried it for [imath]n=9[/imath] and couldn't find one. Added note: I would like to assume [imath]G[/imath] is connected. (thisn comes from an example in a comment) | 927785 | Constructing self-complementary regular graphs
It can be easily shown that if a graph is self-complementary and regular then the number of vertices, [imath]n[/imath], is equal to [imath]4k +1[/imath] for some [imath]k \in \mathbb{Z}[/imath]. But, how to we prove (prove by constructing) that there is a self-complementary regular graph for [imath]n = 4k +1[/imath] |
2374396 | Complex analysis: relationship between a complex number and its reciprocal number
The question is: Let [imath]z_1...z_n[/imath] be non-zero complex numbers such that [imath]\sum_{i=1}^n\frac{1}{z_i}=0[/imath] Prove for any line passing through the origin, the number [imath]z_1...z_n[/imath] cannot all lie in a half-plane on either side of the line. To solve this, I tried using the argument of [imath]z[/imath] and [imath]\frac1z[/imath], but I have not got anything particularly useful. I also thought using vector notation of complex number in a 2d plane could help, but can't find a way to formalize that. Any help is much appreciated. | 185523 | A finite sum of reciprocals of complex numbers cannot be confined to a half-plane
Let [imath]z_1,\dots,z_n[/imath] be non-zero complex numbers such that [imath]\sum_{k=1}^n \frac{1}{z_k} = 0[/imath]. Prove that for any line [imath]ax+by=0[/imath] passing through the origin, [imath]z_1,\dots,z_n[/imath] cannot all lie in either of the half-spaces [imath]\{ ax+by<0\}[/imath], [imath]\{ ax+by >0\}[/imath]. Any help would be greatly appreciated, thanks in advance! |
2337341 | How to go from elliptic coordinates back to cartesian?
I know that the transform form cartesian to elliptical is [imath]x=a \cosh \mu \cos \nu; \;\;\;\; y=a \sinh \mu \sin \nu[/imath] How do I go back to cartesian if I have an expression in elliptical, meaning that I need the relations [imath]\mu = f(x,y);\;\;\;\nu=g(x,y)[/imath]? | 1529551 | Elliptic Coordinates - Inverting the transformation
The standard way to transform elliptic coordinates [imath](\mu, \nu)[/imath] [imath]\ to[/imath] Cartesian coordinates [imath](x,y)[/imath]: [imath]x = a \cosh(\mu) \cos(\nu)[/imath] [imath]y = a \sinh(\mu) \sin(\nu)[/imath] Is there any way to get the transformation [imath](x,y)[/imath] to [imath](\mu,\nu)[/imath]? Meaning is there a way to find: [imath]\mu = f(x,y)[/imath] [imath]\nu = g(x,y)[/imath] I'm guessing that it would involve [imath]\sinh^{-1}[/imath]'s and [imath]\cosh^{-1}[/imath]'s, if it was possible to do this at all. |
2374874 | Does 'All countable unions' include 'all finite unions'?
One of the requirements of a sigma-algebra is that all countable unions are in the algebra i.e. if [imath]A_n\in B[/imath] for [imath]n=0,1,2,...[/imath] ([imath]B[/imath] been our algebra and [imath]A_n\subset U[/imath]) then we require: [imath] \bigcup\limits^\infty_{n=1} A_n\in B[/imath] my question is whether this includes all finite unions i.e. do we also require: [imath]A_1 \cup A_4\in B[/imath] etc? | 2324503 | Finite union sigma field
I'm really struggling to grasp why [imath]\sigma[/imath]-fields are not necessarily closed under finite union. If the above statement is true then; Let [imath](\Omega, \mathcal{F},P) [/imath] be our probability space such that [imath](A,B \in \mathcal{F} \Rightarrow A \cup B \not\in \mathcal{F})[/imath]. Surely [imath]P(A \cup B) [/imath] then cannot be evaluated. Any advice on how I could move forward would be greatly appreciated. |
2375128 | Suppose X and Y are independent random variables and [imath]Z = X + Y[/imath]. Find the density [imath]f_Z(z)[/imath]
Let [imath]X[/imath] and [imath]Y[/imath] be independent random variables defined on the space [imath]\Omega[/imath], with density functions [imath]f_X(x)[/imath] and [imath]f_Y(y)[/imath], respectively. Suppose [imath]Z = X + Y[/imath]. Find the density [imath]f_Z(z)[/imath] if [imath]f_X(x)=f_Y(x)= \begin{cases} \frac{1}2 &, & \text{if $0<x<2$,} \\ 0 &, & \text{otherwise.} \end{cases}[/imath] Attempt: I tried to take the integral [imath]\int_{0}^{z}\frac{1}2,\mathrm dx[/imath] then I eventually did the integral to get [imath](z/2)[/imath] and [imath](1-(z/2)) [/imath]. unfortunately I was unable to figure this out | 574246 | Find the density of the sum of two uniform random variables
Let [imath]X[/imath] and [imath]Y[/imath] be independent and uniform random variables on [imath](0,1)[/imath]. Find the density of [imath]X+Y[/imath]. I know [imath]f_{X+Y}=\int_{-\infty}^{\infty}\,f_X(x)\,f_Y(z-x)dx[/imath] Now when [imath]x\in(0,1)[/imath], I know [imath]f_X(x)=1[/imath]. Assuming [imath]z\in (0,1)[/imath], I need [imath]z>x[/imath] in order for [imath]f_Y(y)=1[/imath]. In that case, [imath]f_{X+Y}=\int_{0}^{z}\,(1)\,(1)dx=z.[/imath] Now when [imath]x\in(0,1)[/imath], [imath]f_Y(y)[/imath] can also equal [imath]1[/imath] if [imath]z \in (1,2)[/imath]. In that case, I need [imath]z-x<1 \implies x>z-1.[/imath] So I have: [imath]f_{X+Y}=\int_{z-1}^{1}\,(1)\,(1)dx=2-z.[/imath] Does this all look correct? |
2374763 | sum of perfect cubes formula proof without induction
is there any proof for the sum of cubes except induction supposition? there are some proofs using induction in below page Proving [imath]1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2[/imath] using induction | 2370264 | Compilation of proofs for the summation of natural squares and cubes
I want to know different proofs for the following formulas, [imath] \sum_{i=1}^n{i^2} = \frac{(n)(n+1)(2n+1)}{6} [/imath] [imath] \sum_{i=1}^n{i^3} = \frac{n^2(n+1)^2}{2^2} [/imath] Please do not mark this as duplicate, since what I specifically want is to be exposed to a variety of proofs using different techniques (I did not find such a compilation anywhere on the net) I am only familiar with two proofs, one which uses expansion of [imath](x+1)^2 - x^2[/imath] and [imath](x+1)^3 - x^3[/imath] and the other which uses induction. I have provided link for the induction proof in a self-answer. I am particularly interested in proof without words and proofs which use a unrelated mathematical concept (higher level math upto class 12 level is acceptable). Also, (1) Don't think I am being rude or anything, it is out of genuine interest that I am asking this question. (2) Someone marked this as a duplicate of Methods to compute [imath]\sum_{k=1}^nk^p[/imath] without Faulhaber's formula My question is different in three ways: (i) I want to focus only on these two summation and not the general case, (ii) Hence, it follows that the proofs which I am looking for a simpler than the ones provided in that link and are simpler (using images, pictures or high school algebra). What I want is to study new proofs. I believe it is a good practice when learning math to so this. (iii) Since the proofs in the link are given for the general case, they are complicated and I am finding it hard to understand them. If someone is able to use the same method to the two cases in my question, then it would probably become much simpler and easier to digest. Appendix Feel free to make use of these topics in your answers, Calculus Basic Binomial Expansion Coordinate Geometry Algebra (upto what 18 year olds learn) Taylor Series Expansions Geometry (18 year old level) basically.............math which 18 year old's learn on Earth. If you want to err, then err on the higher math side:) Answers Compilation List By Newton series By Sterling Numbers By Induction From the book Generatingfunctionology by Herbert Wilf By generalizing the following pattern [imath]\begin{align} &\ \,4\cdot5\cdot6\cdot7\\=&(1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+3\cdot4\cdot5\cdot6+4\cdot5\cdot6\cdot7)\\-&(0\cdot1\cdot2\cdot3+1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+3\cdot4\cdot5\cdot6)\\ =&(1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+3\cdot4\cdot5\cdot4+4\cdot5\cdot6\cdot4)\\ \end{align}[/imath] By Lagrangian Interpolation By Formal Differentiation By the Euler-Maclaurin Summation Formula By Assuming that the expression is a polynomial of degree [imath]2[/imath]. A Proof Without Words for the cube case By integrating and assuming a error term. SimplyBeatifulArt's Personal Approach |
2375579 | If [imath]\cos A+\cos B+\cos C=\frac{3}{2}[/imath], prove [imath]ABC[/imath] is an equilateral triangle
If in [imath]\Delta ABC[/imath] [imath]\cos A+\cos B+\cos C=\frac{3}{2}[/imath]prove that it is an equilateral triangle without using inequalities. I tried using the cosine rule as follows: [imath]\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}[/imath] [imath]\implies[/imath] [imath]a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2)=3abc[/imath] [imath]\implies[/imath] [imath](a+b+c)(a^2+b^2+c^2)=2(a^3+b^3+c^3)+3abc[/imath] Any clues as to how I can take it from here? | 949530 | In [imath] \triangle ABC[/imath] show that [imath] 1 \lt \cos A + \cos B + \cos C \le \frac 32[/imath]
Here is what I did, tell me whether I did correct or not: \begin{align*} y &= \cos A + \cos B + \cos C\\ y &= \cos A + 2\cos\left(\frac{B+C}2\right)\cos\left(\frac {B-C}2\right)\\ y &= \cos A + 2\sin\left(\frac A2\right)\cos\left(\frac {BC}2\right) && \text{since [imath]A+B+C = \pi[/imath]} \end{align*} Now for maximum value of [imath]y[/imath] if we put [imath]\cos\left(\frac {B-C}2\right) = 1[/imath] then \begin{align*} y &\le \cos A + 2\sin\left(\frac A2\right)\\ y &\le 1-2\sin^2\left(\frac A2\right) + 2\sin\left(\frac A2\right) \end{align*} By completing the square we get [imath]y \le \frac 32 - 2\left(\sin\frac A2 - \frac 12\right)^2[/imath] [imath]y_{\max} = \frac 32[/imath] at [imath]\sin\frac A2 = \frac 12[/imath] and [imath]y_{\min} > 1[/imath] at [imath]\sin \frac A2>0[/imath] because it is a ratio of two sides of a triangle. Is this solution correct? If there is a better solution then please post it here. Help! |
2375697 | Spivak's Calculus 8-8(a)
This question is from Calculus by Spivak, Chapter 8 on Least Upper Bounds: Suppose that [imath]f[/imath] is a function such that [imath]f(a) \leq f(b)[/imath] whenever [imath]a<b[/imath]. Prove that [imath]\lim_{x_\to a^-} f(x)[/imath] and [imath]\lim_{x_\to a^+} f(x)[/imath] both exist. Worked on it for a while but didn't get anywhere. Any help would be appreciated. | 968157 | Show that one-sided limits always exist for a monotone function (on an interval)
Show that one-sided limits always exist for a monotone function on an interval [imath][a,b][/imath]. Me attempt: 1) If a function is monotone on an interval [imath][a,b][/imath], then [imath]f(a)\le f(x) \le f(b)[/imath] for [imath]x\in[a,b][/imath]. Therefore if there exists left-hand (right-hand) limit of this function at a given point, then it must be finite. Now we must show that a both left-hand and right=hand limits exist. We use the fact that if we take a sequence [imath](x_n)\rightarrow c^{+} \in [a,b][/imath] then [imath]f(x_n)[/imath] is bounded and monotone and therefore it is convergent. |
1764013 | Which functions are irrelevant?
Call a bijection [imath]f : \mathbb{N} \rightarrow \mathbb{N}[/imath] irrelevant over [imath]\mathbb{R}[/imath] iff for all sequences [imath]a : \mathbb{N} \rightarrow \mathbb{R}[/imath], if [imath]\sum_{i=0}^\infty a_n[/imath] exists, call its value [imath]\lambda[/imath], then [imath]\sum_{i=0}^\infty a_{f(n)}[/imath] also exists, and its limits is also [imath]\lambda[/imath]. (I feel slightly bad about the melodramatic title.) Question. Which bijections [imath]f : \mathbb{N} \rightarrow \mathbb{N}[/imath] are irrelevant over [imath]\mathbb{R}[/imath]? | 1383730 | Rearrangements that never change the value of a sum
Which bijections [imath]f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}[/imath] have the property that for every sequence [imath]\{a_n\}_{n=1}^\infty[/imath], [imath] \lim_{n\to\infty} \sum_{k=1}^n a_k = \lim_{n\to\infty} \sum_{k=1}^n a_{f(k)}, [/imath] where "[imath]=[/imath]" is construed as meaning that if either limit exists then so does the other and in that case then they are equal? It is clear that there are uncountably many of these. Might it just be that [imath]\{f(n)/n : n=1,2,3,\ldots\}[/imath] is bounded away from both [imath]0[/imath] and [imath]\infty[/imath]? These bijections form a group. Can anything of interest be said about them as a group? PS: Here's another moderately wild guess (the one above appears to be wrong): Might it be just the bijections whose every orbit is finite? |
2375175 | Riemann sums upper and lower sums question
I asked a similar question yesterday, but I didn't really get the info I wanted so maybe if I post a question and get an answer I will understand this concept better. Just some background info the topic is using Riemann sums approximation to find upper/lower sums. The question is [imath]f(x)=(x-2)^{2} +1, [a,b]=1,3,[/imath] find the lower sum with [imath]n=3.[/imath] Here is my attempt: [imath]\sum_{1}^{3}[/imath], by Riemann's definition: [imath]\Delta x= (b-a)/n,[/imath] so our [imath]\Delta\,x[/imath] is [imath]2/3.[/imath] [imath]x*k= a+\Delta\,x*k= 2/3k+1[/imath] for [imath]k\in \{0,1,2,3\}[/imath] according to my book, why does it include 0 is my question? So I have my [imath]\Delta x[/imath] and my [imath]x∗k.[/imath] Now I have to check where the function given is decreasing and increasing on the intervals given, correct? So my 3 sub intervals are [imath][1,5/3][/imath] and [imath][5/3,7/3][/imath] and [imath][7/3,3][/imath] If I plug [imath]1[/imath] in [imath](x-2)^{2} +1[/imath] I get [imath]2,[/imath] plugging in [imath]5/3[/imath] I get [imath]10/9.[/imath] so It appears from [imath][1,5/3][/imath] we are decreasing. So for this first interval my lower sum is 5/3. Next interval is from [imath][5/3,7/3][/imath]. Plugging in [imath]5/3[/imath] I get [imath]10/9.[/imath] Plugging in [imath]7/3[/imath] I get [imath]10/9.[/imath] In this interval I have no lower sum, so I add the two and divide by 2, giving me a lower sum of 2. Final interval is from [imath][7/3,3][/imath].Plugging in [imath]7/3[/imath] I get [imath]10/9.[/imath] Plugging in [imath]3[/imath] I get [imath]2.[/imath] In this interval the lower sum is 7/3. Now here is where I get lost. The answer according to me text is 2/3*(((5/3 -2)^2 +1) +((2-2)^2 +1) + ((7/3 -2)^2 +1)). So what I understand what they did is they expanded the summation 3 times, I get that. They plugged x as the lower for each expansion, ie. 5/3 for the first expansion, 2 for the secound expansion, and 7/3 for the last expansion. I get this. Then they multiplied the whole thing by [imath]\Delta x[/imath] My main concern is where did we use 2/3K +1. What was the point of even figuring this out, wouldn't we have been fine with just [imath]\Delta x[/imath]. I thought we would plug in 2/3K +1 where x is in each expansion like right/left sums. EDIT: Can someone answer the questions in my post please, I get how to do most of it | 2376147 | Riemann Lower Sum by definition
I have asked a similar question before, and didn't get the answer I was looking for, so i'll try to be more clear here. If necessary here is the link: Riemann sums upper and lower sums question Okay I am trying to find the lower Riemann sum. The question is [imath]f(x)=(x-2)^{2} +1, [a,b]=1,3,[/imath] find the lower sum with [imath]n=3.[/imath] An answer given by someone: ''' f′(x)=[imath]2(x−2)[/imath] For [imath]n=3[/imath], the partition of [imath][1,3][/imath] is the 3 sub intervals [imath][1,5/3][/imath] and [imath][5/3,7/3][/imath] and [imath][7/3,3][/imath] f is decreasing at [imath][1,2][/imath] and increasing at [imath][2,3][/imath]. in [imath][5/3,7/3][/imath] the minimum of f(x) is f([imath]2[/imath])= [imath]1[/imath] the lower sum is [imath]2/3[/imath]*(([imath](5/3 −2)^2 +1)[/imath]+[imath]1[/imath]+([imath](7/3 −2)^2) +1)[/imath] ''' I have two major questions: I see the derivative was taken, was this to see where f was increasing and decreasing? If so how do I do this, just plug in the partitioned values in the derivative? By definition of a lower sum our delta x was 2/3, which implies our x_k=a+k*(delta x) = 1+(2/3)k. We did not use this x_k anywhere, why is it necessary to calculate this(my text does it), I know in a right sum we would plug this x_k anywhere their is an x, but we didn't do that here. |
2375646 | [imath]\int_{-\infty}^{\infty} x^pf(x)=0[/imath] for all [imath]p\geq 0[/imath], how to prove [imath]f=0[/imath] with the Fourier transform?
I found this as an exercise in Kolmogorov-Fomin, let me state the problem precisely: Let [imath]S_{\infty}[/imath] be the space of functions on the real line for which there exists a colelction of constants [imath]C_{pq}[/imath] for all [imath]p, q[/imath] such that [imath] |x^pf^{(q)}(x)|\leq C_{pq}. [/imath] Is it true that if for all [imath]p\geq 0[/imath] [imath]\int_{-\infty}^{\infty} x^pf(x)=0[/imath], then [imath]f=0[/imath]? The exercise follows a section on the Fourier transform of functions of [imath]S_{\infty}[/imath]. Now, I guess one should prove this exercise (I am fairly sure it is true) using some properties of the Fourier transform, but I cannot see how. So my question is, how is this to be proved with the fundamental properties of the Fourier transform in [imath]S_{\infty}[/imath]? | 1034090 | If [imath]f\in S_\infty[/imath] and [imath]\int_{\mathbb{R}}x^pf(x)d\mu=0[/imath] for all [imath]p\in\mathbb{N}[/imath] then [imath]f\equiv 0[/imath]?
Let [imath]f\in S_\infty\subset L_1(\mathbb{R},\mu)[/imath] with [imath]\mu[/imath] as the Lebesgue linear measure be a Lebesgue-summable function such that [imath]\forall (p,q)\in\mathbb{N}^2_{\ge 0}\quad\exists C_{pq}>0: \Bigg|x^p\frac{d^q}{d x^q}f(x)\Bigg|< C_{pq}[/imath] I was wondering whether, if [imath]\forall p\in\mathbb{N_{\ge 0}}\quad\int_{\mathbb{R}}x^pf(x)d\mu=0[/imath], then [imath]f[/imath] is constantly, or almost everywhere, null. I cannot find a counterexample and therefore I think that the implication might well hold, but I cannot prove it either. Since [imath]x^p f(x)[/imath] belongs to [imath]S_\infty\subset L_1[/imath], thanks to the fact that [imath]f[/imath] belongs to it, and is continuous I think that the Lebesgue integral and the Riemann improper integral [imath]\mathscr{R}\int_{-\infty}^\infty t^pf(t)dt[/imath] are the same. Please correct if I am wrong. Nevertheless I cannot use calculus facts to prove the desired implication... I have also tried using the fact that the Fourier transform induces a bijection [imath]S_\infty\to S_\infty[/imath], but with no result. What do you think about it? Has anybody got a counterexample or proof? Thank you very much! |
2376185 | Understanding when [imath]\lim_\limits{n \to \infty} \frac{a_n}{b_n} = \lim_\limits{n \to \infty} \frac{a_0 + a_1+ \cdots + a_n}{b_0+b_1+ \cdots + b_n}[/imath].
Understanding when [imath]\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_0 + a_1+ \cdots + a_n}{b_0+b_1+ \cdots + b_n}.[/imath] That is, supposing the first limit exists, what properties must be satisfied by the sequences [imath]\{a_n\}_{n \ge 0}, \{b_n\}_{n \ge 0}[/imath] such that we may conclude that the above equality holds? I tried using the inequality [imath]\frac{a_n}{b_n} + \frac{a_{n-1}}{b_{n-1}} < \frac{a_n+a_{n-1}}{b_n + b_{n-1}}[/imath], which holds when our sequences consist of positive real numbers, having in mind the notion of applying induction but didn't get anywhere with that idea. This isn't a homework assignment, I just want to know the answer. | 1137451 | How to understand intuitively the Stolz-Cesaro Theorem for sequences?
I have to give a presentation on the theorem in Real Analysis with a fellow student. While I've looked over the proof and verified that, yes, step B does indeed follow logically from step A, etc. and have internalized the proof to the extent that I can replicate it myself on paper, I still feel that I have made little progress as to why this theorem works the way it does, i.e. what is the intuition behind the theorem, such that it should make sense that it follows the way it does. Therefore I ask: what kind of intuition is there about this theorem? It would also be helpful to understand how the theorem can be used effectively in analysis. I have seen it referred to as a sort of L'Hopital's rule for sequences, which certainly seems to make sense, but I similarly have little intuitive understanding of how that rule works, either. I am asking this question not only for my personal understanding, but also for the sake of being able to present it in an illuminating manner, such that the rest of the class can also come away with the same sort of intuition of how the theorem works and how it is useful. Any help would be greatly appreciated. EDIT: I should point out that the formulation of the theorem we are being tasked to prove is the following: Let [imath]\lbrace a_n \rbrace[/imath], [imath]\lbrace b_n \rbrace[/imath] be sequences, [imath]{b_n}[/imath] strictly increasing and unbounded. Then if [imath]\lim\limits_{n \to \infty} \frac {a_{n+1} - a_n}{b_{n+1} - b_n} = l[/imath] for some [imath]l \in \mathbb R[/imath], then [imath]\lim\limits_{n \to \infty} \frac{a_n}{b_n} = l[/imath] also. |
2376695 | Does [imath]\sum _{n=1}^\infty\frac{\left(-1\right)^{\pi \left(n\right)}}{n}[/imath] converge, where [imath]\pi(x)[/imath] is the prime counting function?
Does [imath]\sum _{n=1}^\infty\frac{\left(-1\right)^{\pi \left(n\right)}}{n}[/imath] converge, where [imath]\pi(x)[/imath] is the prime counting function? The resulting sum should look like [imath]\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}+\frac{1}{13}-\cdots[/imath] So the sign of each fraction in the sequence switches whenever a prime is reached. | 707870 | Convergence or divergence of [imath]\sum\limits_n(-1)^{\pi(n)}\frac1n[/imath] where [imath]\pi(n)[/imath] is the number of primes less than or equal to [imath]n[/imath]
Consider [imath]\sum_{n=1}^{\infty}\frac{(-1)^{\pi(n)}}{n}[/imath] where [imath]\pi(n)[/imath] is the number of primes less than or equal to [imath]n[/imath]. Does this sum converge or does it diverge? Are there any results related to this? |
2376959 | Is [imath]\int^0_0 x^{-1} dx[/imath] defined?
Is an object like [imath]\int^0_0 x^{-1} dx[/imath] defined? Context: when considering functions such as [imath]f(x)=\int^{x^2}_xt^{-1}dt[/imath] would it therefore be necessary to give a piecewise definition of [imath]f[/imath] if we wished to include [imath]0[/imath] in the domain? A little bit of random speculation, but if the object is supposedly defined to be [imath]0[/imath], then we would be allowed to write something like [imath]\int^0_0\sqrt{\text{banana}}\quad d(\text{apple})=0[/imath] ??? | 1785405 | This integral is defined ? [imath]\displaystyle\int_0^0\frac 1x\:dx[/imath]
[imath]\displaystyle\int_0^0\frac 1x\:dx[/imath] This integral is defined ? If it define, what is the value ? Please,step by step |
2377013 | If [imath](-1)^{\gcd(n,F_n)}=1[/imath] where [imath]F_n[/imath] is the [imath]n[/imath]-th Fibonacci number then [imath]n[/imath] is divisible by [imath]6[/imath] ?
Question: If [imath]n[/imath] is any positive integer I write [imath]F_0=0, F_1=1[/imath] and [imath]F_n=F_{n-1}+F_{n-2}[/imath]. The numbers [imath]F_n[/imath] are called the Fibonacci numbers. Let [imath]a_n=(-1)^{\gcd(n,F_n)}[/imath]. Is it true that if [imath]a_n=1[/imath] then [imath]n[/imath] is a multiple of [imath]6[/imath] ? Not much motivation here just an experimental observation I noticed. I am not sure where exactly to start since I do not know a formula for [imath]\gcd(n,F_n)[/imath]. | 488518 | Strong Induction Proof: Fibonacci number even if and only if 3 divides index
The Fibonacci sequence is defined recursively by [imath]F_1 = 1, F_2 = 1, \; \& \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.[/imath] Prove that [imath]2 \mid F_n \iff 3 \mid n.[/imath] Proof by Strong Induction : [imath]\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}[/imath] [imath]2 \mid F_1[/imath] is false. Also, [imath]3 \mid 1[/imath] is false. The implication [False [imath]\iff[/imath] False] is vacuously true. [imath]\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}[/imath] Assume that [imath]2 \mid F_i \iff 3 \mid n[/imath] for every integer [imath]i[/imath] with [imath]1 ≤ i ≤ k[/imath]. [imath]\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;[/imath] To prove: [imath]\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.[/imath] [imath]\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;[/imath] [imath]2 \mid F_2[/imath] is false. Also, [imath]3 \mid 2[/imath] is false. So [False [imath]\iff[/imath] False] is vacuously true. Hence assume that [imath]k + 1 ≥ 3.[/imath] We now consider three cases: [imath]\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}[/imath] Thus [imath]3 \require{cancel}\cancel{\mid} k[/imath] and [imath]3 \require{cancel}\cancel{\mid} (k − 1)[/imath]. By the ind hyp, [imath]3 \require{cancel}\cancel{\mid} k \iff F_k[/imath] odd & [imath]3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}[/imath] odd. Since [imath]F_{k+1} = F_k + F_{k−1}[/imath], thus [imath]F_{k+1}[/imath] = odd + odd = even. [imath]\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}[/imath] Thus [imath]3 | k[/imath] and [imath]3 \require{cancel}\cancel{\mid} (k − 1).[/imath] By the ind hyp, [imath]3 | k \iff F_k[/imath] even & [imath]3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}[/imath] odd. Thus [imath]F_{k+1}[/imath] odd. [imath]\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}[/imath] Thus [imath]3 \require{cancel}\cancel{\mid} k[/imath] and [imath]3 | (k −1).[/imath] By the ind hyp, [imath]3 \require{cancel}\cancel{\mid} k \iff F_k[/imath] odd and [imath]3 \mid (k − 1) \iff F_{k - 1}[/imath] even. Thus [imath]F_{k+1}[/imath] odd. [imath]\blacksquare[/imath] [imath]\Large{1.}[/imath] Does the proof clinch the [imath](\Leftarrow)[/imath] of the [imath](k + 1)[/imath]th case? [imath]\Large{2.}[/imath] Since the recursion contains [imath]n, n - 1, n - 2[/imath], thus the recursion "time lag" is [imath]3[/imath] here. So shouldn't [imath]3[/imath] base cases be checked? [imath]\Large{3.}[/imath] Further to #2, shouldn't "assume [imath]k + 1 \geq \cancel{3} 4[/imath]" instead? [imath]\Large{4.}[/imath] Shouldn't the [imath]n = k + 1 = 2[/imath] case precede the induction hypothesis? I referenced 1. Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al Supplement to peterwhy's Answer: [imath]\Large{1.1.}[/imath] I wrongly believed that all 3 Cases proved the [imath]\Leftarrow[/imath]. I now see that Case 1 is [imath]\Leftarrow[/imath] via a Direct Proof. Cases 2 and 3 are [imath]\Rightarrow[/imath] via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from [imath]3 \mid n[/imath] for both directions of the proof? |
2376886 | Prove [imath]I[/imath] [imath]\in[/imath] [imath]\mathbb{Z}[x][/imath] s.t. [imath]I=\{f(x)\mid f(0) \text{ is even}\}[/imath] is a maximal ideal, and find the number of elements in [imath]\mathbb{Z}[x]/I[/imath]
I have already shown (in a previous question) that [imath]I=\langle x,2\rangle[/imath], and that I is a prime ideal, and I believe that allows me to show that [imath]\mathbb{Z}[x]/I[/imath] is an integral domain, which if it is finite would allow me to conclude that [imath]\mathbb{Z}[x]/I[/imath] is a field, which would imply that I is a maximal ideal. But I don't know how to determine the number of elements in [imath]\mathbb{Z}[x]/I[/imath], and thus do not know how to determine whether or not it is finite. Can someone tell me if this thinking is correct, and if so how to determine the number of elements in [imath]\mathbb{Z}[x]/I[/imath]? Note: we have not covered irreducibility in this course. | 2370724 | order of quotient ring
In [imath]\mathbb{Z}\left [ x \right ][/imath], let [imath]I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}[/imath] In fact, [imath]I=\left \langle x,2 \right \rangle[/imath] How do I show that the order of [imath]\mathbb{Z}\left [ x \right ]/I[/imath] is 2? |
2378337 | Proof that the function [imath]f[/imath] is not injective.
Let [imath]f:U\subset\mathbb{R}^m\rightarrow\mathbb{R}^n[/imath] be a function of class [imath]C^1[/imath] with [imath]m> n[/imath]. Prove that [imath]f[/imath] is not an injection function. I do not know how to solve this problem. Could someone help me with the solution to this problem? | 1086314 | Continuously differentiable map from [imath]\mathbb{R}^{m+n}[/imath] to [imath]\mathbb{R}^n[/imath]
Suppose [imath]m,n>0[/imath], [imath]U[/imath] an open subset of [imath]\mathbb{R}^{m+n}[/imath] and let [imath]f: U \to \mathbb{R}^n[/imath] be continuously differentiable. Is it possible for [imath]f[/imath] to be injective? My thinking is that continuous differentiability+injectivity suggests that this is sort of getting at a converse to the Inverse Function Theorem. We know the Jacobian is not invertible at any point in [imath]U[/imath], and the question is asking if this implies that [imath]f[/imath] cannot be invertible in any neighborhood contained in [imath]U[/imath]. |
1995738 | An inequality involving [imath]n[/imath] blue points and [imath]n[/imath] red points on a line.
Question : Let [imath]n[/imath] blue points [imath]A_i[/imath] and [imath]n[/imath] red points [imath]B_i[/imath] ([imath]i=1,2,...,n[/imath]) be situated on a line. Prove that [imath]\sum_{i,j}A_iB_j\ge \sum_{i<j}A_iA_j+\sum_{i<j}B_iB_j[/imath] I tried inducting on [imath]n[/imath] but cant proceed in the inductive step ( as usual ). Please help, if there is a direct method of approach please give hints. Is this true if all the blue points were on a different line? | 1385354 | An inequality about the sum of distances between points : same color [imath]\le[/imath] different colors?
When I was drawing some points on paper and studied the distances between them, I found that an inequality holds for many sets of points. Suppose that we have [imath]2[/imath] blue points [imath]b_1,b_2[/imath] and [imath]2[/imath] red points [imath]r_1,r_2[/imath] in the Euclidean plane. Then using the triangular inequality, it is easy to see that the following inequality always holds : [imath]\text{the sum of Euclidean distances between points in the same color}\le\text{the sum of Euclidean distances between points in different colors},[/imath] i.e. [imath]d(\color{blue}{b}_1,\color{blue}{b}_2)+d(\color{red}{r}_1,\color{red}{r}_2)\le d(\color{blue}{b}_1,\color{red}{r}_1)+d(\color{blue}{b}_1,\color{red}{r}_2)+d(\color{blue}{b}_2,\color{red}{r}_1)+d(\color{blue}{b}_2,\color{red}{r}_2)\tag1[/imath]where [imath]d(p,q)= \sqrt{(q(x)-p(x))^2 + (q(y)-p(y))^2}[/imath] is the Euclidean distance between two points [imath]p(p(x),p(y))[/imath] and [imath]q(q(x),q(y))[/imath]. However, its generalization is difficult for me. Question : Let [imath]n\ge 3[/imath]. Suppose that we have [imath]n[/imath] blue points [imath]b_1,b_2,\cdots,b_n[/imath] and [imath]n[/imath] red points [imath]r_1,r_2,\cdots,r_n[/imath] in the Euclidean plane. Then, can we say that the following inequality always holds? [imath]\text{the sum of Euclidean distances between points in the same color}\le\text{the sum of Euclidean distances between points in different colors},[/imath] i.e. [imath]\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}d(\color{blue}{b}_i,\color{blue}{b}_j)+\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}d(\color{red}{r}_i,\color{red}{r}_j)\le\sum_{i=1}^{n}\sum_{j=1}^{n}d(\color{blue}{b}_i,\color{red}{r}_j)[/imath] So far, I have not been able to find any counterexample, so it seems that this inequality always holds. However, I've been facing difficulty even in the [imath]n=3[/imath] case. For the [imath]n=3[/imath] case, using the result of the [imath]n=2[/imath] case several times, we can obtain [imath]3\left(\sum_{i=1}^{2}\sum_{j=i+1}^{3}d(\color{blue}{b}_i,\color{blue}{b}_j)+\sum_{i=1}^{2}\sum_{j=i+1}^{3}d(\color{red}{r}_i,\color{red}{r}_j)\right)\le 4\sum_{i=1}^{3}\sum_{j=1}^{3}d(\color{blue}{b}_i,\color{red}{r}_j)\tag2[/imath] But this does not seem to help. Can anyone help? Added 1 : Induction does not seem to help. As I wrote, the result of the case [imath]n=2[/imath] does not seem to help for the case [imath]n=3[/imath]. Added 2 : For the case [imath]n=2[/imath], I used the triangular inequality. (sorry, I didn't write the details because what I'm asking is for [imath]n\ge 3[/imath].) However, it seems that the triangular inequality does not help for [imath]n\ge 3[/imath]. Added 3 : The following is how I obtained [imath](2)[/imath] from [imath](1)[/imath]. From [imath](1)[/imath], we have [imath]d(\color{blue}{b}_1,\color{blue}{b}_2)+d(\color{red}{r}_1,\color{red}{r}_2)\le d(\color{blue}{b}_1,\color{red}{r}_1)+d(\color{blue}{b}_1,\color{red}{r}_2)+d(\color{blue}{b}_2,\color{red}{r}_1)+d(\color{blue}{b}_2,\color{red}{r}_2)[/imath] [imath]d(\color{blue}{b}_1,\color{blue}{b}_2)+d(\color{red}{r}_2,\color{red}{r}_3)\le d(\color{blue}{b}_1,\color{red}{r}_2)+d(\color{blue}{b}_1,\color{red}{r}_3)+d(\color{blue}{b}_2,\color{red}{r}_2)+d(\color{blue}{b}_2,\color{red}{r}_3)[/imath] [imath]d(\color{blue}{b}_1,\color{blue}{b}_2)+d(\color{red}{r}_3,\color{red}{r}_1)\le d(\color{blue}{b}_1,\color{red}{r}_3)+d(\color{blue}{b}_1,\color{red}{r}_1)+d(\color{blue}{b}_2,\color{red}{r}_3)+d(\color{blue}{b}_2,\color{red}{r}_1)[/imath] [imath]d(\color{blue}{b}_2,\color{blue}{b}_3)+d(\color{red}{r}_1,\color{red}{r}_2)\le d(\color{blue}{b}_2,\color{red}{r}_1)+d(\color{blue}{b}_2,\color{red}{r}_2)+d(\color{blue}{b}_3,\color{red}{r}_1)+d(\color{blue}{b}_3,\color{red}{r}_2)[/imath] [imath]d(\color{blue}{b}_2,\color{blue}{b}_3)+d(\color{red}{r}_2,\color{red}{r}_3)\le d(\color{blue}{b}_2,\color{red}{r}_2)+d(\color{blue}{b}_2,\color{red}{r}_3)+d(\color{blue}{b}_3,\color{red}{r}_2)+d(\color{blue}{b}_3,\color{red}{r}_3)[/imath] [imath]d(\color{blue}{b}_2,\color{blue}{b}_3)+d(\color{red}{r}_3,\color{red}{r}_1)\le d(\color{blue}{b}_2,\color{red}{r}_3)+d(\color{blue}{b}_2,\color{red}{r}_1)+d(\color{blue}{b}_3,\color{red}{r}_3)+d(\color{blue}{b}_3,\color{red}{r}_1)[/imath] [imath]d(\color{blue}{b}_3,\color{blue}{b}_1)+d(\color{red}{r}_1,\color{red}{r}_2)\le d(\color{blue}{b}_3,\color{red}{r}_1)+d(\color{blue}{b}_3,\color{red}{r}_2)+d(\color{blue}{b}_1,\color{red}{r}_1)+d(\color{blue}{b}_1,\color{red}{r}_2)[/imath] [imath]d(\color{blue}{b}_3,\color{blue}{b}_1)+d(\color{red}{r}_2,\color{red}{r}_3)\le d(\color{blue}{b}_3,\color{red}{r}_2)+d(\color{blue}{b}_3,\color{red}{r}_3)+d(\color{blue}{b}_1,\color{red}{r}_2)+d(\color{blue}{b}_1,\color{red}{r}_3)[/imath] [imath]d(\color{blue}{b}_3,\color{blue}{b}_1)+d(\color{red}{r}_3,\color{red}{r}_1)\le d(\color{blue}{b}_3,\color{red}{r}_3)+d(\color{blue}{b}_3,\color{red}{r}_1)+d(\color{blue}{b}_1,\color{red}{r}_3)+d(\color{blue}{b}_1,\color{red}{r}_1)[/imath] Adding these gives [imath](2)[/imath]. However, [imath](2)[/imath] does not seem to help. |
2378889 | Estimate of sum of 1/n for n≤x
How does one prove the elementary estimate, [imath]\sum_{n≤x}{1/n}=\log{x}+\gamma+O(1/x)[/imath]? The result is necessary in a proof for E. Landau's estimate of [imath]\sum_{n≤x}{1/\phi(n)}[/imath]. | 596621 | Asymptotic estimate of the sequence of harmonic series [imath]\sum_{k=1}^{n} \frac{1}{k} [/imath]
Asymptotic estimate of the sequence of partial sums, for [imath]n\rightarrow \infty[/imath]: [imath] s_n=\sum_{k=1}^{n} \frac{1}{k} [/imath] |
2375481 | Do the irrationals form a group?
It's true that the irrationals do not form a group under addition or multiplication, but I want to find a binary operation [imath]*[/imath] such that [imath](\mathbb{R}\setminus \mathbb{Q}, *)[/imath] is a group. It is possible by transpose of operation from [imath]\mathbb{R}[/imath], but I want a solid example. | 2378996 | Do Irrationals form a group?
It's true that the irrationals do not form a group under addition or multiplication, but I want to find a binary operation "∗" such that ([imath]\mathbb R \setminus \mathbb Q,∗[/imath]) is a group. It is possible by transpose of operation from [imath]\mathbb R[/imath], but I want a explicit example. |
2379053 | Prove that the product of two infinite cyclic groups is not infinite cyclic
Prove that the product of two infinite cyclic groups is not infinite cyclic. Let [imath]G=<a>[/imath] and [imath]G'=<b>[/imath]. The product group is [imath]\{a^mb^n: m,n\in\mathbb{Z}\}=<a^kb^l>[/imath] Now, for given [imath]m,n\in\mathbb{N}[/imath], we can find [imath]u[/imath] such that [imath](a^kb^l)^u=a^mb^n[/imath]. Then [imath]u=\frac{m}{k}=\frac{n}{l}\Rightarrow \frac{k}{l}=\frac{m}{n}[/imath], which is true for any [imath]m,n\in\mathbb{N}[/imath], which can not be true, since [imath]k,l[/imath] are fixed, so the ratio is. And then I am stuck. What to do next??? Another question from my curiosity: We can take [imath]G=<e^{i\pi \sqrt{2}}>[/imath] and [imath]G'=<e^{i\pi \sqrt{3}}>[/imath]. What is the product group? It should be [imath]<e^{i\pi \theta}>[/imath], where [imath]\theta[/imath] is rational. But what is [imath]\theta[/imath]? | 2379109 | Prove that the product of two infinite cyclic groups is not infinite cyclic.
I am very sorry that I deleted my previous post on this same topic and specially to @lhf, since I forced him/her to delete his/her answer. Prove that the product of two infinite cyclic groups is not infinite cyclic. The part I did: [imath]G=<a>[/imath] and [imath]G'=<b>[/imath]. The product group is [imath]G\times G'[/imath], where mulplication rule is [imath](a^{k_1},b^{l_1}),(a^{k_2},b^{l_2})\rightsquigarrow (a^{k_1+k_2},b^{l_1+l_2})[/imath] where [imath]k_1,k_2,l_1,l_2\in \mathbb{Z}[/imath] Now suppose the cyclic group [imath]G\times G'[/imath] is generated by [imath](a^k,b^l)[/imath] Then for any [imath]m,n\neq 0[/imath], [imath](a^m,b^n)=(a^{ku},b^{lu})\Rightarrow \frac{k}{l}=\frac{m}{n}[/imath], which can not be true since [imath]k,l[/imath] are fixed and [imath]G,G'[/imath] are infinite cyclic group. But then I am stuck. What should be the lat argument to end this proof! |
2378521 | Proving that the connect sum of 3 copies of [imath]\mathbb{C}P^2[/imath] does not have a complex structure
Pardon my total lack of knowledge in complex geometry. How would one go about proving that the connect sum of 3 copies of [imath]\mathbb{C}P^2[/imath] does not admit a complex structure? Note that this manifold does admit an almost complex structure. | 786774 | Complex structure on [imath]\mathbb{C}\mathbb{P}^2\# \dots \# \mathbb{C}\mathbb{P}^2[/imath]
I know the chern classes-related theorem that states that [imath]\mathbb{C}\mathbb{P}^2\# \dots \# \mathbb{C}\mathbb{P}^2[/imath] ([imath]k[/imath] times) has no almost complex structure (hence no complex structure) if and only if [imath]k[/imath] is even. I also know that [imath]\mathbb{C}\mathbb{P}^2\# \mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2[/imath] has no complex structure, hence the almost complex ones you define on it are not integrable. Where can I find a proof of this proposition? How do I see that the connected sum of three (or five, or seven, or every odd number) copies of [imath]\mathbb{C}\mathbb{P}^2[/imath] doesn't admit a complex structure? |
2379275 | Prove that [imath]\{f_n \}[/imath] has a uniformly convergent subsequence on [imath][0,1][/imath].
For each positive integer [imath]n[/imath] let [imath]f_n : [0,1] \to R[/imath] be a continuous function, differentiable on [imath](0,1][/imath], such that [imath]|f_n^{~'} (x)| \leq\frac{1 + |\ln x|}{\sqrt{x}} [/imath]for [imath]0 < x \leq 1[/imath]. and such that [imath]−10 \leq \int^1_0 f_n (x)~\mathrm{d}x \leq 10.[/imath] Prove that [imath]\{f_n \}[/imath] has a uniformly convergent subsequence on [imath][0,1][/imath]. I try to use the Arzelà–Ascoli theorem, which requires me to prove [imath]\{f_n\}[/imath] is uniformly bounded and equicontinuous. From the second inequality, we conclude [imath]\{f_n\}[/imath] is uniformly bounded. However, I stuck with proving they are equicontinuous. I try to use the Lagrange theorem so that it suffices to show the [imath]\{f_n^{~'}\}[/imath] is uniformly bounded. However, the first inequality cannot give us what we want since the function on the right side is not bounded on [imath](0,1][/imath]. So what should I do? | 2281164 | Uniformly Convergent Subsequence
This is almost surely an Arzela-Ascoli question, since it comes from an old exam of which such problems are quite common. Unfortunately, I can't seem to get it though. [imath]\{ f_n \}[/imath] is a sequence of functions [imath][0,1] \to \mathbb{R}[/imath] satisfying [imath]|f_n'(x)| \leq \frac{1 + |\ln (x)|}{\sqrt{x}}[/imath] and [imath]-10 \leq \int_0^1 f_n(x) dx \leq 10[/imath] for every [imath]n[/imath]. The question is to show the existence of a uniformly convergent subsequence. In these kinds of situations, the bounds given usually turn into something nice to force uniform boundedness. Maybe I could hope to establish equicontinuity using the integral somehow, but I'm not told that the [imath]f_n[/imath] are always positive, so the absolute values would probably screw things up anyway. Is it an easy problem I'm missing or is this totally the wrong track? If so, what's the right track? |
2379316 | Does there always exist a surjection from [imath]\Bbb R[/imath] to [imath]\omega_1[/imath]?
I'm pretty sure that, with the axiom of choice, there always exists a surjection from [imath]\Bbb R[/imath] to [imath]\omega_1[/imath] (well-order [imath]\Bbb R[/imath], send the first [imath]\omega_1[/imath] elements to the corresponding ordinals, if there are any reals left over send them to whatever). What happens in the absence of choice? | 407833 | Proving existence of a surjection [imath]2^{\aleph_0} \to \aleph_1[/imath] without AC
I'm quite sure I'm missing something obvious, but I can't seem to work out the following problem (web search indicates that it has a solution, but I didn't manage to locate one -- hence the formulation): Prove that there exists a surjection [imath]2^{\aleph_0} \to \aleph_1[/imath] without using the Axiom of Choice. Of course, this surjection is very trivial using AC (well-order [imath]2^{\aleph_0}[/imath]). I have been looking around a bit, but an obvious inroad like injecting [imath]\aleph_1[/imath] into [imath]\Bbb R[/imath] in an order-preserving way is impossible. Hints and suggestions are appreciated. |
673841 | Show that [imath]\sigma^2[/imath] is a Cycle iff the length of [imath]\sigma[/imath] is Odd
I got this question. I'm totally stumped and I don't know what to do. Let [imath]\sigma[/imath] be a cycle of length [imath]k > 2[/imath]. Show that [imath]\sigma^2[/imath] is a cycle iff [imath]k[/imath] is odd. | 544673 | α² is a cycle if and only if s is odd
let [imath]\alpha[/imath] be a cycle of length [imath]s[/imath], say [imath]\alpha = (a_1, a_2, \ldots, a_s)[/imath] Prove [imath]\alpha^2[/imath] is a cycle if and only if [imath]s[/imath] is odd. Let me start off by saying I am in my 5th week of Group Theory. I often have trouble getting these problems started. This is my first proof based course. I believe [imath]\alpha^2 = (a_2, a_3, \ldots, a_1)[/imath] Any tips on where to go from here would be great. Also...if there are any tips for starting proofs like these in general, I could really use them! My teacher teaches as if a proofing class was a pre-req, which it was not. |
2379541 | Optimum Dimensions of a Closed Cylindrical Can
I was doing math for time-killing, and I stumbled upon this question: "Find the most economical dimensions for a closed cylindrical can containing a quart." My work: I do know that the volume [imath]V[/imath]of cylinder is [imath]V = \pi r^2h.[/imath] I applied differentiation like this:[imath]d(V) = d(\pi r^2h) [/imath] Then: [imath]dV = \pi r^2 + 2\pi rh.[/imath] I set the [imath]dV[/imath] to be zero to get the optimum dimensions: [imath]0 = \pi r^2 + 2\pi rh[/imath] [imath]-\pi r^2 = 2\pi rh[/imath] [imath]-r=2h[/imath] The solution I got seems legitimate because I would just ignore the negative sign, getting the dimensions the radius must equal to twice of the cylinder's height. And that's one fat can we got. In the book I used the answer is the dimensions must be Diameter = Height. How do you get the optimum dimensions Diameter = Height? | 1449563 | Height/Radius ratio for maximum volume cylinder of given surface area
I am a bit confused by this problem I have encountered: A right circular cylindrical container with a closed top is to be constructed with a fixed surface area. Find the ratio of the height to the radius which will maximize the volume. I know the volume to be [imath] \pi{r}^2h[/imath], but I don't see what equation I should be solving for. How can I solve for the ratio? Thanks |
2379746 | How can a basis of a topological space NOT contain the empty set?
By definition any topology must contain the empty set. Also by definition if [imath]B\subset T[/imath] is a basis for topology [imath]T[/imath], then every member of [imath]T[/imath], in particular the empty set, must be a union of some members of [imath]B[/imath]. But how can you get a union of sets equal to the empty set if none of them are empty sets? | 490272 | Base and empty set of a topology
Given space [imath]X = \{a, b, c\}[/imath], [imath]\beta[/imath] is a basis for a topology [imath]\tau[/imath] on X. [imath]\tau = \{ \varnothing, X, \{a\}, \{b\}, \{a,b\}\}[/imath], [imath]\beta = \{\{a\}, \{b\}, X\}[/imath]. [imath]\beta[/imath] can't union its elements to get empty set [imath]\varnothing[/imath] contained in [imath]\tau[/imath] , but the definition of basis require that every open set can be expressed as a union of basis elements. So why [imath]\varnothing[/imath] is not an element of [imath]\beta[/imath] ? |
2380102 | How to calculate the residue of this function
Consider a function [imath]f(x)=\frac{1}{(x-1)^\alpha (x+1)^\alpha}[/imath] where [imath]\alpha[/imath] is a positive small number (not necessary to be an integer). How to calculate the residue at [imath]x=1[/imath]? | 366025 | Calculus of residue of function around poles of fractional order (complex analysis)
The complex function [imath]f(z)=\frac{1}{\sqrt{z^2+r_0z}}[/imath] with [imath]r_0>0[/imath] has two poles (at [imath]z=0[/imath] and [imath]z=-r_0[/imath]). But they are not simple poles. They are poles of fractional order. Am I right? How I can calculate residue of the function at the poles? Please help me. Thanks Vahid |
2380325 | For which values of [imath]x[/imath], [imath]0[/imath]°[imath]≤x<360[/imath]°, is [imath]\sin x > \cos x[/imath]?
For which values of [imath]x[/imath] such that [imath]0[/imath]°[imath]≤x<360[/imath]°, we have: [imath]\sin x > \cos x[/imath]? The answer is "When [imath]45[/imath]° [imath]< x < 225°[/imath], the [imath]y[/imath]-coordinate of the point on the unit circle is greater than the [imath]x[/imath]-coordinate", but I don't understand why this is the answer and how the book got it. | 2375816 | Sine is algebraically less than cosine in given interval.
Prove that Sine is algebraically less than Cosine for any angle between [imath]2n\pi-\dfrac{3\pi}{4}[/imath] and [imath]2n\pi+\dfrac{\pi}{4}[/imath]. My attempts: [imath]\sin \bigg(-\dfrac{3\pi}{4}\bigg)=\cos \bigg(-\dfrac{3\pi}{4}\bigg)=-\dfrac{1}{\sqrt{2}}[/imath]. Now when moved further [imath]\sin[/imath] will starts increasing in magnitude (till [imath]\dfrac{-3\pi}{2}[/imath]) and [imath]\cos[/imath] will decrease, hence because they are in [imath]3^{rd}[/imath] quadrant [imath]\implies\sin<\cos[/imath]. Now if we enter [imath]4^{th}[/imath] quadrant [imath]\cos>0[/imath] and [imath]\sin<0[/imath] till [imath]2\pi[/imath] [imath]\implies\cos>\sin[/imath]. Now if we cross [imath]4^{th}[/imath] quadrant [imath]\sin\uparrow[/imath], from [imath]0[/imath] and [imath]\cos\downarrow[/imath], from [imath]1[/imath], and becomes euqals at [imath]\dfrac{\pi}{4}[/imath], will imply [imath]\cos>\sin[/imath] till [imath]\dfrac{\pi}{4}[/imath]. Hence the results. But is there any way to do this with pure algebra, without these intuitions. Please help. |
2380612 | If [imath]x + \frac{1}{x} = \sqrt{3}[/imath], then find [imath]x^{18}[/imath]
If [imath]x + \frac{1}{x} = \sqrt{3}[/imath], then find [imath]x^{18}[/imath]. This is sorta like If [imath]x^3+\frac{1}{x^3}=18\sqrt{3}[/imath] then to prove [imath]x=\sqrt{3}+\sqrt{2}[/imath]. So my question is, do I solve my question the same way as the question in the link? | 2276746 | Given that [imath]x+\frac{1}{x}=\sqrt{3}[/imath], find [imath]x^{18}+x^{24}[/imath]
Given that [imath]x+\frac{1}{x}=\sqrt{3}[/imath], find [imath]x^{18}+x^{24}[/imath] Hints are appreciated. Thanks in advance. |
2380846 | Need help with the following inequality
I am stuck with the following problem : If [imath]a,b,c[/imath] are positive and not all equal then prove that [imath]\,\,(a+\frac1a)^2+(b+\frac1b)^2+(c+\frac1c)^2 \geq 33\frac13,[/imath] when [imath]a+b+c=1[/imath]. Try: [imath]\,\,(a+\frac1a)^2+(b+\frac1b)^2+(c+\frac1c)^2 \implies a^2+b^2+c^2+\dfrac{1}{ a^2}+\dfrac{1}{ b^2}+\dfrac{1}{ c^2}+6[/imath] and then I did few more steps but could not come up with the result. I will be grateful if someone explains it. Thanks in advance for your time. | 1671937 | [imath]a,b,c[/imath] are real numbers [imath]>0[/imath]. If [imath]a+b+c=1[/imath], show that [imath](a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge\frac{100}{3}[/imath]
[imath]a,b,c[/imath] are real numbers [imath]>0[/imath]. If [imath]a+b+c=1[/imath], show that [imath](a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\geq\frac{100}{3}[/imath] |
2380991 | Almost surely convergence of random variables
Let [imath]X_1(x),X_2(x),\ldots ,X_n(x)[/imath] be iid random variables. They are non-decreasing, piecewise linear, convex and they depend on a parameter [imath]x\in \mathbb{R}[/imath]. We know that for all [imath]x\in \mathbb{R}[/imath] [imath]\mathbf{P}\left(\lim_{n\to\infty}X_n(x)=X(x)\right)=1,[/imath] where [imath]X[/imath] is a constant for each [imath]x\in\mathbb{R}[/imath], meaning it can be thought of as a function [imath]X\colon \mathbb{R}\to\mathbb{R}[/imath]. I want to prove that then [imath]\mathbf{P}\left(\lim_{n\to\infty}X_n(x)=X(x), \ \text{for all }x\in \mathbb{R}\right)=1.[/imath] Problem is that [imath]\mathbb{R}[/imath] is not countable. My thought is to choose a dense countable subset [imath]\mathbb{Q}[/imath] and from countable additivity we get [imath]\mathbf{P}\left(\bigcup_{x\in\mathbb{Q}} \lim_{n\to\infty} X_n(x)\neq X(x)\right)=\sum_{x\in\mathbb{Q}}\mathbf{P}\left(\lim_{n\to\infty}X_n(x)\neq X(x)\right)=0+0+\ldots =0.[/imath] Since the limit of convex functions is continuous, we know that [imath]X[/imath] is continuous. Hence the convergence must also happen in [imath]\mathbb{R}\setminus \mathbb{Q}[/imath]. Is my explanation correct? Any ideas on how to improve it? It is different from Probability of random variables converging as [imath]X_i[/imath] are continuous. | 2380282 | Probability of random variables converging
Let [imath]X_1(x),X_2(x),\ldots ,X_n(x)[/imath] be [imath]n[/imath] iid random variables. I know that [imath]\mathbf{P}(\lim_n X_n(x)=X(x))=1[/imath] for each [imath]x\in \mathbb{R}[/imath]. We also know that the limit random variable [imath]X[/imath] is a.s. constant and considering [imath]X\colon \mathbb{R}\to\mathbb{R}[/imath], it is continuous. I want to show that [imath]\mathbf{P}(\lim_nX_n(x)=X(x)\ \text{ for all }x\in \mathbb{R})=1.[/imath] I thought about using countable additivity of probability on [imath]\mathbb{Q}[/imath] and then applying continuouty to extend the result to the whole [imath]\mathbb{R}[/imath]. Since [imath]\mathbf{P}(\lim_nX_n(x)\neq X(x)\ \text{ for all }x\in \mathbb{Q})=\sum_{x\in \mathbb{Q}}\mathbb{P}(X_n(x)\neq X(x))=0+0+\ldots.[/imath] Is there anything else I should add to this explanation? Ideas? |
2381560 | Without using the proof by induction establish the formula for the simple arithmetic series
How to establish the formula for this without induction? [imath]1 + 2 + 3 + · · · + n =\sum_{j=1}^n j=\frac{n(n + 1)}{2}[/imath] | 2106689 | Different ways to come up with [imath]1+2+3+\cdots +n=\frac{n(n+1)}{2}[/imath]
I am trying to compile a list of the different ways to come up with the closed form for the sum in the title. So far I have the famous Gauss story, the argument by counting doubletons, and using generating functions. Is there any other (significantly) different way to discover the closed form? To clarify, Gauss's method was to take [imath]1+2+3+\dots+n[/imath], write it backwards, and add it up. |
2381660 | [imath]\frac{f(t)}{t}\rightarrow b[/imath] as [imath]t\rightarrow\infty[/imath]
Let [imath]f:(0,\infty)\rightarrow\mathbb{R}[/imath] be differentiable and let [imath]b[/imath] be a real number. Prove that if [imath]f'(t)\rightarrow b[/imath] as [imath]t\rightarrow\infty[/imath], then [imath]\frac{f(t)}{t}\rightarrow b[/imath] as [imath]t\rightarrow\infty[/imath]. Median Value Theorem: If [imath]f[/imath] is continuous and differentiable, then [imath]\exists c[/imath] such that [imath]a<c<b[/imath] and [imath]f'(c)=\frac{f(b)-f(b)}{b-a}[/imath]. So when [imath]c\rightarrow\infty[/imath], [imath]a,b\rightarrow\infty[/imath]. How do I get [imath]\frac{f(t)}{t}[/imath]? | 1044668 | if [imath]f'(x)\rightarrow L[/imath] as [imath] x \rightarrow \infty[/imath], [imath]-\infty \leq L \leq \infty [/imath] then [imath] f(x)/x \rightarrow L [/imath] as [imath]x \rightarrow \infty[/imath]
If [imath]f[/imath] is differentiable on [imath](a,\infty)[/imath], Show that if [imath]f'(x)\rightarrow L[/imath] as [imath] x \rightarrow \infty[/imath], [imath]-\infty \leq L \leq \infty [/imath] then [imath] f(x)/x \rightarrow L [/imath] as [imath]x \rightarrow \infty[/imath]. Deduce further that if [imath]f(x) \rightarrow M[/imath] as [imath] x\rightarrow \infty[/imath] , and [imath] f'(x)\rightarrow L [/imath] as [imath] x \rightarrow\infty [/imath] then [imath] L = 0[/imath] For the first part I tried converting [imath]f(x)[/imath] to [imath]F(\frac{1}{x})[/imath] and [imath]x \rightarrow 0[/imath] and then using L'Hospital's rule but I am getting stuck. How do I proceed with this problem ? |
2382018 | Let [imath]G[/imath] be a group of order [imath]125[/imath]. Then which of the following statements are necessarily true?
Let [imath]G[/imath] be a group of order [imath]125[/imath]. Then which of the following statements are necessarily true? A) [imath]G[/imath] has a non-trivial abelian subgroup B) the center of [imath]G[/imath] is proper subgroup C) center of [imath]G[/imath] has exactly [imath]5[/imath] elements D) there is a subgroup of order [imath]25[/imath] My attempt: Since [imath]5[/imath] is a prime number and order of [imath]g[/imath] is [imath]5^3[/imath] so center of [imath]G[/imath] has exactly [imath]5[/imath] elements. So options A, B, C are correct. | 2380926 | When we can say a group can not be abelian
Let [imath]G[/imath] be a group of order [imath]125[/imath]. Which of the following statements are necessarily true? [imath]G[/imath] has a non-trivial abelian subgroup. The centre of [imath]G[/imath] is a proper subgroup. The centre of [imath]G[/imath] has order [imath]5[/imath]. There is a subgroup of order [imath]25[/imath]. Solution(Partial): Since [imath]G[/imath] has a subgroup of order [imath]5[/imath], [imath]G[/imath] has a non-trivial abelian subgroup. Hence (1) is true. (*)I think (2) is not true, since [imath]G[/imath] can be abelian, then centre may not be proper subgroup. Since (2) is not true it is not necessary to be centre of order [imath]5[/imath]. Hence (3) is not true. (4) is true for the fact that [imath]125=5\times 25[/imath]. Also see here. I think I need better explanation for (*). More generally is there any specific way to argue that ---- [imath]\color{red}{\text{When we can say a group can not be abelian if we only know the order of the group?}}[/imath] |
2381805 | Alternative way to prove [imath]\lim_{n\to\infty}\frac{2^n}{n!}=0[/imath]?
It follows easily from the convergence of [imath]\sum_{n=0}^\infty\frac{2^n}{n!}[/imath] that [imath] \lim_{n\to\infty}\frac{2^n}{n!}=0\tag{1} [/imath] Other than the Stirling's formula, are there any "easy" alternatives to show (1)? | 2478862 | Prove using the basic properties of a sequence that [imath]\lim \frac{n}{2^n} = 0[/imath]
Prove using the basic properties of a sequence that [imath]\lim \frac{n}{2^n} = 0[/imath] I tried to prove this using the standard epsilon-definition of a limit but the math got really hard, so I stopped. Then, I tried to represent the sequence as a quotient or a product of two sequences, but those sequences were not convergent. Thus, I've got no more ideas on how to tackle this problem. Any suggestions would be most appreciated. |
2382554 | If any two indices are equal, then the term is zero
Let [imath]X[/imath] be a topological space. Let [imath]U_i[/imath] be an open cover, with the index [imath]i[/imath] running through some set [imath]I[/imath]. For [imath]i_0[/imath], [imath]i_1[/imath], ..., [imath]i_n \in I[/imath], write [imath]U_{i_0 i_1 \cdots i_n}[/imath] for [imath]U_{i_0} \cap \cdots \cap U_{i_p}[/imath]. [imath]\def\cE{\mathcal{E}}[/imath]Let [imath]F[/imath] be a sheaf of abelian groups on [imath]X[/imath] and [imath]F_{i_0,i_1,...i_n}:=F(U_{i_0 i_1 \cdots i_n})[/imath], we will write the group operation additively. Let [imath]I[/imath] to be a set of indices, and [imath]i_0, i_1, i_2... \in I[/imath]. For each array [imath](i_0,i_1,...i_n)[/imath] associate a set [imath]F_{i_0,i_1,...i_n}[/imath] and take [imath](f_{\large i_0,i_1,...i_n})_{i_0,i_1,...i_n\in I}\in \prod_{\large i_0,i_1,...i_n}F_{\large i_0,i_1,...i_n}.[/imath] Define [imath]d((f_{\large i_0,i_1,...i_n})_{\large i_0,i_1,...i_n\in I})=(\sum_{i=0}^{n+1}(-1)^kf_{\large i_0,...\hat{i_k},...i_{n+1}}|U_{i_0 i_1 \cdots i_{n+1}})_{\large i_0,i_1,...i_{n+1}\in I} [/imath] I wonder if the following statement is true: Assume [imath]d((f_{\large i_0,i_1,...i_n})_{\large i_0,i_1,...i_n\in I})=0[/imath]. If any two of the indices [imath]{k_0,k_1,...k_n}[/imath] are equal, then [imath]f_{\large k_0,k_1,...k_n}=0[/imath] See here for the description pf the cohomology w.r.t. an open cover and this problem N for the original question(from Rick Miranda's book). I have made no constructive approach to this problem so far. One idea is to prove it backwards, first prove that it's zero when all of the indices are equal, then it is zero. But there will be an evenness-oddness problem. Update: I referred to the wrong problem earlier. It should be problem N. | 1269863 | Skew symmetry of indices in cocycles of Cech cohomology
In Cech cohomology, the coboundary operator [imath]\delta:C^p(\underline U, \mathcal F)\to C^{p+1}(\underline U, \mathcal F) [/imath] is defined by the formula [imath] (\delta \sigma)_{i_0,\dots, i_{p+1}} = \sum_{j=0}^{p+1}(-1)^j \sigma_{i_0,\dots, \hat {i_j},\dots i_{p+1}}{\Huge_|}_{U_{i_0}\cap\dots\cap U_{i_{p+1}}}. [/imath] In the book by Griffiths and Harris, the authors claim that [imath]\delta\sigma=0[/imath] implies the skew-symmetry condition [imath] \sigma_{i_0,\dots,i_p} = -\sigma_{i_0,\dots, i_{q-1},i_{q+1},i_q,i_{q+2},\dots,i_p}. [/imath] How can I see this? EDIT: Or is it possible that the skew symmetry is merely an assumption for all cochains? |
2379252 | Does the n'th derivative of a smooth transition have to have more than n+1 roots for some n?
I'm wondering if the space [imath]S=\left\{f\in C^{\infty}([0,1])\ \ :\ \ f(0)=0\bigwedge f(1)=1\bigwedge\forall n\in\mathbb{N},\ f^{(n)}(0)=f^{(n)}(1)=0\right\}[/imath] contains an element [imath]f_0[/imath] such that [imath]f_0^{(n)}[/imath] has at most [imath]n+1[/imath] roots for all [imath]n\in\mathbb{N}_0[/imath]. Intuitively I'm somewhat inclined to think it's false. Do anyone know if this holds, or how I should go on about finding out? If [imath]f_0^{(n)}[/imath] has [imath]n+1[/imath] roots, then by the mean value theorem [imath]f_0^{(n+1)}[/imath] has [imath]n[/imath] distinct roots in the interior of the support in addition to the boundary points being roots. Hence by induction [imath]f_0^{(n)}[/imath] must have [imath]n+1[/imath] roots for all [imath]n\in\mathbb{N}[/imath]. As to whether the lower bound is non-strict, I think maybe the way to go is to assume that [imath](a_i)_{i\in\mathbb{N}}[/imath] is a sequence with [imath]a_i \in S[/imath] and [imath]a_i^{(i)}[/imath] having at most [imath]i+1[/imath] roots. Because if those properties imply that not all derivatives of [imath]a_i[/imath] remain bounded as [imath]i\to\infty[/imath], then no such sequence can have an element of [imath]S[/imath] as limit, and so the lower bound must be strict. I made some numerically experiments with linear programs which indicated the existence of such a sequence where also [imath]\sup_x a_i^{(1)}(x)<2[/imath]. So several derivatives must be considered simultaneously. | 664741 | Prove that there exists [imath]n\in\mathbb{N}[/imath] such that [imath]f^{(n)}[/imath] has at least n+1 zeros on [imath](-1,1)[/imath]
Let [imath]f\in C^{\infty}(\mathbb{R},\mathbb{R})[/imath] such that [imath]f(x)=0[/imath] on [imath]\mathbb{R}[/imath]\ [imath](-1,1)[/imath]. Prove that there exists [imath]n\in\mathbb{N}[/imath] such that [imath]f^{(n)}[/imath] has at least n+1 zeros on [imath](-1,1)[/imath] My attempt : Assume [imath]f[/imath] is strictly positive (if not n=0 works). Let [imath]Z(f^{(n)})[/imath] the number of zeros of [imath]f^{(n)}[/imath] on [imath](-1,1)[/imath]. Then we have: [imath] \forall n\in\mathbb{N}:\;f^{(n)}(-1)=f^{(n)}(1)=0 [/imath] so that iterating the Rolle theorem to the successive derivatives of f we have [imath] \;Z(f^{(k+1)}) \geq Z(f^{(k)}) +1 [/imath] thus, by induction : [imath] \forall n\in\mathbb{N}: \;Z(f^{(n+1)}) \geq Z(f^{(n)}) +1 [/imath] Therefore, [imath] Z(f^{(n)}) \geq n [/imath] How can I find an another zero ? I'm still looking for an another proof which does not use the lemma .., Thank you in advance. |
2383717 | [imath]A[/imath] is an [imath]n\times n[/imath] matrix where [imath]n[/imath] is an even number. If [imath]A^2 = 0[/imath] what must also be true?
The answer choices are: [imath]\operatorname{rank}(A) = 0[/imath] [imath]\operatorname{rank}(A) \leq \frac{n}{2}[/imath] (and possibly [imath]>0[/imath]) [imath]\operatorname{rank}(A) \leq n-1[/imath] (and possibly [imath]> \frac{n}{2}[/imath]) I know that 1. is false, but I don't know about 2 and 3. | 2034296 | Matrix algebra: If [imath]A^2=0[/imath], Proof rank(A) [imath]\le \frac{n}{2}[/imath]
For my matrix algebra class I need to prove the following: If [imath]A^2=0[/imath], Proof rank(A) [imath]\le \frac{n}{2}[/imath]. So if A is nilpotent prove Rank(A) [imath]\le \frac{n}{2}[/imath] I know already how to solve this, but my initial way of solving is false. I am looking for the mistake, but cannot find one. I know there already exists a question where this is asked. I'm just curious about my particular mistake. Proof: [imath]A=\begin{bmatrix}a_1&&a_2&& ...&&a_n\end{bmatrix}[/imath] and [imath]A=\begin{bmatrix}a^T_1\\a^T_2\\...\\a^T_n\end{bmatrix}[/imath] [imath]AA=\begin{bmatrix}a_1 && a_2 && ... && a_n\end{bmatrix}\begin{bmatrix}a^T_1\\a^T_2\\...\\a^T_n\end{bmatrix}[/imath] [imath]=\begin{bmatrix}a_1a^T_1&&...&&a1a^T_n\\...\\a_na^T_1 && ...&& a_na^T_n\end{bmatrix}[/imath] [imath]=\begin{bmatrix} a_1 \cdot a_1&&...&&a_1 \cdot a_n\\...\\a_n \cdot a_1 && ...&& a_n \cdot a_n\end{bmatrix}[/imath] [imath]=0_{nxn}[/imath] So we know the diagonal is zero thus [imath]a_i \cdot a_i = 0[/imath] this equals [imath]\vert\vert{a_i}\vert\vert^2[/imath] The square root of this equals the length, therefore the length is equal to 0. The only vector with this property is the zero vector. Herefore all vectors [imath]a[/imath] must be equal to the zero vector. The last however is not the truth, is there anyone who can spot my mistake? |
2383403 | continuum, cofinality must be uncountable
How does it follow from [imath]\kappa < \kappa^{cf \kappa}[/imath] that continuum [imath]2^{\aleph_0}[/imath] cannot be [imath]\aleph_{\omega}[/imath]? It is mentioned everywhere that it is obvious. | 1511419 | Does this seemingly elementary question require König's theorem?
Question Let [imath]f:\;\mathbb{R}\to\mathbb{N}[/imath] and let [imath]X_n[/imath] be the set of reals mapped to the integer [imath]n[/imath]. Show that for some [imath]n,\;X_n[/imath] has cardinality of the continuum. This is straightforward if we use König's theorem: assuming the contrary, i.e. [imath]\left|X_n\right|<2^{\aleph_0}[/imath] for all [imath]n[/imath]: [imath]2^{\aleph_0}=\left|\bigcup_{\mathbb{N}} X_n\right|=\sum_{\mathbb{N}}\big|X_n\big|\overset{\text{K.T.}}<\prod_{\mathbb{N}}2^{\aleph_0}=\big(2^{\aleph_0}\big)^{\aleph_0}=2^{\aleph_0}[/imath] a contradiction. This feels overkill, but perhaps it is not. Am I missing a more elementary solution to this problem? |
2383989 | Proof of the identity [imath]\sin(x+y) =\sin(x)\cos(y) + \cos(x)\sin(y)[/imath] for all [imath]x[/imath] and [imath]y[/imath]
I don't know if I'm asking for too much, but the proofs I've seen of the statement [imath]\sin(x+y) =\sin(x)\cos(y) + \cos(x)\sin(y)[/imath] consist of drawing a couple of triangles, one on top of each other and then figuring out some angles and lengths until they arrive at the identity. And I agree with the proof, is just that, even by flipping the triangle around, it only proves the identity for the case [imath]x+y<\pi/2[/imath], or if it does prove it for all values of [imath]x[/imath] and [imath]y[/imath], I wouldn't understand why. As to construing a proof by using Euler's identity or the derivatives of sin and cos, I would ask the writer to first prove his/her already accepted formulas without using the addition identity. So that is my humble question. How could one prove that for all the values of [imath]x[/imath] and [imath]y[/imath], the identity [imath]\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)[/imath] holds. Any thoughts/ideas would be really appreciated. | 635077 | Proof of the angle sum identity for [imath]\sin[/imath]
[imath]\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)[/imath] How can I prove this statement? |
2384061 | Integral [imath]\int_0^\infty x^ne^{-\frac{x}{a}} dx[/imath]
Improper integral of [imath]\int_0^\infty x^ne^{-x/a} \, dx[/imath] I tried to do it by using Partial Integral and looked for the trend of [imath]a[/imath] and [imath]n[/imath] but it's confusing and I cannot finish it Is there any other method to solve this? (I think of the Laplace Transform, but cannot find any relation) | 2378736 | Evaluate the integral [imath]\int e^{-x}x^n dx[/imath]
Evaluating the integral [imath]\int_{0}^{\infty} e^{-x}x^n dx[/imath] where n=0,1,2 ... 1st Integration by part: let [imath]a=x^n[/imath], [imath]da= nx^{n-1} dx[/imath], Also [imath]db=e^{-x}[/imath], [imath]b=-e^{-x}[/imath] We have [imath] -x^n e^{-x} + n\int x^{n-1}e^{-x} dx[/imath] 2nd: let [imath]a=x^{n-1}, da= (n-1)x^{n-2}, db= e^{-x}, b= -e^{-x}[/imath] We have [imath]-x^n e^{-x} + n\left[x^{n-1}(-e^{-x}) - \int(n-1)x^{n-2}\cdot -e^{-x}\right][/imath] How would you find the general form? What would it be? |
2383964 | Expressing [imath]e^{e^i}[/imath] in the form [imath]a+bi[/imath]
If Euler's formula states that [imath]e^{it} = \cos(t) + i\sin(t)[/imath], then how would I use it to express [imath]e^{e^i}[/imath]? Edit: I didn't even see that [imath]e^{e^i}[/imath] could be rendered as [imath]e^{cos(1)+isin(1)}[/imath]. I mishandled the problem and thought that [imath]e^{e^i}[/imath] could be rendered as [imath]e^{ie}[/imath], but can now clearly see that isn't the case. Thank you all for your input, this makes a lot of sense now. | 2383871 | Compute [imath]e^{e^z}[/imath], where z is a complex number
I am self-studying the MAT 67 linear algebra course and I got stuck at a problem in Chapter 2: Compute the real and imaginary parts of [imath]e^{e^z}[/imath] for [imath]z[/imath] [imath]\in[/imath] [imath]\Bbb C[/imath]. I tried to expand it using Euler's formula but it seemed endless. Any help? |
2384291 | Is it possible to to solve algebraically: [imath]e^z=z[/imath]?
While solving [imath]e^z=z[/imath], I realise there is no real solution and want to find a complex solution, if any. I searched around the internet only to find functions describing approximations that might be sufficiently close, but I can't find a way to know the exact answer. My question is : DERIVE all solutions to [imath]e^z=z[/imath] (and prove if there are none) :-) I'm NOT interested in the answer, I am interested in the solution | 137927 | Are there any simple ways to see that [imath]e^z-z=0[/imath] has infinitely many solutions?
Joseph Bak and Donald Newman's complex analysis book (p.236) has a proof that the equation [imath]e^z-z=0[/imath] has infinitely many complex solutions: I'm curious if there are any particularly elegant ways to see this, other than that given in the text. |
2384492 | Is multiplicative mod Z9 cyclic?
I'm asked if [imath]\mathrm Z_9[/imath]or [imath]\mathrm Z_{17}[/imath] under multiplication is cyclic or not, proof needed. I understand the rules with addition, but under multiplication i can't make sens of the rules. I tried to manually calculate it with but did not find a generator thus i assumed that it is not cyclic but with [imath]\mathrm Z_{17}[/imath] it's a little bit more excruciating, what is the best way to prove if a group like this is cyclic or not? thanks in advance. | 524019 | group generators of [imath](\mathbb Z_{17}-\{0\},\times)[/imath]
How to find generators of [imath](\mathbb Z_{17}-\{0\},\times)[/imath]? Is there a faster way to find generators than trying every element in the group? I know that for additive group, if a number say m is relatively prime to n (in this case 17), then it's a generator of that group. Do we have something like this for multiplicative group? |
2384093 | [imath]\lim_{x \rightarrow 0} f'(x) = L[/imath] exists, does it follow that [imath]f'(0)[/imath] exists?
The question is the following: Assume [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] is continuous, and for all [imath]x \neq 0[/imath], [imath]f'(x)[/imath] exists. If [imath]\lim_{x \rightarrow 0} f'(x) = L[/imath] exists, does it follow that [imath]f'(0)[/imath] exists? My intuition is that this does not have to hold in general. However, I keep finding counter examples such that [imath]f'(0)[/imath] exists where [imath]\lim_{x \rightarrow 0} f'(x) = L[/imath] does not exist, which is obviously not the original question. | 2989427 | Continuity of [imath]f[/imath] and the existence of [imath]f'(x)[/imath] given [imath]\lim_{x\rightarrow 0} f'(x)[/imath] exists.
Assume that [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] is continuous, and for all [imath]x \neq 0,[/imath] [imath]f'(x)[/imath] exists. If [imath]\lim_{x\rightarrow 0} f'(x)[/imath] exists, does it follow that [imath]f'(0)[/imath] exists? Prove or disprove. Intuitively, it seems to me that it does does. That [imath]f'(x)[/imath] would not exist implies a point discontinuity, since [imath]\lim_{x\rightarrow 0} f'(x)[/imath] exists. Given that [imath]f[/imath] is continuous, [imath]f[/imath] is defined at [imath]x.[/imath] However, I'm not sure how to go about proving this (if I'm correct). Any hints? Thanks :) |
2384764 | On the evaluation of [imath]\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{n \log n}[/imath]
I have tried to evaluate the series [imath]\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{n \log n}[/imath]. I get that it converges to [imath]\frac{1}{2}[/imath] but there must be a flaw in my reason since then result I get does not match that of Wolfram's Alpha. Let me show you my work. One good way to begin with is to recall the known Fourier series \begin{equation} \sum_{n=1}^{\infty} \frac{\cos nx}{n^s} =\frac{1}{2} \bigg({\rm Li}_s \left ( e^{-ix} \right ) + {\rm Li}_s \left ( e^{ix} \right ) \bigg) \end{equation} where [imath]{\rm Li}_s[/imath] is the polylogarithm of order [imath]s[/imath]. Hence \begin{equation} \sum_{n=2}^{\infty} \frac{\cos nx}{n^s} = - \cos x + \frac{1}{2} \bigg({\rm Li}_s \left ( e^{-ix} \right ) + {\rm Li}_s \left ( e^{ix} \right ) \bigg) \end{equation} We differentiate with respect to [imath]s[/imath] , hence \begin{equation} \sum_{n=2}^{\infty} \frac{\cos nx}{n^s \log n} = \frac{e^{ix}}{2} {\rm Li}_{s-1} \left ( e^{-ix} \right ) + \frac{e^{-ix}}{2} {\rm Li}_{s-1} \left ( e^{ix} \right ) \end{equation} For [imath]x=\pi[/imath] and [imath]s=1[/imath] we have that [imath]\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n} = \frac{1}{2}[/imath]. Where have I gone wrong? | 2178873 | On the series [imath]\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{n \log n}[/imath]
The series [imath]\displaystyle \sum_{n=2}^{\infty} \frac{1}{n \log n}[/imath] is known to diverge. This can be seen with comparison to the harmonic series. The other series [imath]\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n}[/imath] converges due to Liebniz's alternating criterion. I see these two series quite often in exams' questions to examine them if they converge or not. The answers are pretty straight forward. Now , out of curiosity , could we find a closed form for the series [imath]\mathcal{S}=\sum_{n=2}^{\infty} \frac{(-1)^n}{n \log n}[/imath] The interesting fact is that if we define the function [imath]f(s)=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^s \log n} , \; s \geq 1[/imath] then taking the derivative with respect to [imath]s[/imath] we get that [imath]f'(s)=-\sum_{n=2}^{\infty} \frac{(-1)^n}{n^s} = \zeta(s) - 2^{1-s} \zeta(s) - 1[/imath] I don't know how to go back and evaluate [imath]f(1)[/imath]. Has anyone done that before. Expressing the value of the series in terms of special functions will suit me. Numerically it is approximately equal to [imath]0.526412[/imath]. That said Wolfram. So, any clues? |
2384723 | Can [imath]\sin(xy)[/imath] be written in terms of trigonometric functions of only [imath]x[/imath] or [imath]y[/imath]?
Can [imath]\sin(xy)[/imath] be written in terms of trigonometric functions of only [imath]x[/imath] or [imath]y[/imath]? I am tempted to say yes, because the double- and half-angle formulae exist, and these would be special cases of [imath]\sin(xy)[/imath]. I first looked at the Taylor series, [imath]\sin(xy) = \sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}y^{2n+1}}{(2n+1)!} = xy-\frac{x^3y^3}{6}+\frac{x^5y^5}{120}-\cdots[/imath] but as far as I know, neither [imath]x[/imath] nor [imath]y[/imath] can come out of the sum since they are not constants. | 1656407 | Is there a formula for [imath]\sin(xy)[/imath]
Can we express a trigonometric function for the product of two angles as a function of trigonometric functions of its factors? For example: Is there a formula for [imath]\sin(xy)[/imath] as a function of [imath]\sin x[/imath] and [imath]\sin y[/imath] or other trigonometric functions of [imath]x[/imath] and [imath]y[/imath]. |
2385012 | Solving [imath]\lim_{x\to0} \frac{e^x-1-x}{x^2}[/imath]
[imath] = \lim_{x\to0} (\frac{e^x-1}{x} * \frac{1}{x}) -\frac{x}{x^2}[/imath] [imath] = \lim_{x\to0} (\frac{e^x-1}{x} * \frac{1}{x}) -\frac{x}{x^2}[/imath] [imath] = 1 * \lim_{x\to0} \frac{1}{x} -\lim_{x\to0}\frac{1}{x}[/imath] [imath] = \infty -\infty[/imath] How do I actually solve this without using L'Hospital. I'm just looking for a hint, not a concrete answer. | 1179383 | [imath]\lim_{x\to0}\frac{e^x-1-x}{x^2}[/imath] using only rules of algebra of limits.
I would like to solve that limit solved using only rules of algebra of limits. [imath]\lim_{x\to0}\frac{e^x-1-x}{x^2}[/imath] All the answers in How to find [imath]\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}[/imath] without using l'Hopital's rule nor any series expansion? do not fully address my question. A challenging limit problem for the level of student who knows that: [imath]\begin{align*} \lim\limits_{x\to +\infty} e^x&=+\infty\tag1\\ \lim\limits_{x\to -\infty} e^x&=0\tag2\\ \lim\limits_{x\to +\infty} \frac{e^x}{x^n}&=+\infty\tag3\\ \lim\limits_{x\to -\infty} x^ne^x&=0\tag4\\ \lim\limits_{x\to 0} \frac{e^x-1}{x}&=1\tag5 \end{align*}[/imath] |
2385874 | For which integers is [imath]i,j \geq 1[/imath] exists an integer [imath]k \geq 0[/imath], such that [imath]\frac{(i+j-1)!}{i!(j-1)!}=2^k[/imath]
Are there integers is [imath]i,j \geq 1[/imath] for which there is an integer [imath]k \geq 0[/imath], such that: [imath]\frac{(i+j-1)!}{i!(j-1)!}=2^k ?[/imath] There are two trivial families of solutions: For [imath]i=1[/imath]: [imath] \frac{j!}{(j-1)!}=j [/imath] and whenever [imath]j[/imath] is a power of 2 we have a solution [imath](1,j)[/imath]. And, for [imath]j=2[/imath]: [imath] \frac{(i+1)!}{i!} = i+1[/imath] and whenever [imath]i+1[/imath] is a power of 2 we have the second family of solutions (i,2). Are there any other solutions? And if not, is there a proof? EDIT: As was pointed out, one can write the equation as a binomial coefficient: [imath]\binom{n}{m}:=\binom{i+j-1}{i}=2^k,[/imath] with [imath]n=i+j-1[/imath] and [imath]m=i[/imath]. Since the product of [imath]m[/imath] consecutive integers greater than [imath]m[/imath] is divisible by a prime [imath]p[/imath] greater than [imath]m[/imath], we have that [imath]\binom{n}{m}= \frac{n(n-1)\ldots (n-m+1)}{m!} [/imath] is divisible by a prime [imath]p[/imath] and can therefore not be a power of 2 (if [imath]m,n-m\geq 2 \Leftrightarrow j,i-1 \geq 2[/imath]). The two families of solutions given above are therefore the only solutions to this equation. See also: Binomial Coefficients that are powers of 2 | 2338488 | Binomial coefficients that are powers of 2
I would like a proof that [imath] {{n}\choose{k}} = \frac{n!}{k!(n-k)!} = 2^m [/imath] for [imath]n,k,m\in \mathbb{N}[/imath], only if [imath]k=1[/imath] or [imath]k=n-1[/imath]. It seems to me that this must be true since for other values of [imath]k[/imath] the numerator contains more factors that are not powers of 2 than the denominator. Furthermore, the numerator also contains larger factors than the denominator and thus can't all be cancelled. Nevertheless, I have been unable to form an elegant, watertight proof. |
2386034 | all analytic bijective map from [imath]\mathbb{C}[/imath] to [imath] \mathbb{C}[/imath]
I am trying to prove that all analytic bijective map from [imath]\mathbb{C}[/imath] to [imath] \mathbb{C}[/imath] has the form of [imath]f(z)=az+b[/imath] where [imath]a[/imath] is nonzero. I tried to approach this problem by using taylor series and Cauchy integral formula, but how do I use the property of bijection? | 307185 | Automorphisms of the complex plane
How can it be shown that [imath]Aut(\mathbb{C})=\{f|f(z)=az+b,a\neq 0\},[/imath] where an automorphism of [imath]\mathbb{C}[/imath] is defined as a bijective entire function with entire inverse? If [imath]f[/imath] is of the form [imath]f(z)=az+b[/imath], with [imath]a\neq 0[/imath], then obviously [imath]f[/imath] is in [imath]Aut(\mathbb{C})[/imath]. How can I prove the converse? |
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