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2206619 | Summation of Binomial Coefficient Proof
Prove that [imath]\sum_{k=0}^n(-1)^k {{n}\choose{k}} = 0[/imath] I have no idea about how to approach this problem. | 1999341 | Working with binomial coefficient [imath]\sum_{k=0}^n (-1)^k \binom nk=0[/imath]
I'm sure there's an identity or some known "trick" to solve this but I couldn't find it (Looked in Wikipedia, Wolfram and here): [imath] \sum_{k=0}^n (-1)^k{n \choose k} =0 [/imath] I want to know how this can be proven. I tried induction but I'm not sure what to do with [imath]{{n+1}\choose k}[/imath]. Thanks! |
2204972 | Partial sum back into series.
If the [imath]n^{th}[/imath] partial sum of a series [imath]\sum a_n[/imath] is [imath]S_n = \frac{n−1}{n+1}[/imath] then find [imath]a_n[/imath] and find the sum of this series. Fully justify your answer. Using the definition. [imath]a_n = S_{n} - S_{n-1} [/imath] [imath]a_n = \frac{n-1}{n+1} - \frac{n-2}{n}[/imath] [imath]= \frac{n(n-1) - (n-2)(n+1)}{n(n+1)}[/imath] [imath]= \frac{n^2-n - (n^2-n-2)}{n(n+1)}[/imath] [imath]= \frac{2}{n(n+1)}[/imath] Hmm, I can use PFD on this to make it. [imath]= \frac{2}{n} - \frac{2}{n+1}[/imath] Is this right? What about n = 0 The sum is 1 right? cus lim at [imath]\infty[/imath] ? | 2198997 | If the [imath]n^{th}[/imath] partial sum of a series [imath]\sum a_n[/imath] is [imath]S_n = \frac{n-1}{n+1}[/imath], find [imath]a_n[/imath] and the sum.
If the [imath]n^{th}[/imath] partial sum of a series [imath]\sum a_n[/imath] is [imath]S_n = \dfrac{n-1}{n+1}[/imath], find [imath]a_n[/imath] and the sum. By definition, the sum of the series is the [imath]\lim n\to\infty[/imath] of it's [imath]n^{th}[/imath] partial sum. [imath]\text{ Sum = } \lim_{n\to\infty} = \dfrac{n-1}{n+1} = 1[/imath] I am asked to find [imath]a_n[/imath]. How do I do this? What is the procedure? |
2206219 | Is [imath]\sqrt{a+b} \leq \sqrt{a} + \sqrt{b}[/imath] when [imath]a,b \geq 0 [/imath] and [imath]a,b \in \mathbb{R}[/imath]?
Is [imath]\sqrt{a+b} \leq \sqrt{a} + \sqrt{b}[/imath] when [imath]a,b \geq 0 [/imath] and [imath]a,b \in \mathbb{R}[/imath]? I am convinced this must be true always but I can't prove it. Is this somehow related to the triangle inequality? | 1650642 | Does [imath]\sqrt{a+b} \le \sqrt a + \sqrt b[/imath] hold for all positive real numbers a and b?
I thought of this a while ago, but can't make up a proof or a counterexample. Does anyone know more about this? [imath]\sqrt{a+b} \le \sqrt a + \sqrt b , \forall a,b \in \mathbb R_+[/imath] Moreover, what happens with more variables? Say: [imath]\sqrt{x_1+x_2+...+x_n} \le \sqrt x_1 +\sqrt x_2 + ... + \sqrt x_n [/imath] with [imath] x_i \in\mathbb R_+ \forall i \in \{1,2,...,n\} [/imath] Or when [imath]i = \mathbb N[/imath] PS: I tried looking on the internet for this but I don't know how I would call this. |
2206851 | How can I prove that a sequence of real numbers always includes a decreasing or an increasing subsequence?
I'm trying to prove that a sequence of real numbers always includes a decreasing or increasing subsequence. At first I came up with the new sequence [imath]y_n:=inf\{x_m|m \ge n\}[/imath]. I figured out [imath]y_n[/imath] is increasing and every member of [imath]y_n[/imath] is an element of R due to the GLBP. However I now realize that the members of [imath]y_n[/imath] need not be an element of [imath]\{x_n\}[/imath], since it's an infinum... I'm stuck here and would appreciate any help. Thanks. | 1161048 | Prove there exists a subsequence of the real numbers such that it is monotonically increasing or decreasing.
Let [imath]x_n[/imath] be a sequence of real numbers. Prove that there exists a subsequence [imath]x_{n_k}[/imath] such that either [imath]x_{n_{k+1}} \le x_{n_k} [/imath] for all [imath]k[/imath] or [imath]x_{n_{k+1}} \ge x_{n_k} [/imath] for all [imath]k[/imath]. Would it suffice to define [imath]\{a_p\}[/imath] as [imath]a_p = x_{n_k}[/imath], [imath]n_k = min\{n \ |\ x_n \gt x_{n_{k-1}}\}[/imath] and [imath]\{b_p\}[/imath] as [imath]b_p = x_{n_k}[/imath], [imath]n_k = min\{n \ |\ x_n \lt x_{n_{k-1}}\}[/imath] |
2207396 | A and B are similar if and only if [imath]\text{rank}(A)=\text{rank}(B)[/imath]?
Suppose that [imath]A[/imath] and [imath]B[/imath] are linear transformations (on the same finite dimensional vector space) such that [imath]A^2=A[/imath] and [imath]B^2=B[/imath]. Is it true that [imath]A[/imath] and [imath]B[/imath] are similar if and only if rank([imath]A[/imath])=rank([imath]B[/imath])? As far as I can tell, it is true. What I have for the forward direction is this: [imath]A=C^{-1}BC[/imath] [imath]CA=BC[/imath] And by a theorem, we can say, rank([imath]CA[/imath])=rank([imath]BC[/imath]). Further, by the same theorem, we can say, rank([imath]A[/imath])=rank([imath]B[/imath]) because [imath]C[/imath] is invertible. I'm stuck on the reverse direction though. I can put an invertible [imath]C[/imath] in there, but past that, I don't think I can make the assumption that [imath]CA=BC[/imath]. Thanks. | 2001157 | Two idempotent matrices are similar iff they have the same rank
Definitions: For [imath]F[/imath] a field, [imath]A,B\in F^{m\times n}[/imath] are equivalent means there exist invertible matrices [imath]Q\in F^{m\times m}[/imath] and [imath]P\in F^{n\times n}[/imath] so that [imath]B=QAP[/imath], [imath]A,B\in F^{n\times n}[/imath] are similar means there exists an invertible matrix [imath]P\in F^{n\times n}[/imath] so that [imath]B=PAP^{-1}[/imath]. [imath][T]_\mathcal{B}[/imath] is the matrix of the linear transformation [imath]T:V\to V[/imath] with respect to an ordered basis [imath]\mathcal B[/imath] of [imath]V[/imath]. I've seen the result that [imath]A,B\in F^{m\times n}[/imath] are equivalent iff they have the same row and column reduced echelon form iff they have the same rank. I'd like to show that if [imath]A,B\in F^{n\times n}[/imath] are idempotent matrices, they are similar iff they have the same rank. I've noted that similarity of two matrices implies their equivalence, and hence that they have the same rank. So it remains to show that [imath]rank(A)=rank(B)[/imath] implies [imath]A[/imath] and [imath]B[/imath] are similar. Observations: If [imath]V[/imath] is a finite-dimensional vector space over field [imath]F[/imath] and [imath]T:V\to V[/imath] is an idempotent linear transformation (i.e. [imath]T^2=T[/imath]), then [imath]V=im T\oplus \ker T[/imath] and if [imath]\mathcal B_1=\{v_1,v_2,...,v_m\},\mathcal B_2=\{v_{m+1},...,v_n\}[/imath] are bases for [imath]im T[/imath] and [imath]\ker T[/imath] respectively, then [imath][T]_{\mathcal B_1\cup \mathcal B_2}\in F^{n\times n}[/imath] is a [imath]1\times 2[/imath] block matrix with [imath]C\in F^{n\times m}[/imath] in the first entry and zero matrix for the other entry, where the [imath]j[/imath]th column of [imath]C[/imath] is given by the scalars [imath]a_{1j},...,a_{nj}[/imath] so that [imath]T(v_j)=\sum_{i=1}^n a_{ij}v_i[/imath] (which uniquely define [imath]T(v_j)[/imath] since [imath]\{v_1,...,v_n\}[/imath] are a basis for [imath]V[/imath]). Let [imath]L_A,L_B[/imath] be the linear transformations representing left-multiplication by [imath]A[/imath] and [imath]B[/imath], respectively. Then [imath]A'=[L_A]_\mathcal B[/imath] and [imath]B'=[L_B]_\mathcal B[/imath] take the block matrix form described above, and since [imath]rank (A)=rank (B)[/imath] by assumption, [imath]A'[/imath] and [imath]B'[/imath] have the same row and column reduced echelon form. So starting from the row and column reduced echelon form that they take, we can multiply on the left and right by invertible matrices (by doing the inverse elementary matrix operations that we did to get the reduced forms) to get back to either matrix, i.e. if [imath]D[/imath] is the row and column reduced echelon form for the two, then [imath]QA'P=D=RB'S[/imath] where [imath]Q,P,R,S[/imath] are invertible [imath]n\times n[/imath] matrices. Hence [imath]B'=R^{-1}QA'PS^{-1}[/imath]. We want to show that the [imath]A,B[/imath] being idempotent gives us that [imath]R^{-1}Q=(PS^{-1})^{-1}[/imath]. This is where I'm stuck. Any help would be greatly appreciated! Thanks in advance |
2208419 | Can the "proves consistent" relation be proven antisymmetric?
Can the "proves consistent" relation be proven antisymmetric? By antisymmetric I mean there is not some chain of theories in which some theory proves other theories consistent which in turn prove it consistent. If [imath]\succ[/imath] is the "proves consistent" relation: [imath]A\succ B\iff (A\models Con(B))\quad\forall A\forall B [/imath] then I think antisymmetry is usually written as follows: [imath]A\succ B\land B\succ A\implies A=B\quad\forall A\forall B[/imath] Can this be proven? This is an inquiry motivated in part by learning that Gentzen's Consistency Proof is neither weaker nor stronger than PA, which suggests an ordering may not necessarily be implied by proving consistency. It would seem to me that by Godel's imcompleteness theorem we're denied reflexivity: [imath]A\succ B\implies A\neq B[/imath] which denies us the usual antisymmetry rule. But it seems this doesn't necessarily rule out antisymmetry provided we have transitivity. Because if [imath]\succ[/imath] were both transitive and symmetric, we might have the following contradiction: [imath]A\succ C\succ A \implies A\succ A\implies A\neq A[/imath] [imath]\therefore\nexists C\mid A\succ C\succ A[/imath] Which fulfils the purpose of antisymmetry when we're denied reflexivity. So the limit of what I can prove is that "proves consistent" is antisymmetric provided it is also transitive - which I have, as yet, no reason to believe it is. | 2076800 | Proofs of consistency for two formal systems
Do there exist two formal systems, [imath]F_1[/imath] and [imath]F_2[/imath], such that [imath]F_1[/imath] proves [imath]F_2[/imath] is consistent and [imath]F_2[/imath] proves [imath]F_1[/imath] is consistent? Would these proofs bypass Gödel's Second Incompleteness Theorem? By "consistent", I mean the formal system does not prove [imath]X[/imath] and NOT [imath]X[/imath]. |
2206415 | Use integration by parts to prove
Let [imath]n \in \mathbb{Z_+},[/imath] use integration by parts to prove [imath]\int_0^{\infty}x^ne^{-x}dx = n![/imath] I know that if you repeatedly differentiate [imath]x^n[/imath] you get [imath]n![/imath], but I don't know how to prove this? A step by step answer would be helpful. Thanks! | 2204941 | Use integration by parts to prove [imath] \int^\infty_0 x^ne^{-x} dx=n![/imath]
[imath] \int^\infty_0 (x^n)(e^{-x}) dx [/imath] and show that it is equal to [imath](n!)[/imath] ? I know that if you differentiate [imath]x^n[/imath] infinitely you get [imath]n![/imath] but I don't know how to prove? |
2207863 | Proving a Binomial Summation with Induction [imath]\sum_{k=0}^n(-1)^k \binom nk^2[/imath]
I'm attempting to prove that [imath]\sum_{k=0}^n(-1)^k{n \choose k}^2=\begin{cases}0&\text{if $n$ is odd}\\(-1)^m{2m \choose m}&\text{if $n$ = 2m}\end{cases}[/imath] via induction. Here's what I have so far: Base Case (odd): [imath]n=1[/imath] [imath]\sum_{k=0}^1(-1)^k{1 \choose k}^2=0[/imath] Base Case (even): [imath]n=2[/imath] [imath]\sum_{k=0}^2(-1)^k{2 \choose k}^2=-2[/imath] Hypothesis: [imath]n=l[/imath] [imath]\sum_{k=0}^l(-1)^k{l \choose k}^2=\begin{cases}0&\text{if $n$ is odd}\\(-1)^m{2m \choose m}&\text{if $n$ = 2m}\end{cases}[/imath] Inductive Step: [imath]n=l+1[/imath] [imath]\sum_{k=0}^{l+1}(-1)^k{l+1 \choose k}^2[/imath] I know that the goal is to get this sum into terms of the hypothesis plus the [imath](l+1)^{th}[/imath] term, but I'm not sure how. I've tried manipulating various identities, but so far I'm not coming up with anything workable. Thanks in advance for any help. | 926937 | find [imath]\sum_{k=0}^{t}(-1)^k\binom{t}{k}^2[/imath] for odd t then for even t
find [imath]\sum_{k=0}^{t}(-1)^k\binom{t}{k}^2[/imath] for [imath]t=2n[/imath] then for [imath]t=2n+1[/imath] I tried by expand [imath](1-x)^n(1-x)^n[/imath], with no result. Any Help ? |
2209081 | Showing that [imath]Aut~Z_p \simeq Z_{p-1}[/imath]
Okay. This is similar to a question I asked a few days ago. I am trying to show that [imath]Aut~ Z_p[/imath] is isomorphic to [imath]Z_{p-1}[/imath], where [imath]Z_p[/imath] denotes the congruence class of integers [imath]\mod p[/imath] and [imath]p[/imath] a prime. I have shown [imath]Aut~ Z_p[/imath] consists of [imath]p-1[/imath] elements, using the fact that a homomorphism is uniquely determined by how it maps [imath][1]_p[/imath], the generator of [imath]Z_p[/imath]; the only element it cannot be mapped to is [imath][0]_p[/imath], otherwise [imath][1]_p \rightarrow [k]_p \neq [0]_p[/imath] extends to an automorphism. Now I am trying to show that [imath]Aut~ Z_p[/imath] is a cyclic, and show that mapping the generator of [imath]Aut~ Z_p[/imath] to [imath]Z_{p-1}[/imath] defines an isomorphism. However, I am having trouble showing this. I could use some hints. Note: At this point, I don't know anything about fields (let alone rings), the order of an element, and related concepts. Edit: I noticed that the question was asked here, but from my cursory reading I didn't see any ideas I could use yet. | 361651 | Aut [imath]\mathbb Z_p\simeq \mathbb Z_{p-1}[/imath]
I'm trying to prove that Aut [imath]\mathbb Z_p\simeq \mathbb Z_{p-1}[/imath] (p prime). I know that Aut [imath]\mathbb Z_p[/imath] has [imath]p-1[/imath] elements because [imath]\mathbb Z_p[/imath] has [imath]p-1[/imath] possiblities of generators, so intuitively I see Aut [imath]\mathbb Z_p\simeq \mathbb Z_{p-1}[/imath], but I couldn't prove it formally. I'm trying to build an isomorphic function, but I don't how to do it. Am I in the right way? Maybe I'm forgetting some trick or something. Thanks |
2209383 | Well Ordering of Sets of Natural Numbers
Subsets of [imath]\Bbb N[/imath] are well ordered. So two subsets of [imath]\Bbb N[/imath], A and B could be compared by comparing the least elements in A\B and B\A; whichever has the lesser is the lower. This relation appears also transitive. It looks to me that this "ordering" will find a "least" in any set of subsets of [imath]\Bbb N[/imath]. But this would produce an effective well ordering of the continuum, which is impossible. So the proposed "order" is either non-transitive or it will fail to find the "least" in some set of subsets of [imath]\Bbb N[/imath]. Is there immediately obvious where the mistake (if any)? PS The proposed relation is not even an ordering, as detailed below. Although the counter example from the other post will work here as well. | 2189342 | Why isn't this a well ordering of [imath]\{A\subseteq\mathbb N\mid A\text{ is infinite}\}[/imath]?
So, to explain the title, I'm referring to the necessity of the axiom of choice in the existence of a well ordering on reals, or any uncountable set. Now, while tweaking some sets, I came across this : We start with the natural numbers, [imath]N[/imath]. We take the power set of the naturals, [imath]P(N)[/imath]. Then we remove all the finite subsets of [imath]N[/imath] from [imath]P(N)[/imath]. Let us call this new set [imath]S[/imath]. This set is the set of all infinite subsets of [imath]N[/imath]. It is easy to show that [imath]S[/imath] has uncountable cardinality, same as that of real numbers. This is because the removal of finite subsets only removes a countable number of elements. (Haven't posted this deduction, for it is very easy, but I may post it if it is not so evident) Now, we seek to find an ordering on the set [imath]S[/imath]. Every set in this set is an infinite subset of natural numbers, so each of these sets are well ordered by the natural ordering of [imath]N[/imath]. Taking any two sets in [imath]S[/imath], say [imath]A[/imath], and [imath]B[/imath], we seek to order them by checking their elements lexicographically. We compare the first two elements in [imath]A[/imath], and [imath]B[/imath]. Let them be [imath]a_1[/imath], and [imath]b_1[/imath] respectively. If [imath]a_1 = b_1[/imath], then we move on to the second elements in the sets, [imath]a_2[/imath], and [imath]b_2[/imath], and so on. If, at any point, [imath]a_n < b_n[/imath], then [imath]A < B[/imath], or if [imath]b_n < a_n[/imath], then [imath]B < A[/imath]. This order seems to be a well ordering of the uncountable infinity of reals. I don't seem to have invoked the axiom of choice anywhere in the construction of this set [imath]S[/imath]. So, why isn't this a well ordering on the uncountable of reals? |
2209470 | How to prove two closed intervals are equinumerous?
Prove that [imath][0,1] \approx [-\pi,e^2][/imath] (The notation [imath]\approx[/imath] is used to denote equinumerous) I know to prove two closed intervals are equinumerous I need to show a bijective function that will map from one interval to the other but I cannot seem to find the correct mapping. | 2032508 | The interval [0,1] andd [3,5] are equivalent. Is my proof correct?
The intervals [imath][0,1][/imath] and [imath][3,5][/imath] are equivalent. My proof goes like this. Proof. To show that the two sets are equivalent, we should show a bijection between them. Consider the function [imath]f:[0,1] \to [3,5][/imath] such that [imath]f(x)=2x+3[/imath]. Since [imath]f[/imath] is linear, it is bijective. Therefore, the intervals [imath][0,1][/imath] and [imath][3,5][/imath] are equivalent. I am not sure if this is right. Assuming that this is correct, is it better to show that [imath]f[/imath] is reflexive, symmetric, and transitive? If this is wrong, what is the right way of proving? Sorry if this is so simple. |
621717 | If [imath]f(0)=0 ,f(1)=1 ,f'(0)=f'(1)=0[/imath], then [imath]|f''(x)|>4[/imath]
Let [imath] f [/imath] be a twice differentiable function with [imath] f (0) = 0, f (1) = 1[/imath] and [imath] f '(0) = f' (1) = 0 [/imath], then [imath] 4 \leq | f'' (x ) | [/imath], for some [imath] x \in [0,1] [/imath]. I tried using the mean value theorem for derivatives with [imath] \dfrac{f '(1)-f' (0)}{f (1)-f (0) } = f '(c) = 0[/imath] so there is a value at which [imath] f ''(c) = 0 [/imath], but that did not help me. Any ideas? Any help is appreciated. | 835042 | [imath]f(0)=f'(0)=f'(1)=0[/imath] and [imath]f(1)=1[/imath] implies [imath]\max|f''|\geq 4[/imath]
Let [imath]f\in C^2(\mathbb [0,1],\mathbb [0,1])[/imath] such that [imath]f(0)=f'(0)=f'(1)=0[/imath] and [imath]f(1)=1[/imath] Prove that [imath]\max_{[0,1]}|f''|\geq 4[/imath] Progress Applying Cauchy mean value theorem three times proves the existence of [imath]\xi\in (0,1)[/imath] such that [imath]f'(\xi)=1[/imath] [imath]\eta\in(\xi,1)[/imath] such that [imath]\displaystyle f''(\eta)=\frac{1}{\xi-1} <0[/imath] [imath]\beta\in(0,\xi)[/imath] such that [imath]\displaystyle f''(\beta)=\frac{1}{\xi}>0[/imath] If [imath]\displaystyle \xi\leq \frac{1}{4}[/imath] or [imath]\displaystyle \xi\geq \frac{3}{4}[/imath], we're done. What about other cases ? I haven't used the continuity of [imath]f''[/imath] yet... |
2208747 | Need help finding the limit of this sequence.
If [imath]x_1< x_2[/imath] are real numbers, and [imath]x_n = \frac{1}{2}(x_{n-2} + x_{n-1})[/imath] for every [imath]n > 2[/imath], how do I show [imath]\lim x_n = \frac{1}{3}x_1 + \frac{2}{3}x_2[/imath]? | 1862616 | If [imath]2a_{n+2} \le a_{n+1}+a_n[/imath], then [imath]\lim \sup a_n \le \frac23 a_2 + \frac13 a_1[/imath]
This is a reformulation of a deleted question: If [imath]a_1 > 0[/imath] and [imath]a_2 > 0[/imath] and [imath]2a_{n+2} \le a_{n+1}+a_n[/imath], show that [imath]\lim \sup a_n \le \frac23 a_2 + \frac13 a_1[/imath]. My proof involves showing that [imath]a_{n+2} \le u_{n} a_{n+1} + (1-u_n)a_n [/imath] where [imath]u_n \to \frac23[/imath], and I wondered if there is a simpler proof. I will post my proof in a couple of days if no one does. |
2210065 | If [imath]a[/imath] is a positive integer of the type [imath]3n+2[/imath],then pove that at-least one prime divisor of [imath]a[/imath] is of the form [imath]3n+2[/imath]
If [imath]a[/imath] is a positive integer of the type [imath]3n+2[/imath],then pove that at-least one prime divisor of [imath]a[/imath] is of the form [imath]3n+2[/imath] My try:If [imath]3n+2[/imath] is prime then we are done since [imath]3n+2[/imath] is of the form [imath]3n+2[/imath]. What do we do in the case it isn't prime? | 320643 | prime divisor of [imath]3n+2[/imath] proof
I have to prove that any number of the form [imath]3n+2[/imath] has a prime factor of the form [imath]3m+2[/imath]. Ive started the proof I tried saying by the division algorithm the prime factor is either the form 3m,3m+1,3m+2. Case 1: [imath]3m[/imath] suppose [imath]3m[/imath] does divide [imath]3n+2[/imath]. That means [imath]3mr=3n+2[/imath] where [imath]r[/imath] is some integer. But we get [imath]mr-n= 2/3[/imath] and since [imath]mr-n[/imath] is an integer that's a contradiction. Thus [imath]3m[/imath] does not work. Case 2: [imath]3m+1[/imath] Same argument here Case 3: [imath]3m+2[/imath] By the fundamental theorem of arithmetic every integer is divisible by a prime. But not sure what to do from here. |
2209985 | Prove [imath]1 + \frac{n}{2} \leq 1+ \frac{1}{2} +\frac{1}{3} +\cdots + \frac{1}{2^n}[/imath] for all natural numbers [imath]n[/imath]
Definitions [imath]H_n = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}[/imath] for all [imath]n \in \mathbb{N}[/imath] The Question Prove [imath]1 + \frac{n}{2} \leq H_{2^n}[/imath] for all [imath]n \in \mathbb{N}[/imath] My Work Base Case: [imath]1+\frac{1}{2} \leq 1+\frac{1}{2} = H_1[/imath] Inductive Hypothesis: [imath]1 + \frac{k}{2} \leq H_{2^k}[/imath] for all [imath]k \in \mathbb{N}[/imath] Induction Step: [imath]1+\frac{k+1}{2} = 1+\frac{k}{2} + \frac{1}{2} \leq H_{2^k}+\frac{1}{2} \leq H_{2^k} + \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}} = H_{2^{k+1}} [/imath] My Problem My problem is actually understanding the [imath]H_{2^k}+\frac{1}{2} \leq H_{2^k} + \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}[/imath] step. I think that's how the proof should finish, but I don't know why. My Question Can someone explain why the inequality under the "My Problem" header is true? Or if it even is true, am I going about this proof the wrong way? | 116011 | Proof by induction of summation inequality: [imath]1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2[/imath]
Prove by induction the summation of [imath]\frac1{2^n}[/imath] is greater than or equal to [imath]1+\frac{n}2[/imath]. We start with [imath]1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2[/imath] for all positive integers. I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used. This is equivalent to [imath]\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.[/imath] Let [imath]P(n)[/imath] be summation shown above. Base case for [imath]n=1[/imath], the first positive integer, [imath]\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,[/imath] so base case is true. Induction step: Assume [imath]P(n)[/imath] is true and implies [imath]P(n+1)[/imath]. Thus [imath]\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.[/imath] This can be written as [imath]\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.[/imath] I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes. |
2209680 | Difference between number of scalars for proving linear transformations
Why did the solution to the first question only have one scalar, and the solution to the second had two scalars? Is it up to the solver's preference? Question 1: Define [imath]T : \mathbb{P}_3 \to \mathbb{R}^4[/imath] by [imath]T(p) = \begin{bmatrix} p(-3) \\ p(-1) \\ p(1) \\ p(3) \end{bmatrix}[/imath] Show that [imath]T[/imath] is a linear transformation. Question 2: Let [imath]T : \mathbb{P}_2[/imath] -> [imath]\mathbb{P}_4[/imath] be the transformation that maps a polynomial [imath]p(t)[/imath] into the polynomial [imath]p(t) + t^2 p(t)[/imath]. Show that [imath]T[/imath] is a linear transformation. | 2209658 | Prove that T is a linear transformation
Does it matter that in the first line it's written [imath]T(\alpha p+ \beta g)[/imath] and not [imath]T(\alpha p(t)+ \beta g(t))[/imath] but at the end it is written with [imath]\alpha T(p(t)) + \beta T(g(t)))[/imath] with the [imath]t[/imath]'s. Define [imath]T : \mathbb{P}_3 \to \mathbb{R}^4[/imath] by [imath]T(p) = \begin{bmatrix} p(-3) \\ p(-1) \\ p(1) \\ p(3) \end{bmatrix}[/imath] Show that [imath]T[/imath] is a linear transformation. |
2210187 | Number of bases of [imath]F_2^3[/imath]
Number of bases of [imath]F_2^3[/imath]: Any ideas in which to express this?. Is the formula [imath]V=F_P^3[/imath] in any way applicable? | 1235852 | Finding the number of bases for [imath] \mathbb{F}_3^2 [/imath] and isomorphisms
Find the number of bases for [imath] \mathbb{F}_3^2 [/imath] and also the amount of Isomorphisms [imath] \mathbb{F}_3^2 \rightarrow \mathbb{F}_3^2 [/imath] Here if [imath](v1,v2)[/imath] is a basis, then [imath](v2, v1)[/imath] is a different basis. I have no idea how to approach this question. Could someone help me out? |
2210263 | Alternative methods for [imath]\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx[/imath]
[imath]I=\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx[/imath] I was recently in a small, friendly integration competition and this problem came up. Eventually, I found a solution using messy long division (shown below). I was wondering of other methods (preferably real-method solutions, but complex methods such as residues (if applicable) would also be nice to see) that could be used here. I spent a lot of time trying to find a nice trigonometric substitution that would work (if you use [imath]x=\tan(u)[/imath], it can be brought to integrating [imath]\int_{0}^{\frac{\pi}{4}}\tan^4(u)(1-\tan(u))^4dx[/imath] which doesn't seem fun/faster and other substitutions I tried ended up similarly). So, that's what I'm hoping to see. There may also be a clever change of variables that that I didn't see (e.g. somehow using [imath]\int_{a}^{b} f(x)dx=\int_{a}^{b}f(a+b-x)dx[/imath] or similar) Long Division Method: Multiply the numerator out to prepare for long division: [imath](1-x)^4=((1-x)^2)^2=(1-2x+x^2)^2=1-4x+6x^2-4x^3+x^4[/imath] hence [imath]x^4(1-x)^4=x^4-4x^5+6x^6-4x^7+x^8[/imath] Then long division shows that [imath]\dfrac{x^4(1-x)^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+4-4\cdot\dfrac{1}{1+x^2}[/imath] Finally, we get that [imath]I=\int_{0}^{1}\left[x^6-4x^5+5x^4-4x^2+4-4\cdot\frac{1}{1+x^2}\right]dx[/imath] [imath]I=\frac{1}{7}x^7-\frac{2}{3}x^6+x^5-\frac{4}{3}x^3+4x-4\arctan(x)\bigg\vert_{0}^{1}[/imath] [imath]I=\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-4\cdot\frac{\pi}{4}-0=\frac{1}{7}+3-\pi=\boxed{\frac{22}{7}-\pi}[/imath] (Yes, I do see the cleverness of [imath]\frac{22}{7}[/imath], an approximation of [imath]\pi[/imath].) | 1450910 | Shortest method for [imath]\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}[/imath]
I don't want to solve by expanding it and all, I tried corollary but denominator becomes messy, also in the options there is [imath]\pi[/imath] so tried several trigonometric substitutions too |
2209815 | [imath]E[/imath] is a splitting field of [imath]f(x)=x^3-3x+1[/imath] over [imath]\mathbb Q[/imath].Then determine the group [imath]G(\frac {E}{\mathbb Q})[/imath]
[imath]E[/imath] is a splitting field of [imath]f(x)=x^3-3x+1[/imath] over [imath]\mathbb Q[/imath].Then determine the group [imath]G(\frac {E}{\mathbb Q})[/imath] MY try : Actually I don't Know how to relate splitting field to group,But I found something like [imath]Gal(E/\mathbb{Q}) \simeq S_3[/imath] | 2017810 | Galois group of irreducible cubic equation
The polynomial [imath]f(x) = x^3 -3x + 1[/imath] is irreducible, and I'm trying to find the splitting field of the polynomial. We've been given the hint by our lecturer to allow an arbitrary [imath]\alpha[/imath] to be a root and then to check that [imath]\alpha[/imath] is a root, then [imath]2-\alpha^2[/imath] is also root. I've done this with no problem. My issue is since this is a polynomial of degree [imath]3[/imath], there must be another root that is somehow related to [imath]\alpha[/imath] so that I can create the splitting field and in turn the Galois group of my polynomial but I have no idea what this last root could be. |
2210803 | Quadratic (Pythagorean?) Diophantine Equation
How do you find all (non-trivial) [imath](x, y)[/imath] such that [imath]x^2 + y^2 = 2017^2?[/imath] I really don't want to go through trial and error, like what I managed to do here: So, we can rewrite the equation as [imath]x^2 = 2017^2 - y^2 \implies x^2 = (2017 + y)(2017 - y).[/imath] From there, I know nothing but guessing values of [imath]y,[/imath] which would be rather painful. Yes, I know the Babylonian method of finding Pythagorean triples, but I figured since [imath]2017[/imath] is prime it would still simplify to [imath]x^2 + y^2 = 2017^2.[/imath] Thanks in advance for your help. Ah, I got my Babylonian method wrong. It is for [imath]a, b, c, m, n \in \mathbb{Z}[/imath] such that [imath]a^2 + b^2 = c^2[/imath] and (WLOG) [imath]m > n,[/imath] \begin{align} a &= k\left(m^2 - n^2\right) \\ b &= 2kmn \\ c &= k\left(m^2 + n^2\right) \end{align} | 657226 | [imath]2017[/imath] as the sum of two squares
Write the prime [imath]2017[/imath] as the sum of two squares [imath]2017[/imath] can be written as the sum of two squares because it is a prime of the form [imath]p\equiv 1\ ([/imath]mod [imath]4)[/imath] Using an appropriate algorithm find the two numbers that, when squared, add to the total of [imath]2017[/imath] |
173113 | Relationship between prime factorizations of [imath]n[/imath] and [imath]n+1[/imath]?
Are there any theorems that give us any information about the prime factorization of some integer [imath]n+1[/imath], if we already know the factorization of [imath]n[/imath]? Recalling Euclid's famous proof for the infinity of the set of prime numbers, I guess we know that if [imath]n = p_1 p_2 p_3[/imath], then [imath]n+1[/imath] cannot have [imath]p_1[/imath], [imath]p_2[/imath], or [imath]p_3[/imath] as factors. But is there any way we could use the information about [imath]n[/imath]'s factorization to determine something more precise about the factorization of [imath]n+1[/imath]? | 2061815 | Prime factorization of [imath]x + 1[/imath] given prime decomposition of [imath]x[/imath]
Suppose you had the prime factorization of [imath]x = \prod_{k=1}^m p_k[/imath]. Are there any algorithms for deciding the prime factorization of [imath]x + 1[/imath]? Are their prime factorizations related in any way at all? EDIT: It's known that [imath]x[/imath] and [imath]x + 1[/imath] are coprime. That means that you get some amount of information about the prime factors of [imath]x + 1[/imath]. Can the information about prime factors of [imath]x[/imath] be used to figure out the prime factors of [imath]x + 1[/imath] faster than just "prime sieve" + "filter out the factors of [imath]x[/imath]"? |
2210192 | Need to prove [imath]f(x)=0[/imath] for all [imath]x \in R[/imath]
[imath]f(x)[/imath] is infinitely differentiable and [imath]∃ L∈\mathbb{R}[/imath] such that [imath]|f^{(k)}(x)|≤L[/imath] for any [imath]k∈\mathbb{N}[/imath]. I need to prove that : If [imath]f(1/n)=0[/imath] then [imath]f(x)=0[/imath] for any [imath]x∈\mathbb{R}[/imath]. | 1086860 | If all derivatives of [imath]f[/imath] are uniformly bounded by a common constant and [imath]f(1/n) = 0[/imath] for all [imath]n[/imath], is [imath]f[/imath] identically zero?
[imath] f: \Re \longrightarrow \Re \ \in C^{\infty} \\ \exists \ L>0: \ \forall x \in \Re, \forall n\in N \\ |f^{(n)} (x)| \le L \\ f(\frac{1}{n})=0 \ \forall n\in N \\ f(x) \equiv 0 [/imath] Good morning, Can you give me an help to make this proof? This is my thought: [imath]\lim_{n \rightarrow +\infty } \dfrac{1}{n}=0=\theta[/imath] [imath]f(\theta)=0[/imath] [imath]f^{(n)} (\theta) =0\\f(x)=\sum_{k=0}^n \dfrac{f^{(k)}(x-\theta)^k}{k!}[/imath] [imath]f[/imath] is analytic, the rest of Taylor's expansion is 0, because the derivatives are bounded, [imath]f[/imath] is null. I have thought this reasoning, but I have difficulties to demonstrate part 3. Can you help me? Is there any mistake? Can I use an easier way to make the proof? Thanks. |
2211073 | Prove that three points are collinear using their corresponding vectors.
Let [imath]a = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}[/imath], [imath]b = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}[/imath], and [imath]c = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}[/imath]. Show that [imath](x_a,y_a,z_a)[/imath], [imath](x_b,y_b,z_b)[/imath], and [imath](x_c,y_c,z_c)[/imath] are collinear if and only if [imath]a \times b + b \times c + c \times a = 0.[/imath] I was thinking of proving that the area of the triangle formed by the three points is 0. I thought the box product, [imath]|(a \times b)\bullet c|[/imath], would be helpful but I don't know how to relate that to the equation. All help is greatly appreciated. | 1339242 | Proof with 3D vectors
Let [imath]{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}[/imath], [imath]{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}[/imath], and [imath]{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}[/imath]. Show that [imath](x_a,y_a,z_a)[/imath], [imath](x_b,y_b,z_b)[/imath], and [imath](x_c,y_c,z_c)[/imath] are collinear if and only if [imath]{a} \times {b} + {b} \times {c} + {c} \times {a} ={0}.[/imath] Hello, Is there any other way to do this problem without bashing? I can't seem to find a nice and slick solution to this. Thanks in advance! |
1474150 | Proving [imath]A \cap (B-A) = \emptyset[/imath] (new to proofs!)
I'm a novice mathie - only calculus courses until this past semester, when I started Foundations of Mathematics (a bit of naïve set theory, introduction to basic proofs.) Up to now, I have been able to complete proofs assigned by using parity, simple modular math, greatest common divisor, etc. That being said, I am stuck on developing a proof for this. Let [imath]A[/imath] and [imath]B[/imath] be sets. [imath]A \cap(B-A) = \emptyset[/imath]. The methods I have at my disposal are direct, contrapositive, and contradiction. I would appreciate a gentle nudge in the right direction. This one has me stumped. | 250500 | Is a contradiction enough to prove a set equality to [imath]\varnothing[/imath]?
An exercise asks me to prove the following: [imath]A \cap (B-A) = \varnothing[/imath] This is what I did: I need to prove that [imath]A \cap (B-A) \subseteq \varnothing[/imath] and [imath]\varnothing \subseteq A \cap (B-A)[/imath] The second one seems to be obvious by definition (not sure if it is fine to say that in a test). But the first one goes like this: [imath]x \in A \cap (B-A)[/imath] [imath]x \in A \land x \in (B-A)[/imath] [imath]x \in A \land (x \in B \land x \notin A)[/imath] [imath](x \in A \land x \notin A) \land x \in B[/imath] [imath](x \in A \land x \notin A)[/imath] Since there is a contradiction... err... the proof... is good. Okay, that's my problem. I think that the procedure is fine, but clearly I am unable to word it out (well, I don't even know if it is valid at all). Is it valid? How can I... "prove that my proof is correct"? I don't really see this contradiction as an obvious indicator of [imath]A \cap (B-A) \subseteq \varnothing[/imath] |
2211317 | How many generators do we need to generate the Rubik's Cube group?
Until today, I was convinced that we needed at least 6 generators to generate the Rubik's Cube group. However, consider the following... [imath]a = (0,5,7,2)(1,3,6,4)(8,16,24,32)(9,17,25,33)(10,18,26,34)[/imath] [imath]b = (5,15,42,24)(6,12,41,27)(7,10,40,29)(16,21,23,18)(17,19,22,20)[/imath] [imath]c = (0,26,47,13)(1,28,46,11)(2,31,45,8)(32,37,39,34)(33,35,38,36)[/imath] [imath]d = (0,39,40,16)(3,36,43,19)(5,34,45,21)(8,13,15,10)(9,11,14,12)[/imath] [imath]e = (2,18,42,37)(4,20,44,35)(7,23,47,32)(24,29,31,26)(25,27,30,28)[/imath] d^{-1}c^{-1}da^{-1}e^{-1}aeb^{2}d^{-1}b^{-1}ab^{-1}e^{-1}b^{-1}d^{-1}baba^{-1}d^{-1}a^{-1}ca^{-1}e^{-1}aeab^{2}db^{-2}e^{-1}d^{-1}c^{-1}d^{-1}cd^{-2}b^{-1} dbd^{-1}ec^{-1}e^{-1}d^{-1}ada^{-1}dcd^{-1}c^{-1}eb^{2}c^{2}d^{-1}b^{-1}c^{-2}ebc^{-1}a^{-1}b^{-2}ca^{-1}e^{-1}c^{-1}eca^{2}b^{-1}d^{-1}ba^{-1}b^{-1}dbd^{-1}ada^{-1}dcd^{-1}c^{-1}e^{-1}c^{-1}d^{-1}eada^{-1}de^{-1}cd^{-1}eacea^{-1}e^{-1}a^{-2}c^{-1}e^{-1}ceac^{-1}ba^{-2}b^{-1}ce^{-2}c^{2}e^{2}d^{2}c^{2}a^{-1}b^{2}c^{2}d^{-2}e^{2}ac^{2}abd^{-1}c^{-2}b^{-2}ec^{2}ba^{-1}c^{-2}a^{-1}ed^{-1}b^{2}e^{-1}da^{-2}eabab^{-1}e^{-1}a^{2}b^{-1}e^{-1}beae^{-1}a^{2}cea^{-1}e^{-1}a^{-2}c^{-1}e^{-1}ceaeba^{-1}c^{-1}bdcb^{-2}e^{-1}c^{-1}bd^{-1}cb^{-1}ae^{2}db^{-1}e^{2}d^{-2}ceb^{2}de^{-1}a^{-2}b^{2}e^{-2}d^{-2}c^{-2}d^{-2}e^{-2}ce^{-1}c^{-1}bab^{-1}a^{-1}eacec^{-1}ba^{-1}b^{-1}a^{-1}e^{-1}a^{-1}b^{-1}ce^{2}bc^{-1}ec^{-2}ed^{-1}a^{-2}de^{-2}a^{-1}c^{2}a^{-2}c^{-2}a^{2}c^{2}d^{-1}a^{-1}de^{-1}cdc^{-1}d^{-1}e = [imath](13,21,29,37)(14,22,30,38)(15,23,31,39)(40,42,47,45)(41,44,46,43)[/imath] This is a lot of information, but bear with me. The permutations [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], [imath]d[/imath] and [imath]e[/imath] correspond to 5 faces of a Rubik's Cube. I then write an expression in these 5 generators that gives me the inverse of what is usually cited as the 6th generator of the group. This expression is large, but can anyone verify it? Note: If someone cares, I am willing to post a You-Tube video that animates the Rubik's Cube and shows that the above sequence actually does rotate the remaining face using only the other faces of the cube. | 1641110 | Proof that the Rubik’s Cube group is 2-generated
Singmaster (1981) writes, on page 32 of his Notes on Rubik’s Magic Cube: Frank Barnes observes that the group of the cube is generated by two moves: \begin{align*} \alpha &= L^2 B R D^{-1} L^{-1} &=(RF,RU,RB,UB,LD,LB,LU,BD,DF,FL,RD)& \\ &&\cdot (FUR,UBR,LDB,LBU,DLF,BDR,DFR)\\ \beta &= UFRUR^{-1}U^{-1}F^{-1} &=(UF,UL)_+(UR)_+(UBR,UFL)_-(URF)_+ \end{align*} Observe that [imath]\alpha^7[/imath] is an [imath]11[/imath]-cycle of edges and [imath]\alpha^{11}[/imath] is a [imath]7[/imath]-cycle of corners, that [imath]\beta[/imath] affects the edge and corner left fixed by [imath]\alpha[/imath], and that [imath]\beta^2 = (UF)_+(UL)_+(UBR)_-(UFL)_-(UFR)_-[/imath] [...] The remaining details are left as an exercise. I hadn't seen this notation before, so I'll explain it here. Notation like [imath](LU, BD, DF)[/imath] means an edge cycle, in which: The [imath]L[/imath]-[imath]U[/imath] edge moves to the [imath]B[/imath]-[imath]D[/imath] edge's place, with the [imath]L[/imath] half ending up on the [imath]B[/imath] face, and the [imath]U[/imath] half ending up on the [imath]D[/imath] face. Similarly [imath]BD \to DF[/imath] and [imath]DF \to LU[/imath]. The notation for corners is similar. Notation like [imath](UF, UL)_+[/imath] is a twisted cycle: again, [imath]UF \to UL[/imath], but now [imath]UL \to FU[/imath]; the final edge gets flipped when cycling back to the first edge. For corners, the notation is similar, but corners rotate, they don't flip. A subscript [imath]+[/imath] means clockwise rotation, a subscript [imath]-[/imath] means counterclockwise rotation. [imath](UR)_+[/imath] means a single edge is flipped. [imath](UBR)_-[/imath] means a single corner is rotated counterclockwise. I would like to show that this is indeed true, by writing each element in [imath]\{F,B,L,R,U,D\}[/imath] as a product of elements in [imath]\{\alpha, \beta, \alpha^{-1}, \beta^{-1}\}[/imath] – preferably, having those product be as short as possible. How would I go about finding them? (I'm okay with using software like GAP – if it is at all computationally possible.) |
2210558 | Is this normed linear space a Banach space?
Let [imath]E[/imath] be a measurable set of finite measure and [imath]1 < p_1 < p_2 < \infty[/imath]. Consider the linear space [imath]L^{p_2} (E)[/imath] normed by [imath]||.||_{p_1}[/imath] . Is this normed linear space a Banach space? | 98889 | Is [imath]L^{p}[/imath] space with alternate norm Banach?
[imath]E[/imath] is a measurable set of finite measure and [imath]1 \le p_{1} < p_{2} < \infty[/imath]. Consider the linear space [imath]L^{p_{2}}(E)[/imath] normed by [imath]||.||_{p_{1}}[/imath]. Is this space Banach? |
2211152 | How do I show that the units of [imath]R[x] = [/imath] the units of [imath]R[/imath] where [imath]R[/imath] is an integral domain?
How do I show that the units of [imath]R[x] = [/imath] the units of [imath]R[/imath] where [imath]R[/imath] is an integral domain? I understand that given [imath]a,b\in R[/imath], [imath]a[/imath] is a unit if [imath]a\cdot b=1[/imath]. But I'm not really sure what this means as far as [imath]R[x][/imath] is concerned. I'm pretty confused so if someone can talk about any of this in simple terms, that would be awesome. | 144956 | Intuition surrounding units in [imath]R[x][/imath]
My lecture notes state that an 'easy' result is If [imath]R[/imath] is an integral domain then an irreducible element of [imath]R[/imath] remains irreducible in [imath]R[x][/imath], and the units in [imath]R[/imath] and in [imath]R[x][/imath] are the same. I can't seem to get my head around why this is the case, and what a unit in [imath]R[x][/imath] means intuitively because I don't see how the units can be the same if [imath]R[/imath] is the coefficients of the polynomials in [imath]R[x][/imath]. I.e. for an unit say [imath]\alpha \in R[/imath] then what is the 'corresponding' unit in [imath]R[x][/imath]? I is it [imath]\alpha x[/imath] or [imath]\alpha x^2[/imath]... or am I getting the wrong end of the stick here? |
2210822 | The system of equation 2
Let [imath]a,b,c\in \mathbb{R}[/imath]. We want to find [imath]x,y,z,w[/imath] in the following equations(according to [imath]a,b,c[/imath]): [imath]\begin{align} x^{2}+y^{2}+z^{2}+w^{2}&=1 \tag{1}\\ x^{2}+y^{2}-z^{2}-w^{2}&=a \tag{2}\\ xw+yz&=2b \tag{3}\\ yw-xz&=2c.\tag{4} \end{align}[/imath] (In my other post, that problem had a condition. But in this post, there is not.) | 2210463 | The system of equations
Let [imath]a,b,c\in \mathbb{R}[/imath] that [imath]a^{2}+b^{2}+c^{2}=1[/imath]. We want to find [imath]x,y,z,w[/imath] in the following equations: [imath]\begin{align} x^{2}+y^{2}+z^{2}+w^{2}&=1 \tag{1}\\ x^{2}+y^{2}-z^{2}-w^{2}&=a \tag{2}\\ xw+yz&=2b \tag{3}\\ yw-xz&=2c.\tag{4} \end{align}[/imath] |
2210681 | Triangular Square insight
I'm trying to find Triangular numbers that are also square. So [imath]m^2=\frac{n(n+1)}{2}[/imath] I don't want the actual answer with the solution but I'd like to solve the problem myself : I need some clues to take me on the right track. | 751316 | Which triangular numbers are also squares?
I'm reading Stopple's A Primer of Analytic Number Theory: Exercise 1.1.3: Which triangular numbers are also squares? That is, what conditions on [imath]m[/imath] and [imath]n[/imath] will guarantee that [imath]t_n=s_m[/imath]? Show that if this happens, then we have: [imath](2n+1)^2-8m^2=1,[/imath] a solution to Pell's equation, which we will study in more detail in Chapter [imath]11[/imath]. I thought about the following: [imath]\begin{eqnarray*} {t_n}&=&{s_n} \\ {\frac{n^2+n}{2}}&=&{m^2} \\ {n^2+n}&=&{m^2} \end{eqnarray*}[/imath] I've solved for [imath]n[/imath] and [imath]m[/imath] but I still have no clue of how to proceed. I've looked at the book's solution and the solution is as follows: [imath]\begin{eqnarray*} {\frac{n(n+1)}{2}}&=&{m^2} \\ {n(n+1)}&=&{2m^2} \\ {\color{red}{4n(n+1)}}&\color{red}{=}&{\color{red}{8m^2}} \\ {4n^2+4n+1-1}&=&{8m^2} \\ {(2n+1)^2-1}&=&{8m^2} \\ \end{eqnarray*}[/imath] In the red line, he multiplies the equation by [imath]4[/imath], I don't understand why to do it nor how the condition is achieved. |
2212100 | For [imath]p\neq q[/imath] odd prime integers, [imath](\mathbb{Z} / pq \mathbb{Z})^*[/imath] is not cyclic.
I am working through Aluffi's Algebra Chapter [imath]0[/imath] and I'm not sure how the author intended us to use the conclusion from exercise [imath]4.9[/imath] in [imath]4.10[/imath] Since [imath]p[/imath] and [imath]q[/imath] are distinct prime numbers, they are relatively prime. So using [imath]4.9[/imath], [imath]C_p \times C_q \cong C_{pq} \cong (\mathbb{Z} / pq \mathbb{Z})^*[/imath]. However [imath]|(\mathbb{Z} / pq \mathbb{Z})^*| \ne pq[/imath]. The order of [imath](\mathbb{Z} / pq \mathbb{Z})^*[/imath] is [imath]pq-p-q+1 = (p-1)(q-1)[/imath]. Since it is assumed that [imath]p[/imath] and [imath]q[/imath] are odd, [imath]gcd(p-1,q-1) \geq 2[/imath] and we can no longer utilize the result of [imath]4.9[/imath] | 354559 | Is my proof that [imath]U_{pq}[/imath] is not cyclic if [imath]p[/imath] and [imath]q[/imath] are distinct odd primes correct?
Prove that [imath]U_{pq}[/imath] is not cyclic if [imath]p[/imath] and [imath]q[/imath] are distinct odd primes. I am a self taught person. I just learned this and tried this on my own and came up with this. [imath]x \equiv 1 \pmod{p}[/imath] and [imath]x \equiv -1 \pmod{q}[/imath] has a solution [imath][a]_{pq}[/imath], since [imath]p[/imath] and [imath]q[/imath] are relatively prime. Because [imath]q[/imath] is an odd prime, [imath][-1]_{pq}[/imath] is not a solution, so [imath][a]_{pq}\neq [-1]_{pq}[/imath]. But [imath]a^2 \equiv 1 \pmod{p}[/imath] and [imath]a^2 \equiv 1 \pmod{q}[/imath], so [imath]a^2 \equiv 1\pmod{pq}[/imath] since [imath]p[/imath] and [imath]q[/imath] are relatively prime, and thus [imath][a]_{pq}[/imath] has order 2. Can someone please tell me if this proof this correct? Please help with proof as I learned it just now as a self taught person. |
2211828 | Similar matrices over a complex field question
Is it true that for all [imath] n \times n[/imath] matrices [imath]A[/imath], [imath]B[/imath] over [imath]\mathbb { C}[/imath], the matrices [imath]AB[/imath] and [imath]BA[/imath] are similar? Give a proof or counterexample. | 176034 | Are the matrix products [imath]AB[/imath] and [imath]BA[/imath] similar?
Given two matrices [imath]A,B.[/imath] On what conditions does [imath]AB \sim BA[/imath] hold? |
2210918 | Help on unclear question determining a probability distribution
The exercise of my homework assignment says "Let [imath]E[/imath], [imath]F[/imath], [imath]G[/imath] be independent events with probability [imath]\frac{1}{2}[/imath], [imath]\frac{1}{4}[/imath], [imath]\frac{1}{5}[/imath]. Let [imath]X[/imath] be the number of events that occur. Determine the random variable [imath]X[/imath], its mean value and variance." My concern is on the first question, and I hope to be able to explain it. Shouldn't I know how many of the event that occur belong to [imath]E[/imath], [imath]F[/imath], [imath]G[/imath]? On a first guess I expect not to be able to write analitically [imath]\Bbb{P}(X=k)[/imath]. My only attempt is by using conditioning, but I guess I should know how many of the three events occurred; defining [imath]X\colon = X_1+\cdots+X_n[/imath] \begin{multline} \Bbb{P}(X = k) = \Bbb{P}(X=k|X_1,\cdots,X_{n_1}\in E)\cdot\Bbb{P}(X_1,\cdots,X_{n_1}\in E) + \\ \Bbb{P}(X=k|X_{n_1+1},\cdots,X_{n_2}\in F)\cdot\Bbb{P}(X_1,\cdots,X_{n_2}\in F) + \\ \Bbb{P}(X=k|X_{n_2+1},\cdots,X_{n_3}\in G)\cdot\Bbb{P}(X_1,\cdots,X_{n_3}\in G) + \\ \Bbb{P}(X=k|X_{n_1 + n_2 + n_3},\cdots,X_{n}\notin E\cup F\cup G)\cdot\Bbb{P}(X_{n_1 + n_2 + n_3},\cdots,X_{n}\notin E\cup F\cup G) \end{multline} where [imath]n_1,\,n_2,\,n_3\,[/imath] are the respective numbers of the events occurring that belong to [imath]E[/imath], [imath]F[/imath], [imath]G[/imath] respectively. Let me know if my question is as unclear as the exercise is to me. | 2207046 | E/Var of a random variable on three events
E, F and G are independent events with probabilities [imath]\frac12,\frac14,\frac15[/imath] of occurring. Let [imath]X[/imath] be the number of events that occur. A. Describe [imath]X[/imath] and calculate its expectation and variance. B. The same, but this time assume the events are mutually exclusive. I really don't know how to proceed. I have tried to set the support of [imath]X[/imath] as [imath]\{0,1,2,3\}[/imath] and then write the probability of each event as [imath]P[E](X=1)=\frac12[/imath] but I don't think it's the real solution. |
2212204 | show that [imath]\lim_{n \rightarrow \infty} \ln(1/n) = -\infty[/imath]
How would you show that [imath]\lim_{n \rightarrow \infty} \ln(1/n) = -\infty[/imath] using the definition of what it means to diverge to [imath]-\infty[/imath]. This is what I'm thinking, but is it right? Let m be a real number. Then there exists [imath]n_0 \in \mathbb{N}[/imath] such that [imath]n_0<m[/imath]. Let [imath]n < n_0[/imath] then [imath]\ln(1/n) < n < n_0 < m[/imath]. Hence, [imath]\ln(1/n) \rightarrow -\infty[/imath] | 2212236 | How to show that a limit diverges to [imath]−∞[/imath]
How would you show that [imath]\lim_{n→∞}\ln(1/n)=−∞[/imath] using the definition of what it means to diverge to [imath]−∞[/imath]. If I plug in different [imath]n[/imath] I see that it goes to [imath]−∞[/imath], but I don't understand how to show this. Could someone please demonstrate? |
2212090 | Connected sum of torus and klein bottle classification
How is connected sum of torus T and klein bottle K classified? What about connected sum of K and K? Let [imath]M_1=mT^2[/imath] and [imath]M_2=nP^2[/imath]? What is the connected sum of [imath]M_1[/imath] and [imath]M_2[/imath]? I am not looking for the connected sum of projection plane. Hence I would appreciate a solution which does not make use of Mobius band?? | 1039819 | Connected sum of projective plane [imath]\cong[/imath] Klein bottle
How can I see that the connected sum [imath]\mathbb{P}^2 \# \mathbb{P}^2[/imath] of the projective plane is homeomorphic to the Klein bottle? I'm not necessarily looking for an explicit homeomorphism, just an intuitive argument of why this is the case. Can we see it using fundamental polygons? |
2212908 | Proving [imath](x_1 - a_1, \ldots, x_n -a_n)[/imath] is a maximal ideal
First of all, I don't think this is the smartest question and I feel quite ashamed about it, but here goes nothing: I wanted to prove that in the polynomial ring [imath]k[x_1, \ldots, x_n][/imath] the ideal [imath]I = (x_1 - a_1, \ldots, x_n - a_n)[/imath] is maximal and tried as follows: I have defined a map [imath]\varphi: k[x_1, \ldots, x_n] \to k: f(x_1, \ldots, x_n) \mapsto f(a_1, \ldots, a_n)[/imath] for which I was able to prove that it is a surjective ring homomorphism. Hence I know from the first isomorphism theorem that [imath]k[x_1, \ldots, x_n]/ \ker(\varphi) \cong k[/imath] showing that [imath]\ker(\varphi)[/imath] is maximal. I see that the ideal [imath]I \subset \ker(\varphi)[/imath] but I am stuck on the other direction: suppose [imath]f \in \ker(\varphi)[/imath], then [imath]f(a_1, \ldots, a_n) = 0[/imath]. How do I show that it must be of the form [imath](x_1-a_1)g_1 + \ldots (x_n - a_n)g_n[/imath] for [imath]g_i \in k[x_1, \ldots, x_n][/imath]? [imath]\textbf{Remark: }[/imath] I know the question on how to prove this is a maximal ideal has been asked many times before, e.g. here and here, but they do not seem to adress my particular question. | 1854588 | [imath]\ker \phi = (a_1, ..., a_n)[/imath] for a ring homomorphism [imath]\phi: R[x_1, ..., x_n] \to R[/imath]
Let [imath]R[/imath] be a commutative ring, [imath]a_1, ..., a_n[/imath] its elements and [imath]\phi: R[x_1, ..., x_n] \to R[/imath] defined by [imath] \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)[/imath] a ring homomorphism. Prove: [imath]\ker \phi = (x_1-a_1, ..., x_n-a_n)[/imath] It is obvious that [imath](x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi[/imath]. I'm not sure how to prove the converse. At this point I don't know any division algorithms for multivariable polynomials, only for the ones in [imath]R[x][/imath](and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for [imath]R[x][/imath]). Though I know this could be solved for [imath]i = 1[/imath] by dividing by [imath]x-a[/imath]: Let [imath]f(x) \in \ker \phi[/imath], divide by [imath]x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0[/imath], so [imath]f(x) = q(x)(x-a) \in (x-a)[/imath]. |
2213174 | How to find [imath] \displaystyle \int_0^{\pi/2} \frac{1}{1+(\tan x)^e}dx [/imath]?
How to find [imath]\displaystyle \int_0^{\pi/2} \frac{1}{1+(\tan x)^e}dx [/imath]Substitution seems not to work. | 1770529 | Finding the value of [imath]\int_0^{\pi/2} \frac{dt}{1+(\tan(x))^{\sqrt{2}}}[/imath]
The problem: Find the value of [imath]\displaystyle\int_0^{\pi/2} \frac{dx}{1+(\tan(x))^{\sqrt{2}}}[/imath] I tried a few different substitutions and the closest I got to an okay looking integral is [imath]\int_0^\infty \frac{t^2 \, dt}{t^{1/\sqrt{2}}(1+\sqrt{t})},[/imath] which still looks scary. How do I approach this integral? |
2213163 | Proving [imath]f(x)=0[/imath]
Let [imath]f[/imath] be a function defined on [imath][a,b][/imath] so that [imath]f[/imath] is continuous on [imath][a,b][/imath], differential on [imath](a,b)[/imath], [imath]f(a)=0[/imath] and there exist a real number [imath]A[/imath] so that [imath]|f'(x)| \leq A |f(x)|[/imath] for all [imath]x \in [a,b][/imath] Prove that [imath]f(x)=0[/imath] for all [imath]x \in [a,b][/imath] | 2046696 | Show that [imath]f(x) = 0[/imath] for all [imath]x \in [a,b][/imath] given [imath]|f'(x)| \leq C|f(x)| [/imath]
Suppose for real numbers [imath]a<b[/imath] one has a function with continuous derivative [imath]f:[a,b] \to \mathbb{R}[/imath] such that [imath]f(a) = 0[/imath] and there exists a real number [imath]C[/imath] with [imath]|f'(x)| \leq C|f(x)|[/imath] for all [imath]x \in [a,b][/imath]. Show that [imath]f(x) = 0[/imath] for all [imath]x \in [a,b][/imath]. Well, since [imath]f(a) = 0[/imath], we have that [imath]|f'(a)| \leq C|0|[/imath], so [imath]|f'(a)| = 0[/imath]. Since [imath]f[/imath] has a continuous derivative, we also know that [imath]f[/imath] is continuous. Since [imath]f[/imath] is continuous on a compact interval, [imath]f[/imath] obtains a maximum, say at [imath]\xi \in [a,b][/imath]. So, [imath]|f'(x)| \leq C|f(\xi)|[/imath]. Since the derivative is bounded, we obtain that [imath]f[/imath] is Lipschitz, so [imath]f[/imath] is also uniformly continuous. Suppose to the contrary that [imath]f(\xi) > 0[/imath]? The above is pretty much everything I could figure out about [imath]f[/imath], so I'm not sure what to try next. This one is also from an old qual and possibly uses methods from beyond our course. I think maybe I should try to show that [imath]|f'(x)| = 0[/imath] for all [imath]x[/imath], but I don't know how. |
2212777 | Prove that for every natural number [imath]n[/imath] and for every real numbers
Prove that for every natural number [imath]n[/imath] and for every real numbers [imath]x\neq \dfrac {k\pi}{2^t }[/imath] [imath](t=0,1,....n;)[/imath] (where [imath]k[/imath] is any integer) [imath]\dfrac {1}{\sin 2x}+\dfrac {1}{\sin 4x}+.....+\dfrac {1}{\sin 2^n x}=\cot x-\cot 2^nx[/imath] | 1987415 | Prove: [imath] \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x [/imath]
Prove [imath] \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x [/imath] where [imath]n \in \mathbb{N}[/imath] and [imath]x[/imath] not a multiple of [imath]\frac{ \pi }{2^k} [/imath] for any [imath]k \in \mathbb{N}[/imath]. My try. If [imath]n=2[/imath], we have [imath]\begin{align} \frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt] &= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt] &= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt] &= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x} \end{align}[/imath] but here I got stuck. I am on the right track? My goal is to ultimately use induction. |
2213773 | Formula obtained by repeated integration by parts.
Tonight I was messing with the integration by parts formula and found the following formula. Starting with the usual [imath]\int udv = uv - \int vdu[/imath] and letting [imath]u = f(x)[/imath], [imath]du = f'(x)dx[/imath], [imath]dv =dx[/imath] and [imath]v=x[/imath] we obtain [imath]\int f(x)dx = xf(x) - \int xf'(x).[/imath] We then apply the apply the integration by parts formula again to the integral on the RHS to obtain [imath]\int f(x)dx = xf(x) - \frac{f'(x)x^2}{2} + \frac{1}{2}\int x^2f''(x)dx[/imath] and again: [imath]\int f(x)dx = xf(x) - \frac{f'(x)x^2}{2} + \frac{f''(x)x^3}{2\cdot 3} - \frac{1}{2\cdot 3}\int x^3f'''(x)dx[/imath] So, in general, it seems (at least to me) safe to say that [imath]\int f(x) dx = xf(x) - \frac{f'(x)x^2}{2} + \frac{f''(x)x^3}{2\cdot 3} - \frac{f'''(x)x^4}{2\cdot 3\cdot 4} + \ldots =\sum_{n=1}^{\infty}\frac{x^nf^{(n)}(x)}{n!}(-1)^{n+1}[/imath] Next, applying the alternating series estimation theorem, we have that [imath]|E_N|=\left|\left(\sum_{n=1}^{\infty}\frac{x^nf^{(n)}(x)}{n!}(-1)^{n+1}\right) - \left(\sum_{n=1}^{N}\frac{x^nf^{(n)}(x)}{n!}(-1)^{n+1}\right)\right| \leq \frac{x^{(N+1)}f^{(N+1)}(x)}{(N+1)!}(-1)^{N+1}[/imath] I then tried to use this estimate to approximate [imath]\int_{a}^{b}f(x)dx[/imath]. That is, since we have [imath]E_N \geq -\frac{x^{(N+1)}f^{(N+1)}(x)}{(N+1)!}(-1)^{N+1}[/imath] we can conclude that [imath]\int_{a}^{b}f(x)dx \approx \left(\sum_{n=1}^{N}\frac{x^nf^{(n)}(x)}{n!}(-1)^{n+1}\right)-\frac{x^{(N+1)}f^{(N+1)}(x)}{(N+1)!}(-1)^{N+1}.[/imath] It then occured to me that to use the alternating series estimation theorem we must have that [imath]0\leq\frac{x^{n+1}f^{(n+1)}(x)}{(n+1)!}\leq \frac{x^{n}f^{(n)}(x)}{(n)!}[/imath] and [imath]\lim_{n \to \infty}\frac{x^nf^{(n)}(x)}{n!} = 0[/imath] So it seems that the above approximation only works if we choose an [imath]f(x)[/imath] so that the above hypothesis are satisfied. Is there a way to estimate the infinite sum that does not rely on the alternating series test so we do not have to worry about satisfying the above restrictions? More generally, I am interested if there is a more interesting route to take with this process of repeated integration by parts. The form of the terms in the infinite series remind me of taylor series. Is there a connection? | 2141794 | Infinite Integration by Parts
A while back, I was explaining to a friend a method to compute the indefinite integral of [imath]\ln(x)[/imath], by taking [imath]\int \ln(x) \, dx[/imath] and setting [imath]u = \ln(x), du = \frac{1}{x} \, dx, dv = 1 \, dx, v = x[/imath], then proceeding with integration by parts. However, I was wondering if this method could be generalized for any infinitely differentiable function [imath]f(x)[/imath], so I attempted the same method. Below is what I computed: [imath]\int f(x) \, dx[/imath], so let [imath]u = f(x), du = f'(x)\,dx, dv = 1\, dx, v = x \implies[/imath] [imath]x f(x) - \int x f'(x) \, dx[/imath], so let [imath]u = f'(x), du = f''(x) \, dx, dv = x \, dx, v = \frac{x^2}2 \implies[/imath] [imath]xf(x) - \frac{x^2} 2 f'(x) + \int \frac{x^2}2 f''(x) \, dx[/imath], so let [imath]u = f''(x), du = f'''(x) \, dx, dv = \frac{x^2} 2 \, dx, v = \frac{x^3} 6 \ldots[/imath] This eventually yields [imath]\int f(x) \, dx= \sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} f^{(n-1)}(x),[/imath] where [imath]f^{(n-1)}(x)[/imath] represents the [imath](n-1)[/imath]-th derivative of [imath]f(x)[/imath] . At first glance, while this looks like a Taylor expansion, it is actually quite different. Firstly, the general Taylor expansion does not contain the [imath](-1)^{n-1}[/imath] term that this sum does. Secondly, Taylor expansions are centered around some point, while this sum is not. Finally, and perhaps most notably, in a Taylor expansion, the derivatives are evaluated at a specific point, while in this expansion, they are left as functions. My questions, then, are the following: 1) Is this something new, or has it been documented before? (I've spent a lot of time looking to see if I could find something similar, and so far this has yielded nothing. But, if it has been done, I'd love to read more about it) 2) If this is novel (or even if it is not), what are some of the interesting consequences of this expression? I really do appreciate all of your input and advice. I realize this is a little open ended, but it's been nagging at me for quite some while, and I'd love a second look. Thanks! EDIT: As Alex Kruckman pointed out in the comments, the claim 'this eventually yields' leads to a loss of precision, specifically as it pertains to the [imath]+C[/imath] term in an indefinite integral. After spending some time thinking about this, I noted several remedies to this issue: 1) Rather than express the integral as an infinite sum, we can represent it as a finite sum added to an 'error' term of sorts. In other words, we can say: [imath]\int f(x)\, dx = \sum_{n=1}^k [\frac{x^n}{n!} (-1)^{n - 1} f^{(n-1)}(x)] + \int \frac{x^k}{k!}(-1)^{k}f^{(k)}(x) \, dx[/imath] 2) Interestingly, when considering the infinite sum, we note that every term has a polynomial component of at least degree one. This implies that if [imath]x = 0[/imath], the sum should evaluate to [imath]0[/imath] as well. Therefore, setting [imath]G(x) = \int_{a}^{x} f(t) \, dt = F(x) - F(a)[/imath], we see that as [imath]G(0) = 0[/imath], [imath]F(0) - F(a) = 0,[/imath] which implies [imath]a = 0[/imath]. In other words, to make my original claim more precise, we can use the definite integral: [imath]\int_{0}^{x} f(t) \, dt= \sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} f^{(n-1)}(x)[/imath] I believe these two edits help to eliminate the problem with the [imath]+C[/imath] term. EDIT 2: I've tried a couple common functions to see how they interact with the formula. Considering [imath]f(x) = e^x[/imath], we have: [imath]\int e^x \, dx= \sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} e^x[/imath] [imath]e^x + C = e^x * \sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} = e^x * (1 - e^{-x}) = e^x - 1[/imath] This implies [imath]C = -1[/imath], a result consistent with the first edit, where we integrate from [imath]0[/imath] to [imath]x[/imath]. Similarly, considering [imath]f(x) = e^{-x}[/imath], we have: [imath]\int e^{-x} \, dx= \sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} e^{-x} * (-1)^{(n-1)} = \sum_{n=1}^\infty \frac{x^n}{n!} * e^{-x}[/imath] [imath]-e^{-x} + C = e^{-x} * \sum_{n=1}^\infty \frac{x^n}{n!} = e^{-x} * (e^x - 1) = 1 - e^{-x}[/imath] This implies [imath]C = 1[/imath], which we would again get if we integrated from [imath]0[/imath] to [imath]x[/imath]. Now, I decided to consider [imath]f(x) = ln(x)[/imath], a decidedly harder function to analyze in this respect, which also is not defined at 0. Here, we have: [imath]\int ln(x) \, dx= \sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} \frac{d^{(n-1)}}{dx^{(n-1)}} ln(x)[/imath] [imath]x*ln(x) - x + C = x * ln(x) + \sum_{n=2}^\infty \frac{x^n}{n!} (-1)^{n - 1} \frac{d^{(n-1)}}{dx^{(n-1)}} ln(x) = [/imath] [imath]x*ln(x) - x + C = x * ln(x) + \sum_{n=2}^\infty \frac{x^n}{n!} (-1)^{n - 1} \frac{(n-2)!}{x^{n-1}} * (-1)^n[/imath] [imath]x*ln(x) - x + C = x * ln(x) - \sum_{n=2}^\infty \frac{x}{n(n-1)} = x*ln(x) - x[/imath] Interestingly, here, we see that [imath]C = 0[/imath]. While [imath]ln(x)[/imath] is not defined at [imath]0[/imath], this intuitively still makes some sense, as we note [imath]\lim_{x\to0} (x*ln(x) - x) = 0[/imath]. |
2213676 | Prove statement using formulas and then using a counting argument.
Prove the following statement first by using formulas and second by using a counting argument. [imath] \binom{n}{k}\binom{n-k}{j} = \binom{n}{j}\binom{n-j}{k} [/imath] | 755989 | explanation for a combinatorial identity involving the binomial coefficient
I am looking for an intuitive explanation for the identity: [imath]\binom{n}{h}\binom{n-h}{k} = \binom{n}{k}\binom{n-k}{h}[/imath] Thanks! |
2213286 | Proving a tricky divisibility by 8
I've been trying to prove by induction that [imath]5^n + 2 * 3^{n + 1} + 1[/imath], for [imath]n > 0[/imath]. My first attempt was trying to prove that the difference [imath](5^n + 2 * 3^{n + 1} + 1) - (5^{n + 1} + 2 * 3^{n + 2} + 1)[/imath] is always a multiple of 8. Working this difference, I've found that the difference between consecutive terms is [imath]4(5^{n + 1} + 3^n)[/imath], which proves that their difference is always a multiple of 4. I could isolate an 8 instead of 4, but the inner calculations wouldn't always be integer, which I think doesn't prove anything. Any tips on how to proceed ? | 2204834 | Why is [imath]5^{n+1}+2\cdot 3^n+1[/imath] a multiple of [imath]8[/imath] for every natural number [imath]n[/imath]?
I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is [imath]5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0[/imath], because it is a multiple of 8, you can say that[imath]5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}[/imath]. |
2213996 | Show that if [imath]\gcd(b,c)=1[/imath] then [imath]\gcd(a,bc) = \gcd(a,b)\cdot \gcd(a,c)[/imath]
everyone I'm stuck on this proof. I know that [imath]\gcd(a,bc)=\gcd(a,b)\cdot \gcd(a,c)[/imath] but I don't know how I use [imath]\gcd(b,c)=1[/imath] to get that [imath]\gcd(a,bc)=\gcd(a,b)\cdot \gcd(a,c)[/imath] | 806096 | Let [imath]a,m,n \in \mathbf{N}[/imath]. Show that if [imath]\gcd(m,n)=1[/imath], then [imath]\gcd(a,mn)=\gcd(a,m)\cdot\gcd(a,n)[/imath].
Let [imath]a,m,n\in\mathbf{N}[/imath]. Show that if [imath]\gcd(m,n)=1[/imath], then [imath]\gcd(a,mn)=\gcd(a,m)\cdot\gcd(a,n)[/imath]. Proof: Let [imath]u,v\in\mathbf{Z}[/imath] such that [imath]\gcd(m,n)=um+vn=1[/imath]. Let [imath]b,c\in\mathbf{Z}[/imath] such that [imath]\gcd(m,a)=ab+cm[/imath]. Let [imath]d,e\in\mathbf{Z}[/imath] such that [imath]\gcd(a,n)=ad+en[/imath]. So [imath]\gcd(a,m)\cdot\gcd(a,n)=a^2bd+cmen+aben+emad[/imath]. Where do I go from here? |
2212604 | Show that [imath]\lvert \mathbb Z^n / N\rvert = \lvert \det (A) \rvert[/imath].
Let [imath]N[/imath] be a rank [imath]n[/imath] submodule of [imath]\mathbb Z^n[/imath], and let [imath]A[/imath] be the matrix with rows being the generators of [imath]N[/imath]. Show that [imath]\lvert \mathbb Z^n / N\rvert = \lvert \det (A) \rvert[/imath]. So this is a homework problem, and I am a little confused. Shouldn't it be the case that [imath]\mathbb Z^n[/imath] is the unique (up to isom.) free module of rank [imath]n[/imath]? Then, wouldn't this imply that [imath]N = \mathbb Z^n[/imath]? I think I am missing something here, but even assuming that this is indeed the case, then we would be trying to prove that [imath]\lvert \det (A) \rvert = 1[/imath] for every invertible matrix [imath]A[/imath] with entries in [imath]\mathbb Z[/imath], which I don't think is true. | 834431 | Quotient group [imath]\mathbb Z^n/\ \text{im}(A)[/imath]
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix with integer coefficients and nonzero determinant. Can we say something about [imath] \mathbb{Z}^n /\ \text{im}( \phi )[/imath] (here [imath]\phi : v \mapsto Av[/imath] )? This problem has arised as I was solving some problem in homology theory. |
1796949 | A funtion and its fourier transformation cannot both be compactly supported unless f=0
Problem : Suppose that [imath]f[/imath] is continuous on [imath]\mathbb{R}[/imath]. Show that [imath]f[/imath] and [imath]\hat f[/imath] cannot both be compactly supported unless [imath]f=0[/imath]. Hint : Assume [imath]f[/imath] is supported in [0,1/2]. Expand [imath]f[/imath] in a Fourier series in the interval [-,1], and note that as a result, f is a trigonometric polynomial. I proved that f is trigonometric polynomial by using hint. But, I don't know how to prove function's fourier transform cannot compactly supported function. Can I get some hints? | 2348402 | [imath]C_c^{\infty}\cap{\cal F}C_c^{\infty}=\{0\}[/imath]
I am trying to show that [imath]C_c^{\infty}\cap{\cal F}C_c^{\infty}=\{0\}[/imath], where [imath]C_c^{\infty}[/imath] denotes the space of all [imath]C^\infty[/imath] functions on [imath]\mathbb{R}^n[/imath] whose support is compact, and [imath]{\cal F}C_c^{\infty}[/imath] denotes the space of all Schwartz functions whose Fourier transform is in [imath]C_c^{\infty}[/imath]. My thought is the following: it is obvious that [imath]C_c^{\infty}\cap{\cal F}C_c^{\infty}\supset\{0\}[/imath]. So I need to only show that [imath]C_c^{\infty}\cap{\cal F}C_c^{\infty}\subset\{0\}[/imath]. Let [imath]f\in C_c^{\infty}\cap{\cal F}C_c^{\infty}[/imath]. Then there exists a compact set [imath]K\subset\mathbb{R}^n[/imath] such that [imath]K={\rm supp}\ {\cal F}f[/imath]. And I want to show that [imath]f=0[/imath], but I could not do that. Any advice will be greatly appreciated. Thank you in advance. |
846522 | If [imath]\gcd(a, b) = 1[/imath], then [imath]\gcd(ab, c) = \gcd(a, c) \cdot\gcd(b, c)[/imath]
How can I prove that if [imath]\gcd(a, b) = 1[/imath], then [imath]\gcd(ab, c) = \gcd(a, c) \times \gcd(b, c)[/imath]? By eea there exists [imath]ax+by=1[/imath] from [imath]\gcd(a,b)=1[/imath] so a and be are co-primes there also exists [imath]dk=a[/imath] and [imath]dj= b[/imath] where [imath]d=\gcd(a,b)=1[/imath] this is all the information I have gathered from the question but I dont know how to approach and solve it. Can anyone help explain to me how to arrive at the answer? Thanks! | 2407391 | show that if gcd(b,c) = 1 , gcd(a,bc) = gcd(a,b)gcd(a,c)
Show that if [imath]\gcd(b,c) = 1[/imath] then [imath]\gcd(a,bc) = \gcd(a,b)\gcd(a,c).[/imath] Everyone, I'm stuck on this proof. I know that: [imath]\gcd(a,bc)=\gcd(a,b)\cdot\gcd(a,c)\gcd(a,bc)\\=\gcd(a,b)\cdot\gcd(a,c),[/imath] but I don't know how I use [imath]\gcd(b,c)=1[/imath] to get that [imath]\gcd(a,bc)=\gcd(a,b)\cdot\gcd(a,c)[/imath] |
2203711 | For which prime numbers [imath]p[/imath] is [imath]\cos(2\pi/p)[/imath] irrational?
I am beginner in field extension theory. But I have not learnt Galois theory. I need to solve the following problem: For which prime numbers [imath]p[/imath] is [imath]\cos\left(\frac{2 \pi}{p}\right)[/imath] irrational? I have spent some time on this question. But I did not make much progress. If someone could provide a full solution, that would be great. But hints will be appreciated as well. Thanks so much. | 1273388 | Calculate the degree of the extension [imath][\mathbb{Q}(\cos(\frac{2\pi}{p})):\mathbb{Q}][/imath]
Calculate the degree of the extension [imath][\mathbb{Q}(\cos(\frac{2\pi}{p})):\mathbb{Q}][/imath] where [imath]p[/imath] is a prime number. My thoughts are: I am lost My intuition says it has to be [imath] \frac{p-1}{2}[/imath] and that the base vectors are probably [imath]\{\cos(\frac{2n\pi}{p}) \mid 0 \le n \le p-1 \}[/imath], notice [imath]\cos(\frac{2n\pi}{p})=\cos(\frac{2(p-n)\pi}{p})[/imath]. I don't know how to prove/calculate it. |
2214878 | How to show that this is a binary operation?
Let [imath]G = \mathbb{Q}\setminus\{1\}[/imath] be the set of all rational numbers other than [imath]1[/imath] and suppose [imath]*[/imath] is defined as [imath]a*b=a+b-ab[/imath]. Show that [imath]*[/imath] is a binary operation on [imath]G[/imath]. | 980196 | Show that [imath]a \star b=a \cdot b+a+b[/imath] is binary operation for the group [imath]\Bbb{ Q} - \{-1\}[/imath]
The group [imath]\left(\Bbb{ Q} - \{-1\},\star\right)[/imath] has as its underlying set the rational numbers different from [imath]-1[/imath] and the operation [imath]\star[/imath] is defined as [imath]a \star b=a \cdot b+a+b[/imath] where multiplication/addition are the usual operations with rational numbers.Show that this is a binary operation. What I did (Please help verify): identity element has to be 0 since [imath]a\star 0 = a+0+a(0)[/imath] so for [imath]a\star b=0[/imath] we have [imath]ab+a+b=0[/imath] [imath]ab+a=-b \\ a(b+1)=-b \\ a=\frac{-b}{b+1}[/imath] so [imath]b[/imath] cant be [imath]-1[/imath] since that solution is not in [imath]\Bbb{Q}[/imath]. Similar question might have been answered here. Is this operation true for all groups? I think so. Thanks for the tips and help. |
2214927 | If two cosets are equal how can we conclude this
Let [imath]aH=bH[/imath] where [imath]H[/imath] is a subgroup of a group [imath]G[/imath] Then [imath]a=bh\quad \text{for}\quad h \in H.[/imath] I couldn't understand why. [imath]a[/imath] and [imath]b[/imath] are any numbers from the group [imath]G[/imath]. How come some number I chose from [imath]G[/imath] satisfies the condition above. Is there any proof behind this, since I couldn't manage to find one. Thanks | 1216088 | Is it true that [imath]aH = bH[/imath] iff [imath]ab^{-1} \in H[/imath]
Let [imath]H[/imath] be a subgroup of a group [imath]G[/imath]. The I know that for [imath]a,b\in G[/imath] we have [imath]aH = bH[/imath] if and only if [imath]a^{-1}b \in H[/imath]. My question is if it is also true that [imath]aH = bH[/imath] if and only if [imath]ab^{-1} \in H[/imath]? Does it matter where the [imath]-1[/imath] goes? I am guessing that this is not true since we keep putting the [imath]-1[/imath] on the left element, but I am also thinking that it might be true because they look like the same. If [imath]G[/imath] is abelian, then it doesn't matter, so a possible counter example would have to involve a non-abelian group. If [imath]H[/imath] is a normal subgroup, then it should also be true because then [imath]a^{-1}bH = Ha^{-1}b[/imath]. |
2214988 | Proving we can partition [imath]\Bbb R^+[/imath] into two sets which are closed under addition and Zorn's lemma
Let us assume [imath]A,B\subseteq \Bbb R^+ [/imath] are disjoint sets and [imath]A\cup B=\Bbb R^+ [/imath]. Furthermore, [imath] \forall_{x,y\in A} \ x+y \in B \ \text{and} \ \ \forall_ {x,y \in B} \ x+y \in A.[/imath] Is it possible to provide an example for such [imath]A[/imath] and [imath]B[/imath]? If not, I would like to know why, and also prove the existence of such [imath]A[/imath], [imath]B[/imath] using Zorn's lemma. However, I couldn't figure out how to do it. Please do not provide full proof as I am only looking for hints. | 244456 | Using Zorn's lemma show that [imath]\mathbb R^+[/imath] is the disjoint union of two sets closed under addition.
Let [imath]\Bbb R^+[/imath] be the set of positive real numbers. Use Zorn's Lemma to show that [imath]\Bbb R^+[/imath] is the union of two disjoint, non-empty subsets, each closed under addition. |
2215135 | If [imath] \lim_{x \to + \infty} f'(x) = + \infty [/imath], then [imath] \lim_{x \to + \infty} f(x) = + \infty [/imath]
[imath] f [/imath] is a differentiable function on the interval [imath] [a, + \infty) [/imath] with [imath] a \in \mathbb{R} [/imath] Prove that if [imath] \lim_{x \to + \infty} f'(x) = + \infty [/imath], then [imath] \lim_{x \to + \infty} f(x) = + \infty [/imath] [imath] f [/imath] is a differentiable function on the interval [imath] [a, + \infty) [/imath], so by the mean value theorem we have: [imath] \exists c \in (a,x): f'(c) = \frac{f(x) - f(a)}{x-a}[/imath] [imath]\implies f(x) = f(a) + (x-a)f'(c)[/imath] [imath]\implies\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty}(x-a)f'(c) + f(a) [/imath] I don't know how to proceed to prove [imath] \lim_{x \to + \infty} f'(c) = + \infty [/imath] because we have: [imath] a < c < x [/imath]. Thank you. | 878894 | If [imath]\lim_{x \to \infty}f'(x)=+\infty[/imath] then [imath]\lim_{x \to \infty}(f(x)-f(x-1))=+\infty[/imath] and [imath]\lim_{x \to \infty}f(x)=+\infty[/imath].
Let [imath]f[/imath] be differentiable and let [imath]\lim_{x \to \infty}f'(x)=+\infty[/imath] prove that: 1) [imath]\lim_{x \to \infty}(f(x)-f(x-1))=+\infty[/imath] and 2) [imath]\lim_{x \to \infty}f(x)=+\infty[/imath]. 1) I'll prove by contradiction. Let [imath]\lim_{x \to \infty}(f(x)-f(x-1))=a[/imath], where [imath]a[/imath] is from and [imath]\Bbb R[/imath]. [imath]\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=a[/imath], where c is from [imath](x-1,x)[/imath] (mean value theorem). As [imath]x[/imath] goes to infinity so does [imath]c[/imath] i.e. we have that [imath]\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=a[/imath]. Which is a contradictition since we have that [imath]\lim_{x \to \infty}f'(x)=+\infty[/imath] . 2) I will also prove this by contradiction. We have that [imath]\lim_{x \to \infty}f(x)=a[/imath] This is equivalent to [imath]\lim_{x \to \infty}f(x)-f(x-1)=0[/imath]. I use mean value theorem and get [imath]\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=0[/imath], where c is from [imath](x-1,x)[/imath]. As [imath]x[/imath] goes to infinity so does [imath]c[/imath] i.e. we have that [imath]\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=0[/imath]. Which is a contractition since we have that [imath]\lim_{x \to \infty}f'(x)=+\infty[/imath] . |
2215231 | Proof of true or false statement
Deciding whether this is true or not. I believe the statement is true just based of the definitions of what [imath]A \cup B[/imath] and [imath]A \cap B[/imath] mean. But I'm not sure how to prove this. For any sets A and B, if [imath](A \cup B)[/imath] \ [imath](A \cap B ) = ∅[/imath] , then [imath]A = B[/imath] | 1951173 | Symmetric difference = Ø
This is my first post on this forum :) I've been trying for a while to solve this problem about symmetric difference of two sets: Given two sets [imath]A[/imath] and [imath]B[/imath], we call symmetric difference of [imath]A[/imath] and [imath]B[/imath] the following: [imath]A\mathrel{\triangle}B = (A\cup B)-(A\cap B)[/imath] Show that if [imath]A\mathrel{\triangle} B = \emptyset[/imath], then [imath](A\subset B) \vee (B\subset A)[/imath]. I don’t know where to start; thank you very much for the help :) |
2215301 | [imath]\operatorname{tr}(A^m)=0\implies A[/imath] is Nilpotent
[imath]\newcommand{\tr}{\operatorname{tr}}[/imath]Let [imath]A[/imath] be a complex [imath]n\times n[/imath] matrix such that [imath]\tr(A^m)=0[/imath] for each [imath]m=1,\ldots,n[/imath]. Show that [imath]A[/imath] is nilpotent. My attempt: Since [imath]A[/imath] is a complex matrix, there exists a diagonal matrix [imath]D[/imath] and a nilpotent matrix [imath]N[/imath] such that [imath]A=D+N[/imath]. Furthermore, this decomposition is unique and [imath]D[/imath] and [imath]N[/imath] commute. So we wean to show [imath]D=0[/imath]. Since [imath]DN=ND[/imath], [imath]A^2=D^2+2DN+N^2\implies \tr(A^2)=\tr(D^2)+2\tr(DN)+\tr(N^2)[/imath]. Since [imath]N[/imath] is nilpotent, [imath]N^2[/imath] is nilpotent [imath]\implies \tr(N^2)=0[/imath]. Also, since [imath]D[/imath] and [imath]N[/imath] commute, [imath]DN[/imath] is nilpotent, so [imath]\tr(DN)=0[/imath]. Hence [imath]0=\tr(D^2)=\lambda_1^2+\cdots+\lambda_n^2[/imath], where [imath]\lambda_1,\ldots,\lambda_n[/imath] are the diagonal entries of [imath]D[/imath]. This implies each [imath]\lambda_i=0[/imath] and so [imath]D=0[/imath], which completes the proof. I'm a little unsure though because I didn't use the fact that [imath]\tr(A^m)=0[/imath] for [imath]m>2[/imath]. Is my argument correct, or does someone see a flaw? | 1798703 | Traces of powers of a matrix [imath]A[/imath] over an algebra are zero implies [imath]A[/imath] nilpotent.
I would like to have a result similar to "Traces of all positive powers of a matrix are zero implies it is nilpotent". Namely: Let [imath]R[/imath] be a commutative [imath]\mathbb{C}[/imath]-algebra, [imath]A \in \mathcal{M}_n(R)[/imath] such that [imath]\mathrm{tr}(A)=\cdots=\mathrm{tr}(A^n)=0[/imath]. Does it follow that [imath]A^n=0[/imath]? I have no idea weather or not the analogy of eigenvalue exists for matrices over algebra so I can't simply generalize considerations from cited post. Any help will be appreciated. |
2214696 | Why is [imath] \emptyset[/imath] considered a set?
My question is short and concise. Here it goes - In my book the definition of a set is given as a well defined collection of things and in mathematicse they are well defined collection of mathematical objects. Then why is [imath]\emptyset[/imath] which has nothing is even considered as a set. Is it merely a mathematica convention or is it that it has a special significance ? Though it is pretty general, I want to know the reason behind it. Thanks for your help . | 1255726 | What is an Empty set?
We define the term "Set" as, A set is a collection of objects. And an "Empty set" as, An empty set is a set which contains nothing. First problem I encountered: How the definition of "Empty set" is consistent with the definition of "sets" if "Empty set" contains nothing and a "set" is a collection of objects. Further we discovered in set theory that every set has a subset that is the Null set. Such as, If [imath]A=\emptyset[/imath] and [imath]B=\{1,2,3\}[/imath] then, [imath]A \subset B[/imath] Second problem I encountered: How and why "No element" is referred and considered as an element as we do in case of null set that is, when we say that every set has a subset that is the Null set? Third and Last one: How can a set possess "some thing" and "nothing" simultaneously that is when we say that every set (containing objects) has a subset which is Null set (contains nothing)? |
2215777 | Construct a set [imath]A[/imath] for which the lower box dimension of [imath]A[/imath] is less than the upper box dimension of [imath]A[/imath]
I'm revising for an exam and would appreciate any answers to this question. Thanks | 1843152 | Minkowski Dimension of Special Cantor Set
As can be seen at the top of the page here (exercise 1), Terry Tao gives an exercise to find the Minkowski Dimension of the Quadnary Cantor Set, and of a special Quadnary Cantor Set. The two sets are: [imath]C := \left\{\sum_{i=1}^{\infty} a_i4^{-i} : a_i \in \{0,3\} \right\}[/imath] and [imath]C' : = \left\{\sum_{i=1}^{\infty} a_i4^{-i} : a_i \in \{0,3\} \, \, \text{if} \, \, (2k)! \leq i \leq (2k+1)! \,\, \text{and arbitrary otherwise} \right\}.[/imath] I managed to find the Minkowksi Dimension of the first one by noting that [imath]C_\delta[/imath], the [imath]\delta[/imath]-fattening of [imath]C[/imath], has [imath]\text{vol}(C_\delta)[/imath] equal to the [imath]n^{th}[/imath] iteration of the Quadnary Cantor set. For the second one, I've found a closed form expression for the volume if [imath]\delta_n = 4^{-(2n)!}[/imath], since I suspect this should be a lim sup, but I'm not sure. The closed form I found is: [imath]\text{vol}(C_{\delta_n}) = \sum_{k=1}^{n-1} \sum_{j=0}^{2k(2k)!} 2^{-(2k-1)!+2j)},[/imath] but I'm not sure how to even begin manipulating the expression [imath] \frac{\log(\text{vol}(C_{\delta_n}))}{-(2n)! \log(4)}. [/imath] In particular, we are to show that the upper MD of [imath]C'[/imath] is [imath]1[/imath], while the lower MD is [imath]\frac{1}{2}[/imath], but it seems to me that the worst sequence we can use is [imath]4^{-(2n+1)!}[/imath], since this would give a volume of [imath]\text{vol}(C_{\delta_n}) = \sum_{k=1}^{n} \sum_{j=0}^{2k(2k)!} 2^{-(2k-1)!+2j)}.[/imath] To note how I got my closed form expression, at the [imath]i=(2k)![/imath] iteration, we are removing [imath]2^{(2k)! + ((2k)! - (2(k-1)+1)!)}[/imath] intervals of length [imath]4^{-(2k)!}[/imath] note that we must remove [imath]((2k)! - (2(k-1)+1)!)[/imath] more intervals, since we did not remove anything for some time and also that [imath](2k+1)! - (2k)! = 2k(2k)!.[/imath] Also it is clear that the lim inf (lower MD) is at least [imath]\frac{1}{2}[/imath], since [imath]C \subset C'[/imath], but it's unclear why it isn't strictly larger. |
2215945 | Prove that [imath]2^{1/n}[/imath] is irrational
Proof by contradiction, Assume [imath]2^{1/n}[/imath] is rational so: [imath]2^{1/n} = \frac ab [/imath] where a,b have no common factors. [imath]2 = \frac{a^n}{b^n}[/imath] [imath]2[/imath] divides LHS, therefore [imath]2[/imath] divides RHS so [imath]2[/imath] divides [imath]a^n[/imath] or [imath]2[/imath] divides [imath]b^n[/imath] which implies [imath]2[/imath] divides [imath]a[/imath] or [imath]2[/imath] divides [imath]b[/imath]. Stuck on what to do next. | 1932602 | How to prove that any (integer)[imath]^{1/n}[/imath] that isn't an integer, is irrational?
Is my proof beneath perfect and complete? I wanted to prove that for any nth root of an integer, if it's not an integer, than it's irrational: [imath]\begin{cases} m,n\in \mathbb{N}\\\sqrt[n]{m}\notin \mathbb{N} \end{cases}\implies \sqrt[n]{m}\notin \mathbb{Q}.[/imath] I start by assuming that [imath]m^{\frac 1n}[/imath] is rational and non-integer. So there exist co-prime integers [imath]a,b[/imath] so that [imath]\sqrt[n]{m}=\frac{a}{b}[/imath] [imath]\implies m=\frac{a^n}{b^n}\in\mathbb{N}.[/imath] But since [imath]a[/imath] and [imath]b[/imath] have no common factor, [imath]a^n[/imath] and [imath]b^n[/imath] also have no common factor. So: [imath]\frac{a^n}{b^n}\notin\mathbb{N},[/imath] a contradiction. |
2216020 | If [imath]g[/imath] is the generator of a group [imath]G[/imath], order [imath]n[/imath], when is [imath]g^k[/imath] a generator?
Say a cyclic group of order [imath]9[/imath] is denoted by [imath]G = \langle g \rangle[/imath]. I have a claim that [imath]\langle g^k \rangle = G [/imath] whenever [imath]k[/imath] is a unit in [imath]\mathbb{Z}_9[/imath]. I'm struggling to understand why this is true. Is this specific to order 9? In general, for a group of order [imath]n[/imath], do we require that [imath]k[/imath]'s multiplicative inverse in [imath]\mathbb{Z}_n[/imath] be [imath]n[/imath] itself? Otherwise the generating group will be of order less than [imath]n[/imath], and thus cannot be equal to the group. | 2155137 | Cyclic Group Generators of Order [imath]n[/imath]
How many generators does a cyclic group of order [imath]n[/imath] have? I know that a cyclic group can be generated by just one element while using the operation of the group. I am having trouble coming up with the generators of a group of order [imath]n[/imath]. Any help would be great! Thanks! |
2216106 | How do I solve [imath]n <8 \log_2(n)[/imath]?
I am several years away from college so I've forgotten my math and I don't know how to solve this one. I've reviewed several logarithmic properties but I'm still stumped. I am at this part: move 8 to other side of equation: [imath]\log_2 n < n/8[/imath] convert to exponential form: [imath]2^{n/8} < n[/imath] Now what? anyone? | 20652 | How to solve [imath]n < 2^{n/8}[/imath] for [imath]n[/imath]?
This is from an exercise (1.2.2) in introduction to algorithms that I'm working on privately. To find at what point a [imath]n \lg n[/imath] function will run faster than a [imath]n^2[/imath] function I need to figure out for what value [imath]n[/imath] [imath]8n^2 > 64n \lg n[/imath] (with lg here being the binary log) after some elementary simplification we get [imath]n > 8\lg n[/imath] Playing around with properties of log I can further get this to [imath]n^8 < 2^n[/imath] or [imath]n < 2^{n/8}[/imath] While I'm sure it's something very elementary I've lost somewhere over the years, after checking out a few logarithm tutorials I'm just not finding how to get any further on this. Any help with solving for [imath]n[/imath] would be appreciated. |
2215743 | schwarz lemma on the unit disc
If [imath]f(z)[/imath] is analytic function and satisfies [imath] \vert f(z)\vert <1[/imath] for [imath]\vert z\vert<1[/imath]. show if [imath]f(z)[/imath] has a zero of order m at [imath]z_o[/imath] , then [imath]\vert z_o \vert^ m > \vert f(0)\vert[/imath]. here is what I know: if we let [imath]\psi = (z-z_o)/ ( 1-\bar z_o z )[/imath] then [imath]\vert f(z) \vert < \vert \psi(z)\vert^m[/imath] but how can I prove the last inequality? i.e. how to prove that [imath]\vert f(z) \vert < \vert \psi (z) \vert^m[/imath]? I just think that Im missing something? Im interested specifically in solving it by using this way [imath]\psi[/imath] thanks | 164772 | Schwarz Lemma - like exercise
There's this exercise: let [imath]\,f\,[/imath] be analytic on [imath]D:=\{z\;\;;\;\;|z|<1\}\,\,,\,|f(z)|\leq 1\,\,,\,\,\forall\,z\in D[/imath] and [imath]\,z=0\,[/imath] a zero of order [imath]\,m\,[/imath] of [imath]\,f\,[/imath]. Prove that [imath]\forall z\in D\,\,,\,\,|f(z)|\leq |z|^m[/imath] My solution: Induction on [imath]\,m\,[/imath]: for [imath]\,m=1\,[/imath] this is exactly the lemma of Schwarz, thus we can assume truth for [imath]\,k<m\,[/imath] and prove for [imath]\,k=m>1\,[/imath] . Since [imath]\,f(z)=z^mh(z)\,\,,\,h(0)\neq 0\,[/imath] analytic in [imath]\,D\,[/imath] , put [imath]g(z):=\frac{f(z)}{z}=z^{m-1}h(z)[/imath] Applying the inductive hypothesis and using Schwarz lemma [imath]\,\,(***)\,\,[/imath] we get that [imath]|g(z)|=\left|\frac{f(z)}{z}\right|=|z|^{m-1}|h(z)|\stackrel{ind. hyp.}\leq |z|^{m-1}\Longrightarrow |f(z)|\leq |z^m|[/imath] and we're done...almost: we still have to prove [imath]\,|g(z)|\leq 1\,[/imath] for all [imath]\,z\in D[/imath] in order to be able to use the inductive hypothesis and this is precisely the part where I have some doubts: this can be proved as follows (all the time we work with [imath]\,z\in D\,[/imath]): [imath](1)\,\,[/imath] For [imath]\,f(z)=z^mh(z)\,[/imath] we apply directly Schwarz lemma and get [imath]|f(z)|=|z|^m|h(z)|\leq |z|\Longrightarrow |z|^{m-1}h(z)|\leq 1[/imath] And since now the function [imath]\,f_1(z)=z^{m-1}h(z)\,[/imath] fulfills the conditions of S.L. we get [imath](2)\,\,[/imath] Applying again the lemma, [imath]|f_1(z)|=|z|^{m-1}|h(z)|\leq |z|\Longrightarrow |z^{m-2}h(z)|\leq 1[/imath]and now the function [imath]\,f_2(z):=z^{m-2}h(z)\,[/imath] fulfills the conditions of them lemma so...etc. In the step[imath]\,m-1\,[/imath] we get [imath]|z||h(z)|\leq |z|\Longrightarrow {\color{red}{\mathbf{|h(z)|\leq 1}}}\,[/imath] and this is what allows us to use the inductive hypothesis in [imath]\,\,(***)\,\,[/imath] above. My question: Is there any way I can't see right now to deduce directly, or in a shorter way, that [imath]\,|h(z)\leq 1\,[/imath] ? |
2217080 | Find Trace [imath]T[/imath] of Matrix [imath]A[/imath]
[imath]A[/imath] is a non singular [imath]n \times n[/imath] matrix with all eigenvalues real. The trace [imath]T(A^2)= T(A^3)= T(A^4)[/imath]. Find [imath]T(A)[/imath]. I have tried using the fact that trace is a linear functional and tried to explore its kernel, but have made no progress. Can I get some help? My knowledge of linear algebra is limited to the first [imath]5[/imath] chapters of Hoffman and Kunze, basic properties of eigenvalues, annihilating polynomials and invariant subspaces. Is it sufficient to do this problem? | 1683869 | [imath]Tr(A^2)=Tr(A^3)=Tr(A^4)[/imath] then find [imath]Tr(A)[/imath]
Let [imath]A[/imath] be a non singular [imath]n\times n[/imath] matrix with all eigenvalues real and [imath]Tr(A^2)=Tr(A^3)=Tr(A^4).[/imath]Find [imath]Tr(A)[/imath]. I considered [imath]2\times 2[/imath] matrix [imath]\begin{bmatrix}a&b\\c&d\end{bmatrix}[/imath] and tried computing traces of [imath]A^2,A^3,A^4[/imath] and ended up with following [imath]Tr(A^2)=Tr(A)^2-2\det(A)[/imath] [imath]Tr(A^3)=Tr(A)^3-3Tr(A)\det(A)[/imath] [imath]Tr(A^4)=Tr(A)^4-4Tr(A)^2\det(A)+2\det(A)[/imath] I have no idea how to proceed from here... |
1951267 | Why is [imath]\{\emptyset\} \not = \emptyset[/imath]?
Why is the following true? [imath]\{ \emptyset \} \neq \emptyset [/imath] Is it just because the Empty set already refers to there being a set present and so it is just the same argument that [imath]\{ \{3\} \} \neq \{ 3\} [/imath]? And is there a proof for the following? For any set[imath]\, X[/imath], [imath]X \cup \emptyset = X \quad \text{and} \quad X \cap \emptyset = \emptyset[/imath] | 2511981 | Why is [imath]\emptyset[/imath] not [imath]\{\emptyset\}[/imath]?
\begin{align} . \end{align} \begin{align} \text{Why is } \emptyset \neq\{\emptyset\}? \end{align} \begin{align} . \end{align} |
2217303 | What does [imath]\frac{dy}{dx}[/imath] mean in calculus?
In calculus, does [imath]\frac{dy}{dx}[/imath] represents the slope of a function? If so, what does [imath]d[/imath] in the numerator represents in [imath]\frac{d}{dx}[/imath]? And why does [imath]\frac{d}{dx} y=\frac{dy}{dx}[/imath]? | 700605 | What exactly does [imath]\frac{dx}{dy}[/imath] mean?
I asked 3 professors at my university and none gave me a clear cut answer, but instead merely told me qualities of this notation. Here is what I understand so far from what they told me: 1)Treat the top variable as as variable when finding the derivative 2)Treat the bottom variable as a constant when finding derivative 3)It it said "Find x with respect to y", but what exactly does that mean? What does it mean for something to be in respect to something else? It seems like [imath]\frac{dx}{dy}[/imath] notation changes according to values in the problem. For instance, If [imath]y = x^3 + 2x[/imath] and [imath]\frac{dx}{dt} = 5[/imath], find [imath]\frac{dy}{dt}[/imath] when [imath]x=2[/imath]. Why do the values of [imath]\frac{dx}{dy}[/imath] change in this problem and how do I solve this? |
2217400 | Prove that the order of the cyclic subgroup [imath]\langle g^k\rangle [/imath] is [imath]n/{\operatorname{gcd}(n,k)}[/imath]
Let [imath]G[/imath] be a cyclic group of finite order [imath]n[/imath]. Then, the order of the cyclic subgroup [imath]\langle g^k\rangle[/imath] is [imath]\frac {n}{\operatorname{gcd}(n,k)}[/imath] Proof: [imath]G[/imath] cyclic [imath]\Rightarrow \exists g \in G: G = \langle g\rangle[/imath] [imath]\Rightarrow \operatorname{ord}(g) = \lvert G\rvert = n [/imath] We know, because every subgroup of a cyclic group is cyclic, that [imath]\lvert\langle g^k\rangle\rvert = \operatorname{ord}(g^k)[/imath]. Therefore, we only need to find the order of [imath]g^k[/imath]. Let [imath]l[/imath] be the order of [imath]g^k[/imath]. Then: [imath](g^k)^l = e \iff g^{kl} = e \iff n|(kl)[/imath] I'm now trying to show that [imath]l = n/{\operatorname{gcd}(n,k)}[/imath], but this seems difficult for me. I tried to assume that there was an element [imath]l' < n/{\operatorname{gcd}(n,k)}[/imath] that satisfied the conditions, in the hope of reaching a contradiction, but this did not work out. Can anyone help? | 1906718 | How to prove [imath]|a^k|=n/\gcd(n,k)[/imath] whenever [imath]|a|=n[/imath]?
This is an exercise from "Contemporary Abstract Algebra" I'm not sure how to solve. Exercise: Let [imath]\langle a\rangle [/imath] be a (cyclic) group of order [imath]n[/imath]. Prove that the order of [imath]a^k=\frac{n}{\gcd(n,k)}[/imath]. Direction: (1) Let [imath]d=\gcd(n,k)[/imath], thus by the Euclidian algorithm we can find [imath]X,Y\in\mathbb{Z}[/imath] s.t. [imath]d=Xn+Yk[/imath], thus, [imath]a^d=a^{Xn+Yk}=a^{Xn}a^{Yk}=(a^n)^X(a^k)^Y=(a^k)^Y[/imath]. What to do from here? (2) We know that [imath]d|n[/imath], thus [imath]\langle a^{n/d} \rangle[/imath] is of order [imath]d[/imath] and [imath]\langle a^d \rangle[/imath] is of order [imath]\frac{n}{d}=\frac{n}{\gcd(n,k)}[/imath]. Is it mean that [imath]d=k[/imath]? Where is my mistake? |
2217516 | Prove that if the [imath]\mathrm{ord}(p)a =3[/imath], then [imath]\mathrm{ord}(p)(a+1) = 6[/imath], [imath]p[/imath] prime
I couldn't answer this question. This only one. It looks simple, but I got stuck. Here is the image of the question. The number [imath]12[/imath]. Sorry for my english. Prove that if the [imath]\mathrm{ord}(p)a=3[/imath], then [imath]\mathrm{ord}(p)(a+1)=6[/imath] Example: [imath]3^3=1(\mathrm{mod}\;13)[/imath] then [imath]4^6=1(\mathrm{mod\;}13)[/imath]. Prove that if the [imath]\mathrm{ord}(p)a=3[/imath], then [imath]\mathrm{ord}(p)(a+1)=6[/imath] | 463010 | Show that if [imath]a[/imath] has order [imath]3\bmod p[/imath] then [imath]a+1[/imath] has order [imath]6\bmod p[/imath].
Show that if [imath]a[/imath] has order [imath]3\bmod p[/imath] then [imath]a+1[/imath] has order [imath]6\bmod p[/imath]. I know I am supposed to use primitive roots but I think this is where I am getting caught up. The definition of primitive root is "if [imath]a[/imath] is a least residue and the order of [imath]a\bmod p[/imath] is [imath]\phi(m)[/imath] then [imath]a[/imath] is a primitive root of [imath]m[/imath]. But I really am not sure how to use this to my advantage in solving anything. Thanks! |
2212948 | Giving the formula of a line through points in projective geometry
I'm having some trouble with giving the formula through 2 (or more) points in projective geometry. For example I have P=(1:0:0), and Q=(1:1:1). How do I find a formula of a line through these points? I thought I had to use [imath]ax_0+bx_1+cx_2=0[/imath], but then you get a line L= {[imath](x_0,x_1,x_2)|x_1=x_2=x_3=0[/imath]} ? Or for two points with for example [imath]P=(1:0:0)[/imath] and [imath]Q=(m_1,m_2,m_3)[/imath]? Would that be [imath](1,x,x)[/imath]? | 1123759 | Line between points in projective space?
I am trying to find the line through the points [imath](0 : 1 : 0)[/imath] and [imath](1 : 1 : 1)[/imath] in [imath]\mathbb P^2[/imath] and [imath](0 :1 : 0: 1)[/imath] and [imath](1: 1: 1: 0)[/imath] in [imath]\mathbb P^3.[/imath] Would the first line be the set of points [imath]\{(b, a+b, b)\}[/imath] for all [imath]a,b \in \mathbb P^1[/imath] and the second, [imath]\{(d, c+d, d,c)\}[/imath] for all [imath]c, d \in \mathbb P^1[/imath]? Thanks |
2218349 | Gaussian curvature and geodesics. Can such a surface have non vanishing curvature?
Consider a surface diffeomorphic to a cylinder (the boundary is 2 curves diffeomorphic to circles). Suppose the boundary curves are geodesic, can the surface have non-vanishing gaussian curvature (i.e. [imath]K \neq 0[/imath] everywhere)? I tried to use the Gauss-Bonnet theorem, but I'm not sure how. | 2207278 | Cylinder Gaussian Curvature
The question is as follows. Imagine we have two closed geodesics on a surface, where these two geodesics bound a cylinder together and the geodesics are diffeomorphic to circles. Can the cylinder have [imath]K>0[/imath] everywhere or [imath]K<0[/imath] everywhere? Here [imath]K[/imath] is Gaussian Curvature. What's confusing me about this question is that a cylinder has Gaussian Curvature of [imath]K=0[/imath] everywhere, so wouldn't the answer to both questions be no? But then I don't get the point of the question with the geodesics. Thank you in advance |
727495 | Prove that if [imath]A[/imath] is transitive, then [imath]\mathcal{P}(A)[/imath] is transitive too.
My book on set theory says Prove that is a set [imath]A[/imath] is transitive, then [imath]\mathcal{P}(A)[/imath] is also transitive, where [imath]\mathcal{P}(A)[/imath] is the power set of [imath]A[/imath]. I don't know understand how. Let [imath]A=\{a,b\}[/imath]. Clearly it is transitive. Now [imath]\mathcal{P}(A)=\{\{a\},\{b\},\{a,b\},\emptyset\}[/imath]. This is clearly not transitive. | 2220647 | Transitive sets and power sets
I'm currently studying set theory and came across this statement here. If a set A is a transitive set, then P(A) (its power set) is also a transitive set. A set [imath]S[/imath] is called a transitive set if the elements of its elements are elements of [imath]S[/imath]. Could anyone give me a hand on proving this? I tried some examples and started with the transitive set [imath]A = \{0, \{0\}\}[/imath]. However the power set of this would be [imath]\mathcal{P}(A) = \{ \{0\}, \{\{0\}\}, \{0, \{0\}\}, \emptyset \}[/imath] and it doesn't seem to be a transitive set which contradicts the statement above. Am I doing anything wrong with my example? Any help would be greatly appreciated |
2219280 | Let [imath]G[/imath] be a group of order [imath]21[/imath] contains contains an element [imath]a[/imath] of order [imath]7[/imath]
Let [imath]G[/imath] be a group of order [imath]21[/imath] contains an element [imath]a[/imath] of order [imath]7[/imath] . Prove that [imath]A[/imath]=([imath]a[/imath]) ,the subgroup generated by [imath]a[/imath] , is normal in [imath]G[/imath] . I'm more concerned with how I can derive the prove of this question | 615514 | Number of normal subgroups of a non abelian group of order [imath]21[/imath]... CSIR December [imath]2013[/imath]
Question is to : Find the number of Normal subgroups of a nonabelian group [imath]G[/imath] of order [imath]21[/imath] other than [imath]\{e\}[/imath] and [imath]G[/imath]. What I have done so far is : As [imath]|G|=21=3\cdot7[/imath] we have : No. of sylow [imath]3[/imath] subgroups [imath]1+3k[/imath] dividing [imath]7[/imath] leaving out possibilities [imath]1[/imath] or [imath]7[/imath] No. of sylow [imath]7[/imath] subgroups [imath]1+7k[/imath] dividing [imath]3[/imath] leaving out only possibilities [imath]1[/imath]. So, we have a unique sylow [imath]7[/imath] subgroup and so it is normal. I remember somehow that any normal group should come from a normal sylow subgroup or something like that. S0, I prefer to conclude there is only one Normal subgroup for a non abelian group of order [imath]21[/imath]. Please let me know if this is true and please help me to fill that gap : I remember somehow that any normal group should come from a normal sylow subgroup or something like that. Thank you. |
2219377 | If G be a non-trivial group with no non-trivial proper subgroup, show that G is a group of prime order.
If G be a non-trivial group with no non-trivial proper subgroup, show that G is a group of prime order. Attempt: Given [imath]G\neq \{e\}[/imath] and let [imath]a\in G, a\neq e[/imath]. Then [imath]H=<a>[/imath] is a subgroup of G. Then by the hypothesis, [imath]H= \{e\} ~or~ G[/imath]. But [imath]H\neq <a>, ~as ~a\neq e[/imath]. So G is generated by [imath]a[/imath]. I can show also that G is finite (if G is infinite, considering group [imath]<a^2>[/imath]). Let if possible, [imath]o(G)=n=composite=kl, ~k,l>1[/imath] are intergers. As [imath]G=<a>[/imath], [imath]o(a)=n[/imath]. As [imath]<a^k>[/imath] is a subgroup of G, [imath](a^k)^l=a^{kl}=a^n=e\implies o(a^k)=l<n[/imath], i.e [imath]<a^k>[/imath] is a subgroup of order [imath]<n[/imath], but G cannot have a non-trivial subgroup. So [imath]l=1 ~or~ l=n[/imath]. How to conclude the rest? The problem is in the shaded portion only. Please correct it. | 1036019 | If [imath]G[/imath] has no proper subgroup, then [imath]G[/imath] is cyclic of prime order
This is something I'm supposed to be able to prove for an upcoming test, but I can't find anything to help me prove this in my notes or the chapter, which is on cosets and Lagrange's theorem. If all I start with is the group having no proper subsets, then that means any subset is the whole group. By Lagrange's theorem, that means the index is [imath]1[/imath], but that doesn't get me anywhere. Where do I start? |
2219492 | If [imath]\textbf{Z}(A_n)>1[/imath], then [imath]n=3[/imath].
Isaacs 6.13: If [imath]\textbf{Z}(A_n)>1[/imath], then [imath]n=3[/imath]. I am unsure where to begin. I know that [imath]\textbf{Z}(A_n)[/imath] is the center of [imath]A_n[/imath]. If we assume that [imath]\textbf{Z}(A_n)>1[/imath], then the center is not the trivial center (i.e. it doesn't only contain the identity element. But how do I proceed from there? | 1878344 | The center of [imath]A_n[/imath] is trivial for [imath]n \geq 4[/imath]
I need to prove that the center of [imath]A_n[/imath] is trivial for [imath]n \geq 4[/imath]. [imath]Z(A_3) = A_3[/imath], since [imath]A_3 = \mathbb{Z}/3\mathbb{Z}[/imath] is commutative. One idea is two use "counting" technique. First of, all we count cojugacy classes of even permutations in [imath]S_n[/imath]. It's the classes correspodning to the types [imath][\lambda_1, ..., \lambda_r][/imath] where [imath]n \equiv r \mod 2[/imath] Now, the conjugacy class in [imath]S_n[/imath] correspoding to the type [imath][\lambda_1, ..., \lambda_r][/imath] splits into two equal-sized classes in [imath]A_n[/imath] if [imath]\lambda_1, ..., \lambda_r[/imath] are distinct odd numbers. If they aren't then it is preserved in [imath]A_n[/imath]. Then we can use "the counting formula"( counting the number of elements in a conjugacy class of [imath]S_n[/imath] ) of elements in each "even" class of [imath]S_n[/imath] and divide one by [imath]2[/imath] if needed. Still, I'm not sure if it can be done. Seems to be a lot of work there. Maybe there are other, he easier ways? Or, maybe, it is the way, in this case, I wouild appreciate any advices. |
2219642 | Determinant of a 4 by 4 matrix
Suppose [imath](a,b,c,d)\in \mathbb R^4 [/imath] is nonzero. [imath] M = \left( {\begin{array}{cc} a & -b & -c &-d \\ b & a & -d & c \\ c &d &a &-b \\ d&-c&b&a \end{array} } \right) [/imath]. Is det[imath](M)[/imath] non-zero? | 2217821 | Sign of determinant when using [imath]det A^\top A[/imath]
We have been given matrix: [imath]A = \begin{pmatrix} a& b& c &d \\ b &−a& d& −c\\ c& −d &−a& b \\ d &c& −b& −a\\ \end{pmatrix} [/imath] ...and have been asked to calculate [imath]\det(A)[/imath] using [imath]AA^T[/imath]. We see that: [imath]AA^T=\begin{pmatrix} a^2+ b^2+ c^2+ d^2& 0& 0&0\\ 0 &a^2+ b^2+ c^2+ d^2& 0& 0\\ 0& 0 &a^2+ b^2+ c^2+ d^2& b \\ 0 &0& 0& a^2+ b^2+ c^2+ d^2\\ \end{pmatrix} [/imath] So, [imath]\det(AA^T)= (a^2+ b^2+ c^2+ d^2)^4[/imath] Now which should I choose: [imath](a^2+ b^2+ c^2+ d^2)^2 [/imath] or [imath] -(a^2+ b^2+ c^2+ d^2)^2[/imath] ? Please explain me which one and why. |
2219103 | Fixed point of closed disk
Let [imath]D = \{(x,y)\in \mathbb{R}^2: x^2 + y^2 ≤ 1 \}[/imath]. Let [imath]A \subset \mathrm{int}D[/imath]. Let [imath]A[/imath] be connected and compact and let [imath]D \setminus A[/imath] be connected. Let [imath]f:A \longrightarrow A[/imath] be a continuous function and let [imath]g:D \longrightarrow D[/imath] be also continuous function such that [imath]g_{|A} = f[/imath].[imath]\\[/imath] Does it imply that there exists [imath]x \in \mathrm{int}D[/imath] such that [imath]g(x)=x[/imath]? | 2180896 | Fixed point of interior of closed disk
Let [imath]D = \{(x,y)\in \mathbb{R}^2: x^2 + y^2 \leq 1 \}[/imath]. Let [imath]A \subset \mathrm{int}D[/imath]. Let [imath]A[/imath] be connected and compact and let [imath]D \setminus A[/imath] be connected. Let [imath]f:A \longrightarrow A[/imath] be a continuous function and let [imath]g:D \longrightarrow D[/imath] be also continuous function such that [imath]g_{|A} = f[/imath].[imath]\\[/imath] Does it imply that there exists [imath]x \in \mathrm{int}D[/imath] such that [imath]g(x)=x[/imath]? |
2219483 | A difficult integration that contains [imath]e^{tx} x^{\alpha - 1}(1-x)^{\beta -1}[/imath] .
How can I integrate this integral: [imath]\int^{1}_{0} e^{tx} x^{\alpha - 1}(1-x)^{\beta -1}dx.[/imath] Could anyone give me a hint ? | 2217409 | Need help calculating a definite integral
I'm trying to calculate the following integral: [imath] \int_0^1 e^{-\lambda(1-x)} (1-x)^{n-1} x^{k-n} dx [/imath] It seems like kind of a combination of Gamma and Beta function. I'm suspecting that it has something to do with both of them, however I can't see the connection. Any hint would be welcome! |
2218696 | Show every prime ideal in [imath]\mathbb{Z}[/imath] is of form [imath]\langle p\rangle[/imath] where [imath]p[/imath] is prime.
An ideal [imath]P[/imath] is called a prime ideal if whenever [imath]ab\in P[/imath] we have either [imath]a \in P[/imath] or [imath]b \in P[/imath]. I am confused about how to connect the definition with with [imath]\langle p\rangle[/imath]. | 626535 | Maximal and Prime Ideal in Z
In the ring [imath]\mathbb{Z}[/imath] the following conditions on a nonzero ideal [imath]I[/imath] are equivalent: (i) [imath]I[/imath] is prime; (ii) [imath]I[/imath] is maximal; (iii) [imath]I=(p)[/imath] with [imath]p[/imath] prime. |
984998 | Combinatorial proof of [imath] k{n \choose k} = n {n-1\choose k-1} [/imath]
I have to prove this using a combinatorial proof [imath] k{n \choose k} = n {n-1\choose k-1} [/imath] What's the standard procedure on doing this? The only thing I managed was to split it into: (by fixing one element) [imath] k{n \choose k} = k\bigg[ {n-1 \choose k} {n-1\choose k-1}\bigg][/imath] But does this help? Or how do I go from here? | 528927 | Help finding a combinatorial proof of [imath]k {n \choose k } = n {n - 1 \choose k -1}[/imath]
Help finding a combinatorial proof of [imath]k {n \choose k } = n {n - 1 \choose k -1}[/imath] I have expanded it this far: [imath]\frac{k \cdot n!}{k!(n-k)!} = \frac{n \cdot (n-1)!}{(k-1)!(n-k)!} [/imath] but then I am stuck from where to go from there. |
2219968 | Prove [imath]a^n \ge n^a[/imath]
I was reading a little from a book on Real Analysis (which I'm not that far into yet), and I came across a problem: prove [imath]2^n \ge n^2, \forall n >4\in \Bbb N[/imath]. Though "challenging", I was able to prove this problem with induction without much struggle. I then considered proving this same problem for any integer '[imath]a[/imath]' in replace of [imath]2[/imath]. I.e. prove: [imath]a^n \ge n^a[/imath] where [imath]a,n \in \Bbb N[/imath]. The first thing I decided to do was look for the smallest '[imath]n[/imath]' that would actually work for a given '[imath]a[/imath]'. I did this by simply solving for [imath]n[/imath]: [imath]\begin {align} a^n \ge& n^a\\ n^{1 \over n} \ge& a^{1 \over a}\\ {1 \over n}\ln n \ge& \ln a^{1 \over a}\\ \ln n e^{-\ln n} \ge& \ln{a^ {1 \over a}}\\ -\ln n e^{-\ln n} \ge& -\ln{a^ {1 \over a}}\\ -\ln n \ge& W_0(-\ln{a^{1 \over a}})\\ n \ge& e^{-W_0(-\ln a^{1 \over a})} \end {align}[/imath] where [imath]a \in \Bbb N[/imath] and [imath]W_0(x)[/imath] is Lambert's [imath]W[/imath] Function. My first question is: "Is this a valid constraint to put on [imath]n[/imath]?", and my second and third questions are: "Is this a valid method for proving the inequality for values of [imath]n[/imath] larger than or equal to what I just derived? And if it isn't, how do I prove it using induction?" | 507744 | threshold of n to satisfy [imath]a^n [/imath]
How to find the minimum of [imath]n[/imath] when we know [imath]a[/imath], to satisfy: [imath]a^n<n^a[/imath] [imath]a^m>m^a[/imath] for each [imath]m>n[/imath] [imath]n[/imath] and [imath]m[/imath] are natural numbers. |
2220530 | Is there an integral domain [imath]R[/imath] with ideal [imath]I[/imath] such that [imath]I^2 = I[/imath] and [imath]I[/imath] is nontrivial?
Where [imath]I[/imath] being nontrivial means [imath]I \neq 0[/imath] and [imath]I \neq R[/imath]. I have found that such a ring must not be Noetherian (since otherwise, roughly, you could take some nonzero nonunit [imath]x_1 \in I[/imath] and recursively take [imath]x_i \in I[/imath] such that [imath]x_i^2 = x_{i-1}[/imath] to create a strictly increasing chain of ideals), and if we don't require our ring to be an integral domain, the ring [imath]\mathbb{Z}/6\mathbb{Z}[/imath] with ideal [imath]I = (2)[/imath] works. | 246083 | Can an ideal in a commutative integral domain be its own square?
If [imath]I[/imath] is a non-zero proper ideal of a commutative integral domain, is it possible for [imath]I[/imath] to be its own square? |
2218417 | the price of the European call option
For the Black-Scholes market model where the price of the riskless asset (bond) satisfies [imath]dB_t=rB_tdt, B_0 = 1[/imath] for some [imath]r>0[/imath] and the stock price evolves according to [imath]dS_t = µS_tdt + σS_tdW_t, S_0 = 1,[/imath] where [imath]µ, σ > 0[/imath] constants and [imath]W_t[/imath] is a (standard) Brownian motion. With fixed time horizon [imath]T > 0[/imath], and fixed a constant [imath]K>0[/imath]. How can we find the price of the European call option [imath]G=f(S_T ) [/imath] where [imath]f(x) = (x − K)_+[/imath]. I was wondering if anyone could help me? Thanks. | 411780 | Easy proof of Black-Scholes option pricing formula
I use this Book to read the option pricing in Black-Scholes model in pages 93-99, The proof of the formula given by [imath]c(s,t)= N(d_1(s,t)- Ke^{-rT}N(d_2(s,t)))[/imath] where [imath]d_{1,2}=\frac{\ln(s/K)+(r\pm \frac{1}{2}\sigma^2)t}{\sigma \sqrt{t}},[/imath] seem for me more long to read. Where do I find a short demonstration with adequate assumptions? |
2219321 | Prove that [imath]k\binom{n}{k}=n\binom{n-1}{k-1}[/imath]
I'm struggling to prove [imath]k\binom{n}{k}=n\binom{n-1}{k-1}[/imath] Here's what I'm doing: [imath]k\binom{n}{k}=\frac{k\cdot n!}{k!(n-k)!}=\frac{k\cdot n\cdot(n-1)!}{k(k-1)!(n-k)!}=n\frac{(n-1)!}{(k-1)!(n-k)!}[/imath] I see the problem that in the denominator we have [imath](n-k)![/imath] I guess it should be [imath](n-(k-1))!=(n-k+1)![/imath] to make the whole fraction equal [imath]n\binom{n-1}{k-1}[/imath] What's the problem? | 2047226 | Proof of [imath]k {n\choose k} = n {n-1 \choose k-1}[/imath] using direct proof
I've seen many posts regarding a combinatorial proof of the following question. But for a non-combinatorial proof would the following method work? Also... is this the easiest way to arrive at a proof? It seems to be rather verbose. Show the formula [imath]k {n\choose k} = n {n-1 \choose k-1}[/imath] is true for all integers [imath]n,k[/imath] with [imath]0\le k \le n[/imath]. My answer: Observe that [imath]\begin{align*} {n\choose k}&= \frac {n}{k} {n-1 \choose k-1}\\ &= \frac {n}{k}\left[ {n-2 \choose k-2} + {n-2 \choose k-1}\right]\\ &= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-2-(k-2))!} + \frac {(n-2)!}{(k-1)!(n-2-(k-1))!}\right]\\ &= \frac {n}{k}\left[ \frac{(n-2)!}{(k-2)!(n-k)!} + \frac {(n-2)!}{(k-1)!(n-k-1)!}\right]\\ &= \frac{(n-1)!}{(k-1)!(n-k)!} + \frac {(n-1)!}{(k)!(n-k-1)!}\\ &= \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} + \frac {(n-1)!}{(k)!(n-1-k)!}\\ &={n-1 \choose k-1} + {n-1 \choose k}={n \choose k}. \end{align*}[/imath] |
2219307 | Proving Commutative ring with Conditions
Let [imath]R[/imath] be a ring such that [imath]{x^2}-x[/imath] [imath]\in[/imath] [imath]C(R)[/imath] [imath]\forall[/imath] [imath]x[/imath] [imath]\in[/imath] [imath]R[/imath]. Prove that [imath]R[/imath] is commutative. I have tried like this : Let [imath]x[/imath], [imath]y[/imath] [imath]\in[/imath] [imath]R[/imath]. So , [imath]{x^2}-x[/imath] , [imath]{y^2}-y[/imath] [imath]\in[/imath] [imath]C(R)[/imath]. Now, we have to prove that [imath]xy[/imath] [imath]=[/imath] [imath]yx[/imath]. Now, [imath]y({x^2}-x)[/imath] [imath]=[/imath] [imath]({x^2}-x)y[/imath] , [imath]x({y^2}-y)[/imath] [imath]=[/imath] [imath]({y^2}-y)x[/imath]. From these , I got that [imath]x(x-y)y[/imath] [imath]=[/imath] [imath]y(x-y)x[/imath] . Now I am stucked. Please help. | 2221409 | Prove that [imath]R[/imath] is commutative
Let [imath]R[/imath] be a ring such that [imath] x^2-x \in C(R) \quad \forall x \in R,[/imath] where [imath]C(R)[/imath] is the center of [imath]R[/imath]. Prove that [imath]R[/imath] is commutative. I cannot find any route to proceed. How can I start? Please give hints instead of full solution. |
2221553 | "Every vector space over [imath]\mathbb{Q}[/imath] has a basis" [imath]\Leftrightarrow[/imath] Axiom of choice?
It is known that "every vector space has a basis" [imath]\Leftrightarrow[/imath] Axiom of choice but do we have the same thing with a vector space over [imath]\mathbb{Q}[/imath]? | 994632 | Bases of complex vector spaces and the axiom of choice
In Zermelo-Fraenkel set theory [imath]ZF[/imath] consider the following statement defined for every field [imath]K[/imath]: [imath]B_K[/imath] : Every vector space over [imath]K[/imath] has a basis. It is well-known that [imath]AC \Rightarrow \forall K (B_K)[/imath]. A. Blass proved the converse. In his paper from 1984 he said that for a fixed field [imath]K[/imath], the problem of proving [imath]B_K \Rightarrow AC[/imath] is open. What is the current status? Specifically, I would like to know: Does [imath]B_{\mathbb{Q}}[/imath] imply [imath]AC[/imath]? Does [imath]B_{\mathbb{C}}[/imath] imply [imath]AC[/imath]? |
2221292 | I've got this Integral to evaluate , a really messy one
[imath] I=\displaystyle \int_{0}^{\infty} \dfrac{\sin (\pi x^{2})}{\sinh^{2} (\pi x)} ~\mathrm{d}x [/imath] Found this on the cover of a book called "Integral Kokeboken" written in some language that i've never seen and btw the ans is [imath] I=\dfrac{2-\sqrt{2}}{4}[/imath] | 61605 | tough integral involving [imath]\sin(x^2)[/imath] and [imath]\sinh^2 (x)[/imath]
I ran across this integral I get no where with. Can someone suggest a method of attack?. [imath]\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}[/imath] I tried series, imaginary parts, and so forth, but have made no progress. Thanks very much. |
2221751 | Proofs using cardinality [imath]c[/imath] of set [imath]S[/imath]
Suppose [imath]S[/imath] is a subset of cardinality [imath]c[/imath]. Given two elements [imath]x,y \in S[/imath], prove that there exist two disjoint subsets [imath]S_1[/imath] and [imath]S_2[/imath] of [imath]S[/imath] each of cardinality [imath]c[/imath] such that [imath]x \in S_1, y \in S_2[/imath]. For two sets [imath]S[/imath] and [imath]T[/imath], prove that [imath]|S| ≤ |T|[/imath] implies [imath]|\mathcal{P(S)}|≤|\mathcal{P(T)}|[/imath]. Let [imath]\mathcal{P_0}(S)[/imath] denote the collection of all countable subsets of [imath]S[/imath]. Given that [imath]|S| = |T| = c[/imath], show that [imath]|\mathcal{P_0}(S)| = |\mathcal{P_0}(T)|[/imath]. Can someone please prove this for me? I'm really having a tough time with cardinality and don't know where to begin with. I'm assuming everyone is familiar with the [imath]\mathcal{P}[/imath] notation for the power set. | 2219649 | Question about Cardinality
The question says: [imath]S[/imath] is a set of cardinality [imath]c[/imath]. Given two distinct elements [imath]x, y ∈ S[/imath], prove that there exist two disjoint subsets [imath]S_1[/imath] and [imath]S_2[/imath] of [imath]S[/imath] and each of cardinality [imath]c[/imath] such that [imath]x ∈ S_1[/imath] and [imath]y ∈ S_2[/imath]. Here is what I am done so far: Since we know that [imath]|[0,1]|=|\mathbb{R}|=c[/imath]. Let [imath]S=[0,1][/imath], then pick [imath]x=1/2, y=7/10[/imath]. Define [imath]S_1=[0,1/2][/imath] and [imath]S_2=[3/5,1][/imath], where [imath]x∈S_1[/imath], [imath] S_1 \subset S[/imath] and [imath]y∈S_2[/imath], [imath] S_2 \subset S[/imath]. Since [imath]|[0,1/2]|=c[/imath] and [imath]|[3/5,1]|=c[/imath]; therefore, we can conclude that there exist two disjoint subsets [imath]S_1[/imath] and [imath]S_2[/imath] of [imath]S[/imath] and each of cardinality [imath]c[/imath] such that [imath]x ∈ S_1[/imath] and [imath]y ∈ S_2[/imath]. Can someone give me some suggestions on my proof? Thanks! |
2221961 | How do I find the equation of the circumcircle?
Tangents are drawn from the point P(1,8) to the circle [imath]x^2+y^2-6x-4y-11=0[/imath] touch the circle at points A and B.So what will be the equation of the circumcircle of [imath]\Delta[/imath] PAB?I am not asking you to to solve the question and tell me the answer.Just tell me how to do it please. | 912228 | Circles and tangents and circumcircles
Question: Tangents drawn from the point [imath]P(1, 8)[/imath] to the circle [imath]x^2 + y^2 -6x -4y -11=0[/imath] touch the circle at the points [imath]A[/imath] and [imath]B[/imath]. What is the equation of the circumcircle of the triangle [imath]PAB[/imath]? I tried drawing a diagram, and found the length of the tangent. However, I lost as to where to proceed from there. Please help me! |
1576231 | [imath]G[/imath] a group s.t. every non-identity element has order 2. If [imath]G[/imath] is finite, prove [imath]|G| = 2^n[/imath] and [imath]G \simeq C_2 \times C_2 \times\cdots\times C_2[/imath]
Let [imath]G[/imath] be a group s.t. every non-identity element has order 2. If [imath]G[/imath] is finite, prove [imath]|G| = 2^n[/imath] and [imath]G \simeq C_2 \times C_2 \times\cdots\times C_2[/imath] I know G is abelian since [imath]ab = (ab)^{-1} = b^{-1} a^{-1} = ba[/imath] for all non-trivial [imath]a,b \in G[/imath] so I have several questions remaining: How do I prove [imath]|G| = 2^n[/imath]? I'd like to say we use induction to prove this but I'm at a loss as to where I would start. Why is [imath]G \simeq C_2 \times C_2 \times\cdots\times C_2[/imath]? I've read several answers to question similar to this yet unfortunately most of them involve Galois Fields and vector spaces, both concepts I'm unfamiliar with. I'd greatly appreciate an intuitive proof. | 2431492 | [imath] G \cong F_{2} [/imath]?
Consider a group , where any element has [imath]ord(g)=2[/imath]. Could we say that [imath] G \cong F_{2} [/imath]? My idea was : consider all generating elements and say [imath]g_{1} = e_{1} \dots[/imath] am I right? |
311892 | Show that this entire function is polynomial.
Let [imath]f[/imath] be an entire function such that [imath] |f(z)| \to \infty[/imath] as [imath]|z| \to \infty[/imath]. Prove that [imath]f[/imath] is a polynomial. | 2256658 | [imath]f(z)[/imath] entire and [imath]\lim_{|z|\rightarrow \infty} |f(z)|=\infty[/imath] then [imath]f(z)[/imath] is a polynomial, without using the Casorati-Weierstrass theorem
I am trying to prove the following: If [imath]f(z)[/imath] is an entire function on [imath]\mathbb{C}[/imath] and [imath]\displaystyle\lim_{|z|\rightarrow \infty} |f(z)|=\infty[/imath] then [imath]f(z)[/imath] is a polynomial. Let [imath]a\in \mathbb{C}\setminus \{0\}[/imath]. Since [imath]f(z)[/imath] is analytic at [imath]a[/imath], it can be expressed as the power series [imath]f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (z-a)^n.[/imath] I have not learned anything about poles in my complex analysis class, so I am expected to solve this without any theorems involving poles. I am trying to work with the function [imath]g(z):=f(1/z)[/imath]. This function would have a singularity at zero, but I won't see how to use the function to prove the claim. Is there a way I can show that there exists some [imath]N[/imath] such that [imath]\frac{f^{(n)}(a)}{n!}=0[/imath] for all [imath]n\geq N[/imath]? Also, prior to this I showed that if [imath]f(z)[/imath] is entire and [imath]|f(z)|\leq A+B|z|^n[/imath] for all [imath]z \in \mathbb{C}[/imath] then [imath]f(z)[/imath] is a polynomial. Perhaps I can invoke this result? Any help is greatly appreciated, thank you. |
2219244 | Another solution to the problem of finding the interior points and limit points of [imath]A=\{(x,y):x^2+2y^2<1\}[/imath]
interior points and limit points of [imath]\{(x,y): x^2+2y^2 < 1\}[/imath] Find out all interior points and limit points of [imath]A=\{(x,y):x^2+2y^2<1\}[/imath] ([imath](x,y) \in \mathbb R^2[/imath]) I've got an answer from original post but as i'm not familiar with this subject, I cannot understand the solution. Is there any other solution? | 2218984 | interior points and limit points of [imath]\{(x,y): x^2+2y^2 < 1\}[/imath]
Find out all interior points and limit points of [imath] A =\{(x,y): x^2+2y^2 < 1\}[/imath] I understand this problem graphically, but i'm not quite sure how to prove the answer rigorously using mathematical word. My try : Let k be an element of A, and suppose t is an element of [imath]\{(x,y):x^2+2y^2 =1\}[/imath], which makes minimum [imath]||t-k||[/imath]. Then [imath]N(k,\delta)\subset A[/imath] when [imath]\delta = \||t-k||[/imath]. I suppose this will work, but I want more clear and detailed proof. Thank you in advance. |
2222482 | On finding a limit value
What is the value for [imath]\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{n}\ ?[/imath] According the answers given in the book, it should be [imath]e[/imath], but I can't solve it. Can someone help me please? | 666704 | [imath]\lim (1+\frac{1}{n+1})^n = e[/imath]
Assuming the definition [imath]e = \lim (1+\frac{1}{n})^n[/imath], I want to show that one also has [imath]e = \lim (1+\frac{1}{n+1})^n.[/imath] My idea was to try and squeeze [imath](1+\frac{1}{n+1})^n[/imath] between two sequences which we know converge to [imath]e[/imath], i.e., something like [imath] \left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n+1}\right)^n \leq \left(1+\frac{1}{n}\right)^n,[/imath] but the equality on the left sadly isn't true. Is there a way to adapt this idea into a working proof? Thanks! |
2222117 | Tangent Points to Ellipse
First: This is not a question for an assignment! I have an ellipse given by the equation \begin{equation} \frac{x^2}{2^2} + \frac{y^2}{1^2} = 1 \end{equation} and the intersection point [imath](0,4)[/imath] of two tangents to this ellipse. How do I determine the two tangent points to the ellipse for the aforementioned tangents? | 72982 | How to find points of tangency on an ellipse?
The problem I have to solve is: If tangent lines to ellipse [imath]9x^2+4y^2=36[/imath] intersect the y-axis at point [imath](0,6)[/imath], find the points of tangency. |
2222659 | Does Champernowne's constant converge to the digits of [imath]\pi[/imath]?
Does Champernowne's constant converge to the digits of [imath]\pi[/imath]? Clearly if you go far enough into Champernowne's constant you will find a string of the first [imath]n[/imath] digits of [imath]\pi[/imath] for any finite [imath]n[/imath]. For every finite set of [imath]n[/imath] digits of [imath]\pi[/imath] contained within the constant, [imath]n+1[/imath] digits are subsequently represented [imath]a(n+1)[/imath] digits later, for some [imath]a\leq b[/imath] where [imath]b[/imath] is the base. By induction there is no string of digits which is not, ultimately, represented. If it contains the digits of [imath]\pi[/imath], it cannot contain anything after them since they never end, so if it contains them, it must contain them at its end. Could it converge to the digits of [imath]\pi[/imath]? However it would seem we can never reach them if it did, since they would occur at a length greater than the length of the digits of [imath]\pi[/imath]. Does this mean they are not represented? | 1314350 | Can we show that the decimal expansion of [imath]\pi[/imath] doesn't occur in the decimal expansion of the Champernowne constant?
This question was inspired by an answer and some comments to this question. Recall that the Champernowne constant is obtained by concatenating all natural numbers written in base 10 and then put [imath]0.[/imath] in front, that is, [imath]C_{10}=0.123456789101112131415161718192021222324\cdots[/imath] The question is does the decimal expansion of [imath]\pi[/imath] occur as a tail of this number? By which I mean is there some [imath]n[/imath] such that [imath]\pi=C_{10}\cdot10^{n+1}-10\cdot\lfloor C_{10}\cdot10^n\rfloor[/imath] Now it is obvious that any finite initial string of [imath]\pi[/imath] occurs in [imath]C_{10}[/imath]. Not only that, it also occurs infinitely often (that's just because any finite string of numbers occurs infinitely often in [imath]C_{10}[/imath]). Hagen von Eitzen (who's answer inspired this question) also notes that proving that [imath]\pi[/imath] does occur as the tail of [imath]C_{10}[/imath] would imply that [imath]\pi[/imath] is base-[imath]10[/imath]-normal which is an open question see e.g. here, so it is very unlikely we can prove that. He also notes "that if such a position exists [one where [imath]C_{10}[/imath] starts giving the digits of [imath]\pi[/imath]] (somewhere in the middle of an [imath]n[/imath]-digit integer, say) then the first [imath]10^{n/2}n[/imath] or so digits of [imath]\pi[/imath] turn out nearly regular and this should give rise to an unusually good rational approximation." I tried to consider if there might be some relationship with relative algebraicity, but even in that direction little seems to be known since it is unknown even whether [imath]\pi[/imath] is algebraic over [imath]e[/imath]. In conclusion it seems intuitively extremely unlikely that [imath]\pi[/imath] would occur in [imath]C_{10}[/imath], but I can't think of a proof that it doesn't. A good answer would also be a reduction of this problem to some known open problem (note that we need to reduce that [imath]\pi[/imath] is not the tail of [imath]C_{10}[/imath]). PS If someone can come up with better tags please have a go at it. |
2222755 | Finding a basis over P2
Let [imath]V=P_2[/imath] be the vector space of polynomials of degree [imath]\leq 2[/imath] with real coefficients, and let [imath]W[/imath] be the subset of polynomials [imath]p(x)[/imath] in [imath]P_2[/imath] such that: [imath]\int_{-2}^0 p(x) \,\mathrm{d}x = 4 \cdot \int_0^2 p(x) \,\mathrm{d}x[/imath] B) Find a basis for [imath]W[/imath], and compute [imath]\dim(W)[/imath]. Doing an example with just [imath]a[/imath] or [imath]a+bx+cx^2[/imath] tells me that [imath]a=0[/imath], but where do I go from here? Does this mean that [imath]b[/imath] and [imath]c[/imath] are free variables and the dimension is [imath]2[/imath]? Does running through examples [imath](a, ax, a+bx, a+bx+cx^2, bx+cx^2)[/imath] prove that [imath]b[/imath] and [imath]c[/imath] are free? If they are free, what does the basis look like? | 2222616 | Proving [imath]W[/imath] is a subspace of [imath]P_{2}[/imath]
Let [imath]V = P_{2}[/imath] be the vector space of polynomials of degree less than or equal to [imath]2[/imath] with real coefficients, and let [imath]W[/imath] be the subset of polynomials [imath]p(x)[/imath] in [imath]P_{2}[/imath] such that: [imath]\int_{-2}^{0}p(x)\,dx = 4\int_{0}^{2}p(x)\,dx.[/imath] [imath]b)[/imath] Find a basis for [imath]W[/imath], and compute [imath]\dim(W)[/imath]. For [imath]b)[/imath], I know after plugging in an example p(x)=a and p(x)=a+bx+c^2 I get that a = 0, where do I go from here? |
2213855 | Homotopic functions in [imath]n[/imath]-sphere
Can anyone give me a hint in this problem? Please. Let [imath]X[/imath] be a topological space. If [imath]f,g[/imath] are continuous function of [imath]f,g \colon X\to S^n[/imath] which [imath]f(x)\neq -g(x)[/imath] for all [imath]x \in X [/imath], show [imath]f[/imath] and [imath]g[/imath] are homotopics. | 2209652 | Homotopy functions
I'm new here. Can anyone give me a hint in this problem? Please. Let [imath]X[/imath] be a topological space. If [imath]f,g[/imath] are continuous function of [imath]f,g\colon X\to S^n[/imath] which [imath]f(x)\neq -g(x)[/imath] for all [imath]x[/imath] in [imath]X[/imath], show [imath]f[/imath] and [imath]g[/imath] are homotopics. I try to show a homotopy using [imath]q(x) = \frac{f(x)+g(x)}{\|f(x)+g(x)\|}[/imath] it's well defined, but i'm stuck. |
2222966 | Ruler and compass construction
Given the three line segments below, of lengths a, b and 1, respectively: construct the following length using a compass and ruler: [imath]\frac{1}{\sqrt{b+\sqrt{a}}} \ \ \text{and} \ \ \ \sqrt[4]{a} [/imath] Make sure to draw the appropriate diagram(s) and describe your process in words. We are also to use the following axioms and state where they are used: Any two points can be connected by a line segment, Any line segment can be extended to a line, Any point and a line segment define a circle, Points are born as intersection of lines, circles and lines and circles Can someone please guide me or show me as to how to construct this? I know if we draw a triangle whose base(let's suppose this is [imath]a+1[/imath]) is the diameter of a semi-circle, then the line perpendicular to this base leading to the top of the semi-circle will divide the trianlge into two smaller triangles with the bases resulting in [imath]a[/imath] and [imath]1[/imath]. I don't know how to end up with [imath]\sqrt{a}[/imath] from there. But with it, the process can be repeated to end up with [imath]\sqrt[4]{a}[/imath]. Can someone explain or show me? I will then be able to tackle a whole lot of other questions. | 2221802 | Compass and ruler construction
Given the three line segments below, of lengths a, b and 1, respectively: construct the following length using a compass and ruler: [imath]\frac{1}{\sqrt{b+\sqrt{a}}} \ \ \text{and} \ \ \ \sqrt[4]{a} [/imath] Make sure to draw the appropriate diagram(s) and describe your process in words. We are also to use the following axioms and state where they are used: Any two points can be connected by a line segment, Any line segment can be extended to a line, Any point and a line segment define a circle, Points are born as intersection of lines, circles and lines and circles |
783836 | Apply the intermediate value theorem to the intersection of two functions
Prove the following using the intermediate value theorem: [imath]f[/imath], [imath]g[/imath] are continuous on [imath][a,b][/imath], [imath]f(a)< g(a)[/imath] but [imath]f(b) >g(b)[/imath] Prove: [imath]f(c)=g(c)[/imath] for some [imath]c \in (a,b)[/imath] I am not sure if i am correct: [imath]g[/imath] is continuous: using IVT [imath]g(c)=j[/imath], [imath]g(a)< j <g(b)[/imath] , c belongs to (a,b) [imath]f[/imath] is continuous: using IVT [imath]f(c)=k[/imath], [imath]f(a)< k < f(b)[/imath] , therefore by the given inequality: [imath]f(a)< j <f(b)[/imath] Hence, by IVT [imath]j= f(c)= g(c)[/imath] | 348363 | Proving that [imath]f(x) = g(x)[/imath] for some [imath]x \in [a,b][/imath] if [imath]f,g[/imath] continuous, [imath]f(a) < g(a)[/imath] and [imath]f(b) > g(b)[/imath]
Suppose [imath]f[/imath] and [imath]g[/imath] are continuous on [imath][a,b][/imath] and that [imath]f(a) < g(a)[/imath] but [imath]f(b) > g(b)[/imath]. Prove that [imath]f(x) = g(x)[/imath] for some [imath]x \in [a,b][/imath]. I did try and actually did get as far as coming up with [imath]f-g[/imath]. Would I set [imath]h(x)= f(x)- g(x)[/imath] and say [imath]h(a) <x< h(b)[/imath]? |
2223565 | Discrete Mathematics how to prove [imath](A\setminus B)\setminus C ⊆ A\setminus (B\setminus C)[/imath]
I can't figure out how to prove this one. Help please how to prove [imath](A\setminus B)\setminus C ⊆ A\setminus(B\setminus C)[/imath] in which prove system to choose? | 488834 | If [imath]A, B, C[/imath] are sets prove that [imath](A\setminus B)\setminus C\subseteq A\setminus(B\setminus C)[/imath]
If [imath]A, B, C[/imath] are sets prove that [imath](A\setminus B)\setminus C\subseteq A\setminus(B\setminus C)[/imath]. Find a description of when it is that we have equality, and give an example where the inclusion is strict. I'm not sure how to do this one I know that to prove that two sets [imath]X[/imath] and [imath]Y[/imath] are equal, we need to show that if [imath]x[/imath] is any element of [imath]X[/imath], then [imath]x\in Y[/imath], and that if [imath]x[/imath] is any element of [imath]Y[/imath], then [imath]x\in X[/imath]: the first of these means that [imath]X\subseteq Y[/imath] and the second that [imath]Y\subseteq X[/imath], so the two together establish that [imath]X=Y[/imath]. I'm not sure how to use this for a question and how to finish it. |
2223778 | What will be Summation [imath]\sum_{n=1}^{\infty}\frac{n^2}{n!}[/imath]
I break it down into two parts the numerator one into [imath]\dfrac{n(n+1)(2n+1)}{ 6}[/imath] but what should be summation of [imath]n![/imath] ? | 2213518 | Show that [imath]\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1[/imath]
Show that: [imath]\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1[/imath] First I will re-define the sum: [imath]\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}[/imath] Bow I will define e: [imath]e^2 = 1+ \frac{2}{1!} + \frac{x^2}{2!} + ... + \infty[/imath] [imath]e' = 1 + \frac{1}{1!} + \frac{1}{2!} + ... + \infty[/imath] [imath](e'-2) = \sum^\infty_{n=1} \frac{1}{(n+1)!}[/imath] Now I need help. |
2225202 | Show that [imath] (\sum_{i=1}^n x_iy_i)^2\leq \sum_{i=1}^n x_i^2\sum_{i=1}^n y_i^2 [/imath] with induction
Let [imath]x_i,y_i\in\mathbb{R},\quad i=1,\dots,n[/imath]. Show that [imath](\sum_{i=1}^n x_iy_i)^2\leq \sum_{i=1}^n x_i^2\sum_{i=1}^n y_i^2[/imath] where the equality holds if and only if [imath]x_i=\lambda y_i[/imath] where [imath]\lambda_i\in\mathbb{R}[/imath] If I think I already saw it in some place but I don't remember how to proof For [imath]n=1[/imath] [imath](x_1y_1)^2=x_1^2y_1^2\leq x_1^2\times y_1^2[/imath] Assuming that is true for [imath]n=k[/imath] for any [imath]k\in\mathbb{Z}[/imath]. Then [imath](\sum_{i=1}^k x_iy_i)^2\leq \sum_{i=1}^k x_i^2\sum_{i=1}^k y_i^2[/imath] Then we need to show that for [imath]n=k+1[/imath] it is true too [imath](\sum_{i=1}^{k+1} x_iy_i)^2\leq \sum_{i=1}^{k+1} x_i^2\sum_{i=1}^{k+1} y_i^2[/imath] I'm stuck now, I thought it was something related to moving the series but I got nowhere. | 918580 | Proving the Cauchy-Schwarz inequality by induction
I ran across this problem in some old notes, and I frustratingly can't figure out how to do it Let [imath]a_i[/imath] and [imath]b_i[/imath] be sequences of natural numbers, use induction to show [imath]\sum_{i=1}^n (a_ib_i)^{1/2} \le (\sum_{i=1}^n a_i)^{1/2}(\sum_{i=1}^n b_i)^{1/2} [/imath] Obviously this is trivial to show for n=1. I can't make much progress on n+1. I've tried various tactics, squaring both sides etc. Any hint or help would be appreciated. Thanks. |
2223628 | k litres of Milk delivery with minimum cost
Note: This is the exact same as a question asked 12 days ago but I do not understand the answer to that and I could not comment on the question so here we are. This is the link to that question. I think the answer is checking all the routes and the costs and then choosing the best one. Is that the best algorithm? Assume that you are the CEO of a milk delivery company and want to deliver [imath]k[/imath] litres of milk to one of your customers. You can buy milk from another company. Then you should use an arbitrary path to reach your customer. These paths are given to you as a graph [imath]G=(V,E)[/imath]. At first, you are at vertex [imath]p[/imath]. The objective is to reach vertex [imath]q[/imath]. But there are some rules. Assume that [imath]u[/imath] and [imath]v[/imath] are two arbitrarily chosen vertices. When you go from vertex [imath]u[/imath] to [imath]v[/imath], You should pay [imath]C_{uv} \gt 0[/imath] per litre of milk. Also, for example if before going from [imath]u[/imath] to [imath]v[/imath], the amount of milk was [imath]m[/imath] litres, after going from [imath]u[/imath] to [imath]v[/imath], this amount becomes [imath]m \times Y_{uv}[/imath] such that [imath]0 \lt y_{uv} \le 1[/imath]. Assume that the cost of buying [imath]1[/imath] litre of milk is [imath]\alpha[/imath]. Notice that at the end, there should be [imath]k[/imath] litres of milk left that can be delivered to the customer. Question: Find an algorithm to deliver [imath]k[/imath] litres of milk to the customer, with the minimum cost. Note1 (The meaning of cost): Assume that you want to go from vertex [imath]a[/imath] to vertex [imath]b[/imath], [imath]\alpha=2[/imath] and the amount of milk you want to deliver is [imath]1[/imath] litre. There are [imath]2[/imath] paths from [imath]a[/imath] to [imath]b[/imath]: 1. Going directly from [imath]a[/imath] to [imath]b[/imath]: You should buy [imath]1[/imath] litre of milk, which costs [imath]2[/imath] units. Then you should go from [imath]a[/imath] to [imath]b[/imath], which costs [imath]C_{ab}=10[/imath]. Also, if [imath]Y_{ab}=1[/imath], the amount of milk doesn't change. So, the total cost becomes [imath]2+10=12[/imath]. 2. Going from [imath]a[/imath] to [imath]c[/imath], and then from [imath]c[/imath] to [imath]b[/imath]: You should buy [imath]3[/imath] litres of milk, which costs [imath]6[/imath] units. Then, you should go from [imath]a[/imath] to [imath]c[/imath]. If we assume that [imath]Y_{ac}=0.5[/imath], then the amount of milk becomes [imath]1.5[/imath] litres. Also, if [imath]C_{ac}=5[/imath], then we should pay [imath]5 \times 3 = 15[/imath] units for going from [imath]a[/imath] to [imath]c[/imath]. Then we go from [imath]c[/imath] to [imath]b[/imath]. When we reach [imath]c[/imath], there are [imath]1.5[/imath] litres of milk left. Also, If [imath]C_{cb}=6[/imath], then the price of going from [imath]c[/imath] to [imath]b[/imath] will be [imath]6 \times 1.5 = 9[/imath] units. Also, [imath]Y_{cb}= \frac{2}{3}[/imath]. Thus, [imath]1[/imath] litre of milk remains at the end. So, the total cost will be [imath]6+15+9=30[/imath] units. | 2203824 | [imath]k[/imath] litres of Milk delivery with minimum cost
Assume that you are the CEO of a milk delivery company and want to deliver [imath]k[/imath] litres of milk to one of your customers. You can buy milk from another company. Then you should use an arbitrary path to reach your customer. These paths are given to you as a graph [imath]G=(V,E)[/imath]. At first, you are at vertex [imath]p[/imath]. The objective is to reach vertex [imath]q[/imath]. But there are some rules. Assume that [imath]u[/imath] and [imath]v[/imath] are to arbitrarily chosen vertices. When you go from vertex [imath]u[/imath] to [imath]v[/imath], You should pay [imath]C_{uv} \gt 0[/imath] per each litre of milk. Also, For example if before going from [imath]u[/imath] to [imath]v[/imath], the amount of milk was [imath]m[/imath] litres, after going from [imath]u[/imath] to [imath]v[/imath], this amount becomes [imath]m \times Y_{uv}[/imath] such that [imath]0 \lt y_{uv} \le 1[/imath]. Assume that the cost of buying [imath]1[/imath] litre of milk is [imath]\alpha[/imath]. Notice that at the end, there should be [imath]k[/imath] litres of milk remained that can be delivered to the customer. Question: Find an algorithm to deliver [imath]k[/imath] litres of milk to the customer, with the minimum cost. Note1 (The meaning of cost): Assume that you want to go from vertex [imath]a[/imath] to vertex [imath]b[/imath], [imath]\alpha=2[/imath] and the amount of milk you want to deliver is [imath]1[/imath] litre. There are [imath]2[/imath] paths from [imath]a[/imath] to [imath]b[/imath]: 1. Going directly from [imath]a[/imath] to [imath]b[/imath]: You should buy [imath]1[/imath] litre of milk, Which costs [imath]2[/imath] units. Then you should go from [imath]a[/imath] to [imath]b[/imath] which costs [imath]C_{ab}=10[/imath]. Also, if [imath]Y_{ab}=1[/imath], The amount of milk doesn't change. So, the total cost becomes [imath]2+10=12[/imath] 2. Going from [imath]a[/imath] to [imath]c[/imath], and then from [imath]c[/imath] to [imath]b[/imath]: You should buy [imath]3[/imath] litres of milk, Which costs [imath]6[/imath] units. Then, you should go from [imath]a[/imath] to [imath]c[/imath]. If we assume that [imath]Y_{ac}=0.5[/imath], Then the amount of milk becomes [imath]1.5[/imath] litres. Also, If [imath]C_{ac}=5[/imath], then we should pay [imath]5 \times 3 = 15[/imath] units for going from [imath]a[/imath] to [imath]c[/imath]. Then we go from [imath]c[/imath] to [imath]b[/imath]. When we reach [imath]c[/imath], there are [imath]1.5[/imath] litres of milk remained. Also, If [imath]C_{cb}=6[/imath], Then the price of going from [imath]c[/imath] to [imath]b[/imath] will be [imath]6 \times 1.5 = 9[/imath] units. Also, [imath]Y_{cb}= \frac{2}{3}[/imath]. Thus, [imath]1[/imath] litre of milk remains at the end. So, the total cost will be [imath]6+15+9=30[/imath] units. Note2 (What my problem is): First, i thought of labeling the edges with their costs. I thought that Kruskal's algorithm would solve the problem. But, There is a huge difference here! We are not sure about the costs! They depend on the amount of milk remained after each choice of vertices! That's where i'm stuck! Its like finding a minimum spanning tree, In a graph with edges having dynamic weights! |
2204425 | Determine the automorphism of ring
Determine the group [imath]\text{Aut}(\Bbb{Q}(\sqrt{2})[/imath],where [imath]\Bbb{Q}(\sqrt{2})=\left\{a+b\sqrt{2}:a,b\in \Bbb{Q}\right\}[/imath]. First, identity and conjugation which is [imath]a+ib\mapsto a-bi[/imath]. But I don't know whether there are some other elements in [imath]\text{Aut}(\Bbb{Q}(\sqrt{2}))[/imath]. If we can prove that there is no other element, then [imath]\text{Aut}(\Bbb{Q}(\sqrt{2}))\cong \Bbb{Z}_{2}[/imath]. | 2225485 | Finding all ring homomorphisms from a field to itself
I am currently working to find all ring homomorphisms [imath]\phi : \mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}(\sqrt{2})[/imath]. My work so far: Since [imath]\mathbb{Q}(\sqrt{2})[/imath] is a field extension of [imath]\mathbb{Q}[/imath], a field, it is obviously a field as well. Therefore, its only ideals are [imath]\{0\}[/imath] and itself. Since every homomorphism's kernel is an ideal, we have that either [imath]Ker(\phi)=\{0\}[/imath] or [imath]\mathbb{Q}(\sqrt{2})[/imath]. In the latter case, we have the trivial homomorphism. In the former case, we have that the homomorphism is injective (and therefore bijective). My question: I think it is true that in the former case [imath]\phi(1)=1[/imath], but am not sure why. If this is the case, can I prove inductively that [imath]\phi(x)=x, \forall x\in \mathbb{Q}(\sqrt{2})[/imath]? Thank you in advance for your help. |
2225331 | Factorial system
Factorial system, hi how can I convert decimal number to "factorial system" like this: [imath](100)_{10}[/imath] [imath]= 4⋅4! + 0⋅3! + 2⋅2! + 0⋅1! [/imath] = [imath](4020)_{!}[/imath] Multipliers of consecutive positions are defined by the force of consecutive positive natural numbers. | 868774 | Calculating a Factorial Base Representation
My friend thought of a system in which each number [imath]n[/imath] (I will first restrict my question to positive integers [imath]n[/imath]) is represented by a digit string [imath](d_l,...,d_1)[/imath] as follows [imath]\forall n \in \mathbb{N}, \exists l \in \mathbb{N}: n = (d_l, d_{l-1}, ..., d_2, d_1) = \sum_{i=1}^l (d_i*(i!))[/imath] where [imath]\forall i \in \mathbb{N}, d_i \in \mathbb{N}: d_i \leq i[/imath] and specifically [imath]\forall i>l, d_i = 0[/imath] ; note that each integer [imath]d_i[/imath] would be represented in decimal with at least 1 digit (in general, these factorial digits can be quite long in decimal representation)- but this is not an issue at the moment [because his system uses a finite alphabet via balanced quinary for each [imath]d_i[/imath]]. [In this post, any number shown is in decimal.] For example, [imath]1729=2*6! + 2*5! + 2*4! + 1=(2,2,2,0,0,1)[/imath] and [imath]1729^3 = (10*12!+9*11!+5*10!+3*9!+6*8!) + 1729 = (10, 9, 5, 3, 6, 0,2,2,2,0,0,1)[/imath]. My question is: how would this digit string [imath](d_l,...,d_1)[/imath] be computed most efficiently for general positive integer n? I have mostly just been trying to guess at the largest [imath]l \in \mathbb{N}: l*(l!) \leq n, (l+1)! > n[/imath] and then (once I have found it), I just go through digit-by-digit, working my way down in values of candidate [imath]d_i[/imath] (mostly by squeeze theorem with the goal of finding out that [imath]d_i + 1[/imath] is too large first and then next testing [imath]d_i[/imath] and finding that it works). This process can be quite arduous. What would you recommend doing, algorithmically and efficiently? Bragging rights for Mathematica code that automates the process. (Using Stirling's formula might be helpful, but it still requires a lot of playing around.) What modifications would need to be made for general real numbers (including nonintegers)? After the radix point, it works similarly, but with [imath](i!)^{-1}[/imath] (where, I think, [imath]d_i<i \forall i \in \mathbb{N}[/imath] and where the first such base is [imath]\frac{1}{2!}[/imath]). More generally and philosophically, are there any spiritual problems with this representation? It is meant to be rather compact and alien. What are some cool properties that it exhibits? |
2225285 | Prove that two paths connected from opposite sides of a square must interesect.
Lef [imath]f,g: I \rightarrow I^2[/imath] be continuous and [imath]f(0) = (a,0), f(1) = (b,1)[/imath] and [imath]g(0) = (0,c), g(1) = (1,d)[/imath]. Prove that there exist [imath]t_1, t_2 \in I[/imath] such that [imath]f(t_1)=g(t_2)[/imath] The book is hinting to use (if [imath]h : D^n \rightarrow D^n[/imath] continuously then there exists a fixed point) to find a function from [imath]I^2 \rightarrow I^2[/imath], but I can't find anything that works even if I replace [imath]D^n[/imath] with [imath]I^n[/imath] by homeomorphism. | 804150 | Prove that two paths on opposing corners of the unit square must cross.
I'm looking for a simple argument to the following: Given two (continuous) paths on the unit square, one from (0,0) to (1,1) and the other from (1,0) to (0,1), prove that the paths cross at some point [imath](x_0, y_0)[/imath]. I have constructed a topological argument for why this is true using compactness (and a proof by contradiction), but the person I'm working with seems to think there is a "simple" and "well-known" argument that says the two paths must cross. I haven't been able to find such a result. Does anyone know of one? Thanks! |
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