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1255778 | Is [imath]\{\frac{1}{n}:n\in\mathbb{N}\}[/imath] nowhere dense in [imath][0,1][/imath]?
Is [imath]\{\frac{1}{n}:n\in\mathbb{N}\}[/imath] nowhere dense in [imath][0,1][/imath] for the metric induced from the Euclidean metric on [imath]\mathbb{R}[/imath]? I think that yes, it is nowehere dense because Cl[imath](\{\frac{1}{n}:n\in\mathbb{N}\})=\{\frac{1}{n}:n\in\mathbb{N}\}\cup \{0\}[/imath] and then Int[imath](\{\frac{1}{n}:n\in\mathbb{N}\}\cup \{0\})=\emptyset[/imath]. Am I right or am I mising something? Thanks. | 144003 | [imath]A =\{ 1/(n+1): n \in \mathbb N \} [/imath] is a nowhere dense subset
Prove that the set [imath]A =\displaystyle \left \{ \frac{1}{n+1} : n \in \mathbb N \right \} [/imath] is a nowhere dense subset of [imath]\displaystyle{ \mathbb R }[/imath]. I have think two ways but I can't finish it. Here it is I tried to prove that [imath] \text{int} (\bar{A}) = \emptyset[/imath]. For this it is enough to prove that [imath] \bar A =\mathbb Q [/imath] but I can't show this. I tried to prove that every interval of [imath] \mathbb R[/imath] contains a subinterval whose intersection with [imath]A[/imath] is the empty set. Any help? Thank's in advance! |
1255891 | The formula for a summation for [imath]\sum_{i=0}^n i^n,\,[/imath] for arbitrary [imath]n[/imath]?
[imath]\text{If }\,\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}[/imath] [imath]\text{and }\,\sum_{i=0}^n i^3 = \frac{[n(n+1)]^2}{4},[/imath] [imath]\text{is there a formula for }\,\sum_{i=0}^n i^n\;?[/imath] | 660041 | Is there a general product formula for [imath]\sum\limits_{k=1}^{n} k^p[/imath]
I'm familiar with Faulhaber's formula to express this sum as a much simpler one, but it appears that for any [imath]p[/imath] there's a product formula in [imath]n[/imath] for the sum e.g.: [imath]\begin{align} & \sum\limits_{k=1}^{n} k^1=\frac{n(n+1)}{2} \\ & \sum\limits_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6} \\ & \sum\limits_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4} \end{align}[/imath] ...and so forth. Is there a general product formula in [imath]p[/imath] and [imath]n[/imath] for this sum? |
752589 | Is independence a transitive property?
If the events [imath]A[/imath] and [imath]B[/imath] are independent and the events [imath]B[/imath] and [imath]C[/imath] are independent, does this necessarily mean events [imath]A[/imath] and [imath]C[/imath] are independent? I used coin tosses to try to model this with [imath]A = H[/imath], [imath]B = T[/imath], and [imath]C = H[/imath] in separate fair tosses. I get that they are independent, but am not sure if I am correct. I also have problems determining if [imath]B[/imath] is independent to [imath]A\cap C[/imath] and also [imath]A \cup C[/imath]. | 581649 | Is statistical dependence transitive?
Take any three random variables [imath]X_1[/imath], [imath]X_2[/imath], and [imath]X_3[/imath]. Is it possible for [imath]X_1[/imath] and [imath]X_2[/imath] to be dependent, [imath]X_2[/imath] and [imath]X_3[/imath] to be dependent, but [imath]X_1[/imath] and [imath]X_3[/imath] to be independent? Is it possible for [imath]X_1[/imath] and [imath]X_2[/imath] to be independent, [imath]X_2[/imath] and [imath]X_3[/imath] to be independent, but [imath]X_1[/imath] and [imath]X_3[/imath] to be dependent? |
685735 | showing that [imath]\Phi:\Pi_{1}(X,x_{0})\rightarrow [S^{1},X][/imath] is onto if [imath]X[/imath] is path connected.
We can regard [imath]\Pi_{1}(X,x_{0})[/imath] as the set of basepoint-preserving homotopy classes of maps [imath](S^{1},s_{0})\rightarrow(X,x_{0})[/imath] . Let [imath][S^{1},X][/imath] be the set of homotopy classes of maps [imath]S^{1}\rightarrow X[/imath] , with no conditions on basepoints. Thus there is a natural map [imath]\Phi:\Pi_{1}(X,x_{0})\rightarrow [S^{1},X][/imath] obtained by ignoring basepoints. Show that [imath]\Phi[/imath] is onto if [imath]X[/imath] is path-connected, and that [imath]\Phi([f])= \Phi([g])[/imath] iff [imath][f][/imath] and [imath][g][/imath] are conjugate in [imath]\Pi_{1}(X,x_{0})[/imath] . Hence [imath]\Phi[/imath] induces a one to one correspondence between [imath] [S^{1},X][/imath] and the set of conjugacy classes in [imath]\Pi_{1}(X)[/imath] when [imath]X[/imath] is path-connected. this is problem 6 from hatcher page 38. for showing that it is onto,my Idea was to take [imath]f[/imath] in [imath][S^{1},X][/imath],and consider a point say [imath]x\in f(S^{1})[/imath],because of [imath]X[/imath] is path connected we can have a path from [imath]x[/imath] to [imath]x_{0}[/imath] call it [imath]\alpha[/imath],I know that it is not exact but I feel [imath]f(S^{1})\cup \alpha[/imath] is the element in [imath]\Pi_{1}(X,x_{0})[/imath] which we need. for second part I have no Idea. 1.please help me to make them complete and exact. 2.I don't underestand this part well,when it say :"We can regard [imath]\Pi_{1}(X,x_{0})[/imath] as the set of basepoint-preserving homotopy classes of maps [imath](S^{1},s_{0})\rightarrow(X,x_{0})[/imath]" . please give me good imagination about it. 3.can you guide to how I must have a good vision to solve algebraic topology problems?I know by practicing it will be achieved but sometimes I feel that there is something that is more like a gap in my mind. I will be very thankful for your consideration. | 1200799 | Question on showing a bijection between [imath]\pi_1(X,x_0)[/imath] and [imath][S^1, X][/imath] when X is path connected.
I am trying to do this question taken from Hatchers algebraic topology and I am struggling to understand the notation and the concepts. As far as I know [imath]\pi_1(X,x_0)[/imath] is the set of end point preserving homotopy classes of loops in X based at [imath]x_0[/imath] and a loop is just a path [imath]f:I \rightarrow X[/imath] with [imath]f(0)=f(1)[/imath]. The question says we can regard [imath]\pi_1(X, x_0)[/imath] as the set of basepoint preserving homotopy classes of maps [imath](S^1, s_0) \rightarrow (X, x_0)[/imath]. This confuses me, does it mean the set of basepoint preserving homotopy class on maps [imath]g:S^1 \rightarrow X[/imath] which map loops in [imath]S^1[/imath] based at [imath]s_0[/imath] to loops in X based at [imath]x_0[/imath]? Then how can [imath]\pi_1(X,x_0)[/imath] be regarded as this set? If someone could explain this it would be really useful |
934595 | dot product vs inner product?
For [imath]V_n[/imath] where [imath]x=(x_1, x_2, \ldots, x_n)[/imath] and [imath]y=(y_1,y_2,\ldots,y_n)[/imath], the dot product is defined by [imath]x_1y_1+x_2y_2+ \cdots+x_ny_n[/imath]. In Apostol's calculs vol 2 It says that if [imath]x=(x_1,x_2)[/imath] and [imath]y=(y_1,y_2)[/imath] are any two vectors in [imath]V_2[/imath], define [imath](x \cdot y)[/imath] by the formula [imath](x \cdot y)= 2x_1y_1+x_1y_2+x_2y_1+x_2y_2[/imath] Why is that? Is there difference between dot and inner product? I thought that following the dot product, [imath](x \cdot y)[/imath] should be [imath]x_1y_1+x_2y_2[/imath]. | 476738 | difference between dot product and inner product
I was wondering if a dot product is technically a term used when discussing the product of [imath]2[/imath] vectors is equal to [imath]0[/imath]. And would anyone agree that an inner product is a term used when discussing the integral of the product of [imath]2[/imath] functions is equal to [imath]0[/imath]? Or is there no difference at all between a dot product and an inner product? |
1252328 | Is a topological space [imath]X[/imath] the colimit of an open cover [imath]\cup U_i[/imath] in this way?
Let [imath]X[/imath] be a topological space space and [imath]X=\cup_{i\in I} U_i[/imath] a covering of [imath]X[/imath] by open subsets [imath]U_i\subseteq X[/imath]. Is it true that [imath] \operatorname{colim}\left(\coprod_{(i,j)\in I\times I} U_i\cap U_j \rightrightarrows\coprod_{i\in I} U_i\right) \cong X [/imath] where the two maps are the inclusions of either [imath]U_i\cap U_j[/imath] into [imath]U_i[/imath] or [imath]U_i\cap U_j[/imath] into [imath]U_j[/imath]? If this is not true in general, is it at least true for [imath]X[/imath] a CW-complex and [imath]\cup_{i=0}^n U_i[/imath] a finite good cover? | 1081949 | Categorical Pasting Lemma
If I'm not mistaken, the pasting lemma for a two-element open cover [imath]X=A\cup B[/imath] of a topological space is equivalent to saying that the following square is a pushout in [imath]\mathsf{Top}[/imath]: [imath]\begin{matrix} A\cup B & \longleftarrow & B \\ \uparrow && \uparrow \\ A & \longleftarrow & A\cap B\end{matrix}[/imath] where the arrows are the canonical embeddings. Is my statement true, and if so, how can I categorically formulate the pasting lemma for general open covers [imath] \left\{ U_i\right\} _{i\in I}[/imath]? |
1256417 | [imath]X_n\rightarrow X[/imath] in probability, but [imath]\mathbb{E}(X_n)[/imath] does not converge to [imath]\mathbb{E}(X)[/imath]
What is an example of a sequence [imath]X_1,X_2,...[/imath] such that [imath]X_n\rightarrow X[/imath] in probability, but [imath]\mathbb{E}(X_n)[/imath] does not converge to [imath]\mathbb{E}(X)[/imath]? | 477942 | Does convergence in probability implies convergence of the mean?
Let [imath]\{X_n\}_{n=1}^\infty[/imath] be a sequence of random variables converging to a random variable [imath]X[/imath]. Does this fact implies that [imath]\lim\limits_{n\to \infty} EX_n = EX[/imath]? If not, is there any other sufficient condition? |
1256821 | Show [imath]f(x) = x\sin{x}[/imath] is not uniformly continuous on [imath]\mathbb{R}[/imath]
I know it's not Lipschitz continuous because its derivative is unbounded but I'm not sure if not Lipschitz continuous implies not uniformly continuous. Thanks | 372766 | Uniform Continuity of [imath]x \sin x[/imath]
How does one go about proving that the function [imath]x \sin x[/imath] is not uniformly continuous on the set of real numbers ? Any method that uses the sequential criterion for discontinuity would be preferred. Thanks in advance ! |
951977 | Geometry of Riemann Stieltjes integration
What is the geometrical interpretation of Riemann Stieltjes Integration ? We know that for Riemaan integration [imath]\int_{a}^b f(x)dx[/imath] represents the area bounded by the curve [imath]y=f(x)[/imath]& the straight lines [imath]x=a[/imath] & [imath]x=b[/imath]. But when we integrate [imath]f(x)[/imath] with respect to another function [imath]g(x)[/imath] then which area represents that integration geometrically ? | 295383 | Integration of a function with respect to another function.
What is the intuition/idea behind integration of a function with respect to another function? Say [imath]\int f(x)d(g(x)) \;\;\;\;\;?[/imath] or may be a more particular example [imath]\int x^2d(x^3)[/imath] My concern is not at the level of problem solving. To solve we could simply substitute [imath]u=x^3[/imath] and then [imath]x^2=u^{2/3}[/imath]. My concern is rather about what meaning physical/geometrical does this impart? ADDED if you ask what kind of meaning I seek, integration of a function w.r.t a variable gives the area under the curve and above x-axis. |
1255342 | Suppose [imath]S_1 =\{ u_1 , u_2 \}[/imath] and [imath]S_2 = \{ v_1 , v_2 \}[/imath] are each independent sets of vectors in an n-dimensional vector space V..
Let us assume that every vector in S_2 is a linear combination of vectors in S_1. Question: Does that mean that S_1 and S_2 are bases for the same subspace of V? I know that the answer to this question is yes, the subspace spanned by both S_1 and S_2 are the same, but how do i show that? | 1255188 | Suppose [imath]S_1 =\{ u_1 , u_2 \}[/imath] and [imath]S_2 = \{ v_1 , v_2 \}[/imath] are each independent sets of vectors in an n-dimensional vector space V.
Let us assume that every vector in [imath]S_2[/imath] is a linear combination of vectors in [imath]S_1[/imath]. Question: Does that mean that [imath]S_1[/imath] and [imath]S_2[/imath] are bases for the same subspace of [imath]V[/imath]? I know that the answer to this question is yes, subspaces spanned by both [imath]S_1[/imath] and [imath]S_2[/imath] are the same. Let's call them [imath]W_1[/imath] and [imath]W_2[/imath] respectively. How do we prove [imath]W_1=W_2[/imath]? Equal subspaces when regarded as sets, must have the same elemnts. How can we show that EVERY vector in one subspace is also in the other subspace? |
1257083 | Vector Spaces and Linear Transformations ([imath]T^2 = 0 \iff R(T) \subseteq N(T)[/imath]).
Let [imath]V[/imath] be a vector space over a field [imath]F[/imath]. Let [imath]T: V\to W[/imath] be a linear transformation. a. Prove that [imath]T^2=0[/imath] if and only if [imath]R(T)[/imath] is contained in [imath]N(T)[/imath]. (Here we denote [imath]T^2[/imath] as the linear transformation such that [imath]T^2(v)=T(T(v))[/imath] for all [imath]v[/imath] in [imath]V[/imath]) b. Assume that [imath]V[/imath] is finite-dimensional and [imath]T^2=0[/imath]. Prove that [imath]\operatorname{nullity}(T)[/imath] is greater than or equal to [imath]\dim(V)/2[/imath]. | 374956 | [imath]T\circ T=0:V\rightarrow V \implies R(T) \subset N(T)[/imath]
Question Let [imath]T:V \rightarrow V[/imath] be a linear map. How do I prove that [imath]T \circ T = T_0[/imath] ( the zero linear map) iff [imath]R(T) \subset N(T)[/imath]? Attempt \begin{eqnarray} T\circ T=T(T(v))&=&T(T(v-0))\\ &=&T(T(v)+T(-0))\\ &=&T(T(v))+T(T(-0))\\ &=&T(T(v))+T(-T(0))\\ &=&T(T(v))-T(T(0))&=&0, \end{eqnarray} so this last bit gives \begin{eqnarray} T(T(v))=T(T(0)), \end{eqnarray} which means [imath]T(v)=T(0)=0~\forall~v\in V[/imath], so [imath]R(T)=0[/imath], this zero of which is in the null space for all linear maps [imath]T[/imath]. |
1246343 | Prove that if [imath]3^\frac{F_n-1}{2} \equiv -1 \pmod {F_n}[/imath], then [imath]F_n[/imath] is prime
[imath]F_n = 2^{2^n} + 1[/imath] is a Fermat Number. Here is my attempt. We square each side of the congruences we get [imath]3^{F_n-1} \equiv 1 \pmod {F_n}[/imath] Now I already know that whenever [imath]m \neq n[/imath] then [imath] \gcd(F_n,F_m) = 1[/imath] and so for all [imath]n > 0[/imath] we do have that [imath]\gcd(11,F_n)=1[/imath] and so [imath]ord(3,F_n) \mid F_n-1[/imath] and we do know that [imath]F_n-1 = 2^{2^n}[/imath] and so [imath]ord(3,F_n) \mid 2^{2^n}[/imath] But however, since we have that [imath]3^\frac{F_n-1}{2} \equiv -1 \pmod{F_n}[/imath] so [imath]ord(3,F_n) \nmid \frac{F_n-1}{2} = \frac{2^{2^n}}{2}[/imath] Now the [imath]ord(3,F_n)[/imath] must be a factor of [imath]2^{2^n}[/imath], so it must be one of the following [imath]2^{2^n},\frac{2^{2^n}}{2},\frac{2^{2^n}}{3},........,\frac{2^{2^n}}{2^n} = 1[/imath] However, since [imath]ord(3,F_n) \neq \frac{2^{2^n}}{2}[/imath] so it must be greater than [imath]\frac{2^{2^n}}{2}[/imath], This only leaves us one option so [imath]ord(3,F_n)= 2^{2^n}[/imath] Now where do I go from here to conclude that [imath]F_n[/imath] is prime ? I want to show that [imath]\phi(F_n) = F_n -1[/imath] and hence I can conclude that [imath]F_n[/imath] is prime. I am missing the connection from the order to this step. We also know that [imath]3[/imath] is a primitive root [imath]\pmod {F_n}[/imath] since [imath]ord(3,F_n) = F_n -1 = \phi(F_n-1)[/imath] ?? | 32940 | [imath]n = 2^k + 1[/imath] is a prime iff [imath]3^{\frac{n-1}{2}} \equiv -1 \pmod n[/imath]
Let [imath]k \geq 2[/imath] be a positive integer and let [imath]n=2^k+1[/imath]. How can I prove that [imath]n[/imath] is a prime number if and only if [imath]3^{\frac{n-1}{2}} \equiv -1 \pmod n.[/imath] Fixed. |
1002459 | Do we have negative prime numbers?
Do we have negative prime numbers? [imath]..., -7, -5, -3, -2, ...[/imath] | 1626120 | Unique prime factorization
We all know that [imath]15=3 \times 5[/imath] And [imath]15 =(-3) \times(-5)[/imath] Since [imath]3 \neq -3[/imath] and [imath]5 \neq -5[/imath] , we have two different prime factorizations ! Is this wrong ? If this is wrong , then there are no negative primes ! |
843763 | For [imath]x\in\mathbb R\setminus\mathbb Q[/imath], the set [imath]\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}[/imath] is dense on [imath][0,1)[/imath]
Let [imath]x\in \mathbb{R}[/imath] an irrational number. Define [imath]X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}[/imath]. Prove that [imath]X[/imath] is dense on [imath][0,1)[/imath]. Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof. I spend part of the day studying this question Positive integer multiples of an irrational mod 1 are dense and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer. This answer has some mistakes, he use that [imath][(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha][/imath] which is not true. Consider [imath]k_1=3, k_2=1, \alpha=\sqrt{2}[/imath] we have [imath][(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] [/imath]. We only can assure that [imath][k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha][/imath]. Who answered said something interesting about additive subgroups of [imath]\mathbb{R}[/imath], but unfortunately the set [imath]X=\{nx-[nx] : n\in \mathbb{N} \}[/imath] is not a subgroup. Considering the additive subgroup [imath]G=\langle X \rangle[/imath], if we prove the part (a) of the link, we get that indeed [imath]G[/imath] is dense on [imath]\mathbb{R}[/imath] but we can not conclude that [imath]X[/imath] is dense on [imath][0,1)[/imath]. I think this problem has not been solved. Thanks! | 2529636 | pigeonhole principle questions centered
We were given yesterday couple of pigeonhole principle questions, i did solve most of them but these 3 i could not, i don't know even where to start. 1) 20 people are sitting around round table, on it there is a big pizza (20 slices) 10 of them with olives and 10 without, some of people like the pizza with olives while other don't, show that one can rotate the pizza such that at least half of the people(at least 10) are happy ? 2) Given a graph with 6 vertices prove that there is at least 3 vertices that have no edge between every two of them or that they all are connected ? 3) Given [imath]x \in \mathbb{R}[/imath] and [imath]x \not \in \mathbb{Q}[/imath] prove that for all [imath]\epsilon >0[/imath] there is [imath]n>0[/imath] such that [imath]\{ n x \} < \epsilon[/imath] ? any help is appreciated, even if you know how solve one question, please leave a comment showing me how to prove it, or posting an answer, Thank you. |
1257411 | Why does [imath]\lim_{n \to \infty} x^{1+1/(2n-1)}=|x|[/imath]?
Consider the following limit for [imath]x \in [-1,1][/imath] and [imath]n \in \mathbb N[/imath]: [imath]\lim_{n \to \infty} x^{1+\frac{1}{2n-1}}=|x|[/imath] I can't see where does the absolute value come from. Why isn't the answer simply [imath]x[/imath]? This limit is available on page 156 of Abbott's Understanding Analysis. EDIT: Sorry i found a dublicate here. | 767959 | Limit of [imath]h_n(x)=x^{1+\frac{1}{2n-1}}[/imath]
[imath]\lim_{n\to\infty}h_n(x) = x\lim_{n\to\infty}x^{\frac{1}{2n-1}}[/imath] where [imath]h_n(x)=x^{1+\frac{1}{2n-1}}[/imath]. I understand that [imath]\lim_{n\to\infty}x^{\frac{1}{2n-1}}[/imath] goes to one but what I don't understand is how did our limit become [imath]h_n(x)=|x|[/imath]? I'm just having hard time wrapping my head around the appearance of absolute value. Note. This is an example (Chapter 6, Section 2) from Understanding Analysis by Abbott. |
1257527 | Help solving probability problem
A box contains [imath]200[/imath] balls and [imath]10[/imath] of them are red. I need to pick up [imath]20[/imath] balls. What is the probability that [imath]4[/imath] of them are red? | 422414 | Probability of selecting q red balls from m red balls and n blue balls
Suppose there are [imath]m[/imath] red balls and [imath]n[/imath] blue balls in an urn. We randomly choose [imath]p:m<p<n[/imath] balls uniformly from the urn. What is the probability that exactly [imath]q[/imath] red balls are chosen? Note:- Normally the answer would be [imath]\frac{{m}\choose{q}}{{m+n}\choose{q}}[/imath]. However, since the number of balls that are chosen are provided, it is confusing me out. Any hints for the answer will also be appreciated. |
1257198 | Show that only one prime can be expressed as [imath]n^3-1[/imath] for some positive integer [imath]n[/imath]
Show that only one prime can be expressed as [imath]n^3-1[/imath] for some positive integer [imath]n[/imath] [imath]n^3-1=(n-1)(n^2+n+1)[/imath]. If [imath]n^3-1[/imath] is prime then [imath]n-1=1[/imath] or [imath]n=2[/imath] so [imath]n^3-1=7[/imath]. But I need a concrete proof of it, not from intuition. | 671076 | Is [imath]7[/imath] the only prime followed by a cube?
I discovered this site which claims that "[imath]7[/imath] is the only prime followed by a cube". I find this statement rather surprising. Is this true? Where might I find a proof that shows this? In my searching, I found this question, which is similar but the answers seem focused on squares next to cubes. Any ideas? |
1258051 | how to combine angle rotations along different axes into one rotation along a single vector
So, lets say I have some rotation a about the x-axis(vector:[imath](1, 0 ,0)[/imath]) and some other rotation about y-axis(vector [imath](0, 1, 0)[/imath]) and a rotation about the z-axis(vector: [imath](0,0,1)[/imath]). How would I combine these rotations so they have an equivalent rotation about a single vector? | 22437 | Combining Two 3D Rotations
Every rotation in 3D space can be defined by a rotation axis and an angle. Now let's say we have two rotations [imath]R_1 (\text{(axis)}_1, \text{(angle)}_1)[/imath], [imath]R_2 (\text{(axis)}_2, \text{(angle)}_2)[/imath]. I remember that Rotation operator is closed under composition, so [imath]R_1 (R_2(\text{object}))[/imath] will be a rotation again. Does anybody know how to find the new axis and angle for [imath]R_1R_2[/imath]? |
1258129 | what i want to know is how to compute the powers of [imath]\theta[/imath] in [imath]F_2[/imath] and also how many powers am i looking to compute. How can i find such powers
This a new chapter that we are learning and the teacher is flying through it and this are also new concept that i have just learn and i was wondering if i can have some guidance in this problem. Show that [imath]x^3 + x + 1[/imath] is irreducible over [imath]F_{2}[/imath] and let [imath]\theta[/imath] be a root. Compute the powers of [imath]\theta[/imath] in [imath]F_{2} (\theta)[/imath]. Proof: [imath]F_2[/imath] has only two element [imath]0[/imath] and [imath]1[/imath] thus to show that the polynomial is irreducible i did [imath]0^3+0+1 = 1 \not=0[/imath] and [imath] 1^3+1+1 = 3 = 1 \not=0[/imath] since there is no root therefore the polynomial is irreducible. Now my question is how to you go about computing the powers of [imath]\theta[/imath] | 768730 | [imath]x^3+x+1[/imath] irreducible over [imath]\mathbb{F}_2[/imath], [imath]\theta[/imath] root, compute powers of [imath]\theta[/imath]
Problem statement: Show that [imath]x^3+x+1[/imath] is irreducible over [imath]\mathbb{F}_2[/imath] and let [imath]\theta[/imath] be a root. Compute the powers of [imath]\theta[/imath] in [imath]\mathbb{F}_2[/imath]. I am having trouble computing the powers of [imath]\theta[/imath]. [imath]x^3+x+1[/imath] is irreducible, since any cubic polynomial over a field is reducible if and only if it has a root in the field. Since [imath]\mathbb{F}_2=\{0,1\}[/imath], we see that none of these is a root of [imath]x^3+x+1[/imath]. To compute the powers of [imath]\theta[/imath], note that [imath]\theta^3+\theta+1=0[/imath]. Hence [imath]\theta^3=-(\theta+1)[/imath]. Is this a good enough answer for [imath]\theta^3[/imath]? To compute [imath]\theta^2[/imath], note that [imath]\theta = -(\theta^3+1)[/imath]. Therefore, [imath]\theta^2=(\theta^3+1)^2 = \theta^6 + 2\theta^3 + 1 = (-\theta-1)^2-2\theta-2+1,[/imath] which just reduces to [imath]\theta^2[/imath]. Any ideas for computing these powers? Thanks! |
1257111 | Holder Inequality when [imath]0 < p < 1[/imath]
If [imath]0 < p < 1[/imath], [imath]f \in L^p[/imath], and [imath]\int \lvert g \rvert^q < \infty[/imath], show that [imath]\int \lvert fg \rvert \ge (\int \lvert f \rvert^p)^{\frac{1}{p}}(\int \lvert g \rvert^q)^{\frac{1}{q}}[/imath] My feeling is that I can use [imath]\frac{p-q}{p}[/imath] and [imath]\frac{q-p}{q}[/imath] as a conjugate pair, and apply Holder Inequality to reach such a conclusion. But I tried a lot and failed to get the desired inequality. I was considering to prove [imath](\int \lvert fg \rvert )(\int \lvert g \rvert^q)^{-\frac{1}{q}} \ge (\int \lvert f \rvert^p)^{\frac{1}{p}}[/imath], then [imath](\int \lvert fg \rvert )^{p-q}(\int \lvert g \rvert^q)^{-\frac{p-q}{q}} \ge (\int \lvert f \rvert^p)^{\frac{p-q}{p}}[/imath], i.e. [imath](\int \lvert fg \rvert )^{p-q}(\int \lvert g \rvert^q)^{\frac{q-p}{q}} \ge (\int \lvert f \rvert^p)^{\frac{p-q}{p}}[/imath]. Seems like dead ends to me.. | 1226863 | How can I prove Holder inequality for [imath]0?[/imath]
[imath]0<p<1[/imath] [imath]\dfrac{1}{p}+\dfrac{1}{p'}=1[/imath] If [imath]f \in L^{p}[/imath] and [imath]0<\int_{\Omega}\vert g(x) \vert^{p'}dx < \infty[/imath] then [imath]\int_{\Omega}\vert f(x)g(x) \vert dx \geq (\int_{\Omega}\vert f(x) \vert^{p}dx)^{\frac{1}{p}}(\int_{\Omega}\vert g(x) \vert^{p}dx)^{\frac{1}{p'}} [/imath] |
174828 | A (probably trivial) induction problem: [imath]\sum_2^nk^{-2}\lt1[/imath]
So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that [imath]\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1[/imath] for all [imath]n[/imath]. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither. | 1615204 | Induction on inequalities: [imath]\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}<2[/imath]
I am trying to solve this inequality by induction. I just started to learn induction this week and all the inequalities we had been solved were like an equation less than another equation (e.g. [imath]n! \geq 2^n[/imath]) So I am confused how to prove an inequality that is less than a particular value like the following problem? Thanks in advance. [imath]\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}<2[/imath] |
1259685 | Prove a set is a subset of another.
I need to prove [imath]A⊆B[/imath] where A and B are defined as: [imath]{A =\{x | x = 2n + 1}\}[/imath] [imath]{B =\{x | x = 2m - 21}\}[/imath] where [imath]n,m∈\mathbb{Z}[/imath] I know that I need take an arbitrary element from A and show that it works in B, but I'm stuck on how to do it in this example. | 1252640 | Proving Two Sets are Equal - Infinite Sets - Example
Let [imath]A = \{x | x = 2n+1, n\in\mathbb{Z}\}[/imath] and [imath]B = \{x | x = 2m-21, m\in\mathbb{Z}\}.[/imath] I am trying to prove [imath]A =B.[/imath] I understand that I need to prove [imath]A\subseteq B[/imath] and [imath]B\subseteq A[/imath]; But my difficulty is how to prove it. Any help would be appreciated! |
1259837 | Proof by induction
[imath]\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} +\cdots+\frac{1}{n^2} < 2 - \frac{1}{n}[/imath]. Proof: Let [imath]p(n)[/imath] be a proposition. [imath]p(n):\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} +\cdots+\frac{1}{n^2} < 2 - \frac{1}{n}.[/imath] Basis step: Let [imath]n=2[/imath], then [imath]\frac{1}{1^2} + \frac{1}{2^2} < 2 - \frac{1}{2}[/imath] we now obtain [imath]\frac{5}{4}< \frac{3}{2}[/imath] Therefore P(2) is true. Inductive step: We now assume that [imath]p(k)[/imath] is true. we will now show that [imath]p(k+1)[/imath] is also true. [imath]p(k):\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} +\cdots+\frac{1}{k^2} < 2 - \frac{1}{k}[/imath] [imath]P(k+1):\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} +\cdots+\frac{1}{k^2}+ \frac{1}{(k+1)^2} < 2 - \frac{1}{k} + \frac{1}{(k+1)^2}[/imath] So now we have to show that [imath]2 - \frac{1}{k} + \frac{1}{(k+1)^2} = \frac{2k+1} {k+1}.[/imath] Right here i am not so sure how to approach to finish this problem help anyone | 1259471 | Proof verification for proving [imath]\forall n \ge 2, 1 + \frac1{2^2} + \frac1{3^2} + \cdots + \frac1{n^2} < 2 − \frac1n[/imath] by induction
Prove by mathematical induction: [imath]\forall n \ge 2, 1 + \frac1{2^2} + \frac1{3^2} + \cdots + \frac1{n^2} < 2 − \frac1n[/imath] Basis Step: (We want to show, [imath]P(2)[/imath], which is 1 + [imath]\frac1{2^2}<2-\frac12[/imath]). [imath]\frac1{2^2}=\frac14[/imath], and [imath]2-\frac12=\frac32[/imath] So, [imath]\frac1{2^2}<2-\frac12[/imath]. Inductive step: (We want to show, [imath]\forall n\ge2,\mathrm P(n)\to\mathrm P(n+1)[/imath]. Let [imath]k\ge2[/imath] be an integer, arbitrary & fixed. (We want to show, [imath]\mathrm P(k)\to\mathrm P(k+1)[/imath]). (I.H.) Assume [imath]1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{k^2}<2-\frac1k[/imath] ("[imath]\mathrm P(k)[/imath]") (We want to show [imath]\mathrm P(k+1)[/imath], which is [imath]1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{(k+1)^2}<2−\frac1{k+1}[/imath]) [imath]\begin{align} 1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{(k+1)^2}&<2−\frac1{k+1}+\frac1{(k+1)^2}\tag{by I.H.}\\ &=2-\frac1k+\frac1{k^2+2k+1} \end{align}[/imath] (Thus, [imath]\mathrm P(k)\to\mathrm P(k+1)[/imath] for all [imath]\mathrm k\ge2[/imath] Also, [imath]\mathrm P(2)[/imath], so by mathematical induction, [imath]\forall k\ge2,\mathrm P(k)[/imath]) Can someone look at my solution and see if it is correct? I'm not 100% confident with it. |
1259862 | Prove by induction [imath]\frac{a_{1} + a_{2} + a_{3} +...+ a_{n}}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot ....a_n}[/imath].
Let [imath]a_1[/imath] [imath]a_2[/imath],..., [imath]a_n[/imath] be positive numbers. Prove that [imath]\frac{a_{1} + a_{2} + a_{3} +...+ a_{n}}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot ....a_n}[/imath]. Mine is about trying to understand how can i use induction to make some justifications so this has nothing to do with solving inequalities | 691807 | Proofs of AM-GM inequality
The arithmetic - geometric mean inequality states that [imath]\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}[/imath] I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some of those), proofs that use theory that doesn't link to this inequality at first sight (e.g. differential equations …)? Induction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics (yeah, I know, it's rather some physical argument that this theorem might be true, not really a proof), convexity, … are some of the proofs I know. |
1259950 | Relationship between Ideals and subgroups ??
I am just wondering if Product of two Ideals is again an Ideal. Let [imath]I[/imath] and [imath]J[/imath] be ideal of Ring, then is [imath]IJ=\{ij: i\in I, j\in J\}[/imath] still an ideal? | 290229 | Explaining the product of two ideals
My textbook says that the product of two ideals [imath]I[/imath] and [imath]J[/imath] is the set of all finite sums of elements of the form [imath]ab[/imath] with [imath]a \in I[/imath] and [imath]b \in J[/imath]. What does this mean exactly? Can you give examples? |
1259627 | Isomorphism inbetween a factor ring and Cartesian product of factor rings
Let R be principal ideal ring and [imath]a_1, ..., a_n \in R[/imath], with [imath]gcf(a_i, a_j) = 1[/imath], for all [imath]i, j \in \{1, ..., n\}[/imath] (with [imath]gcf[/imath] being the greatest common factor of [imath]a_i, a_j[/imath]). Show that the transformation: [imath]\phi: R/(a_1) \cdot ... \cdot (a_n) \to R/(a_1) \times ... \times R/(a_n)[/imath] [imath]r + (a_1) \cdot ... \cdot (a_n) \mapsto (r + (a_1), ..., r + (a_n))[/imath] is an isomorphism. Now I thought that this might be an application of the Chinese remainder theorem, but I'm not so sure on how this could be used. Thanks in advance. | 202113 | proof of chinese remainder theorem for ring
Let [imath]R[/imath] be a ring (not necessary having "1"), and let [imath]I,J[/imath] be ideals of [imath]R[/imath] such that [imath]I+J=R[/imath]. I want to prove that, for any [imath]r, s \in R[/imath], there is a [imath]x\in R[/imath] such that [imath]x\equiv r ({\rm mod} I) \quad x \equiv s ({\rm mod} J)[/imath] My proof: Since [imath]I+J=R[/imath], we can write [imath]r=r_i+r_j,\quad s=s_i+s_j \quad \mbox{for some}~~~r_i,s_i\in I,\quad r_j,s_j\in J[/imath] Let [imath]x=r_j+s_i[/imath]. then [imath]x-r=r_i-s_i\in I, \quad x-s=r_j-s_j\in J[/imath] Thus, [imath]x\equiv r ({\rm mod} I) \quad x \equiv s ({\rm mod} J)[/imath] Is this proof wrong?? I can't look for a mistake. |
1260356 | Do all prime numbers satisfy [imath]p \mid (p-1)! + 1[/imath]?
If [imath]m > 1[/imath], [imath]m \mid (m-1)! + 1[/imath], then we can get the conclusion that [imath]m[/imath] is a prime number. But if we have a prime number [imath]p[/imath], can I get [imath]p \mid (p-1)! + 1[/imath]? (I verify it when [imath]p < 100000[/imath], and there's no errors.) In other words, can [imath]m \mid (m-1)!+1[/imath] generates all prime numbers? | 883111 | Wilson's Theorem textbook proof question
I'm trying to understand this proof from Stein's Elementary Number Theory, and I understand the pairing of inverses but not the other direction. I have two questions: [imath]1).[/imath] When the proof says, [imath]l[/imath] a prime divisor of [imath]p[/imath], and so [imath]l<p[/imath] and [imath]l \mid (p-1)![/imath], is this because [imath]l \mid p \Rightarrow l \mid p!=p*(p-1)![/imath] and so [imath]l \mid (p-1)![/imath] (by Euclid)? [imath]2.)[/imath] Why is it that [imath]p \mid ((p-1)!+1)[/imath]? |
1260960 | Union of definable sets is a definable set
I tried to prove this question but without a success: Let [imath]K_1 \text{and } K_2[/imath] be definable sets, prove that [imath]K_1∪K_2[/imath] is definable. What I tried to do is to assume: [imath]K_1=\text{Ass}(X)=\{v\mid v\models X\}[/imath] and [imath]K_2=\text{Ass}(Y)=\{v\mid v\models Y\}[/imath] and to say that [imath]K1\cup K2=\text{Ass}(X\cap Y)[/imath]. But I couldn't prove that: [imath]K_1\cup K_2\supseteq\text{Ass}(X\cap Y)[/imath]. I would like to get help with this proof | 1260521 | Union of definable sets
I tried to prove this question but without a success: Let [imath]K_1[/imath] and [imath]K_2[/imath] be definable sets, prove that [imath]K_1\cup K_2[/imath] is definable. What I tried to do is to assume: [imath]K_1=Ass(X)=\{ v|v \vDash X \}[/imath] and [imath]K_2=Ass(Y)=\{ v|v \vDash Y \}[/imath] and to say that [imath]K_1\cup K_2=Ass(X \cap Y)[/imath]. but I couldn't prove that: [imath]K_1\cup K_2 \supseteq Ass(X \cap Y)[/imath] . I would like to get help with this proof |
1261031 | Sufficient conditions for [imath]\lim_{m \to \infty}\lim_{n \to \infty} a_{mn} = \lim_{m \to \infty}\lim_{n \to \infty} a_{mn}[/imath]
Recently I came across with some problems concerning the following fundamental problem in the advanced calculus: [imath]\lim_{m \to \infty}\lim_{n \to \infty} a_{mn} = \lim_{m \to \infty}\lim_{n \to \infty} a_{mn}[/imath] Is there any sufficient(or necessary and sufficient, if there is) conditions for this limit equality? I know from the Fubini's theorem that [imath]\sum^{\infty}_{m=1}\sum^{\infty}_{n=1}a_{mn}=\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}a_{mn}[/imath] under some conditions for [imath]a_{mn}[/imath] | 15240 | When can you switch the order of limits?
Suppose you have a double sequence [imath]\displaystyle a_{nm}[/imath]. What are sufficient conditions for you to be able to say that [imath]\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}[/imath]? Bonus points for necessary and sufficient conditions. For an example of a sequence where this is not the case, consider [imath]\displaystyle a_{nm}=\left(\frac{1}{n}\right)^{\frac{1}{m}}[/imath]. [imath]\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}}=\lim_{n\to \infty}{\left(\frac{1}{n}\right)^0}=\lim_{n\to \infty}{1}=1[/imath], but [imath]\displaystyle \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}=\lim_{m\to \infty}{0^{\frac{1}{m}}}=\lim_{m\to \infty}{0}=0[/imath]. |
684811 | Is there a closed form for [imath]\Gamma(i)[/imath]?
I know that [imath]\Gamma(z)\cdot\Gamma(z^*)=|\Gamma(z)|^2\tag{1}[/imath] and [imath]\Gamma (z)\cdot \Gamma (1-z) =\frac{\pi }{\sin{\pi z}}\tag{2}[/imath] but I still can't find closed form for [imath]\Gamma(i)[/imath] | 153719 | What is the value of [imath]\Gamma(\mathrm{i})[/imath] ?
What is the value of [imath]\Gamma(\mathrm{i})[/imath] ? [imath]\Gamma(z)[/imath] is Gamma function. Here [imath]\mathrm{i}^2=-1[/imath].Can you help me with this problem ? |
1261477 | Complete metric space of sequence of positive integers
Let [imath](A,d)[/imath] be the space [imath]\mathbb{N}^{\mathbb{N}}[/imath] of sequences of positive integers where [imath]d((a_i)_i, (b_i)_i)= \frac{1}{n}[/imath] where [imath]n[/imath] is the least coordinate at which [imath](x_i)_i[/imath] and [imath](y_i)_i[/imath] disagree (so [imath]a_j = b_j[/imath] for [imath]j<n[/imath]). Show [imath](A,d)[/imath] is complete metric space. Show [imath]A[/imath] is not a countable union of compact sets. My progress: For (1) I was able to show two properties for [imath](A,d)[/imath] to be a metric space, but for the inequality one, I don't see how this can be true from the given definition of the distance function [imath]d((a_i)_i, (b_i)_i)[/imath]. About the completeness part, I also have the same problems when trying to prove any Cauchy sequence with [imath]d((a^{(m)}_i)_i, (a^{(n)}_i)_i) < \epsilon[/imath] for every [imath]m, n > M[/imath] for some [imath]M>0[/imath]. For (2), I couldn't make much progress. | 709742 | Is the Baire space [imath]\sigma[/imath]-compact?
Is the Baire space [imath]\sigma[/imath]-compact? The Baire space is the set [imath]\mathbb{N}^\mathbb{N}[/imath] of all sequences of natural numbers under the product topology taking [imath]\mathbb{N}[/imath] to be discrete. It is a complete metric space, for example with the metric [imath]d ( x , y ) = \frac{1}{n+1}[/imath] where [imath]n[/imath] is least such that [imath]x(n) \neq y(n)[/imath]. A topological space [imath]X[/imath] is called [imath]\sigma[/imath]-compact if it is the countable union of compact subsets. |
1261774 | Reduction of matrix [imath]A[/imath] to [imath]B[/imath] to find eigenvalues by Power method
How to reduce matrix [imath]A[/imath] to [imath]B[/imath] such that it has all eigenvalues and eigenvectors of [imath]A[/imath] but the dominant eigenvalue (eigenvalue with largest magnitude) is replace by [imath]0[/imath] ? I am using Power method to find the eigenvalues and eigenvectors. The article says that by using above method, one can find not only the dominant eigenvalue but also the other eigenvalues. | 1114777 | Approximate the second largest eigenvalue (and corresponding eigenvector) given the largest
Given a real-valued matrix [imath]A[/imath], one can obtain its largest eigenvalue [imath]\lambda_1[/imath] plus the corresponding eigenvector [imath]v_1[/imath] by choosing a random vector [imath]r[/imath] and repeatedly multiplying it by [imath]A[/imath] (and rescaling) until convergence. Once we have the first eigenpair, is there a similar way to estimate the second eigenpair? |
1255596 | "Distance" of iid gaussian variables
Take two i.i.d. Gaussian R.V.s [imath]X[/imath] are [imath]Y[/imath] both of which are [imath]~N(0,a\sigma)[/imath]. Define a new R.V. [imath]D = \sqrt{X^2 + Y^2}[/imath]. What's the expected value [imath]E(D)[/imath]? In researching this I'm seeing references to chi-squared distributions and chi distributions (of different degrees of freedom?), but I haven't learned about those yet. Do they apply here? What are their means and how does the original mean/variance carry through? | 746127 | Given [imath]X[/imath] and [imath]Y[/imath] are independent N(0,1) random variables and [imath]Z = \sqrt{X^2+Y^2}[/imath] from the marginal pdf of [imath]Z[/imath]
Let [imath]X[/imath] and [imath]Y[/imath] be independent [imath]N(0; 1)[/imath] random variables. Let [imath]Z = \sqrt{X^2+Y^2}[/imath]. (a) Derive the marginal pdf of [imath]Z[/imath] and then using the marginal pdf to compute [imath]{\rm E}[Z^2][/imath] (b) Can you propose a different way other than that in (a) to compute [imath]{\rm E}[Z^2][/imath] (c) Compute [imath]{\rm E}[Z][/imath]. This is the whole question that I was asked. I can do (c) but I don't know how to find the marginal pdf of [imath]Z[/imath] with the given information and can't seem to find any formulas. Any help would be appreciated. |
1253524 | Convergence in measure, almost everywhere and almost uniformly.
I want to find counterexamples for this: [imath]f_n\to f[/imath] in measure implies [imath]f_n\to f[/imath] almost uniformly. [imath]f_n\to f[/imath] in measure implies [imath]f_n\to f[/imath] almost everywhere. I proved that almost uniformly implies almost everywhere, so it suffices to show a counterexample for the second point. I mean, to show a sequence such that [imath]f_n\to f[/imath] in measure but [imath]f_n[/imath] does not converge to [imath]f[/imath] almost everywhere. I know that the sequence of functions of the form [imath]f_n=\chi_{A_n}[/imath], where [imath]\mu(A_n)\to 0[/imath], converges to [imath]0[/imath] in measure. So I wanted to find [imath]A_n[/imath] such that [imath]f_n[/imath] doesn't converge to zero a.e. Any hint? Thank you. | 505014 | Example of Converge in measure, but not converge point-wise a.e.?
Can anyone give an exam of Converge in measure, but not converge point-wise a.e.? And also for the converse part, professor asks us to prove "pointwise a.e. implies converge in measure", but think about this function: [imath]f_n(x)= \chi_{[n,\infty)}[/imath] It converge to [imath]f(x)=0[/imath] pointwise, but it seems that the difference measure between [imath]f(x)[/imath] and [imath]f_n(x)[/imath] is always infinity. |
1261540 | Why does the midpoint method have error [imath]O(h^2)[/imath]
In solving an ode [imath] y'(t) = f(t, y(t)), \quad y(t_0) = y_0 [/imath] the midpoint method estimates [imath]y_{n+1} = y_n + hf\left(t_n+\frac{h}{2},y_n+\frac{h}{2}f(t_n, y_n)\right)[/imath] But why is the error [imath]O(h^2)[/imath]? The estimate for Euler's method is straightforward since it's just the Taylor approximation, but here it doesn't look like a Taylor approximation. | 1207414 | Numerical Approximation of Differential Equations with Midpoint Method
I want to proof that the local truncation error of the Midpoint Method is [imath]d_{k+1}=O\left(h^{3}\right)[/imath] Approach The local truncation error is defined as: [imath]d_{k+1}=u(t_{k+1})-\underset{u_{k+1}}{\underbrace{\left[u(t_{k})+\triangle t\cdot\phi\right]}}[/imath] [imath]u(t_{k+1})\rightarrow[/imath] exact solution of the differential equation [imath]u'(t)=f(u,t)[/imath], which is being approximated [imath]\Phi\rightarrow f\left(u(t_{k+1})+\frac{1}{2}\left(\triangle t\right)\cdot f\left(u(t_{k}),t_{k}\right),t_{k}+\frac{1}{2}\left(\triangle t\right)\right)[/imath] incremental Function called the Midpoint Method The first step is to write the term [imath]u(t_{k+1})[/imath] as its Taylor expansion: [imath]u(t_{k+1})=u(t_{k})+\frac{1}{1!}\cdot h^{1}\cdot u^{(1)}(t_{k})+\frac{1}{2!}\cdot h^{2}\cdot u^{(2)}(t_{k})+\frac{1}{3!}\cdot h^{3}\cdot u^{(3)}(t_{k})+O\left(h^{4}\right)[/imath] Then the second step is to write the incremental function in a fashion that the elements can cancel out with the previously introduced Taylor expansion of [imath]u(t_{k+1})[/imath]. I found a solution to the second step, but i do not understand how it is formed. And i would be happy if someone could explain in detail and write out the rules which lead to it. The solution goes as follows: [imath]\Phi\rightarrow f\left(u(t_{k+1})+\frac{1}{2}\left(\triangle t\right)\cdot f\left(u(t_{k}),t_{k}\right),t_{k}+\frac{1}{2}\left(\triangle t\right)\right)[/imath] What happened here? [imath]=f\left(u\left(t_{k}\right),t_{k}\right)+\frac{1}{2}\cdot h\cdot f_{t}\left(u\left(t_{k}\right),t_{k}\right)+\frac{1}{2}\cdot h\cdot f\left(u\left(t_{k}\right),t_{k}\right)\cdot f_{y}\left(u\left(t_{k}\right),t_{k}\right)[/imath] [imath]+\frac{1}{2}\cdot\left(\frac{1}{2}\cdot h\right)^{2}\cdot f_{tt}+\left(\frac{1}{2}\cdot h\right)^{2}f\cdot f_{tu}+\frac{1}{2}\left(\frac{1}{2}\cdot h\right)^{2}\cdot f^{2}f_{uu}+O\left(h^{3}\right)[/imath] I understand that there is some kind of differentiating going on, but i cant figure it out. Can someone provide the steps involved? Edits I found out that total derivatives are used somehow, but i do not have written out the exact transformation steps. |
1258089 | Show that for a gradient system [imath]\bf\dot x= f(x)[/imath], [imath]\frac{\partial f_i}{\partial x_j}-\frac{\partial f_j}{\partial x_i}=0[/imath] for [imath]1 \leq i, j \leq d[/imath]
The dynamical system [imath]{\bf \dot x} = {\bf f}({\bf x})[/imath] is called a gradient system if there exists a function [imath]V({\bf x})[/imath] such that [imath] {\bf f}({\bf x}) = - \nabla V({\bf x}) [/imath] Show that if [imath]{\bf \dot x} = {\bf f}({\bf x})[/imath] is a gradient system, then [imath] \frac{\partial f_i}{\partial x_j} - \frac{\partial f_j}{\partial x_i} = 0 [/imath] for [imath]1 \leq i, j \leq d[/imath] So let [imath]{\bf f}({\bf x}) = (f_1({\bf x}), f_2({\bf x}), \dots , f_n({\bf x}))[/imath] and from the definition of a gradient system, we have [imath] (f_1({\bf x}), f_2({\bf x}), \dots , f_n({\bf x})) = - \left ( \frac{\partial V({\bf x})}{\partial x_1}, \frac{\partial V({\bf x})}{\partial x_2}, \dots, \frac{\partial V({\bf x})}{\partial x_n} \right ) [/imath] but then I don't see how to follow this to produce the required equation. | 153552 | [imath]\frac{\partial f_i}{x_j}=\frac{\partial f_j}{x_i}\implies(f_1,\ldots,f_n)[/imath] is a gradient
I was reading a solution when I came across this statement. So [imath]\frac{\partial f_i}{x_j}=\frac{\partial f_j}{x_i}.[/imath] Then there exists a differentiable function [imath]g[/imath] on [imath]\mathbb{R}^n[/imath] such that [imath]\frac{\partial g}{\partial x_i}=f_i[/imath]. Why is this true? |
1263568 | How to calculate the integral [imath]I=\int\limits_0^1\frac{x^n-1}{\ln(x)} \,\mathrm dx[/imath]
How can we calculate this integral: [imath]I=\int\limits_0^1\frac{x^n-1}{\ln(x)}\,\mathrm dx[/imath] I believe that integral is equal to [imath]\ln(n+1)[/imath], but I don't lnow how to prove it. | 1254548 | An integration question to be solved without using differentiation under the integral sign.
[imath]I(\alpha)=\int_0^1 \frac{x^\alpha-1}{\ln x}dx.[/imath] As the title says, if someone could solve this without using the differentiation under the integral sign technique, I would be very grateful. |
1261532 | Compute the splitting field and the Galois group of [imath]x^4 - 5[/imath] over [imath]\mathbb{Q} (\sqrt{5})[/imath].
I believe the splitting field is easily found by the following. [imath]x^4 - 5 = (x^2 - \sqrt{5})(x^2 + \sqrt{5})[/imath], so the splitting field is [imath]\mathbb{Q}(\sqrt[4]{5},i)[/imath], or is this incorrect? Once I obtain this, how to do I find the Galois group of this polynomial over [imath]\mathbb{Q}(\sqrt{5})[/imath]? Thanks! | 629460 | Galois group of [imath]x^4-5[/imath]
How do I find the Galois group of [imath]x^4-5[/imath] over [imath]\mathbb{Q}(i)[/imath], [imath]\mathbb{Q}(\sqrt5)[/imath] and [imath]\mathbb{Q}(\sqrt{-5})[/imath]? I've managed to do so over [imath]\mathbb{Q}[/imath] but I don't know how to find the others. I'd appreciate any help. Thanks! |
1261736 | If [imath]\lim\limits_{x \rightarrow \infty} f'(x)^2 + f^3(x) = 0[/imath] , show that [imath]\lim \limits_ {x\rightarrow \infty} f(x) = 0[/imath]
If [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] is a differentiable function and [imath]\lim\limits_{x \rightarrow \infty} f'(x)^2 + f^3(x) = 0[/imath] , show that [imath]\lim\limits_{x\rightarrow \infty} f(x) = 0[/imath]. I really have no clue how to start, I tried things like MVT and using definition of derivatives but I really can't figure this out. | 881706 | How prove that [imath]\lim\limits_{x\to+\infty}f(x)=\lim\limits_{x\to+\infty}f'(x)=0[/imath] if [imath]\lim\limits_{x\to+\infty}([f'(x)]^2+f^3(x))=0[/imath]?
Question: Let [imath]f[/imath] be differentiable on [imath][0,+\infty)[/imath], such as[imath]\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0[/imath]show that [imath]\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0[/imath] I think this problem is similar to this link and (the background) For my problem I guess that if [imath]\lim_{x\to+\infty}\left([f'(x)]^{2m}+f^{2n+1}(x)\right)=0,m,n\in N^{+},n,m>1[/imath] then we have [imath]\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0[/imath]. |
1264833 | Give an example of a pair of real-valued function f and g defined on R for which f is not differentiable at 0, g is not differentiable at f(0)...
Give an example of a pair of real-valued function f and g defined on [imath]\mathbb{R}[/imath] for which f is not differentiable at 0, g is not differentiable at f(0) but for which [imath]g \circ f[/imath] is differentiable at 0. I honestly can't even wrap my mind around this. So some hints would be helpful! Thank you! | 1263931 | Finding a pair of functions with properties
I need to find a pair of functions [imath]f[/imath], [imath]g[/imath] such that [imath]f[/imath] is not differentiable at [imath]x = 0[/imath] [imath]g[/imath] is not differentiable at [imath]f(0)[/imath] [imath]g \circ f[/imath] is differentiable at [imath]x = 0[/imath] I've tried a lot of functions but just can't seem to find ones that work. Any help would be appreciated, thank you. |
1264417 | Prove [imath](\ker T)^\bot = \operatorname{Im}(T^*)[/imath]
If [imath]T[/imath] is a linear operator. Prove the following: [imath] (\ker T)^\bot = \operatorname{Im}(T^*) [/imath] where [imath]T^*[/imath] is the adjoint operator of the [imath]T[/imath]. | 963298 | Proof that [imath]\mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} [/imath]
Where [imath] f: V \rightarrow V [/imath] is a linear operator on an inner product space [imath]V[/imath], I'm trying to show that [imath] \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} [/imath] i.e. the Kernel of the adjoint is equal to the orthogonal complement of the image of f. I taking the inner product of [imath]f(v)[/imath], where [imath]v \in V[/imath] and an arbitary element of the kernel, [imath]k \in \mathrm{ker}(f^*) [/imath] [imath]\langle f(v), k\rangle = \langle v, f^*(k)\rangle = \langle v, 0\rangle = 0[/imath] And then taking the inner product of [imath]f(v)[/imath] with [imath]w[/imath], where [imath]w \in \mathrm{im}(f)^{\perp}[/imath] By definition of the orthogonal complement to the image of f, [imath] \langle f(v), w\rangle = 0 [/imath] My question is, now that we have [imath] \langle f(v), k\rangle = \langle f(v), w\rangle = 0 [/imath] What can we conclude? I would like to conclude that because [imath]w[/imath] and [imath]k[/imath] are arbitary elements of their respective subspaces, that these must be equal, i.e. [imath] \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} [/imath] However, I am not sure this is mathematically sound, is that conclusion valid or are there any technical barriers or additional statements that would have to be made to make that watertight? |
1264870 | If [imath] f [/imath] is injective and [imath] g [/imath] is injective, then [imath] f \circ g [/imath] is surjective.
I can prove that if [imath] f [/imath] and [imath] g [/imath] are both injective, then [imath] f \circ g [/imath] is injective, but I don’t know how to prove that [imath] f \circ g [/imath] is surjective. | 235548 | Proving Functions are Surjective
Prove the following are surjective, or disprove with a counter-example: [imath]f\colon \mathbb{Q} \to \mathbb{Q}[/imath], [imath]f(x) = 1 + 2x[/imath]. [imath]f\colon \mathbb{Z} \to \mathbb{N}\cup\{0\}[/imath], [imath]f(x) = |1 - x|[/imath]. [imath]f\colon \mathbb{Q} \to \mathbb{Q}[/imath], [imath]f(x) = 4 - 2x^3[/imath]. [imath]f\colon \mathbb{Z}^2 \to \mathbb{Z}[/imath], [imath]f(x, y) = x + y[/imath]. Please show me the most effective method to lay out such proofs. These come from a manual on Set Theory, which I am trying to reach to myself. Please be understanding! |
1264473 | Any finite index subgroup of [imath]\mathbb Z_p[/imath] is open
I'm trying to show that every finite index subgroup [imath]H[/imath] of [imath]\mathbb Z_p[/imath] is open. Since [imath]H[/imath] has finite index, it is equivalent (and perhaps easier) to show that it is closed. But I've tried showing that, and I can't work out how to use the finite index condition at any point. Another thing I tried was showing that any non-closed subgroup [imath]K[/imath] of [imath]\mathbb Z_p[/imath] must have infinite index, by using non-convergent sequences in [imath]K[/imath] to construct infinitely many cosets. But I don't think it's obvious even that [imath]\mathbb Z[/imath] has infinite index in [imath]\mathbb Z_p[/imath] (apart from using a cardinality argument that might not generalize to general non-closed subgroups). As far as I'm aware, this is a property that is not shared by general profinite groups. So any proof must use some other part of the structure of [imath]\mathbb Z_p[/imath] in an essential way. | 1092974 | Finite index subgroup [imath]G[/imath] of [imath]\mathbb{Z}_p[/imath] is open.
Suppose [imath][\mathbb{Z}_p:G] = n <\infty[/imath]. Write [imath]n = p^km[/imath] with [imath]p\nmid m[/imath]. The idea is to show that [imath]p^k\mathbb{Z}_p = n\mathbb{Z}_p \subseteq G[/imath], after which I am done, since for any [imath]x\in G[/imath] we have that the open ball [imath]x + p^k\mathbb{Z}_p\subseteq G[/imath], which makes [imath]G[/imath] open. Unfortunately I am stuck at how to prove that [imath]n\mathbb{Z}_p\subseteq G[/imath]. Could someone lend a hand with a hint? |
996330 | A question about sublinear functionals
Could you please give me hints may leads to prove the following: Let [imath]X[/imath] be a real vector space, [imath]\,p_1,p_2:X\to\mathbb R\,[/imath] be two sublinear functionals, and [imath]\,f:X\to\mathbb R\,[/imath] be a linear functional satisfying [imath] f(x)\le p_1(x)+p_2(x), \quad\text{for all $\,x\in X$.} [/imath] Prove that there exist two linear functionals [imath]\,f_1,f_2:X\to\mathbb R[/imath], such that [imath]\,f=f_1+f_2\,[/imath] and [imath]\,f_i(x)\le p_i(x),\,[/imath] for [imath]i\in\{1,2\}[/imath] and [imath]x\in X[/imath]. I think we should use Hahn-Banach Theorem. Thanks in advance. | 1184199 | Prove that there exist linear functionals [imath]L_1, L_2[/imath] on [imath]X[/imath]
Let [imath]X[/imath] be a linear space, [imath]p, q[/imath] sublinear functionals on [imath]X[/imath], and [imath]L[/imath] a linear functional on [imath]X[/imath] such that [imath]|L(x)| ≤ p(x) + q(x),[/imath] for all [imath]x ∈ X[/imath]. Prove that there exist linear functionals [imath]L_1, L_2[/imath] on [imath]X[/imath] such that [imath]L(x) = L_1(x) + L_2(x),[/imath] and [imath]|L_1(x)| ≤ p(x), |L_2(x)| ≤ q(x),[/imath] for all [imath]x ∈ X.[/imath] My Work: First I thought to use Hahn Banach Theorem. But since there is no known subspace it was useless. Then I tried to make [imath]L(x)[/imath] as [imath]L(x)=\frac{L(x+\lambda)+L(x-\lambda)}{2}[/imath] for some scalar [imath]\lambda[/imath] but failed to find suitable [imath]L_1[/imath] and [imath]L_2[/imath]. I think this problem is little bit tricky. I want to try it myself and I only need a hint to start. Can somebody please give me a hint? |
638025 | Definite Integration problem
I was working on a small Integration problem, where i was needed to solve another integral: [imath]I= \int_0^{\pi} \frac{1}{1+\cos^2x} dx [/imath] Working: [imath]\int_0^{\pi} \frac{1}{1+\cos^2x} dx=\int_0^{\pi} \frac{\sec^2x}{\sec^2x+1} dx[/imath] [imath]=\int_0^{\pi} \frac{\sec^2x}{2+\tan^2x} dx[/imath] Substituting, [imath]\tan x=t, \sec^2x dx=dt[/imath] [imath]\implies \frac{\sec^2x}{2+\tan^2x}dx=\frac{dt}{2+t^2}[/imath] which gives us [imath]I= \left[ \begin{array}{cc|c}{\frac{1}{\sqrt2}\tan^{-1}\frac{\tan x}{\sqrt2}} \end{array} \right]_0^\pi[/imath] I think the answer should be zero. After seeing the answer at WolframAlpha, i think i am doing a serious mistake. Ay help will be appreciated. | 170885 | A little integration paradox
The following integral can be obtained using the online Wolfram integrator: [imath]\int \frac{dx}{1+\cos^2 x} = \frac{\tan^{-1}(\frac{\tan x}{\sqrt{2}})}{\sqrt{2}}[/imath] Now assume we are performing this integration between [imath]0[/imath] and [imath]2\pi[/imath]. Hence the result of the integration is zero. On the other hand when looking at the integrand, [imath]\displaystyle \frac{1}{1+\cos^2 x}[/imath], we see that it is a periodic function that is never negative. The fact that it is never negative guarantees that the result of the integration will never be zero (i.e., intuitively there is a positive area under the curve). What is going on here? |
1266426 | Uncountable chain of subsets of [imath]\mathbb N[/imath]
Denote by [imath]\mathbb{N}[/imath] the set of natural numbers and by [imath]2^{\mathbb N}[/imath] the set of all subsets of [imath]\mathbb N[/imath]. Let [imath]E[/imath] be some subset of [imath]2^{\mathbb N}[/imath] such that for every pair of elements in [imath]E[/imath], one is a subset of the other. Clearly, [imath]E[/imath] could be countable by considering [imath]E=\big\{\{1,\cdots, n\}, n\in\mathbb N\big\}[/imath]. My question is the following: could we find an example that [imath]E[/imath] is uncountable? Many thanks for the answer! | 1182145 | Finding an uncountable chain of subsets the integers
How to find an uncountable subset of [imath]P(\mathbb{N})[/imath] such that every two elements of it can be compared. In fact, give an uncountable subset of [imath]P(\mathbb{N})[/imath] such that has totality property. We mean by [imath]P(\mathbb{N})[/imath], the powerset of natural numbers set [imath]\mathbb{N}[/imath]. |
1266495 | How to prove that [imath]I+A^{T}A[/imath] is invertible
Let [imath]A[/imath] be any [imath]m\times n[/imath] matrix and [imath]I[/imath] be the [imath]n\times n[/imath] identity. Prove that [imath]I+A^{T}A[/imath] is invertible. | 363962 | Prove that [imath]A^*A + I[/imath] is invertible
Let [imath]A[/imath] be an [imath]m\times n[/imath] matrix. Prove that [imath]A^*A + I[/imath] is invertible. I'm not sure what to do because everything I try ends up at [imath]A^*A[/imath] again and not [imath]AA^*[/imath]. |
1266333 | A nice set of squares.
Are there integers [imath]a, b, c, d[/imath] such that [imath]a^2+b^2=c^2[/imath] [imath]a^2-b^2 = d^2?[/imath] I have tried by showing that [imath]a^2 = b^2 + d^2[/imath] and thus [imath]a^2+ b^2 = 2b^2+d^2 = c^2[/imath] But how do I show that there are no integers [imath]b, c, d[/imath] such that [imath]2b^2 + d^2 = c^2[/imath] (I think)? I would like an answer other than the one given in the duplicate, which is at Proving that the sum and difference of two squares (not equal to zero) can't both be squares. A possible method maybe by using modulus? Or geometry? | 1246469 | Proving that the sum and difference of two squares (not equal to zero) can't both be squares.
I have the following task: Prove that the sum and the difference of two squares (not equal to zero) can't both be squares. For the sum, I thought about Pythagorean triples: [imath]x^2+y^2=z^2[/imath] works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well? I tried to write down this: Assume that [imath]a^2>b^2[/imath]. Then [imath]a^2+b^2=c^2[/imath] and [imath]a^2-b^2=d^2[/imath]. Summing the left and the right equation, we get [imath]2a^2=c^2+d^2[/imath]. I am just stuck proving why this can't occur. Any ideas? Thanks for your help! |
1266163 | How to find this integral from Gaussian integral?
How to find [imath]\int_0^\infty e^{-ax^2-\frac{b}{x^2}}dx[/imath] using gaussian integral? I tried complete the square: [imath]-ax^2-\frac{b}{x^2}=-\left(\sqrt{a}x+\frac{\sqrt{b}}{x}\right)^2+2\sqrt{ab}[/imath], but what next? I tried integration by parts [imath]u=e^{-ax^2-\frac{b}{x^2}}, dv=dx[/imath], but what next? | 496088 | How to evaluate [imath]\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx[/imath] for [imath]a,b>0[/imath]
How can I evaluate [imath]I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx[/imath] for [imath]a,b>0[/imath]? My methods: Let [imath]a,b > 0[/imath] and let [imath]I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.[/imath] Then [imath]I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,dx.[/imath] What the other methods that can I use to evaluate it? Thank you. |
1267305 | Let [imath]p[/imath] be a prime. Write down the solutions of equation [imath]\frac{1}{x} +\frac{1}{y} =\frac{1}{p}[/imath]
Let [imath]p[/imath] be a prime. Consider the equation [imath]\frac{1}{x} +\frac{1}{y} =\frac{1}{p}[/imath] with [imath]x[/imath] and [imath]y[/imath] positive integers. Write down the complete set of distinct solutions, and prove that your list is complete. | 1260889 | Let p be a prime. Consider the equation [imath]\frac1x+\frac1y=\frac1p[/imath]. What are the solutions?
Write down the set of distinct solutions and prove your list is complete. [imath]x[/imath] and [imath]y[/imath] are positive integers. I have rewritten it as [imath]\frac{(x+y)}{xy} = \frac1p[/imath], but I don't understand where to go from here? |
1263584 | Euler Fermat with double exponent
I have to calculate [imath] 3^{{2014}^{2014}} \pmod {98} [/imath] (without calculus). I want to do this by using Euler/Fermat. What I already have is that the [imath]\gcd(3, 98) = 1[/imath] so I know that I can use the Euler Fermat formula. Then I know that [imath]\varphi(m = 98) = 42 [/imath] Then I can say that [imath]3^{{2014 }^{2014} \pmod {42}} \pmod {98}[/imath] Now I don't know how to progress. Any ideas/hints? | 1259663 | Euler's theorem: [3]^2014^2014 mod 98
Calculate without a calculator: [imath]\left [ 3 \right ]^{2014^{2014}}\mod 98[/imath] I know I have to use Euler's Theorem. As a hint it says I might need to use the Chinese Remainder theorem too. I know how both of these work theoretically. For example I can calculate [imath]5^{256} \mod 13[/imath] with Euler's Theorem. But these large numbers throw me off, especially the double power of 2014. I started of like this for the Chinese Remainder: [imath]98 = 2*7^{2}\\ \left [3 \right ]^{2014^{2014}} \mod 2 \\ \left [3 \right ]^{2014^{2014}} \mod 49 \\[/imath] Now calculate both: [imath]\left [3 \right ]^{2014^{2014}\mod \phi_{(2)} = 2} \equiv \left [3 \right ]^{0} \equiv 1 \mod 2[/imath] That was lucky. I fail at mod 49. Applying Euler's Theorem on 2014^2014 (since ([3]^2014)^2014 seems even more impossible) gives the following: [imath]\phi _{(49)} = 42\\ 2014 = 47*42+40\\ 2014^{2014} = 2014^{42^{47}}*2014^{40}\equiv 2014^{40} \mod 49[/imath] (That doesn't really work without a calculator either.) And what now, applying Euler's Theorem again doesn't work, and I can't do it in my head either. Is my approach right? Did I do something wrong? Please point me in the right direction. I'm also open to entirely different solutions. |
1267349 | Let [imath]f(x) = x^2 + x + 41[/imath]. Show that [imath]f(n)[/imath] is prime for [imath]0 \le n \le 39[/imath], but [imath]f(40)[/imath] is composite.
[imath]40 \cdot 40 + 40 + 41 = 40(40 + 1) + 41 = 40 \cdot 41 + 41 = 41(40 + 1) = 41^2[/imath], so [imath]f(40)[/imath] is composite. Suppose [imath]f(n) = n^2 + n + 41[/imath] is prime for [imath]0 \le n \le 38[/imath]. But [imath]f(n + 1)[/imath] is also prime: [imath]39^2 +39 + 41 = 1601[/imath]. So, [imath]f(n)[/imath] is prime for [imath]0 \le n \le 39[/imath]. Does the solution to the problem look something like that? edit: like some users noted I might have to check all nonnegative [imath]n < 40[/imath] manually. The other post didn't solve my problem because I have no idea what's going on there. So, all I want to do is to see how strong induction fails here. | 289338 | Is the notorious [imath]n^2 + n + 41[/imath] prime generator the last of its type?
The polynomial [imath]n^2+n+41[/imath] famously takes prime values for all [imath]0\le n\lt 40[/imath]. I have read that this is closely related to the fact that 163 is a Heegner number, although I don't understand the argument, except that the discriminant of [imath]n^2+n+41[/imath] is [imath]-163[/imath]. The next smaller Heegner number is 67, and indeed [imath]n^2+n+17[/imath] shows the same behavior, taking prime values for [imath]0\le n\lt 16[/imath]. But 163 is the largest Heegner number, which suggests that this is the last such coincidence, and that there might not be any [imath]k>41[/imath] such that [imath]n^2+n+k[/imath] takes on an unbroken sequence of prime values. Is this indeed the case? If not, why not? Secondarily, is there a [imath]k>41[/imath], perhaps extremely large, such that [imath]n^2+n+k[/imath] takes on prime values for [imath]0<n<j[/imath] for some [imath]j>39[/imath]? |
1267371 | show [imath]p[/imath] is divisible by [imath](x^2 +y^2 +1)[/imath]
Show that, for any prime [imath]p[/imath], there are integers [imath]x[/imath], and [imath]y[/imath] such that [imath]p[/imath] is divisible by [imath](x^2+y^2+1)[/imath] Can you show me what to start with? do I prove [imath]p[/imath] is divisible by [imath]x^2[/imath] and [imath]y^2[/imath] separately? | 1260928 | Show that for any prime [imath] p [/imath], there are integers [imath] x [/imath] and [imath] y [/imath] such that [imath] p|(x^{2} + y^{2} + 1) [/imath].
So we obviously we want [imath] x^{2} + y^{2} + 1 \equiv 0 ~ (\text{mod} ~ p) [/imath]. I haven’t learned much about quadratic congruences, so I don’t really know how to go forward. I suppose you can write it as [imath] x^{2} \equiv -1 - y^{2} ~ (\text{mod} ~ p) [/imath], but again, I’m not sure where to go. I know about the Legendre symbol and quadratic residues, which I know are involved in this question, but I don’t know how to apply what I know. :( |
1267490 | Prove that every convergent sequence has a monotone subsequence
So if a certain sequence [imath]a_n[/imath] is convergent then its bounded.So from Bolzano-Weierstrass [imath]a_n[/imath] has a convergent sub-sequence, but where do I continue from here? | 716461 | Proof Verification - Every sequence in [imath]\Bbb R[/imath] contains a monotone sub-sequence
Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions. I also looked up these related questions - (1), (2), (3) - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated. Show that every sequence in [imath]\Bbb R[/imath] either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence. Let [imath](x_n)[/imath] be a sequence in [imath]\Bbb R[/imath]. Suppose [imath](x_n)[/imath] is not bounded. Without loss of generality we may assume that [imath](x_n)[/imath] is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let [imath]x_{n_1}[/imath] be any member of the sequence. There is [imath]x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}[/imath]. For [imath]i \gt 1[/imath] let [imath]x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}[/imath] then [imath](x_{n_k})[/imath] forms a monotone subsequence of [imath](x_n)[/imath]. Now suppose instead that [imath](x_n)[/imath] is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence [imath](y_n)[/imath] of [imath](x_n)[/imath] which converges to a limit [imath]y[/imath]. Without loss of generality there are infinitely many distinct values in [imath](y_n)[/imath] that are unequal to [imath]y[/imath]. Let [imath]y_{k1}[/imath] be the first such element. Let [imath]y_{k2}[/imath] be any element in [imath]\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}| \}[/imath]. For [imath]i \gt 1[/imath] let [imath]y_{ki} \in \{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}| \}[/imath]. Such [imath]y_{ki}[/imath] exists for every [imath]i \in \Bbb N[/imath] since [imath] \lim (y_n) = y [/imath]. Now let [imath](y_{kn})[/imath] be the sub-sequence of [imath](y_n)[/imath] thus formed. At least one of the two following sets must contain infinitely many elements. [imath]\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}[/imath] [imath]\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}[/imath] The one which does forms a monotone subsequence. |
164864 | Existence of Consecutive Quadratic residues
For any prime [imath]p\gt 5[/imath],prove that there are consecutive quadratic residues of [imath]p[/imath] and consecutive non-residues as well(excluding [imath]0[/imath]).I know that there are equal number of quadratic residues and non-residues(if we exclude [imath]0[/imath]), so if there are two consecutive quadratic residues, then certainly there are two consecutive non-residues,therefore, effectively i am seeking proof only for existence of consecutive quadratic residues. Thanks in advance. | 1269337 | Show for all primes [imath]p>11[/imath] there are two consecutive quadratic residues
I am supposed to use this fact to help prove it. If [imath]p[/imath] is an odd prime, then at least one of the numbers [imath]2,5,10[/imath] is a quadratic residue mod [imath]p[/imath] I can prove this by saying let [imath](\frac{10}{p}) = 1[/imath] Then [imath](\frac{2}{p}) = -1[/imath] [imath](\frac{5}{p}) = -1[/imath] let [imath](\frac{10}{p}) = -1[/imath] Then [imath](\frac{2}{p})[/imath] or [imath](\frac{5}{p}) = 1[/imath] So to start off we know that p is a odd prime [imath]>11[/imath] [imath]2,5[/imath] or [imath]10[/imath] is a quadratic residue of p I am assuming the best way to start this is to say Let [imath](\frac{2}{p}) = 1[/imath] where [imath]p>11[/imath] Then try to show that [imath](\frac{3}{p}) = 1[/imath] where [imath]p>11[/imath] is a quadratic residue Then do the same for the other two. But i am unsure about how to go about this |
1267244 | Sum of series [imath]\sin \theta+\sin 2 \theta+\sin 3\theta+\dots[/imath]
I need to prove sum of series: [imath]\sin \theta+\sin 2 \theta+\sin 3\theta+\dots=\sum_{n=1}^\infty \sin n\theta[/imath] by using in the first place the complex numbers. | 297452 | Sines and cosines of angles in arithmetic progression
Prove that if [imath]\phi[/imath] is not equal to [imath]2k\pi[/imath] for any integer [imath]k[/imath], then [imath]\sum_{t=0}^{n} \sin{(\theta + t \phi)}=\frac{\sin({\frac{(n+1)\phi}2})\sin{(\theta+\frac{n \phi}2)}}{\sin{(\frac{\phi}2)}}[/imath] Find a similar formula for [imath]\sum_{t=0}^{n}\cos{(\theta+t\phi)}[/imath] where the functions sin and cos appear on the right-hand side. Find, for all [imath]\theta[/imath], the values of [imath]\sum_{t=0}^{n}\cos^{2}{(2t\theta)}[/imath] and [imath]\sum_{t=0}^{n}\sin^{2}{(2t\theta)}[/imath] |
1267578 | [imath]trc(A)=0[/imath].why [imath]A=M+N[/imath] where [imath]M[/imath] and [imath]N[/imath] are nilpotent matrices?
Let [imath]trc(A)=0[/imath].why [imath]A=M+N[/imath] where [imath]M[/imath] and [imath]N[/imath] are nilpotent matrices?([imath]A \in {M_n}[/imath]) | 1243650 | Let [imath]trcA=0[/imath].why [imath]A=M+N[/imath] where [imath]M[/imath] and [imath]N[/imath] are nilpotent matrices?
Let [imath]A \in {M_n}[/imath] and [imath]trcA=0[/imath].why [imath]A=M+N[/imath] where [imath]M[/imath] and [imath]N[/imath] are nilpotent matrices? |
1268026 | How to calculate the eigenvalue of the following general matrix
Let the [imath]n\times n[/imath] matrix [imath]Z[/imath] with [imath](i,j)[/imath]-element defined by [imath]Z_{i,j}=i+j[/imath]. How to calculate the eigenvalue of [imath]Z.[/imath]? I have used Matlab to calculate it. I find no matter how bigger n is, there are always two nonzero eigenvalues. But how to get the formula for eigenvalues of [imath]Z[/imath]? | 1261077 | Symmetric matrix eigenvalues
Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix, with [imath]A_{ij}=i+j[/imath]. Find the eigenvalues of [imath]A[/imath]. A student that I tutored asked me this question, and beyond working out that there are 2 nonzero eigenvalues [imath]a+\sqrt{b}[/imath] and [imath]a-\sqrt{b}[/imath] and [imath]0[/imath] with multiplicity [imath]n-2[/imath], I'm at a bit of a loss. |
1268570 | Prove that [imath]A[/imath] and [imath]A^t[/imath] have the same eigenvalues.
Let [imath]A[/imath] be a square matrix. Prove that [imath]A[/imath] and [imath]A^t[/imath] have the same eigenvalues. The solution my lecturer uses is: Consider the characteristic polynomial \begin{align} P_{A^t}(x) &= \det{\left(xI-A^t\right)} \\ &= \det{\left((xI-A)^t\right)} \\ &= \det{(xI-A)} = P_{A}(x). \end{align} Since the characteristic polynomials are equal and the eigenvalues are the roots of the characteristic polynomial, the eigenvalues will be the same. However some of the steps he has taken seem to have come out of nowhere. How do we know [imath](xI-A^t) = (xI-A)^t[/imath], also how do we know [imath](xI-A)^t=(xI-A)[/imath]? | 123923 | A matrix and its transpose have the same set of eigenvalues
Let [imath] \sigma(A)[/imath] be the set of all eigenvalues of [imath]A[/imath]. Show that [imath] \sigma(A) = \sigma(A^T)[/imath] where [imath]A^T[/imath] is the transpose matrix of [imath]A[/imath]. |
1268572 | How to prove or disprove this statement about convergent sub-sequences?
I need to prove through a proof or disprove through a counter example that if [imath]a_n[/imath] is a sequence which has three sub-limits ( [imath]a_n[/imath] has three sub-sequences that each one of them have different limit)and [imath]b_n[/imath] is another sequence that has two sub-limits( [imath]b_n[/imath] has two sub-sequences that each one of them have different limit) then could it be that there is a sequence [imath]c_n[/imath] that happens to be [imath]c_n[/imath]=[imath]a_n[/imath]+[imath]b_n[/imath] and is convergent ? (convergent here means that [imath]c_n[/imath] has only one limit!!) any piece of advice is appreciated!! | 1264359 | Prove or Disprove if [imath]a_n[/imath] has 3 partial limits and [imath]b_n[/imath] has 2 partial limits then could it be that [imath]a_n+b_n[/imath] has one limit?
I think it has to do with [imath]\sin[/imath] and [imath]\cos[/imath] because most of them have three limits but I couldn't find a good example that it is wrong although I think that it is indeed wrong. |
1268469 | Find the range of the function : [imath]\frac{1}{\pi}(\sin^{-1}x+\tan^{-1}x) + \frac{x+1}{x^2+2x+5}[/imath]
Problem : Find the range of the function : [imath]\frac{1}{\pi}(\sin^{-1}x+\tan^{-1}x) + \frac{x+1}{x^2+2x+5}[/imath] My approach : Let [imath]g(x) = (\sin^{-1}x+\tan^{-1}x)[/imath] and [imath]h(x)=\frac{x+1}{x^2+2x+5}[/imath] and domain of [imath]g(x)[/imath] is [imath][-1,1][/imath] [imath]\therefore[/imath] maximum value of [imath]g(x)[/imath] is [imath]g(1) = \frac{3}{4}[/imath] and minimum value of [imath]g(x)[/imath] is at [imath]g(-1) = -\frac{3}{4}[/imath] Now how to find the maximum minimum value of [imath]h(x) [/imath] Let [imath]h(x)=y=\frac{x+1}{x^2+2x+5}[/imath] [imath]\therefore y(x^2+2x+5)=x+1[/imath] [imath]\Rightarrow x^2y +x(2y-1)+5y-1=0[/imath] .......(i) Can we find the range of [imath]h(x)[/imath] using (i) Please guide on this thanks.. or there is any other alternative method using concept of maxima minima i.e. using differentiation. Thanks. | 1259110 | Find the range of the given function [imath]f[/imath]
Find the range of the following function. [imath]\cfrac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) + \cfrac{x+1}{x^2 + 2x + 5} [/imath] where [imath]\sin^{-1}x \ [/imath] and [imath] \ \tan^{-1}x [/imath] are inverse trigonometric functions. This is what I've tried yet. Let : [imath]\cfrac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) = g(x)[/imath] and [imath] \cfrac{x+1}{x^2 + 2x + 5} = h(x)[/imath] . Therefore, [imath]f(x) = g(x) + h(x)[/imath] I can write [imath]h(x)[/imath] as : [imath]h(x) = \cfrac{x+1}{(x+1)^2 + 4} = \cfrac{1}{(x+1) + \cfrac{4}{x+1}}[/imath] So, the function [imath]f(x)[/imath] becomes: [imath]f(x) = \cfrac{1}{\pi} \left(\sin^{-1} x +\tan^{-1} x\right) + \cfrac{1}{(x+1) + \cfrac{4}{x+1}}[/imath] Domain of [imath]g(x)[/imath] is [imath][-1,1] \cap \mathbb{R} = [-1,1][/imath] Domain of [imath]h(x)[/imath] is [imath]\mathbb{R} - \{-1\}[/imath] Therefore, [imath]dom(f(x)) = (-1,1][/imath] Now, finding maximum value of [imath]h(x)[/imath] [imath]\begin{align} \cfrac{dh(x)}{dx} =& \cfrac{d}{dx} \left( (x+1) + \cfrac{4}{x+1} \right) \\ = & 1 - \cfrac{4}{(x+1)^2}\end{align}[/imath] Putting [imath]h'(x)[/imath] equal to zero to find critical points, I got: [imath]\begin{align} 1 - \cfrac{4}{(x+1)^2} =& 0 \\ (x+1) =& \pm 2 \\ \boxed{x = 1 \ \text{OR} \ x = -3} \rightarrow \textsf{CRITICAL POINTS} \end{align}[/imath] How to go further from here? Is this method the perfect one for solving this type of questions? Or is there any other trick to go with it? |
1268795 | How to obtain a closed form for summation over polynomial ([imath]\sum_{x=1}^n x^m[/imath])?
What is the method for obtaining the polynomial equal to \begin{equation*} \sum^{n}_{x=1}x^m \end{equation*} for unknown [imath]n[/imath], and systematically for various values of [imath]m[/imath]? I know it should be a polynomial in [imath]n[/imath] of degree [imath]m+1[/imath], but what are the coefficients? Wolfram Alpha seems to implement this method, at least for [imath]m[/imath] up to around 40, but it is not clear to me how to reproduce it. Thanks. | 449834 | Explicit formula for Bernoulli numbers by using only the recurrence relation
It is not hard to show, by induction on [imath]m\in\mathbb N[/imath], that there exist a sequence [imath](B_n)_{n\geq0}[/imath] of rational numbers such that [imath]\sum_{k=1}^nk^m=\frac1{m+1}\sum_{k=0}^m\binom{m+1}kB_k\,n^{m+1-k},\ \style{font-family:inherit;}{\text{for all}}\ n\geq1\,.[/imath] These are the Bernoulli numbers of the second kind, that is, [imath]B_1=+1/2[/imath]. A corollary of the proof (by induction) of the fact above is a recurrence formula for such numbers [imath]B_n[/imath], which are known as Bernoulli numbers: [imath]\sum_{k=0}^{n-1}\binom nkB_k=0\,.\tag{$\ast$}[/imath] On the other hand, there are a number of explicit formulae for [imath]B_n[/imath], which are obtained using the equality [imath]\frac x{e^x-1}=\sum_{n=0}^\infty B_n\,\frac{x^n}{n!}[/imath]; see here and here. For example: [imath]B_n=\sum_{k=0}^n\frac1{k+1}\sum_{r=0}^k(-1)^r\binom krr^n\,.\tag{$\ast\ast$}[/imath] Is there some clever way to manipulate the recurrence [imath](\ast)[/imath] in order to obtain [imath](\ast\ast)[/imath]? |
1268134 | Linear Programming 3 decision variables (past exam paper question)
This is an exam question I was practising. I have the general understanding of Linear programming, but how would you go about finding the Decision Variables, Objective function and Constraints for this? My Opinions: Decision variables: [imath]M_1[/imath]- # of Model 1 to be produced [imath]M_2[/imath] and [imath]M_3[/imath] (similar definition to [imath]M_1[/imath]) Objective function: I'd Multiple the Cost to make in house by cost to buy from subcontractor to get Minimise: [imath]3900M_1+7500M_2+27300M_3[/imath] Constraints: [imath]3M_1+1.5M_2+2M_3 \geq 0[/imath] [imath]1.25M_1+2.5M_2+0.75M_3 \geq 0[/imath] [imath]M_1,M_2,M_3 \geq 0[/imath] No idea where number order would go? Do I use it in the Constraints and multiply by the Hours? Now, am I on the right track? | 1268129 | Linear Programming 3 decision variables (past exam paper question)
This is an exam question I was practising. I have the general understanding of Linear programming, but how would you go about finding the Decision Variables, Objective function and Constraints for this? My Opinions: Decision variables: [imath]M_1[/imath]- # of Model [imath]1[/imath] to be produced [imath]M_2[/imath] and [imath]M_3[/imath](similar definition as [imath]M_1[/imath]) Objective function: I'd Multiple the Cost to make in house by cost to buy from subtractor to get Minimise: [imath]3900M_1+7500M_2+27300M_3[/imath] Constraints: [imath]3M_1+1.5M_2+2M_3\ge0[/imath] [imath]1.25M_1+2.5M_2+0.75M_3\ge8500[/imath] [imath]M_1,M_2,M_3\ge 5500[/imath] No idea where number order would go? Do I use it in the Constraints and multiply by the Hours? Now, am I on the right track? |
1236838 | Let [imath]I = \{a+ ib \in \Bbb Z[i] : 2 \mid a-b\}[/imath] then [imath]I[/imath] is a maximal ideal of [imath] \Bbb Z[i][/imath].
Let [imath]I = \{a+ ib \in \Bbb Z[i] : 2 \mid a-b\}[/imath] then [imath]I[/imath] is a maximal ideal of [imath] \Bbb Z[i][/imath]. We consider an ideal [imath]J[/imath] such that [imath]I \subset J\subset\Bbb Z[i] [/imath]. So there exists an element [imath]p \in J[/imath] but not in [imath]I[/imath]. Note that [imath]a \not\mid a-b[/imath] when [imath]a,b[/imath] are of odd parity i.e. one is odd and other is even. Without loss of generality we assume that [imath]p = c+ id[/imath] where [imath]c[/imath] is odd and [imath]d[/imath] is even. Then we have [imath]q = (c-1)+id \in I[/imath] since both [imath]c-1,d[/imath] are even and hence [imath]q = (c-1)+id \in J[/imath]. Now since [imath]J[/imath] is an ideal, so [imath]p-q = 1 \in J[/imath]. Thus [imath]J =\Bbb Z[i] [/imath]. Hence [imath]I[/imath] is a maximal ideal of [imath] \Bbb Z[i][/imath]. Is the proof correct?? | 531554 | Prove that [imath]I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}[/imath] is an maximal ideal of [imath]ℤ[i][/imath].
I'm making some exercises to prepare for my ring theory exam: Prove that [imath]I= \{a+bi \in ℤ[i] : a≡b \pmod{2}\}[/imath] is an maximal ideal of [imath]ℤ[i][/imath]. I know that [imath]a+2l=b[/imath] with [imath]l\in ℤ[/imath] (or should I say [imath]l \in ℤ[i][/imath])? Therefore I can write [imath]a+(a+2l)i=a+ai+2li[/imath]. So I don't have [imath]0+i[/imath] in my ideal. For even real parts I have even complex parts, and for odd real parts I have odd complex parts. But I don't see what I can conclude now. |
1269171 | A proof for a theorem related to rank and matrix product.
For all matrix [imath]\mathbf{M} \in \mathbb{R}^{m,n}[/imath] and [imath]\mathbf{N} \in \mathbb{R}^{n,p}[/imath], the inequality [imath]\operatorname{rank}\mathbf{M} + \operatorname{rank}\mathbf{N} - n \leq \operatorname{rank}(\mathbf{M}\mathbf{N}) \leq \min(\operatorname{rank}\mathbf{M},\operatorname{rank}\mathbf{N})[/imath] holds. What would be a proof of this theorem? | 269474 | prove that [imath]\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n.[/imath]
If [imath]A[/imath] is a [imath]m \times n[/imath] matrix and [imath]B[/imath] a [imath]n \times k[/imath] matrix, prove that [imath]\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n.[/imath] Also show when equality occurs. |
1268725 | convergence on sequences using cauchy's criterion
I have the following sequence [imath]x_{n+1}=\frac12\left(x_n+\frac2{x_n}\right),~n>1,~x_1=1[/imath] How can I prove the sequence is Cauchy, and get the limit? | 549428 | Convergence of [imath]x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).[/imath]
Let [imath]x_1=1[/imath] and [imath]x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).[/imath] Prove or disprove [imath](x_n)[/imath] is convergent and show the limit. When I tried working on it I found the sequence was bounded by square root of 2 and it is was monotone. But apparently the sequence is not bounded by square root of two and is not monotone. But I have no idea why. Any help would be greatly appreciated! Thanks! |
1268705 | Prove the existence or the non-existence of a couple of numbers ([imath]n[/imath],[imath]m[/imath]) such that [imath]n^2=m![/imath]
In recent days, while I was doing exercises on combinatorics, I thought if a number [imath]m![/imath] could be a perfect square. I proved to demonstrate it through the prime factorization. My attempt: [imath]k=m!=p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}....\cdot p_n^{a_n}[/imath] with [imath]a_1,a_2,..,a_n[/imath] that have to be even. Now if there exists a prime number [imath]p_j[/imath] between [imath]\frac {k}{2}<p_j<k[/imath] this prime number in the prime factorization the exponent ([imath]a_j[/imath]) relative to the prime number [imath]p_j[/imath] is equal to [imath]1[/imath] ([imath]a_j=1[/imath]) because the number [imath]k[/imath] isn't a multiple of [imath]p_j[/imath]; for example [imath]12![/imath] isn't a perfect square because the prime numbers [imath]7[/imath] and [imath]11[/imath] [imath]6<7,11<12[/imath] aren't divisors of [imath]12[/imath]. My question There exist a prime number [imath]p_j[/imath] between [imath]k[/imath] and [imath]\frac {k}{2}[/imath] and how can I prove if there exist or not exist a number [imath]m!=n^2[/imath]? | 31973 | [imath]n![/imath] is never a perfect square if [imath]n\geq2[/imath]. Is there a proof of this that doesn't use Chebyshev's theorem?
If [imath]n\geq2[/imath], then [imath]n![/imath] is not a perfect square. The proof of this follows easily from Chebyshev's theorem, which states that for any positive integer [imath]n[/imath] there exists a prime strictly between [imath]n[/imath] and [imath]2n-2[/imath]. A proof can be found here. Two weeks and four days ago, one of my classmates told me that it's possible to prove that [imath]n![/imath] is never a perfect square for [imath]n\geq2[/imath] without using Chebyshev's theorem. I've been trying since that day to prove it in that manner, but the closest I've gotten is, through the use of the prime number theorem, showing that there exists a natural number [imath]N[/imath] such that if [imath]n\geq N[/imath], [imath]n![/imath] is not a perfect square. This isn't very close at all. I've tried numerous strategies over the past weeks and am now trying to use the Sylow theorems on [imath]S_n[/imath] to somehow show that [imath]|S_n|[/imath] can't be square (I haven't made any progress). Was my classmate messing with me, or is there really a way to prove this result without Chebyshev's Theorem? If it is possible, can someone point me in the right direction for a proof? Thanks! |
1269377 | Showing [imath]\sum |\hat{f}(n)| \leq C \cdot \int_{0}^{2\pi} |f(t)| \ dt[/imath]
If [imath]f \in L^{1}[0,2\pi][/imath] define [imath]\hat{f}(n)[/imath] for [imath]n \in\mathbb{Z}[/imath] by [imath]\hat{f}(n) = \frac{1}{2\pi} \int_{0}^{2\pi} f(t) \cdot (\cos(nt) -i\sin(nt)) \ dt[/imath] Suppose [imath]M[/imath] is a closed linear subspace of [imath]L^{1}[0,2\pi][/imath] such that [imath]\sum |\hat{f}(n)| < \infty[/imath] for each [imath]f \in M[/imath].Then how can one show that there is a constant [imath]C[/imath] such that [imath]\sum_{n} |\hat{f}(n)| \leq C \int_{0}^{2\pi} |f(t)| \ dt[/imath] for each [imath]f \in M[/imath]. Can anyone help me out with a clear solution. | 288934 | there is [imath]M<\infty[/imath] such that [imath]\sum_{n} |\hat{f}(n)|\le M\int_{0}^{2\pi}|f(t)|dt[/imath] for each [imath]f\in X[/imath]
for [imath]f\in L^1[0,2\pi][/imath] define [imath]\hat{f}(n)=\int_{0}^{2\pi} f(t)e^{-int} dt[/imath] for [imath]n\in\mathbb{Z}[/imath], [imath]X[/imath] is a closed linear subspace of [imath]L^1[0,2\pi][/imath] such that [imath]\sum_{n} |\hat{f}(n)|<\infty[/imath] for each [imath]f\in X[/imath], we need to show that there is [imath]M<\infty[/imath] such that [imath]\sum_{n} |\hat{f}(n)|\le M\int_{0}^{2\pi}|f(t)|dt[/imath] for each [imath]f\in X[/imath] please tell me how to solve this one, I have no clue. |
1270057 | Combinatory: how many differents queues are possible...
Problem: [imath]50[/imath] people go to teather. The price of the ticket is [imath]5[/imath]\[imath]. [/imath]25[imath] of these people have only a [/imath]5[imath]\[/imath] note in their wallet, while the other [imath]25[/imath] only have a [imath]10[/imath]\[imath] banknote. The cashier doesn't have any cash at the beginning of the queue. I would like to know how many queues exist such that the cashier always has the possibility to give the rest.[/imath] For instance: if the first person pays with a 10[imath]\[/imath] note, the queue does not work, so we know that the first person who pays, needs to pay with a [imath]5[/imath]\[imath] note, because if the first one comes with [/imath]5[imath]\[/imath], then the second one can pay with [imath]10[/imath]\[imath], since in this case the cashier would have the rest.[/imath] So, how many different queues are possible? My thoughts: I thought i would have found the solution by finding the number of arrangements of 50[imath] elements by using only [/imath]\{0,1\}[imath], where [/imath]0[imath] stands for a person who pays with a [/imath]5[imath]\[/imath] note, while [imath]1[/imath] stands for a person who pays with a [imath]10[/imath]\[imath] note. In this way I know that the number of [/imath]0[imath]s must always be bigger or equal to the number of [/imath]1[imath]s, if you read the arrangement from left to right.[/imath] I also found that this number must be smaller than \dfrac{1}{2}\begin{pmatrix} 50 \\ 25 \end{pmatrix}$, so a half of total number of arrangements. Can somebody help me? | 22999 | How many ways can we let people into a movie theater if they only have half-dollars and dollars?
My interest in combinatorics was recently sparked when I read about the many things that the Catalan numbers count, as found by Richard Stanley. I picked up a copy of Brualdi's Combinatorics, and while browsing the section on counting sequences I found a nice little puzzle that has definitely puzzled me. Let [imath]m[/imath] and [imath]n[/imath] be nonnegative integers with [imath]n\geq m[/imath]. There are [imath]m+n[/imath] people in line to get into a theater for which admission if [imath]50[/imath] cents. Of the [imath]m+n[/imath] people, [imath]n[/imath] have a [imath]50[/imath]-cent piece and [imath]m[/imath] have a [imath]\[/imath] 1$ dollar bill. The box offices opens with an empty cash register. Show that the number of ways the people can line up so that change is available when needed is [imath] \frac{n-m+1}{n+1}\binom{m+n}{m}. [/imath] I first noted that the first person to enter must be one of the $n[imath] with a half-dollar. Now the register has a half-dollar change. The second person can be either a person with a half-dollar or a dollar. In the first case, the register will now have two half-dollars, in the second case, the register will now have one dollar bill. So it seems to me that when one of the [/imath]n[imath] people with a half-dollar enters, the number of half-dollars in the register increases by [/imath]1[imath], and when one of the [/imath]m[imath] people with a bill enters, the number of half-dollars decreases by [/imath]1[imath] but the number of bills increases by [/imath]1[imath]. [/imath] I tried to model this by looking at paths in \mathbb{Z}^2[imath]. The [/imath]x[imath]-axis is like the number of half-dollars, and the [/imath]y[imath]-axis is the number of bills. You start at [/imath](0,0)[imath], and you can take steps forward [/imath](1,0)[imath] or backwards diagonally [/imath](-1,1)[imath] corresponding to who enters, but you must always stay in the first quadrant of the plane without crossing over the axes. The goal is to make [/imath]m+n[imath] moves, and I figured maybe the number of such paths is counted by [/imath]\frac{n-m+1}{n+1}\binom{m+n}{m}$, but I'm not sure how to show this. I don't know if this observation simplifies the problem at all, as I don't know how to finish up. I'd be happy to see how this problem is done, thank you. |
1270149 | Show that the polynomial [imath]x^8 -x^7 +x^2 -x +15[/imath] has no real root.
Please check if my method is correct. Solution : Let [imath]f(x) = x^8 - x^7 + x^2 -x +15 [/imath] Now, let [imath]g(x)= x^8 -x^7=x^7(x - 1)[/imath] and [imath]h(x)= x^2 -x=x(x-1)[/imath]. Thus, [imath]f(x)=g(x) + h(x) + 15[/imath] On analyzing [imath]g(x)[/imath] and [imath]h(x)[/imath] we see that both the functions are negative between [imath]0[/imath] and [imath]1[/imath] and the maximum negative value of both the functions do not exceed [imath]-1[/imath]. Thus, the minimum value of [imath]f(x)[/imath] is not less than [imath]13[/imath]. Hence, [imath]f(x)[/imath] has no real root. Q.E.D | 335653 | Finding real roots of [imath] P(x)=x^8 - x^7 +x^2 -x +15[/imath]
Let [imath] P(x)=x^8 - x^7 +x^2 -x +15 [/imath], Descartes' Rule of Signs tells us that the polynomial has 4 positive real roots , but if we group the terms as [imath] P(x)= x(x-1)(x^6+1) +15 [/imath] we find that [imath] P(x) [/imath] does not have any positive roots. Where did I err ? |
1269526 | Find all pairs of positive integers [imath](m,n)[/imath] such that [imath]2^{m+1}+3^{n+1}[/imath] is a perfect square
Find all pairs of positive integers [imath](m,n)[/imath] such that [imath]2^{m+1}+3^{n+1}[/imath] is a perfect square My attempt so far Any perfect square is [imath]0,1[/imath] in mod 4, so [imath]n+1[/imath] must be even : [imath]2^{m+1}+3^{2r}=k^2[/imath] Rearranging and factoring [imath]2^{m+1}=(k+3^r)(k-3^r)[/imath] That gives two equations [imath]2^a=k+3^r\\2^b=k-3^r\\a+b=m+1;a>b[/imath] However I am not able to conclude, subtracting/adding... these equations is not giving me anything useful. Any help is appreciated... Thanks! | 1142313 | When is [imath]2^x+3^y[/imath] a perfect square?
If [imath]x[/imath] and [imath]y[/imath] are positive integers, then when is [imath]2^x+3^y[/imath] a perfect square? I tried this question a lot but failed. I tried dividing cases into when [imath]x,y[/imath] are even/odd, but still have no idea what to do when they are both odd. I tried looking into when is [imath]2^x+3^y[/imath] is of the form [imath]6k+1[/imath] and so on .. but couldn't succeed. Can anyone help? |
1270971 | Prove [imath]|f'(\frac{1}{2})|\leq \frac{1}{4}[/imath]
Let [imath]f:[0,1]\to \mathbb{R}[/imath] be a function whose second derivative [imath]f'(x)[/imath] is continuous on [imath][0,1][/imath]. Suppose [imath]f(0)=f(1)=0[/imath] and that [imath]|f''(x)|\leq 1[/imath] for all [imath]x\in [0,1][/imath]. Prove that [imath]|f'(\frac{1}{2})|\leq \frac{1}{4}[/imath] I have played around with Taylor's theorem but it's not going anywhere. help? | 1070561 | If [imath]f(0) = f(1)=0[/imath] and [imath]|f'' | \leq 1[/imath] on [imath][0,1][/imath], then [imath]|f'(1/2)|\le 1/4[/imath]
Let [imath]f : [0,1] \rightarrow \mathbb{R}[/imath] be a function whose second order derivative [imath]f''(x)[/imath] is continuous on [imath][0,1][/imath]. Suppose that [imath]f(0) = f(1)=0[/imath] and that [imath]|f''(x)| \leq 1[/imath] for any [imath]x \in [0,1][/imath]. Prove that [imath]|f'(\frac{1}{2})| \leq \frac{1}{4}[/imath]. I think we need to use the mean value theorem, and I have proven that [imath]|f(x)| \leq \frac{1}{8}[/imath] for any [imath]x \in [0,1][/imath]. I'm not sure how to proceed, though. Could someone please help? Thanks! |
1271343 | Linear Algebra Problem on a Symmetric Matrix
Let [imath]n[/imath] be an odd integer and [imath]A[/imath] an [imath]n \times n[/imath] symmetric matrix with integer entries such that all diagonal entries are zero. Show that [imath]detA[/imath] is even. My Attempt Since all the entries of [imath]A[/imath] are integers its characteristic polynomial [imath]p(x)\in Z[x][/imath] with the constant term of [imath]p(x)[/imath] being the determinant of [imath]A[/imath]. If I can show that 2 is an eigenvalue of [imath]A[/imath] then we are done. But I can't exhibit such an eigenvector. Please Help. | 789438 | Determinant of a Special Symmetric Matrix
If [imath]A[/imath] is a symmatric matrix of odd order with integer entries and the diagonal entries [imath]0[/imath] then [imath]A[/imath] has determinant value even. I can prove the result if I can show that the eigenvalues of [imath]A[/imath] are integers,but I am unable to do that. Thanks for any help. |
1271399 | Proof that every Principal ideal domain is Noetherian
I would like to know if my logic is sound. We know that in every principal ideal domain, every ideal is multiplicatively generated. Thus, for [imath]a \in R[/imath] we have: [imath]aR = [/imath]{[imath]ra: r \in R[/imath]} Thus every ideal has a limit on size, right? So eventually our ascending chain of ideals will have to be the same for some ideal eventually? Is that correct? It seems a bit too easy | 409466 | Every principal ideal domain satisfies ACCP.
Every principal ideal domain [imath]D[/imath] satisfies the ACCP. Proof. Let [imath](a_1) ⊆ (a_2) ⊆ (a_3) ⊆ · · ·[/imath] be a chain of principal ideals in [imath]D[/imath]. It can be easily verified that [imath]I = \displaystyle{∪_{i∈N} (a_i)}[/imath] is an ideal of [imath]D[/imath]. Since [imath]D[/imath] is a PID, there exists an element [imath]a ∈ D[/imath] such that [imath] I = (a)[/imath]. Hence, [imath]a ∈ (a_n)[/imath] for some positive integer [imath]n[/imath]. Then [imath]I ⊆ (a_n) ⊆ I[/imath]. Therefore, [imath]I = a_n[/imath]. For [imath]t ≥ n[/imath], [imath](a_t) ⊆ I = (a_n) ⊆ (a_t)[/imath]. Thus, [imath](a_n) = (a_t)[/imath] for all [imath]t ≥ n[/imath]. I have prove [imath]I[/imath] is an ideal in the following way:- Let [imath] x,y\in I[/imath]. Then there exist [imath]i,j \in \mathbb{N}[/imath] s.t. [imath]x \in (a_i)[/imath] & [imath]y \in (a_j)[/imath]. Let [imath]k \in \mathbb{N}[/imath] s.t [imath]k>i,j[/imath]. Then [imath]x \in (a_k)[/imath] & [imath]y \in (a_k)[/imath]. as [imath](a_k)[/imath] is an ideal [imath]x-y \in (a_k)\subset I[/imath] and [imath]rx,xr \in (a_k)\subset I[/imath]. So [imath]I[/imath] is an ideal. Is it correct? |
1269909 | Elementary substructures and eventually constant variable assignments
One proof of the Downward Löwenheim Skolem Theorem is via consideration of elementary substructures. In a discussion of this theorem, Christopher Leary writes the following: "Suppose that [imath] \mathfrak{L}[/imath] is a countable language and [imath] \mathfrak{B}[/imath] is an [imath] \mathfrak{L}[/imath]-structure. Then [imath] \mathfrak{B}[/imath] has a countable elementary substructure... assume [imath]\textit{B}[/imath] (the underlying set of [imath] \mathfrak{B}[/imath]) is uncountable. As the language [imath] \mathfrak{L}[/imath] is countable, there are only countably many [imath] \mathfrak{L}[/imath]-formulas, and thus only countably many formulas of the form [imath]\exists x \alpha[/imath]. Let [imath]A_o[/imath] be any nonempty countable subset of [imath]\textit{B}[/imath]. We show how to build [imath]A_1[/imath] such that [imath]A_o \subseteq A_1[/imath] , and [imath]A_1[/imath] is countable. The idea is to add to [imath]A_o[/imath] witnesses for the truth (in the structure [imath] \mathfrak{B}[/imath]) of existential statements. Notice that as [imath]A_o[/imath] is countable, there are only countably many functions [imath]s'[/imath]: Vars → [imath]A_o[/imath] that are eventually constant. (This is a nice exercise for those of you who have had a course in set theory...) Also, if we are given any φ and any [imath]s[/imath] : Vars → [imath]A_o[/imath], we can find an eventually constant [imath]s'[/imath] : Vars → [imath]A_o[/imath] such that [imath]s[/imath] and [imath]s'[/imath] agree on the free variables of φ" What does Leary mean by "eventually constant" in this context? Why is the variable assignment [imath]s'[/imath] eventually constant in this case? What might be a proof of the "nice exercise" that Leary refers to? I would be extremely happy if you have answers to my questions. | 1271131 | Eventually constant variable assignments
One proof of the Downward Löwenheim Skolem Theorem is via consideration of elementary substructures. In a discussion of this theorem, Christopher Leary writes the following: "Suppose that [imath] \mathfrak{L}[/imath] is a countable language and [imath] \mathfrak{B}[/imath] is an [imath] \mathfrak{L}[/imath]-structure. Then [imath] \mathfrak{B}[/imath] has a countable elementary substructure... assume [imath]\textit{B}[/imath] (the underlying set of [imath] \mathfrak{B}[/imath]) is uncountable. As the language [imath] \mathfrak{L}[/imath] is countable, there are only countably many [imath] \mathfrak{L}[/imath]-formulas, and thus only countably many formulas of the form [imath]\exists x \alpha[/imath]. Let [imath]A_o[/imath] be any nonempty countable subset of [imath]\textit{B}[/imath]. We show how to build [imath]A_1[/imath] such that [imath]A_o \subseteq A_1[/imath] , and [imath]A_1[/imath] is countable. The idea is to add to [imath]A_o[/imath] witnesses for the truth (in the structure [imath] \mathfrak{B}[/imath]) of existential statements. Notice that as [imath]A_o[/imath] is countable, there are only countably many functions [imath]s'[/imath]: Vars → [imath]A_o[/imath] that are eventually constant. (This is a nice exercise for those of you who have had a course in set theory...) Also, if we are given any φ and any [imath]s[/imath] : Vars → [imath]A_o[/imath], we can find an eventually constant [imath]s'[/imath] : Vars → [imath]A_o[/imath] such that [imath]s[/imath] and [imath]s'[/imath] agree on the free variables of φ" What does Leary mean by "eventually constant" in this context? Why is the variable assignment [imath]s'[/imath] eventually constant in this case? What might be a proof of the "nice exercise" that Leary refers to? I would be extremely happy if you have answers to my questions. |
1271309 | Is there an intuitive, not-too-mathematical way of thinking about limit points?
so I know this question has been asked sooo many times. But I just have a few questions in particular, which despite searching, I haven't found an answer to. I appreciate any help. Book's definition: A point [imath]x[/imath] is a limit point of a set [imath]A[/imath] if every [imath]\epsilon[/imath]-neighborhood [imath]V_{\epsilon}(x)[/imath] of [imath]x[/imath] intersects the set [imath]A[/imath] in some point other than [imath]x[/imath]. Particularly, wanted clarification with the bolded part of this sentence. If we take our set [imath]A = [1,4), A \in \mathbb{R}[/imath], then would the point [imath]\{2\}[/imath] be a limit point of the set [imath]A[/imath]? Since every epsilon neighborhood of [imath]2[/imath] intersects [imath]A[/imath] in a point other than [imath]2[/imath]. Or does that bolded sentence mean to imply that, the limit point itself, cannot belong to the set [imath]A[/imath]? So in this case, that would mean to imply, only [imath]4[/imath] is a limit point of the set [imath]A[/imath] ? Is it possible for someone to give a diagrammatic example of what is and isn't a limit point? Edit: I am trying to get a "non-mathematical" idea of a limit point, and thus I will opt to not mark this question as a duplicate. I mean, technically one could envision and understand this definition with the [imath]\epsilon[/imath] definition, but for people like me who are mathematically challenged, it's nice to have a different way of thinking about it (they are the exact same idea, I know, but this helps me to link and understand the concept). MY question has been solved, now, however. | 211144 | Why the definition of limit have a strict inequality rather than not?
My question is why the definition of limit of sequence is for [imath]\epsilon >0[/imath], there exist [imath]N[/imath] st for [imath]n\ge N[/imath] that [imath]|x_n-L|<\epsilon[/imath] instead of [imath]|x_n-L|\le \epsilon[/imath] |
1271839 | [imath]5^x \equiv 1520 \pmod {9797}[/imath]
How do you solve this? What does mod mean and how will I solve it? I understand that it can be solved but how? 5 to some exponent equals the (mod of 9797) what is the answer to this? | 131127 | Algorithms for solving the discrete logarithm [imath]a^x \equiv b\pmod{n}[/imath] when [imath]\gcd(a,n) \neq 1[/imath]
The general discrete logarithm problem is to find [imath]x[/imath] given [imath]a, b[/imath] and [imath]n[/imath] such that [imath]a^x \equiv b\pmod{n}.[/imath] Normally one can use the "baby-steps giant-steps" algorithm to solve it fairly quickly. But when [imath]\gcd(a,n) \neq 1,[/imath] it is not possible to compute [imath]a^{-1}\pmod{n},[/imath] which the algorithm requires. What algorithms are available to solve this problem in this case (apart from brute force of course)? Are there any theoretical limits on how fast such an algorithm could run? |
1272834 | Is [imath]\phi[/imath] convex if it is continuous and [imath]\phi \Big(\frac{x+y}{2}\Big)\le \frac{1}{2}\phi (x) +\frac{1}{2} \phi (y)[/imath]
Assume that [imath]\phi[/imath] is continuous real function on [imath](a,b)[/imath] such that [imath]\phi\Big(\frac{x+y}{2}\Big)\le \frac{1}{2}\phi (x) +\frac{1}{2} \phi (y)[/imath] for all [imath]x,y \in (a,b)[/imath]. Prove that [imath]\phi[/imath] is convex. We need to show that for [imath]\lambda \in [0,1][/imath] we have [imath]\phi(\lambda x+ (1-\lambda)y) \le \lambda \cdot\phi(x)+(1-\lambda)\cdot \phi (y) \quad\forall x,y \in (a,b)[/imath]. I had drawn some figures and it was coming out to be convex, for rigorous proof I was thinking of proving this for rationals [imath]x,y[/imath] then using continuity argument we can prove it for all [imath]x,y[/imath] but I'm not able to do so. | 1081111 | Jensen's Inequality implies convexity?
Suppose [imath]\phi[/imath] is a real function on [imath]\mathbb{R}[/imath] such that [imath]\phi \Bigl(\int_0^1f(x)\,dx\Bigr)\leq \int_0^1 \phi(f)\,dx[/imath] for every real bounded measurable function [imath]f[/imath]. Prove that [imath]\phi[/imath] is then convex. This one is in Rudin's Real and Complex Analysis, page 74, and, to be quite honest, I am unsure what my first step to solve this problem should even be. To help think of a general methodology, I attempted this similar question, from page 71 of the same book: Assume [imath]\phi[/imath] is a continuous real valued function on [imath](a,b)[/imath] such that [imath]\phi\Bigl(\frac{x+y}{2}\Bigr)\leq\frac{\phi(x)+\phi(y)}{2}[/imath] for all [imath]x, y\in (a,b).[/imath] Prove that [imath]\phi[/imath] is then convex (the conclusion does not follow if continuity is omitted from the hypothesis). I take it that the latter question is just a special case of the former (assuming [imath]\phi[/imath] is bounded on [imath](a,b)[/imath]), so I attempted to solve the latter problem first. My attempt was to note that \begin{align*} \phi\big(\lambda x + (1-\lambda)y\big) &\leq \phi\Bigl(\frac{2\lambda x + 2(1-\lambda)y}{2}\Bigr) \\ &\leq \frac{1}{2}\Bigl(\phi\bigl(2\lambda x\bigr)+\phi \bigl(2(1-\lambda)y\bigr)\Bigr) \\ & \leq \phi\bigl(\lambda x\bigr)+\phi\bigl((1-\lambda)y\bigr) \end{align*} with the latter inequality justified simply by applying the hypothesis again twice to the previous expression. From here, however, I am stuck; I cannot justify "popping out" the [imath]\lambda[/imath] and the [imath](1-\lambda)[/imath] in the last inequality via some bound afforded by the hypothesis and/or the continuity of [imath]\phi[/imath]. |
1272838 | every nonempty compact, locally path-connected and connected metric space is path-connected
I wanna prove that if [imath]M[/imath] is nonempty compact, locally path-connected and connected metric space then it is path connected. I think to prove this the best way is to show that between every to points of [imath]M[/imath] like [imath]p_1[/imath] and [imath]p_2[/imath] there is a path. (Actually this is the definition of path-connectedness). But I don't know how to prove it. | 332108 | Showing that every connected open set in a locally path connected space is path connected
I'm trying to figure out whether my proof is correct for a question I'm trying to tackle in Topology by James R. Munkres. Task: Let [imath]X[/imath] be locally path connected. Show that every connected open set in [imath]X[/imath] is path connected. My attempt at a proof: Well, every open subset of the locally path connected space [imath]X[/imath] is locally path connected. In addition, the path components and components of [imath]X[/imath] are the same in view of Theorem [imath]25.5[/imath] (p. [imath]162[/imath]), which states that if a space [imath]X[/imath] is locally path connected, then the components and the path components of [imath]X[/imath] are the same. Altogether, this implies that then every connected open set in [imath]X[/imath] is path connected. Am I right, or do I need to make any changes? Please provide your input, thanks. |
1273175 | Finite power series
I'm a student and I'm looking for a solution for the following finite power series: [imath] \sum_{n=0}^m \frac{1}{n!} x^n [/imath] By "solution" I meant expansion of the series and finding a closed form representation of it (without the sum). Thanks to everyone for the replies! The solution with the Taylor remainder was what I needed. | 1283 | Closed form of a partial sum of the power series of [imath]\exp(x)[/imath]
I am looking for a closed form (ideally expressed as elementary functions) of the function [imath]\exp_n(x) = \sum_{k=0}^n x^k / k![/imath]. I am already aware of expressing it in terms of the gamma function. Background / Motivation When counting combinations of objects with generating functions, it is useful to be able to express the partial sum [imath]1 + x + \cdots + x^n[/imath] as [imath]\frac{1-x^{n+1}}{1-x}[/imath]. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function [imath]f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)[/imath]. By using the partial sum identity, we can express it as [imath]f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)[/imath]. Simplify, express as simpler product of series, and find the coefficient of the [imath]x^5[/imath] term. I want to be able to do the same for a generating function in the form [imath]g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}[/imath] The easiest way to extract the coefficient of a given term [imath]x^p / p![/imath] would be to use a similar closed form expression for [imath]\exp_n(x)[/imath] and a similar technique to [imath]f[/imath]. Attempted Solutions Differential equation Recall that the way to prove the identity [imath]1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}[/imath] is to define [imath]S = 1 + x + x^2 + \cdots + x^n[/imath] and notice that: [imath]S - Sx = 1 - x^{n+1}[/imath]. Likewise, notice that [imath]y(x) = \exp_n(x)[/imath] satisfies [imath]y - y' = x^n/n![/imath]. Via SAGE, the solution is [imath]y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x[/imath]. Our initial condition [imath]y(0) = 0[/imath] so [imath]c=0[/imath]. By (2), [imath]\Gamma(n+1,x) = n! e^{-x} \exp_n(x)[/imath] so we have [imath]y(x) = \exp_n(x)[/imath]. Recurrence Relation Notice that [imath]\exp_n(x) = \exp_{n-1}(x) + x^n/n![/imath]. Using the unilateral Z-Transform and related properties, we find that [imath]\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)[/imath]. Therefore, [imath]\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz[/imath]. [imath](z^n e^{x/z})/(z-1)[/imath] has two singularities: [imath]z = 1[/imath] and [imath]z = 0[/imath]. The point [imath]z = 1[/imath] is a pole of order one with residue [imath]e^x[/imath]. To find the residue at [imath]z = 0[/imath] consider the product [imath]z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)[/imath]. The coefficient of the [imath]z^{-1}[/imath] term is given when [imath]n - m + j = -1[/imath]. The residue of the point [imath]z=0[/imath] is then [imath]-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m![/imath]. Let [imath]C[/imath] by the positively oriented unit circle centered at the origin. By Cauchy's Residue Theorem, [imath]\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)[/imath]. Finite Calculus I've tried to evaluate the sum using finite calculus, but can't seem to make much progress. |
1271017 | Property of simple bipartite graphs
I'm trying to solve the following exercise from the book A Textbook of Graph Theory by R. Balakrishnan and K. Ranganathan Show that for a simple bipartite graph, [imath]m\leq \frac{n^2}{4}[/imath] [imath]m[/imath] is the size of the graph. [imath]n[/imath] number of vertices. I was considering the following cases: Let [imath]X[/imath] and [imath]Y[/imath] be a partition of the set of vertices of [imath]V(G)[/imath] and I tried to verify some cases accordingly to the cardinality of the each set. If [imath]|V(G)|=n[/imath] and [imath]|X|=|Y|=\frac{n}{2}[/imath] it is easy to see, but I got stuck trying to show the case when the cardinalities are [imath]|X|<|Y|[/imath] Any help will be greatly appreciated! Thanks! | 1098587 | Maximum number of edges in a bipartite graph
Prove that for a bipartite graph [imath]G[/imath] on [imath]n[/imath] vertices the number of edges in [imath]G[/imath] is at most [imath]\frac{n^2}{4}[/imath]. I used induction on [imath]n[/imath]. Induction hypothesis: Suppose for a bipartite graph with less than [imath]n[/imath] vertices the result holds true. Now take a bipartite graph on [imath]n[/imath] vertices.Let [imath]x,y[/imath] be two vertices in [imath]G[/imath] where an edge exist between [imath]x[/imath] and [imath]y[/imath]. Now remove these two vertices from [imath]G[/imath] and consider this graph [imath]G'[/imath]. [imath]G'[/imath] has at most [imath]{(n-2)^2}\over4[/imath]. Add these two vertices back. Then the number of edges [imath]G[/imath] can have is at most [imath]|E(G')|+d(x)+d(y)-1[/imath] My question is in my proof I took [imath]d(x) + d(y) \le n[/imath], where [imath]d(x)[/imath] denotes the degree of vertex [imath]x[/imath]. Can I consider [imath]d(x)+d(y) \leq n[/imath]? I thought the maximum number of edges is obtained at the situation [imath]K_{\frac n 2,\frac n 2}[/imath] |
1273905 | If sequence [imath]x_n[/imath] converges to [imath]x[/imath], prove that [imath]\sqrt x_n[/imath] converges to [imath]\sqrt x[/imath]
If sequence [imath]x_n[/imath] converges to [imath]x[/imath], prove that [imath]\sqrt x_n[/imath] converges to c. I know I have to estimate [imath]| \sqrt x_n - \sqrt x |[/imath]. But I cannot start. Thanks | 99345 | Show that if [imath]x_n \to x[/imath] then [imath]\sqrt{x_n} \to \sqrt{x}[/imath]
Show that if [imath]x_n \to x[/imath] then [imath]\sqrt{x_n} \to \sqrt{x}[/imath] I have been stuck on this for a while. I tried [imath]|\sqrt{x_n} - \sqrt{x}| = \frac{|x_n - x|}{|\sqrt{x_n} + \sqrt{x}|},[/imath] and then I at least can get the top to be as small as I want, so I have [imath]\frac{\epsilon}{|\sqrt{x+\epsilon} + \sqrt{x}|},[/imath] but I get stuck here at choosing the N, and I don't know if my first step in breaking down the absolute value is legitimate. Please help. |
1271724 | Closed formula for [imath]\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)[/imath]
Denote [imath](v_1, \ldots, v_n)[/imath] the matrix that has columns [imath]v_1,\ldots, v_n\in \mathbb{R}^n[/imath]. Let [imath]A\in \mathcal{M}_{n\times n}(\mathbb{R})[/imath]. Is there a clever way (without expanding LHS and doing calculations) to show that [imath]\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)=\text{tr}(A)\det(v_1, \ldots, v_n)\ ?[/imath] | 1258290 | Showing that [imath]f_0 (x_1, \ldots, x_m) \mathrm tr A =\displaystyle{ \sum_{i=1}^n} f_0(x_1, \ldots, Ax_i,\ldots, x_m)[/imath]
Question: Consider [imath]f: (-\epsilon, \epsilon) \to \mathbb R^{m^2}[/imath] a differentiable path of matrices [imath]m \times m[/imath] such that [imath]f(0) = I_m[/imath] and the function [imath]g: I \to \mathbb R[/imath] is defined by [imath]g(t) = \det f(t)[/imath] Show that [imath]g'(0) = \mathrm{tr} \,a [/imath] (trace of the matrix [imath]a[/imath]), where [imath]a = f'(0)[/imath]. Attempt: First we use that [imath]\det f(t) = \det (f_1(t), \ldots, f_m(t)) [/imath], where [imath]f_i[/imath] is the [imath]i[/imath]-th row of the matrix [imath]f(t)[/imath] [imath]m \times m[/imath]. Considering the operator [imath]': \mathbb R^m \to \mathbb R^m[/imath], [imath]f(t) \mapsto f'(t)[/imath] With this in mind [imath]\begin{align}g'(t) = \det'f(t) &= \sum_{i=1}^{m} \det (f_1(t),\ldots,f'_i(t),\ldots, f_m(t)) \\&\underbrace{=}_{(*)}\det(f_1(t), \ldots, f_m(t))\, \mathrm tr f'(t) \end{align}[/imath] For [imath]t = 0[/imath] we have [imath]\begin{align}g'(0) &=\det(e_1, \ldots, e_m) \,\mathrm tr f'(0) \\&= 1 \dot\, \mathrm tr \,a \\& = \mathrm tr \, a\end{align}[/imath] I'm trying to come up with a proof using the fact that [imath]\det[/imath] is a [imath]m[/imath]-linear form. In order to prove [imath](*)[/imath] I need the following equality, which according to this answer holds. If [imath]A: V \to V[/imath] is a linear operator and [imath]f_0[/imath] is a alternated [imath]m[/imath]-linear form, such that [imath]f_0 (e_1, \ldots, e_m) = 1[/imath] then [imath]f_0 (x_1, \ldots, x_m) \mathrm tr A = \sum_{i=1}^m f_0(x_1, \ldots, Ax_i,\ldots, x_m) [/imath] Could someone give a proof to this? |
1230173 | Elementary proof for [imath]\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})[/imath] where [imath]p_i[/imath] are different prime numbers.
Take [imath]p_1, p_2, \ldots, p_n, p_{n+1}[/imath] be [imath]n+1[/imath] prime numbers in [imath]\mathbb{P} \subseteq \mathbb{N}[/imath]. [imath]\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})[/imath] seems to be quite clear, but still need a proof. I know some proofs are involved with Galois theory, which is not I want. | 2495126 | Does [imath]\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})[/imath] has degree [imath]2^k[/imath] over $\mathbb{Q}?
Let [imath]m_1 ,...,m_k[/imath] be [imath]k[/imath] distinct squarefree integers [imath]\neq 1[/imath] which are pairwise coprime. Then is [imath][\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k}):\mathbb{Q}]=2^k[/imath]? Intuitively, I guess that the statement is true, by observing the result when [imath]k=1,2[/imath], but I have trouble actually proving this for the general case. What I know is that [imath]\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})[/imath] is Galois over [imath]\mathbb{Q}[/imath], so [imath][\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k}):\mathbb{Q}]=|\mathrm{Gal}(\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})/\mathbb{Q})|[/imath] and every element of [imath]G=\mathrm{Gal}(\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})/\mathbb{Q})[/imath] sends [imath]\sqrt{m_i}[/imath] to either itself or [imath]-\sqrt{m_i}[/imath] for each [imath]i=1,...,k[/imath], so [imath]G[/imath] has at most [imath]2^k[/imath] elements. I think the key of the proof is to prove that if [imath]\phi(\sqrt{m_i})[/imath] is given, then there is always an automorphism [imath]\phi[/imath] of [imath]\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})[/imath] having such value of [imath]\phi(\sqrt{m_i})[/imath], and here is where I stuck. Does anyone have ideas? Any advice or comments will be helpful! If it is hard to give the answer here directly, then any bibliography for reference is also acceptable. |
1239524 | Prove [imath]\mathbb{P}(\sup_{t \geq 0} M_t > x \mid \mathcal{F}_0)= 1 \wedge \frac{M_0}{x}[/imath] for a martingale [imath](M_t)_{t \geq 0}[/imath]
Let [imath]M[/imath] be a positive, continuous martingale that converges a.s. to zero as [imath]t[/imath] tends to infinity. I now want to prove that for every [imath]x>0[/imath] [imath] P\left( \sup_{t \geq 0 } M_t > x \mid \mathcal{F}_0 \right) = 1 \wedge \frac{M_0}{x}. [/imath] My approach: I thought that rewriting the conditional probability to an expectation would help so we obtain that we must prove: [imath] \mathbb{E} \left[ 1_{\sup_{t \geq 0 } M_t > x} \mid \mathcal{F}_0 \right] = 1 \wedge \frac{M_0}{x}.[/imath] A hint to me was given that I should consider stopping the process when it gets above [imath]x[/imath]. Thus a stopping time that would do this is [imath]\tau = \inf\{t\geq 0 : M_t>x\}[/imath]. But now I'm stuck as I want to apply optional sampling results but we have an indicator which complicates things. How could I proceed from this? Any help is appreciated! | 375449 | Stuck on proof of a martingale equality (similar to doob's inequality)
The question is: Let [imath]M[/imath] be a positive, continuous martingale that converges a.s. to zero as [imath]t[/imath] tends to infinity. Prove that for every [imath]x > 0[/imath] [imath]P\{\sup_{t\geq 0} M_t > x \,| \,F_0\} = 1\wedge \frac{M_0}{x}[/imath] almost surely. From Doob's martingale inequality we may conclude that [imath]P\{\sup_{t\geq 0} M_t > x \,| \,F_0\} \leq 1\wedge \frac{M_0}{x}[/imath], but right now I'm stuck on showing that [imath]P\{\sup_{t\geq 0} M_t > x \,| \,F_0\} \geq 1\wedge \frac{M_0}{x}.[/imath] On this part I'm completely clueless right now. Any tips? P.S. We were given the tip "stop the martingale when it gets above [imath]x[/imath]", but I don't see what we can do with such a stopped martingale. |
1274087 | spectral theory of Laplacian on [imath]\mathbb R^n[/imath]
Can you describe the spectrum of the Laplacian [imath] \Delta : H^2(\mathbb R^n) \subset L^2(\mathbb R^n) \rightarrow L^2(\mathbb R^n)[/imath]? I am interested for which values [imath]z \in \mathbb C[/imath] the equation [imath]\Delta u + z u = f[/imath] has a solution [imath]u \in H^2(\mathbb R^n)[/imath] for every [imath]f \in L^2(\mathbb R^n)[/imath]. I think that the point spectrum is empty, but is every [imath]z[/imath] in the resolvent set? | 766479 | What is spectrum for Laplacian in [imath]\mathbb{R}^n[/imath]?
I know very well that Laplacian in bounded domain has a discrete spectrum. How about Laplacian in [imath]\mathbb{R}^n[/imath]?(not in some fancy-shaped unbounded domain, but the whole domain) Where can I find such results? Moreover, is there a counterpart of Hilbert-Schmidt theorem for Laplacian in [imath]\mathbb{R}^n[/imath]? Hilbert-Schmidt asserts there is a countable set of eigenfunctions [imath]\phi_n[/imath] so that [imath]x=\sum \langle x,\phi_n\rangle \phi_n,\forall x\in H[/imath]. Is there a similar theorem saying [imath]x=\int_0^\infty \langle x,\phi_\lambda\rangle \phi_\lambda\,\mathrm{d}\lambda,\forall x\in H[/imath] where [imath]\phi_\lambda[/imath] is the eigenfunction of Laplacian to spectral value [imath]\lambda[/imath]? |
1274800 | Suppose [imath]X[/imath] and [imath]Y[/imath] are independent Poisson random variables. Find the conditional probability mass function [imath]P(X=k\mid X+Y=m)[/imath]
Suppose [imath]X[/imath] and [imath]Y[/imath] are independent Poisson random variables with parameters [imath]\lambda[/imath] and [imath]\mu[/imath], respectively. Find the conditional probability mass function [imath]P(X=k\mid X+Y=n)[/imath]. Don't know how to approach this. Additionally, the question asks me to find [imath]E[X\mid X+Y=n][/imath]. What is the distinction? Does it have a different distribution? | 936828 | Poisson random variables and Binomial Theorem
I'm working on a problem from Casella and Berger's Statistical Inference. X is distributed as Poisson[imath](\theta)[/imath] and Y is distributed as Poisson[imath](\lambda)[/imath], with X and Y being independent. We let U = X + Y and V = Y, and the conditional pdf of V|U is: [imath]\hspace{15mm}\large f(v|u) = \large \frac{f(u,v)}{f(u)} = \huge \frac{\frac{\theta^{u-v}}{(u-v!)}*\frac{\lambda^ve^{-\lambda}}{v!}}{\frac{(\theta+\lambda)^ue^{-(\theta+\lambda)}}{u!}}[/imath] Apparently this simplifies to [imath]\hspace{15mm}\Large {u \choose v} (\frac{\lambda}{\theta+\lambda})^v(\frac{\theta}{\theta+\lambda})^{u-v}[/imath] I don't see how the [imath]\Large(\frac{\lambda}{\theta+\lambda})^v[/imath] is possible. I keep ending up with [imath]\Large(\frac{\lambda}{\theta})^v[/imath] when I simplify the problem. |
1274356 | Integration of [imath]\int \frac{e^x}{e^{2x} + 1}dx[/imath]
I came across this question and I was unable to solve it. I know a bit about integrating linear functions, but I don't know how to integrate when two functions are divided. Please explain. I'm new to calculus. Question: [imath]\int \frac{e^x}{e^{2x} + 1}dx[/imath] Thanks in advance. | 162238 | How to integrate [imath]\int \frac{e^x dx}{1\,+\,e^{2x}}[/imath]
Ok, I give up, I have tried with [imath]u[/imath]-substitution and integration by parts but I can't solve it. The integral is: [imath]\int{\frac{e^x dx}{1+e^{2x}}}[/imath] I have tried [imath]u=e^x[/imath], [imath]u=e^{2x}[/imath] and also integration by parts but I can't solve it. The result should be: [imath]\arctan(e^x)[/imath] |
1275325 | Prove that the complement of an open ball in [imath]\mathbb{R^n}[/imath] has exactly one unbounded component
Question: Let [imath]B^n \subset \mathbb{R^n}[/imath] be open ball in the Euclidean metric. Prove that the complement of [imath]B^n[/imath] in [imath]\mathbb{R^n}[/imath] has exactly one unbounded component (components of a set are class partitions defining largest connected sets...). This is an exercise of the book 'C. Adam - Topology'. Obviously, it does not hold for [imath]n=1[/imath], since [imath]\mathbb{R}-(-a,a)[/imath] is disconnected and is made of TWO unbounded components. But for the case of [imath]n=2,3[/imath] it is easy to prove that the conjecture holds (each 'side' of the open balls is homeomorphic to [imath]\mathbb{R^n}[/imath] and has non-empty intersection with its neighbour-side...) How to elevate the conjecture for the case of [imath]n\ge 4[/imath], since it is not possible by intuition? Thank you. EDIT - This is not dublicate, since my question is about complement of an open ball not a bounded set in general. I read here before I wrote my question; the answer doesn't prove [imath]\mathbb{R^n}−B^n[/imath] is connected, which I need to prove. | 793562 | Is the complement of an open ball in a Banach space connected?
Let [imath]B[/imath] be a real Banach Space whose dimension is at least [imath]2[/imath], and let [imath]S[/imath] be a subset of [imath]B[/imath] that is an open ball. Is the complement of [imath]S[/imath] (with respect to [imath]B[/imath]) always connected? Idea One could then perhaps make use of the facts that every [imath]2[/imath]-dimensional Banach space is homeomorphic to the Euclidean plane and that - for the Euclidean plane - my question has an obvious "YES" answer (since every open ball is the interior of a circle). |
1275861 | Let [imath]R[/imath] a left artinian ring. Show that if [imath]a \in R[/imath] is not a right zero divisor, then [imath]a[/imath] is a unit in [imath]R[/imath].
Let [imath]R[/imath] a left artinian ring. Show that if [imath]a \in R[/imath] is not a right zero divisor, then [imath]a[/imath] is a unit in [imath]R[/imath]. Comments: I am tryed to do so: See the R-module [imath]_{R}R[/imath] and consider the function [imath]f: _{R}R \longrightarrow _{R}R[/imath] defined by [imath]f(r)= ra[/imath]. As [imath]R[/imath] is artinian then [imath]_{R}R[/imath] has a composition series. I do not know if it's out there the way. | 845488 | Associative ring with identity, inverses, divisors of zero and Artinianity
How to prove the following? [imath]R[/imath] is an associative ring with identity. [imath]R[/imath] contains element [imath]r[/imath]. The element is not invertible on the right and is not a left divisor of zero. Then the ring [imath]R[/imath] cannot be Artinian on the right. |
1275581 | Prove that every prime ideal of [imath]R[/imath] is maximal.
Let [imath]R[/imath] be a commutative ring with identity.For each [imath]a\in R[/imath] there exist [imath]n(>1)\in \mathbb N[/imath] such that [imath]a^n=a[/imath]. Prove that every prime ideal of [imath]R[/imath] is maximal. My try Let [imath]I[/imath] be a prime ideal of [imath]R[/imath]. Then [imath]R/I[/imath] is an integral domain. If I can show using the hypothesis that [imath]R/I[/imath] is a field then we are done. Any hints to show this? | 210940 | Show that any prime ideal from such a ring is maximal.
Let R be a commutative ring with an identity such that for all [imath]r\in[/imath] R, there exists some [imath]n>1[/imath] such that [imath]r^n = r[/imath]. Show that any prime ideal is maximal. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 1, Exercise 7.) Any hints? |
1276620 | A sequence of functions where the integral converges to 0, but the sequence does not
Is there a sequence of functions where [imath]\int_0^1|f_n(x)|->0[/imath] as n approaches infinity, but the sequence of functions is also pointwise divergent over every x in [imath][0,1][/imath]? Initially I thought cos(nx) could be an answer but that is not pointwise divergent at x=0. My other solutions also did not satisfy the integral due to the absolute value. | 1276515 | Analysis Constructing a Sequence
I'm looking for a sequence of functions that is continuous and absolutely integrable, but pointwise divergent for every [imath]z[/imath] [imath]\in [0,1][/imath]. In other words, [imath] \int_0^1 |f_n(z)| dz \rightarrow 0[/imath] as [imath]n \rightarrow \infty[/imath], but [imath](f_n(z))_n[/imath] pointwise diverges for every [imath]z[/imath] [imath]\in [0,1][/imath]. I'm having no luck coming up with an example, and would appreciate any sort of insight on how to go about constructing such a sequence. |
1276212 | dimension of a direct sum is the sum of the separate dimensions
Suppose [imath]U[/imath] is a finite dimensional vector space and [imath]U[/imath] can be expressed as a direct sum [imath]U=U_1\oplus\dots\oplus U_n[/imath]. Then [imath]\dim(U) = \dim(U_1) + \dots + \dim(U_n)[/imath]. I don't really have any idea how to prove this, any tips? | 561481 | Proof that $V = \sum_k U_k$ and $\dim V=\sum_k \dim U_k$ implies $V=\oplus_k U_k$
Question regarding "Linear algebra done right", proposition [imath]2.19[/imath]: Proposition [imath]2.19[/imath]: Suppose [imath]V[/imath] is finite dimensional and [imath]U_1,...,U_m[/imath] are subspaces of [imath]V[/imath] such that [imath]V = U_1 + \dots + U_m\tag{2.20}[/imath] and [imath]\dim V=\dim \, U_1 + \dots + \dim \, U_m.\tag{2.21}[/imath] Then [imath]V = U_1 \oplus \dots \oplus U_m.[/imath] Proof: Choose a basis for each [imath]U_j[/imath]. Put these bases together in one list, forming a list that spans [imath]V[/imath] (by [imath]2.20[/imath]) and has length [imath]\dim V[/imath] (by [imath]2.21[/imath]). Thus this list is a basis of [imath]V[/imath] (by [imath]2.16[/imath]), and in particular it is linearly independent. Now suppose that [imath]u_1 \in U_1,...,u_m \in U_m[/imath] are such that [imath]0=u_1+ ··· +u_m[/imath]. We can write each [imath]u_j[/imath] as a linear combination of the basis vectors (chosen above) of [imath]U_j[/imath]. Substituting these linear combinations into the expression above, we have written [imath]0[/imath] as a linear combination of the basis of [imath]V[/imath] constructed above. Thus all the scalars used in this linear combination must be [imath]0[/imath]. My question Given [imath](v_1, \ldots, v_n)[/imath] as the basis vectors: \begin{align} u_1 &= a_{11}v_1 + \ldots + a_{1n}v_n \\ & \ \vdots \\ u_m &= a_{m1}v_1 + \ldots + a_{mn}v_n. \end{align} It seems the proof only proves [imath]0 = (a_{11} + \ldots + a_{m1}) = \ldots = (a_{1n} + \ldots + a_{mn})[/imath]. But how do we know that each [imath]a_{ij}[/imath] is [imath]0[/imath] too? Thanks. |
1276370 | conditional probability of joints in bayesian net
I have been staring at a bayesian net for an hour and can't understand why this is correct to write: [imath]P(W|B,E)\cdot P(E)\cdot P(R|E)= P(W,R,E|B)[/imath] Note that the joint probability of [imath]P(A,B,E,W,R)[/imath] can be decomposed as follows according to the bayesian net structure: [imath]P(A,B,E,W,R) = P(B)\cdot P(E)\cdot P(A|B,E)\cdot P(R|E)\cdot P(A|B,E)\cdot P(W|A)[/imath] | 1275702 | conditional probability of joints
I have been staring at a bayesian net for an hour and can't understand why this is correct to write: [imath]P(A|B,E)\cdot P(W|A) = P(W,A|B,E)[/imath] Note that the joint probability of [imath]P(A,B,E,W,R)[/imath] can be decomposed as follows according to the bayesian net structure: [imath]P(A,B,E,W,R) = P(B)\cdot P(E)\cdot P(A|B,E)\cdot P(R|E)\cdot P(A|B,E)\cdot P(W|A)[/imath] |
1277462 | [imath](x+y)^a \geq x^a + y^a[/imath]
Take [imath]x,y \in (0,\infty), a \in (1, \infty)[/imath]. For [imath]a \in \mathbb N [/imath] by binomial theorem it holds [imath](x+y)^a \geq x^a + y^a.[/imath] Trying with some numbers it seems true also for [imath]a \in (1, \infty)[/imath]. How can I prove it? | 1275631 | A possible inequality related to binomial theorem (or, convex/concave functions)
Let [imath]x, \ y, \ p[/imath] be any real numbers with [imath]x>0[/imath], [imath]y>0[/imath], and [imath]p>1[/imath]. The question is about (most probably) an elementary inequality: Is it always true that [imath]x^p+y^p\leq (x+y)^p[/imath] ? Note that if [imath]p[/imath] is any positive integer, then the above inequality is obviously correct. What about if the number [imath]p \ (\text{with} \ p>1)[/imath] is any non-integer real number? I guess that (by my intuition) the answer should be positive. But how can we proceed to prove this inequality? |
1276973 | How to show [imath]c_n=\frac11 + \frac12 + \cdots + \frac1n - \ln n[/imath] is a sequence of positive numbers?
For [imath]n \in \mathbb{N}[/imath] let [imath]c_{n}[/imath] be defined by [imath]c_{n}=\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n[/imath] We have to prove that [imath]c_{n}[/imath] is a decreasing sequence of positive numbers. I've already shown the first part, that it is a decreasing sequence by considering the difference of [imath]c_{n+1}-c_{n} = \ln \left(1- \frac{1}{n+1} \right) +\frac{1}{n+1} [/imath] and then using the expansion of [imath]\ln (1-x)[/imath] for [imath]-1\leq x \leq 1[/imath]. But I'm having some trouble in showing the second part that all terms in the sequence are positive. I tried using first form of induction but but stuck in the inductive step, can somebody please suggest explain that to me? or better suggest some other way to prove that part? Any sort of welcome as log as it leads to the solution, thanks in advance. | 701050 | Showing that [imath]\sum_{i=1}^n \frac{1}{i} \geq \log{n}[/imath]
I have been trying to prove this by induction on [imath]n\in \mathbb{N}[/imath], but this approach seemed to get me nowhere. I have a suspicion it might be necessary to express [imath]\log{n}[/imath] as [imath]\int_1^n 1/x\text{ }dx[/imath], but I could not build a rigorous argument from that. Any help greatly appreciated! |
1276775 | Ideals in [imath]\mathbb{Z}[X][/imath] with three generators (and not with two)
It is well-known that in [imath]\mathbb{Z}[X][/imath] we do have non-principal ideals, for example [imath](2,x)[/imath]. This is an ideal with two generators. Now I was wondering if there exists an ideal with three generators, which cannot be generated by two elements. (And of course, if so, if we can find ideals with [imath]n[/imath] generators which cannot be generated by [imath]n-1[/imath] elements). I do have one suggestion: [imath](8, 4x, 2x^2)[/imath], found by trial-and-error. My question is twofold: Is this an ideal as described above? Is there a more constructive way to think about this question? | 110458 | Construct ideals in [imath]\mathbb Z[x][/imath] with a given least number of generators
How do you construct, for each [imath]n\geq 1[/imath], an ideal in [imath]\mathbb Z[x][/imath] of the form [imath](a_1,a_2,\dots,a_n)[/imath] with [imath]a_i\in \mathbb Z[x][/imath] such that it is impossible to have [imath](b_1,b_2,\dots,b_m)=(a_1,a_2,\dots,a_n)[/imath] with [imath]m<n[/imath] and [imath]b_j\in\mathbb Z[x][/imath]? Thank you! |
1277033 | Divergence of the sum of some elements of some uncountable set
I came up with the following conjecture, but I'm not sure how to prove it or whether it is true at all. Conjecture: let some subset of the real numbers [imath]S \subset \mathbb{R}[/imath] have an uncountable amount of elements, all of which are strictly positive, then it is possible to construct some countable subset of this set that diverges when summed. My intuition says that this is probably true since when you create such an uncountable subset of the real numbers, you somehow have to smuggle in a range of real numbers or some set that is already divergent itself (like the irrationals). However, this is nowhere near a proof and I'm not sure whether I have even learned about the tools to rigorously prove this conjecture (or disprove). Therefore I would greatly appreciate it if someone could point me in the right direction. Thanks in advance for your time and trouble. | 359669 | arbitrary large finite sums of an uncountable set.
How to prove that given an uncountable subset [imath]A[/imath] of [imath]R^+[/imath], there exits a finite number of elements in A whose sum is arbitrarily large? |
1278229 | Given [imath]z_n[/imath] = {[imath] x_{1},y_1,x_2,y_2,x_3,y_3 ...[/imath]} Prove that [imath]z_n[/imath] is convergent if [imath]x_n[/imath] and [imath]y_n[/imath] both converge to same limit
I f [imath]x_n[/imath] and [imath]y_n[/imath] be the two sequences such that [imath]z_n[/imath] = {[imath] x_{1},y_1,x_2,y_2,x_3,y_3 ...[/imath]} Prove that [imath]z_n[/imath] is convergent if [imath]x_n[/imath] and [imath]y_n[/imath] both converge to same limit ATTEMPT Let us take that given [imath]x_n[/imath] and [imath]y_n[/imath] converge to different limits [imath]\ell[/imath] and [imath]\ell '[/imath]; with [imath]\ell < \ell'[/imath] . After certain stage [imath]n>n_0[/imath] all elements of [imath]x_n[/imath] lie in [imath](x-\epsilon,x+\epsilon)[/imath] and after [imath]n>n_1[/imath] all elements of [imath]y_n[/imath] lies in interval [imath](y-\epsilon,y+\epsilon)[/imath]. Let us take [imath]N = \max \{n_0 , n_1\}[/imath]. So after that stage elements of sequence [imath]z_n[/imath] lies in both of intervals which is contradiction to our assumption. Is this fine? Isn't this theorem a new way of asking for whether a sequences converge to same limit? Thanks | 1275913 | If [imath]x_n \to a[/imath] and [imath]x'_n \to a[/imath], then [imath]\{x_1, x'_1, x_2, x'_2, ...\} \to a[/imath]
I know that if all subsequences of [imath]\{x_1, x'_1, x_2, x'_2, ...\}[/imath] converge to [imath]a[/imath], then [imath]\{x_1, x'_1, x_2, x'_2, ...\}[/imath] converges to [imath]a[/imath], but I only know two subsequences of [imath]\{x_1, x'_1, x_2, x'_2, ...\}[/imath] that converge to [imath]a[/imath], namely [imath]x_n[/imath] and [imath]x'_n[/imath]. How do I show that there don't exist any other subsequences that converge to something other than [imath]a[/imath]? |
1278127 | Compact Sets in [imath]\mathbb{R^{n^2}}[/imath]
I have a question of multivariable analysis and I don't know how to resolve this. The [imath]n \times n[/imath] orthogonal matrices form a compact subset of [imath]\mathbb{R^{n^2}}[/imath]? I will be very grateful for the help. | 18653 | Compactness of the set of [imath]n \times n[/imath] orthogonal matrices
Show that the set of all orthogonal matrices in the set of all [imath]n \times n[/imath] matrices endowed with any norm topology is compact. |
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