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1234529 | Is GRP a subcategory of SET, or not?
This is the notion of a subcategory [imath]\mathscr{D}[/imath] of a given category [imath]\mathscr{C}[/imath] which I use: it consists of a subcollection of the collection of objects of [imath]\mathscr{C}[/imath] and a subcollection of the collection of morphisms of [imath]\mathscr{D}[/imath] such that: 1) if the morphism [imath]f:x\longrightarrow y[/imath] is in [imath]\mathscr{D}[/imath], then so are [imath]x[/imath] and [imath]y[/imath] 2) if [imath]f:x\longrightarrow y[/imath] and [imath]g:y\longrightarrow z[/imath] are in [imath]\mathscr{D}[/imath], so is the composite [imath]gf:x\longrightarrow z[/imath] 3) if [imath]x[/imath] is in [imath]\mathscr{D}[/imath], then so is the identity morphism [imath]1_X[/imath] Following this definition, the category of groups should be a subcategory of the category of sets. But I have just read on "Universal Algebra" (author Cohn) that: "...it [the category of groups] is not a subcategory [of sets] because distinct groups may have the same carrier." So how can I decide if it is a subcategory or not? | 495447 | Category of Grp is not a subcategory of Set
I am trying to read the book Rings and categories of modules by Anderson and Fuller. On page 7 it is stated that the category of Groups is not a subcategory of a category of Sets, some explanations are given there. Can any one please give me some more details about? especially why [imath]mor_G((G,\circ),(H,\circ))\not\subset Map((G,\circ),(H,\circ))[/imath] ? |
1237520 | If [imath]a, b[/imath] are transcendental then [imath]a+b[/imath] is transcendental or [imath]ab[/imath] is transcendental
I have to prove the following: If [imath]a, b[/imath] are transcendental then [imath]a+b[/imath] is transcendental or [imath]ab[/imath] is transcendental, or both. I don't have any idea on how to solve this. I already proved this: If [imath]z \in \mathbb{C}[/imath], [imath]a_1, a_2[/imath] are algebraic then: [imath]z^2 + a_1z +a_2 = 0[/imath] [imath]\Rightarrow[/imath] [imath]z[/imath] is algebraic But I don't know if this could help me. Could you help me with this problem? Thanks in advance | 779162 | Sum and product of two transcendental numbers can't be both algebraic
Suppose [imath]a[/imath] and [imath]b[/imath] are complex numbers and both transcendental over [imath]\mathbb Q[/imath]. I am wondering why [imath]ab[/imath] and [imath]a+b[/imath] can not both be algebraic. Thanks for any help. |
1237786 | Prove that [imath]\mathbb{Z}_{mn}[/imath] has atleast four idempotent elements.
Suppose [imath]m, n > 1[/imath] are positive integers which are relatively prime. Prove that [imath]\mathbb{Z}_{mn}[/imath] has atleast four idempotent elements. Two of them are [imath][0], [1][/imath], how will I find the other two? | 1064781 | [imath]m,n>1[/imath] are relatively prime integers , then are there at-least four idempotent (w.r.t. multiplication) elements in [imath]\mathbb Z_{mn}[/imath] ?
If [imath]m,n>1[/imath] are integers such that [imath]g.c.d.(m,n)=1[/imath] then is it true that there are at-least four elements in [imath]\mathbb Z_{mn}[/imath] such that [imath]x^2=x[/imath] ( i.e. idempotent ) ? |
397520 | how prove [imath]\phi(n)\ge \frac{n}{6\log \log (n)} [/imath] [imath]\forall n\ge5 [/imath]
How to prove[imath]\forall n\ge5 [/imath] [imath]\phi(n)\ge \frac{n}{6\log \log (n)} [/imath] [imath]\phi[/imath] is Euler function Thanks in advance | 101654 | Lower bound for [imath]\phi(n)[/imath]: Is [imath]n/5 < \phi (n) < n[/imath] for all [imath]n > 1[/imath]?
Is it true that : [imath]\frac {n}{5} < \phi (n) < n[/imath] for all [imath]n > 1[/imath] where [imath]\phi (n)[/imath] is Euler's totient function . Since [imath]\phi(n)[/imath] has maximum value when [imath]n[/imath] is a prime it follows that maximum value of [imath]\phi(n)[/imath] in term of [imath]n[/imath] is [imath]n-1[/imath] , therefore [imath]\phi(n)< n[/imath] for all [imath]n[/imath]. What is the best lower bound for [imath]\phi(n)[/imath]? |
1237936 | Axiom of glueing: direct limit of sheaves in a noetherian topological space.
I'm trying to prove that in a noetherian topological space the following property is satisfied: Consider a direct system of sheaves and morphisms [imath]\{ \cal{F}_t, f_{ij} \}_t[/imath]. Consider the presheaf given by [imath] U \mapsto \varinjlim \cal{F}_t (U)[/imath] I want to prove that this presheaf is a sheaf. I managed to prove that the first axiom os sheaves is satisfied, using the idea given in the comment of my previous question (First axiom of sheaves: in noetherian topological spaces the direct limit presheaf is a sheaf.). I thought that the second axiom would be similar, but when I tried to write it using similar argument I got stuck. These are my attemps and my ideas: Let [imath]U[/imath] an open set and [imath]\{ V_j \}_j[/imath] an open covering of [imath]U[/imath]. Being [imath]X[/imath] noetherian it is compact, so we can suppose [imath]\{ V_j \}_j = \{ V_1, \dots , V_n \}[/imath]. For each [imath]j[/imath] consider [imath]s_j \in \varinjlim \cal{F}_t (V_j)[/imath]. Suppose they have the glueing property: [imath]{s_i}_{|V_i \cup V_j}={s_j}_{|V_i \cup V_j} \quad \forall i, j[/imath]. We want to show that there exists [imath]s \in \varinjlim \cal{F}_t (U)[/imath] such that [imath]s_{|V_j}=s_j[/imath]. For each [imath]j \quad \exists k_j, s'_{k_j} \in \cal{F}_{k_j} \quad s.t. f_{k_j}(V_j)(s'_{k_j})=s_{k_j}[/imath]. We have [imath]{s_i}_{|V_i \cup V_j}={s_j}_{|V_i \cup V_j}[/imath] so [imath]f_{k_i}(V_i \cup V_j)({s'_{k_i}}_{V_i \cup V_j})=f_{k_j}(V_i \cup V_j)({s'_{k_j}}_{V_i \cup V_j})[/imath]. If I could define [imath]s''_j \in \cal{F}_n(V_j)[/imath] such that [imath]s_j=f_n(V_j)(s''_j)[/imath] and verifying the glueing property, then I could work with the [imath]s''_j[/imath] and the [imath]s_j[/imath] and prove the glueing property for the [imath]s_j[/imath]. EDIT: The difference between the proof of the second axiom of sheaves and the first one is the following: in the first axiom one can use that, if [imath]f_i[/imath] is a morphism of the direct limit such that [imath]f_i(x_i)=0[/imath], then there exists some [imath]j \geq i[/imath] such that [imath]f_{ij}(x_i)=0[/imath], being [imath]f_{ij}[/imath] a morphism of the direct system. This property allows to put together the information of the representative elements of the original [imath]s_{V_i}[/imath], using that the original cover of the open set [imath]U[/imath] can be considered finite, since [imath]X[/imath] is compact. | 1234545 | First axiom of sheaves: in noetherian topological spaces the direct limit presheaf is a sheaf.
Consider a topological space [imath]X[/imath] and a direct limit of sheaves and morphisms [imath]\{ \cal{F}_i, f_{ij}\}[/imath]. Define the direct limit presheaf by [imath]U \to \varinjlim \cal{F}_i [/imath]. In general this is just a presheaf. If [imath]X[/imath] is noetherian it is a sheaf, and this is the result I want to prove. My attemp: I tried to prove the first axiom of sheaves (I suspect that the second will need similar arguments). Let be an open set [imath]U[/imath] and an open cover [imath]\{ V_i \}_i[/imath] of [imath]U[/imath]. Since [imath]X[/imath] is noetherian it can be can supposed that the cover is finite. Let [imath]s \in \varinjlim \cal{F_i(U)}[/imath] such that [imath]s|_{V_i}=0 \quad \forall i[/imath]. The goal is to show that then [imath]s=0[/imath]. Being [imath]s \in \varinjlim \cal{F_i(U)}[/imath], we know that there exists some [imath]j / \quad s=f_j(U)(t_j)[/imath] for some [imath]t_j \in \cal{F}_j(U)[/imath] ([imath]f_i(U)[/imath] are the morphisms of the direct limit of the abealian groups [imath]\cal{F}_i(U)[/imath] ). I can consider the germs [imath]{t_j}_{V_i}[/imath]. If I could show that [imath]s|_{V_i}=0 \Rightarrow {t_j}_{V_i}=0[/imath] then the problem would be done, because the [imath]\cal{F}_i[/imath] are sheaves, in particular they verify the first axiom of sheaves, so [imath]t_j=0[/imath] and then [imath]s=f_j(t_j)=0[/imath]. |
1238307 | Prove [imath] \sum \frac{cos(n)} { \sqrt n}[/imath] is bounded
Prove [imath] \sum \frac{\cos(n)} { \sqrt n}[/imath] is bounded. How can I prove [imath]\sum \cos(n)[/imath] is bounded. Can It be done using [imath]\cos[/imath] as real part of complex number as I showed in question ? | 1222484 | Prove [imath] \sum \frac{\cos n} { \sqrt n}[/imath] converges
How to Prove [imath] \sum \frac{\cos n} { \sqrt n}[/imath] converges Using Abel’s theorem ? I think it can be done using [imath]\cos n = Re(e^{in})[/imath] { Real Part of Complex Number } How to proceed ? |
1238554 | Existence of a closed and open set in 0-dimensional Hausdorff space
Given [imath]T = 0-[/imath]dimensional (i.e, T has a basis of sets which are both open and closed), Lindelof Hausdorff space, with closed subsets [imath]A[/imath] and [imath]B[/imath] such that [imath]A\cap B = \emptyset[/imath]. Prove that [imath]\exists[/imath] a closed and open set (clopen) set [imath]D[/imath] with [imath]A\subset D[/imath] and [imath]B\cap D =\emptyset[/imath]. I have thought of this problem for a while, but haven't been able to make any good progress. Can someone please give some detailed help, if possible, on this tricky problem? | 1233527 | Disjoint closed sets in a second countable zero-dimensional space can be separated by a clopen set
I want to prove the following: Let [imath]X[/imath] be second countable zero-dimensional space. If [imath]A,B \subseteq X[/imath] are disjoint closed sets, there exist is a clopen set [imath]C[/imath] such that [imath]A\subseteq C[/imath] and [imath]B\cap C = \emptyset [/imath]. (A topological space [imath]X[/imath] is zero-dimensional if it is Hausdorff and has basis consisting of clopen sets.) |
1238434 | If two subsets [imath]S,T\subseteq G[/imath] have sum of cardinalities greater than [imath]|G|[/imath], then [imath]S+T=G[/imath]
Let [imath]S[/imath] and [imath]T[/imath] are two subset of a finite group [imath](G,+)[/imath] so that [imath]|S|+|T|>|G|[/imath], then Prove that [imath]S+T=G[/imath], where [imath]S+T=\{s+t:s\in S ,t\in T\}[/imath] My effort: It is clear that [imath]S+T\subseteq G[/imath] as elements of [imath]S[/imath] and [imath]T[/imath] are elements of [imath]G[/imath] also and [imath]G[/imath] is closed under addition.Also [imath]|S+T|<|S||T|[/imath]. Then how to show elements of [imath]G[/imath] also in [imath]S+T[/imath]? Does the condition [imath]|S|+|T|>|G|[/imath] say it? I don't understand? | 1091365 | If [imath]G[/imath] is a finite group and [imath]|G| < |A| + |B|[/imath], then [imath]G=AB[/imath].
Let [imath]G[/imath] be a finite group. Suppose that [imath]A[/imath] and [imath]B[/imath] are to subsets of [imath]G[/imath]. If [imath]|G|<|A|+|B|[/imath] prove that [imath]G=AB.[/imath] |
1238238 | Sum of digits of [imath]11\dots 11^2[/imath] where [imath]11\dots 11[/imath] is a 1992 digit number with all digits [imath]1[/imath]
I read this on a non-math forum where the OP says this is a question for Grade 6 elementary school students. Grade 6 elementary school level is somehow ambiguous but clearly this means no advanced math tool can be used. (Maybe some elementary modulo arithmetic is allowed?) I tried in the most dumb way: Since we know that [imath]11\dots 1^2[/imath] of [imath]n\leq 9[/imath] digits of [imath]1[/imath]'s is [imath]123\dots n \dots 321[/imath] because [imath]11\dots 1^2 = 11\dots 1 \times \sum^{n-1}_{m=0}10^m[/imath] (which explains why the answer is [imath]1[/imath] goes to [imath]n[/imath] then goes back to [imath]1[/imath]). So we can say [imath]11\dots 11^2[/imath] is to add up (digit number) of [imath]1[/imath], put it on that digit, and sum them up. Then we apply on [imath]n>9[/imath] and observe the process. This seems promising or at least manageable. You count the number of [imath]1[/imath]'s that are involved in computing number on the digit, and add carries from lower digits. For example, on the 102nd digits this will be [imath]102+11+1=114[/imath], so the carry is [imath]11[/imath], number of 102nd digit is [imath]4[/imath]. Finally sum all digits up. The work involved seems way too immense and is not beautiful. Anyone has some clever ideas about this question? (The result is [imath]17910[/imath] and the original post is in Chinese, so I don't want to put the link here) | 112932 | Sum of digits of repunits
I have a very interesting number theory problem. Let [imath] S_n [/imath] be a number consisting of only [imath]n[/imath] ones. For instance, [imath]S_1=1\\S_2=11 \\ S_4=1111[/imath] The problem is to prove that the sum of the digits of [imath]S^2_n[/imath] can be calculated using the formula [imath]81\cdot \left( \left\lfloor \frac{n}{9} \right\rfloor + \left( \frac{n}{9} - \left\lfloor \frac{n}{9} \right\rfloor \right)^2 \right)[/imath] I would be very grateful for help. I don't even know how to start... |
528753 | An example of a bounded pseudo Cauchy sequence that diverges?
Harmonic series diverges and pseudo Cauchy however it's not bounded. So how can I find such a sequence? A sequence [imath](s_n)[/imath] is pseudo-Cauchy if, for all [imath]\xi>0[/imath], there exists an [imath]N[/imath] such that if [imath]n ≥ N[/imath], then [imath]|s_{n+1}−s_n| < ξ[/imath]. | 768404 | Boundedness and Cauchy Sequence: Is a bounded sequence such that [imath]\lim(a_{n+1}-a_n)=0[/imath] necessarily Cauchy?
If I have a sequence {[imath]a_n[/imath]} that has the property of [imath]\lim(a_{n+1}-a_n)=0[/imath], does that make it a Cauchy Sequence. I think it doesn't because [imath]a_n = \sum_{k=1}^n \frac{1}{k}[/imath] is a counter example. However, by definition, there exists a [imath]M[/imath] such that if [imath]n \geq M[/imath] then [imath]|a_{n+1}-a_n| < \frac{\epsilon}{m-n}[/imath] Hence, we have [imath]|a_m - a_{m-1}|+.....+|a_{n+1}-a_n|<|a_m -a_n| <\epsilon[/imath] This proof doesn't work because I cannot be sure I can find a fixed [imath]M[/imath] which might change according to n. But I wonder if I have an additional condition that says [imath]a_n[/imath] is bounded, I think then the proof works and that I should be able to find a fixed [imath]M[/imath]. However, I don't know how to justify this. Maybe I am wrong. Can someone kindly help me figure out this problem. Thanks |
1239090 | prove that [imath]x^2 + 5 =y^3[/imath] has no solutions for [imath]x,\ y \in \mathbb{Z}[/imath]
So the question is completely stated by the title. My own thoughts: I can prove that [imath]x^2 + 1 = y^3[/imath] has no solutions for [imath]x,y \in \mathbb{Z}[/imath] by using the factorization: [imath] y^3 = (x-i)(x+i) [/imath] in [imath]\mathbb{Z}[i][/imath], using the fact that [imath]\mathbb{Z}[i][/imath] is a UFD and that [imath](x-i)[/imath] and [imath](x+i)[/imath] are coprime we obtain that they are both cubes and it follows that this is impossible. Now I would like to apply similar reasoning for this equation, we factor it as follows: [imath] y^3 = (x-\sqrt{-5})(x+\sqrt{-5}) [/imath] in [imath]\mathbb{Z}[\sqrt{-5}][/imath] but this time [imath]\mathbb{Z}[\sqrt{-5}][/imath] is not a UFD. However, since [imath]\mathbb{Q}(\sqrt{-5})/\mathbb{Q}[/imath] is a finite seperable field extension we find that the integral closure of [imath]\mathbb{Z}[/imath] in [imath]\mathbb{Q}(\sqrt{-5})[/imath] is a Dedekind domain, this integral closure is exactly our ring [imath]\mathbb{Z}[\sqrt{-5}][/imath] (which follows from the fact that [imath]-5 = 3 \neq 1 \text{ mod } 4[/imath]). In a Dedekind domain we know that we write ideals uniquely as a product of maximal ideals, we can apply this to the ideal [imath](y)[/imath], [imath](x-\sqrt{-5})[/imath] and [imath](x+\sqrt{-5})[/imath] for this to be useful we must first prove that the ideals [imath](x-\sqrt{-5})[/imath] and [imath](x+\sqrt{-5})[/imath] are coprime, for this it suffices to show that there is no prime ideal [imath]P \subseteq \mathbb{Z}[\sqrt{-5}][/imath] that contains [imath](x-\sqrt{-5})[/imath] and [imath](x+\sqrt{-5})[/imath]. Let [imath]P[/imath] be a prime ideal s.t. [imath]x-\sqrt{-5},x+\sqrt{-5} \in P[/imath], then it follows that also: [imath] 2\sqrt{-5} = x+\sqrt{-5} - (x-\sqrt{-5}) \in P [/imath] and thus by the given equality: [imath] -20 = (2\sqrt{-5})^2 \in P^2 \subseteq (y)^3 [/imath] so we find an element [imath]a+b \sqrt{-5} \in \mathbb{Z}[\sqrt{-5}][/imath] s.t. [imath]-20 = (a+b\sqrt{-5}) \cdot y^3[/imath] using the norm: [imath] N(r + s \sqrt{-5}) := r^2 +s^2 5 [/imath] we obtain from this equality thath: [imath] 400 = (a^2 + 5b^2) \cdot y^6 [/imath] but as [imath]400 = 2^4 \cdot 5^2[/imath] we find that [imath]y[/imath] must be equal to [imath]1[/imath] and thus we find that: [imath] x^2 + 5 = 1 [/imath] which is clearly impossible as for [imath]x \in \mathbb{Z}[/imath] we have [imath]x^2 \geq 0[/imath]. From this we can conclude that [imath](x-\sqrt{-5})[/imath] and [imath](x+\sqrt{-5})[/imath] are indeed coprime so their decomposition in a product of maximal ideals is disjoint (i.e. we find maximal ideals [imath]\mathfrak{p}_1,\dots,\mathfrak{p}_r[/imath], [imath]\mathfrak{q}_1,\dots,\mathfrak{q}_s[/imath] and natural numbers [imath]e_1,\dots,e_r,f_1,\dots,f_s[/imath] s.t.: [imath](x-\sqrt{-5}) = \mathfrak{p}_1^{e_1}\dots\mathfrak{p}_r^{e_r}, \qquad (x+\sqrt{-5}) = \mathfrak{q}_1^{f_1}\dots\mathfrak{q}_r^{f_r}[/imath] where all [imath]\mathfrak{p}_i[/imath] and [imath]\mathfrak{q}_j[/imath] are different. For [imath](y)[/imath] we also find such a decomposition, which implies that the decompositions of [imath](x-\sqrt{-5})[/imath] and [imath](x+\sqrt{-5})[/imath] must be of the form: [imath] (x-\sqrt{-5}) = \mathfrak{p}_1^{3e_1}\dots\mathfrak{p}_r^{3e_r}, \qquad (x+\sqrt{-5}) = \mathfrak{q}_1^{3f_1}\dots\mathfrak{q}_r^{3f_r} [/imath] which implies that there is some ideal [imath]I[/imath] in [imath]\mathbb{Z}[\sqrt{-5}][/imath] s.t. [imath](x+\sqrt{-5}) = I^3[/imath] so we find elements of [imath]I[/imath]: [imath]a_i+ b_i \sqrt{-5} \in I[/imath] for which: [imath] x+\sqrt{-5} = \prod_{i=1}^3(a_i+ b_i \sqrt{-5}) [/imath] and then I would like that this is impossible, but at this point I'm stuck. I'm sorry if my own idea is absolutaly worthless but this is the first time I try to solve this kind of exercise. | 1161843 | [imath]x^3-9=y^2[/imath] find integral solutions
Find all integral solutions [imath]x^3-9=y^2[/imath] I tried many times but still no idea how to solve it. I will be grateful for any help. |
1239281 | If a matrix A square is 0, does it follow that A = 0?
Let A be a square matrix. If [imath]A^2 = 0[/imath], then it follows that [imath]A = 0[/imath]. Is there a counterexample for this? If there isn't, what kinds of explanation can I make to justify this statement? | 1236230 | If [imath]A[/imath] is a square matrix and [imath]A^2 = 0[/imath] then [imath]A=0[/imath]. Is this true? If not, provide a counter-example.
This is a proof question and I am not sure how to prove it. It is obviously true if you start with [imath]A = 0[/imath] and square it. I was thinking: If [imath] A^2 = 0 [/imath] then [imath] A A = 0 [/imath] [imath] A A A^{-1} = 0 A^{-1}[/imath] [imath]I\,A = 0 [/imath] but the zero matrix is not invertible and that it was not among the given conditions. Where's a good place to start? |
1239500 | Prove or disprove: If [imath]\lim_{n\to\infty} (a_{2n} - a_n) =0[/imath], then [imath]a_n[/imath] has a limit (not infinity).
I need to prove if this is true or false.If true then I need to Prove and if false I need to provide an example that disproves the statement.I tried many times but it didnt work. Note:- Im new so I dont know how to write in the math format so if you know how to please tell me | 1237946 | If [imath](y_{2n}-y_n) \to 0[/imath] then [imath]\lim_{n\to \infty} y_n[/imath] exists
Assume [imath]\lim_{n\to \infty} (y_{2n} - y_n)=0[/imath] then [imath]\lim_{n\to \infty} y_n[/imath] exists. I know it's not true, and I can see a sequence that disprove that [imath](1,1,2,1,3,2,4,1,5,3,...)[/imath] but I want a sequence that has formula. hint/help? thanks |
1239459 | Rank of matrices, prove inequality
Today I'm having hard time with linear algebra problems; this is one: [imath]\forall A,B\in M_n(\mathbb{K})[/imath], [imath]\mathrm{rank}(A)+\mathrm{rank}(B)\le \mathrm{rank}(AB)+n[/imath] [imath]M_n(\mathbb{K})[/imath] is the space of square matrices in the field [imath]\mathbb{K}[/imath], for instance [imath]\mathbb{C}[/imath] or [imath]\mathbb{R}[/imath]. . If I consider A and B as linear mappings from [imath]\mathbb{K}^n[/imath] to [imath]\mathbb{K}^n[/imath] I have: [imath]\mathrm{rank}(A)=\mathrm{dim}(\mathrm{Im}(A))=n-\mathrm{dim}(\mathrm{Ker}(A))[/imath] [imath]\mathrm{rank}(AB)=n-\mathrm{dim}(\mathrm{Ker}(AB))[/imath] so, the inequality becomes: [imath]\mathrm{dim}(\mathrm{Ker}(A))+\mathrm{dim}(\mathrm{Ker}(B))\ge\mathrm{dim}(\mathrm{Ker}(AB))[/imath] Is this correct? Now, how can I conclude? . I thought that if [imath]\mathbb{K}=\mathbb{C}[/imath] then A and B are for sure triangularizable and then, maybe in some way developing the products I can show directly that the inequality holds?? . Thank you. | 298836 | Sylvester rank inequality: [imath]\operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n[/imath]
If [imath]A[/imath] and [imath]B[/imath] are two matrices of the same order [imath]n[/imath], then [imath] \operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n. [/imath] I don't know how to start proving this inequality. I would be very pleased if someone helps me. Thanks! Edit I. Rank of [imath]A[/imath] is the same of the equivalent matrix [imath]A' =\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}[/imath]. Analogously for [imath]B[/imath], ranks of [imath]A[/imath] and [imath]B[/imath] are [imath]r,s\leq n[/imath]. Hence, since [imath]\operatorname{rank}AB = \min\{r,s\}[/imath], then [imath]r+s\leq \min\{r,s\} + n[/imath]. (This is not correct since [imath]\operatorname{rank} AB \leq \min\{r,s\}[/imath]. Edit II. A discussion on the rank of a product of [imath]H_f(A)[/imath] and [imath]H_c(B)[/imath] would correct this, but I don't know how to formalize that [imath]\operatorname{rank}H_f(A) +\operatorname{rank}H_c(B) - n \leq \operatorname{rank}[H_f(A)H_c(B)][/imath]. |
1231677 | In the card came "Projective Set", show that 7 cards do always contain a set.
In the game of Projective Set, it turns out that any seven cards contain a projective set. How can one prove this? And for fewer than 7 cards, how can we determine the probability that one or more sets exist (in terms of the number of cards)? For those not familiar with Projective Set, here's a description from the wikipedia page linked above: A Projective Set card has six binary attributes, or bits, generally represented by colored dots. For each color of dot, each card either has that dot or does not. There is one card for each possible combination of dots except the combination of no dots at all, making [imath]2^6 - 1 = 63[/imath] cards total. Three cards are said to form a "set" if the total number of dots of each color is either 0 or 2. Similarly, four or more cards form a "set" if the number of dots of each color is an even number. A card and itself could be said to form a two-card set, but as the cards in the deck are all distinct, this does not arise in actual gameplay. | 1231936 | In the card game "Projective Set": Compute the probability that [imath]n[/imath] cards contain a set
In the game of Projective Set, it turns out that any seven cards contain a projective set. For fewer than 7 cards, how can we determine the probability that one or more sets exist (in terms of the number of cards)? For those not familiar with Projective Set, here's a description from the wikipedia page linked above: A Projective Set card has six binary attributes, or bits, generally represented by colored dots. For each color of dot, each card either has that dot or does not. There is one card for each possible combination of dots except the combination of no dots at all, making [imath]2^6 - 1 = 63[/imath] cards total. Three cards are said to form a "set" if the total number of dots of each color is either 0 or 2. Similarly, four or more cards form a "set" if the number of dots of each color is an even number. A card and itself could be said to form a two-card set, but as the cards in the deck are all distinct, this does not arise in actual gameplay. |
1237549 | Calculate the ring of integers of quadratic number field [imath]\mathbb{Q}(\sqrt{d})[/imath]
Calculate the ring of integers of quadratic number field [imath]\mathbb{Q}(\sqrt{d})[/imath] Solution: Let [imath]F[/imath] be an algebraic number field. Then an element [imath]b\in F[/imath] is integral iff it monic irreducible polynomial has integer coefficients. For example, [imath]\sqrt{d}[/imath] for integer [imath]d[/imath] is integral. If [imath]d\equiv 1 \mod 4[/imath], then the monic irreducible polynomial of [imath](1+\sqrt{d})/2[/imath] over [imath]\mathbb{Q}[/imath] is [imath]x^2 -x + (1-d)/4 \in \mathbb{Z}[x][/imath], so [imath](1+\sqrt{d})/2[/imath] is integral. Thus the integral closure of [imath]\mathbb{Z}[/imath] in [imath]\mathbb{Q}(\sqrt{d})[/imath] contains the subring [imath]\mathbb{Z}[\sqrt{d}][/imath], and the subring [imath]\mathbb{Z}[(1+\sqrt{d})/2][/imath] if [imath]d\equiv 1 \mod 4[/imath]. We show that there are no other integral elements. An element [imath]a+b\sqrt{d}[/imath] with rational [imath]a[/imath] and [imath]b\neq0[/imath] is integral iff its monic irreducible polynomial [imath]x^2 -2ax +(a^2 -db^2)[/imath] belongs to [imath]\mathbb{Z}[x][/imath]. Therefore, [imath]2a[/imath],[imath]2b[/imath] are integers. If [imath]a=(2k+1)/2[/imath], for [imath]k\in\mathbb{Z}[/imath], then it is easy to see that [imath]a^2 - db^2 \in \mathbb{Z}[/imath] iff [imath]b=(2l+1)/2[/imath] for some [imath]l\in\mathbb{Z}[/imath], and [imath](2k+1)^2 - d(2l+1)^2[/imath] is divisible by 4. The latter implies that d is quadratic residue modulo 4, i.e. [imath]d\equiv 1 \mod 4[/imath]. In turn, if [imath]d\equiv 1 \mod 4[/imath] then every element [imath](2k+1)/2 +(2l+1)\sqrt{d}/2[/imath] is integral. Thus, integral elements of [imath]\mathbb{Q}(\sqrt{d})[/imath] are equal to [imath]\mathbb{Z}[\sqrt{d}][/imath] if [imath]d\not \equiv 1 \mod 4[/imath], and [imath]\mathbb{Z}[(1+\sqrt{d})/2][/imath] if [imath]d\equiv 1 \mod 4[/imath]. Can someone explain why [imath]\sqrt{d}[/imath] for integer [imath]d[/imath] is integral, and why the monic irreducible polynomial of an integral element [imath]a+b\sqrt{d}[/imath] is of the form [imath]x^2 -2ax +(a^2 -db^2)[/imath]? | 1198188 | Why is quadratic integer ring defined in that way?
Quadratic integer ring [imath]\mathcal{O}[/imath] is defined by \begin{equation} \mathcal{O}=\begin{cases} \mathbb{Z}[\sqrt{D}] & \text{if}\ D=2,3\ \pmod 4\\ \mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]\ & \text{if}\ D=1\pmod 4 \end{cases} \end{equation} where [imath]D[/imath] is square-free. I understand [imath]\mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right][/imath] is not closed under multiplication if [imath]D=2,3\pmod 4[/imath]. But still, isn't it more natural to define [imath]\mathcal{O}=\mathbb{Z}[\sqrt{D}][/imath] for all square-free [imath]D[/imath]? (in that case, it really seems like 'quadratic integer') I wonder what is the motivation of this definition. |
1192228 | How to prove the trigonometric equality [imath]\tan(\pi/7) \tan(2\pi/7) \tan(3\pi/7)= \sqrt 7[/imath]?
I want to prove this equality holds: [imath]\tan\big(\frac{\pi}7\big) \tan\big(\frac{2\pi}7\big) \tan\big(\frac {3\pi}7\big)= \sqrt 7.[/imath] Please help me. Thanks. | 1182103 | Trigonometry problem on product of trig functions.
Questions: [imath]\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=?[/imath] [imath]\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}=?[/imath] [imath]\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}=?[/imath] Approach: If [imath]x=\dfrac{\pi}{7}\implies 7x=\pi\implies 4x=\pi-3x\implies \sin 4x = \sin 3x[/imath].Upon expansion, [imath]8cos^3x-4cos²x-4cosx+1=0[/imath] "[imath]\cos \dfrac{\pi}{7}, \cos \dfrac{3\pi}{7}, \cos \dfrac{5\pi}{7}[/imath]will satisfy this cubic." Note: I'm not able to understand the above line. Why do these values satisfy this cubic and how did we get these values? These are therefore the roots of the cubic, and their product is: [imath]\cos\dfrac{\pi}{7}\cos\dfrac{3\pi}{7}\cos\dfrac{5\pi}{7}=\dfrac{-1}{8}[/imath] Finally, [imath]\cos\dfrac{5\pi}{7} = \cos\left(\pi-\dfrac{2\pi}{7}\right)=-\cos \dfrac{2\pi}{7}[/imath]. Replacing this in (1) gives [imath]\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac{1}{8}[/imath] How can we apply this concept to find the values of [imath]\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}?[/imath] [imath]\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}?[/imath] I'm not able to form the equations for these two problems. Please Help. . |
953974 | Proving [imath]9[/imath] divides [imath]n^3 + (n+1)^3 + (n+2)^3[/imath]
I'm trying to prove by MI. I have already distributed n+1, but now I'm stuck on how I can show 9 divides the RHS since [imath]42n[/imath] and [imath]3n^3[/imath] does not divide evenly. [imath](n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36[/imath] | 6805 | Simple Proof by induction: [imath]9[/imath] divides [imath]n^3 + (n+1)^3 + (n+2)^3[/imath]
I'm trying to prove using induction that [imath]9[/imath] divides [imath]n^3 + (n+1)^3 + (n+2)^3[/imath] whenever [imath]n[/imath] is a non-negative integer. So far, I have: Base case: [imath]P(1) = (1) + (8) + (27) = 36, 36[/imath] can be divided by [imath]9[/imath] so the base case is valid Inductive step: let [imath]P(n)[/imath] be the statement [imath]9[/imath] divides [imath]n^3 + (n+1)^3 + (n+2)^3[/imath]. Assume [imath]P(k)[/imath] is true, so [imath]9[/imath] divides [imath]k^3 + (k+1)^3 + (k+2)^3[/imath]. And this is where I'm stuck. I'm not sure how to demonstrate that [imath]9[/imath] divides [imath]P(n)[/imath] when [imath]n = k+1[/imath]. If someone could step me in the right direction that would be awesome. Thanks! |
1239580 | Show all final 2-digit numbers of the decimal expansions of squares are to be found among those of [imath]0^2, 1^2,...25^2[/imath]
I'm not really sure where to begin. The first part of the question states that "every positive integer has a unique representation in the form [imath]50k+l[/imath], with [imath]-24\lt l \le 25[/imath]," which isn't even true, so... | 1239538 | Show that every positive integer has a unique representation in the form [imath]50k+l[/imath]...?
with -24 [imath]\lt l \le[/imath] 25. Then I need to conclude that all final 2-digit numbers of the decimal expansion of squares are to be found among those of [imath]0^2, 1^2, 2^2,...., 25^2[/imath]. I'm thinking that I could maybe rewrite the division theorem proof for this but I don't know how/what to do with that. Any ideas? |
1240303 | A question about primes, number theory
I tried to solve this question but without a success: Let [imath]p[/imath] be a prime number,and [imath]p^2+2[/imath] is also prime, prove that [imath]p=3[/imath]. I tried to show [imath]p^2+2[/imath] as a product of numbers and then to show that [imath]p=3[/imath] is the only option that allows it to be prime. but I didn't find that presentation. I would like to get help with this question, thanks | 628115 | For which primes p is [imath]p^2 + 2[/imath] also prime?
Origin — Elementary Number Theory — Jones — p35 — Exercise 2.17 — Only for [imath]p = 3[/imath]. If [imath]p \neq 3[/imath] then [imath]p = 3q ± 1[/imath] for some integer [imath]q[/imath], so [imath]p^2 + 2 = 9q^2 ± 6q + 3[/imath] is divisible by [imath]3[/imath], and is therefore composite. (1) The key here looks like writing [imath]p = 3q ± 1[/imath]. Where does this hail from? I cognize [imath]3q - 1, 3q, 3q + 1[/imath] are consecutive. (2) How can you prefigure [imath]p = 3[/imath] is the only solution? On an exam, I can't calculate [imath]p^2 + 2[/imath] for many primes [imath]p[/imath] with a computer — or make random conjectures. Matt E's edited answer - E.g. any square is either [imath]\color{purple}0[/imath] or [imath]\color{teal}1[/imath] [imath]\begin{cases}\mod 3, & \text{ depending on whether or not [/imath]3[imath] divides [/imath]x[imath]} \\ \mod 4, & \text{depending on whether or not [/imath]2[imath] divides the number being squared} \end{cases}[/imath], and [imath]0,[/imath] [imath]1[/imath], or [imath]4[/imath] mod [imath]8[/imath] (depending on whether or not [imath]2[/imath] or [imath]4[/imath] divide the number being squared). Thus, when you see [imath]p^2 + 2[/imath], you should think: [imath]\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if [/imath]3[imath] divides [/imath]p[imath]} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if [/imath]3 \not| p[imath]} \end{cases}[/imath]. (3) Can someone please clarify why I'd prefigure or think about mod 3, mod 4, mod 8? Why not consider mod of some random natural number? (4) The last paragraph considers mod 3. How can I prefigure this? |
1240366 | What is an example of [imath]E/F,L/E[/imath] are normal but [imath]L/F[/imath] is not.
Let [imath]E/F,L/E[/imath] be normal field extensions. What would be an example such that [imath]L/F[/imath] is not normal? | 618629 | Example of composition of two normal field extensions which is not normal.
If [imath]F\supset E[/imath] and [imath]E\supset k[/imath] are normal extensions. I want a counter-example where [imath]F\supset k[/imath] is not normal |
1240338 | Derivative [imath](1-x)^{-2}[/imath]
I'm getting answer of [imath]\frac{2}{(1-x)^3}[/imath] , but online calculators suggest it's [imath]\frac{-2}{(1-x)^3}[/imath]. I've tried it as [imath](1-x)^{-2}[/imath], which results [imath]-2*(1-x)*-1 = \frac{2}{(1-x)^3}[/imath]. Same result with [imath](\frac{f}{g})'=\frac{(f'*g-f*g')}{g^2}[/imath] . Any help please? | 1210768 | Derivative of [imath](1-x)^{-2}[/imath]
Wolfram Alpha is telling me that the answer is [imath]-\frac2{(x-1)^3}.[/imath] But I thought that by using the chain rule you multiply the front by [imath]2[/imath] then subtract the exponent by [imath]1[/imath] then multiply by the derivative of the inside ([imath]-1[/imath]) to get [imath]\frac2{(1-x)^3}.[/imath] Wolfram Alpha is never wrong so please tell me what I'm missing its giving me a real headache. |
1240353 | Cyclic group Zp
How to show if [imath]p[/imath] is prime, then group [imath]Z_{p}^{*}[/imath] is cyclic. Tip Let [imath]g[/imath] and [imath]h[/imath] of a commutative group [imath]G[/imath] have orders [imath]n[/imath] and [imath]m[/imath] respectively. There exists and element [imath]x \in G[/imath] of order [imath]LCM (n,m)[/imath] In any field [imath]\mathbb{K}[/imath] a polynomial [imath]f \in \mathbb{K} [x][/imath] of degree [imath]n[/imath] has at most [imath]n[/imath] different roots. I have no idea. | 290427 | Is [imath]\mathbb Z _p^*=\{ 1, 2, 3, ... , p-1 \}[/imath] a cyclic group?
In undergraduate course, the two groups which are most frequently used may be [imath]\{ 0, 1, 2, ... , p-1\}[/imath] and [imath]\{ 1, 2, ... , p-1\}[/imath] where [imath]p[/imath] is a prime. The first one is a group under addition and in addition it is a cyclic group whose generator is [imath]p-1[/imath]. Also we can describe it by the solution set of [imath]x^p=1[/imath] in [imath]{\bf C}[/imath]. The latter is a group under multiplication. Fermat's theorem implies that [imath]a^{p-1} =1~~~ (\text{mod}\;\;p)[/imath] for [imath]a\in \{ 1, ... , p-1\}[/imath]. But this is not sufficient for [imath] \{ 1, ... , p-1\}[/imath] to be cyclic. My question is : [imath] \{ 1, ... , p-1\}[/imath] is cyclic ? Thank you in advance. |
1240591 | What is [imath]\int_0^1 \frac{\log(x+1)}{x^2+1}dx[/imath]
It's so deceptively simple and none of the usual techniques are working. Any and all insights are welcome. | 322229 | How to evaluate [imath]\int_0^1 \frac{\ln(x+1)}{x^2+1} dx[/imath]
This problem appears at the end of Trig substitution section of Calculus by Larson. I tried using trig substitution but it was a bootless attempt [imath]\int_0^1 \frac{\ln(x+1)}{x^2+1} dx[/imath] |
1240482 | A question in Number Theory - prove there exist m>2010 s.t f(m) is not prime
Let [imath]f(x)=\sum_{i=0}^n a_nx^n[/imath] be a polynomial with [imath]a_n \in Z,n>0,a_n\neq0[/imath] Prove that there exists some natural number [imath]m>2010[/imath] such that [imath]|f(m)|[/imath] is not a prime number. I tried to look at f(m) mod m, and i assumed m is relatively prime to a0, so f(m) is reversible mod m. but I didn't know how to continue to prove it is not a prime number. I would like to get help with that. Thanks | 1240790 | Prove there exists [imath]m > 2010[/imath] such that [imath]f(m)[/imath] is not prime
Let [imath]f(x) = \sum_{i = 0}^n a_ix^i[/imath] be a polynomial with [imath]a_i \in \mathbb Z, n > 0, a_n \neq 0[/imath]. Prove that there exists some natural number [imath]m>2010[/imath] such that [imath]|f(m)|[/imath] is not a prime number. I tried to look at [imath]f(m) \bmod m[/imath], and I assumed [imath]m[/imath] is relatively prime to [imath]a_0[/imath], so [imath]f(m)[/imath] is reversible [imath]\mod m[/imath], but I didn't know how to continue to prove it is not a prime number. I would like to get help with that. Thanks. |
297818 | Show that [imath]d_2[/imath] defined by [imath]d_2(x,y)=\frac{|x-y|}{1+|x-y|}[/imath] is a metric
I am pretty sure that this is such a stupid, stupid question, but how do you prove that [imath]d_2(x,y)={|x-y|\over {1+|x-y|}}[/imath] satisfying the third condition to be a metric, which is the triangle inequality. I know that I should really be able to do this, but I just can't, I keep seeing a contradiction. Please help me. | 1109799 | Metric triangle inequality [imath]d_2(x,y):= \frac{d(x,y)}{d(x,y)+1}[/imath]
[imath](X,d)[/imath] is a metric space. [imath]x,y,z \in X[/imath] Now I have to proof that [imath](X,d_2)[/imath] is also a metric space. To show that [imath]d_2(x,y)=0 \leftrightarrow x=y [/imath] and [imath]d_2(x,y) = d_2(y,x)[/imath] are correct was quite trivial. But with the 'triangle inequality' ([imath]d_2(x,y) \leq d_2(x,z)+ d_2(z,y)[/imath]) I have my problems. I have started like this: [imath] d_2(x,y) = \frac{d(x,y)}{d(x,y)+1} = 1-\frac{1}{d(x,y)+1} \leq 1-\frac{1}{1+d(x,z)+d(z,y)}[/imath] But I don't know how to get further. Hints are welcome :) |
1241586 | Prove that for all positive integers [imath]n[/imath], [imath]2^1+2^2+2^3+...+2^n=2^{n+1}-2[/imath]
I want to prove that for all positive integers [imath]n[/imath], [imath]2^1+2^2+2^3+...+2^n=2^{n+1}-2[/imath]. By mathematical induction: 1) it holds for [imath]n=1[/imath], since [imath]2^1=2^2-2=4-2=2[/imath] 2) if [imath]2+2^2+2^3+...+2^n=2^{n+1}-2[/imath], then prove that [imath]2+2^2+2^3+...+2^n+2^{n+1}=2^{n+2}-2[/imath] holds. I have no idea how to proceed with step 2), could you give me a hint? | 123855 | Prove by induction. How do I prove [imath]\sum\limits_{i=0}^n2^i=2^{n+1}-1[/imath] with induction?
[imath]\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1[/imath]. I don't understand induction so I could use some help. |
1030628 | A number theoretical problem
How many pairs [imath](A,B)[/imath] are there up to [imath]n[/imath] such that [imath]\gcd(A,B)=B[/imath] and [imath]B^2\neq A[/imath]? If we consider [imath]n=5[/imath], we have [imath]25[/imath] possible pairs, [imath](1,1)[/imath], [imath](1,2)[/imath], [imath](1,3)[/imath], [imath](1,4)[/imath], [imath](1,5)[/imath], [imath](2,1)[/imath], [imath](2,2)[/imath], [imath](2,3)[/imath], [imath](2,4)[/imath], [imath](2,5)[/imath], [imath]\dots[/imath], [imath](5,5)[/imath]. Eight of these satisfy the above condition. My main objective is to find out the number of such pairs if [imath]n[/imath] is given, with a efficient method. So far I know I have to count those pairs [imath](A,B)[/imath] where [imath]A[/imath] is divisible by [imath]B[/imath]. Hence if I find out the number of divisors of all the numbers up to [imath]N[/imath], we will have the number of pair [imath]\gcd(A,B)=B[/imath]. For example,for [imath]n=10[/imath], [imath]1[/imath] has one divisor, [imath]2[/imath] has two divisors, ..., and [imath]10[/imath] has four divisors. So the total number of "candidate" pairs is [imath]1+2+2+3+2+4+2+4+3+4=27[/imath]. We now remove the pairs with [imath]B^2=A[/imath], of which there are three, so our desired result is [imath]24[/imath]. One of my friends showed me a method but he didn't state the reasoning behind it. Here, I'm presenting his method for [imath]n=10[/imath]. [imath]\begin{align*} 2\sum_{i=1}^{\lfloor\sqrt{n}\rfloor} \big(\lfloor n/i\rfloor-i\big) &=2\big(\lfloor 10/1\rfloor - 1 + \lfloor10/2\rfloor - 2 + \lfloor10/3\rfloor - 3\big)\\ &= 2\big(10 - 1 + 5 - 2 + 3 - 3\big) \\ &= 24\,. \end{align*}[/imath] Why does this give the right answer? Can anyone clearly explain the reasoning behind the above method in detail? | 993445 | Number of pairs [imath](A,B)[/imath] with [imath]\gcd(A,B)=B, A \ne B^2[/imath] with [imath]A,B \le n[/imath]
How many pairs [imath](A,B)[/imath] of integers up to [imath]n[/imath] are there such that [imath]\gcd(A,B)=B[/imath], not counting those pairs where [imath]B^2=A[/imath]? If we consider [imath]n = 5[/imath] we have [imath]25[/imath] possible pairs. They are [imath](1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(2,5),\dotsc,(5,5)[/imath] Of them, [imath]8[/imath] pairs satisfy the above condition. My main objective is to find the number of such pairs if [imath]n[/imath] is given, with a efficient method. One of my friends showed me a method but he didn't state the reasoning behind it. Here, I'm presenting his method for [imath]n=10[/imath]. [imath] \begin{align} & 2 \cdot \left(\left(\left\lfloor\frac{10}{1}\right\rfloor - 1\right) + \left(\left\lfloor \frac{10}{2} \right\rfloor - 2\right) + \left(\left\lfloor \frac{10}{3} \right\rfloor - 3\right)\right) \\ =& 2 \cdot ((10 - 1) + (5 - 2) + (3 - 3)) \\ =& 2 \cdot (9 + 3 + 0) \\ =& 24 \end{align} [/imath] Can anyone explain the reasoning behind the above method in details and clearly? |
1241893 | complex numbers- how do I prove the following statement?
given: [imath] \left | z_{1}\right |=\left | z_{2}\right |=...=\left | z_{n}\right |=1 [/imath] How do I prove: [imath] (1+\frac{z_{2}}{z_{1}})(1+\frac{z_{3}}{z_{2}})*...*(1+\frac{z_{n}}{z_{n-1}})(1+\frac{z_{1}}{z_{n}}) \in \mathbb{R} [/imath] ? | 1184048 | Prove that the complex expression is real
Let [imath]|z_1|=\dots=|z_n|=1[/imath] on the complex plane. Prove that: [imath] \left(1+\frac{z_2}{z_1}\right) \left(1+\frac{z_3}{z_2}\right) \dots \left(1+\frac{z_n}{z_{n-1}}\right) \left(1+\frac{z_1}{z_n}\right) \in\mathbb{R} [/imath] I have tried induction and writing every "subexpression" as [imath](1+e^{i(\theta_n-\theta_{n-1})})[/imath]. Any ideas? |
1067014 | Fermat's Challenge
This is a question French mathematician Pierre de Fermat posed to the English mathematicians of his time. "Prove that x=5 and y=3 are the only positive integer values for which [imath]x^2 +2=y^3[/imath]." I have been trying to come up with an answer for days using modular arithmetic to no avail. Any such method would therefore be appreciated as well as other insights. | 874226 | Proof that [imath]26[/imath] is the one and only number between square and cube
[imath]x^2 + 1 = z = y^3 - 1[/imath] Why [imath]z = 26 [/imath] and only [imath]26[/imath] ? Is there an elementary proof of that ? |
528351 | Show that the order of 5 mod [imath]2^k[/imath] is [imath]2^{k-2}[/imath]
I have that [imath]5^{2^{k-2}}\equiv 1 \mod 2^{k}[/imath] but I am unsure how to show that [imath]2^{k-2}[/imath] is the least integer that has this property. I thought perhaps I could show that [imath]2^{k-2}|\operatorname{ord}_{2^k}(5)[/imath] since I already have (from above) [imath]\operatorname{ord}_{2^k}(5)|2^{k-2}[/imath] but I have been stuck with this. Any help appreciated :) | 74061 | Divisibility question
Let [imath]r[/imath] be an integer greater than [imath]2[/imath]. Is there a simple way of showing that [imath]2^r[/imath] divides [imath]\left(\begin{array}{c} {2}^{r-2} \\ k \end{array}\right) 2^{2k}[/imath] but it does not divide [imath]\left(\begin{array}{c} 2^{r-l} \\ k \end{array}\right) 2^{2k}[/imath] for [imath]l>2[/imath] and for all [imath]k[/imath] greater than [imath]0[/imath] and less or equal than [imath]2^{r-2}[/imath]? Equivalently, is there a simple way of showing that the element [imath]5[/imath] has order [imath]2^{r-2}[/imath] in the group of units of the ring [imath]\mathbb{Z}_{2^r}[/imath]? The two questions are connected by writing [imath]5 = 4 + 1[/imath] and using the binomial expansion formula to study the order of [imath]5[/imath]. |
1241952 | Find spectrum for matrix [imath]A[/imath]
Let [imath]A = \left[ \begin{array}{*{20}{c}} 0&b&0&0&0&0\\ c&0&b&0&0&0\\ 0&c&0&b&0&0\\ 0&0&c&0&b&0\\ 0&0&0&c&0&b\\ 0&0&0&0&c&0 \end{array} \right]_{n \times n}[/imath]. Why does matrix [imath]A[/imath] have spectrum [imath]\sigma (A) = \left\{ 2\sqrt {bc} \cos \left(\frac{\pi k}{n + 1}\right):k = 1,2,\ldots,n \right\}[/imath]? | 1240577 | find spectrum matrix A
Let [imath]A = \left[ \begin{array}{*{20}{c}} 0&b&0&0&0&0\\ c&0&b&0&0&0\\ 0&c&0&b&0&0\\ 0&0&c&0&b&0\\ 0&0&0&c&0&b\\ 0&0&0&0&c&0 \end{array} \right]_{n \times n}[/imath]. Why does matrix [imath]A[/imath] have spectrum [imath]\sigma (A) = \left\{ 2\sqrt {bc} \cos \left(\frac{\pi k}{n + 1}\right):k = 1,2,\ldots,n \right\}[/imath]? |
1241424 | Explaining why proof by induction works
I am learning math proofs for the first time. So I cannot discern the reason for all the details in a proof. Here's the statement of mathematical induction: For every positive integer [imath]n[/imath], let [imath]P(n)[/imath] be a statement. If: (1). [imath]P(1)[/imath] is true and (2). If [imath]P(k)[/imath], then [imath]P(k+1)[/imath] is true for every positive integer [imath]k[/imath] then [imath]P(n)[/imath] is true for every positive integer [imath]n[/imath]. Here's the proof the author gave. Assume that the theorem is false. Then conditions (1) and (2) are satisfied, but there exist some positive integers [imath]n[/imath] for which [imath]P(n)[/imath] is a false statement. Let [imath]S=\{n\in {\rm N}~{\rm :}P\left(n\right)~is~false\}[/imath]Since [imath]S[/imath] is a non-empty subset of [imath]{\rm N}[/imath], it follows by the well-ordering principle that [imath]S[/imath] contains a least element [imath]s[/imath]. Since [imath]P(1)[/imath] is true, [imath]1\notin S[/imath]. Thus [imath]s\ge 2[/imath] and [imath]s-1\in {\rm N}[/imath]. Therefore, [imath]s-1\notin S[/imath] and so [imath]P(s-1)[/imath] is a true statement. By condition (2), [imath]P(s)[/imath] is also true and so [imath]s\notin S[/imath]. This however contradicts our assumption that [imath]s\in S[/imath]. My questions are: Where did [imath]s-1\in {\rm N}[/imath] and [imath]s-1\notin S[/imath] pop from ? Why is the well-ordering principle important for this proof ? | 1139579 | Why is mathematical induction a valid proof technique?
Context: I'm studying for my discrete mathematics exam and I keep running into this question that I've failed to solve. The question is as follows. Problem: The main form for normal induction over natural numbers [imath]n[/imath] takes the following form: [imath]P(1)[/imath] is true, and for every [imath]n, P(n-1)\to P(n).\qquad\qquad\qquad\text{[Sometimes written as [/imath]P(n)\to P(n+1)[imath]]}[/imath] If both 1 and 2 are true, then [imath]P(n)[/imath] is true for every [imath]n[/imath]. The question is to prove the correctness of the form above. My work: My idea was to make a boolean statement and if it's a tautology in a true false table. That means the statement is always correct. I've tried many times but I've failed. Here is the original question in Hebrew: n טענה כלשהי לגבי מםפר טבעי [imath]P(n)[/imath] תהי אם מתקימיים שני התנאים הבאים: 1- הטענה [imath]P(0)[/imath] נכונה 2- לכל n>0, [imath]P(n)[/imath] נכונות הטענה [imath]P(n-1)[/imath] גוררת את נכונות הטענה אז [imath]P(n)[/imath] נכונה לכל מספר טבעי n הוכיחו את נכונות משפט זה |
1242735 | Riemann Integrable Functions to prove [imath]f(x) =0[/imath]
Suppose that [imath]f[/imath] is continuous on [imath][a,b][/imath], that [imath]f(x) \geq 0[/imath] for all [imath]x \in [a,b][/imath] and that [imath]\int_a^b fdx = 0[/imath]. Prove that [imath]f(x) = 0[/imath] for all [imath]x \in [a,b][/imath]. | 1223367 | If [imath]f \geq 0[/imath] is continuous and [imath]\int_{a}^{b} f(x) \, dx = 0[/imath], then [imath]f =0[/imath]
Just wanted to confirm that this is a correct solution: Proof: Suppose [imath]f(x_0) > 0[/imath] for some [imath]x_0 \in [a,b][/imath]. Then, by continuity of [imath]f[/imath], for [imath]\epsilon < f(x_0)[/imath], there exists [imath]\delta > 0[/imath] so that, if [imath]|x-x_0| < \delta[/imath], we have [imath]|f(x) - f(x_0)| < \epsilon[/imath]. That is to say, [imath]f(x)[/imath] is non-zero in a neighborhood [imath]B_\delta (x_0)[/imath]. Now, \begin{eqnarray*} U(f,P) &=& \sum_{P([a,b])} M_i \Delta_i \\ &=& \sum_{P(B_\delta(x_0))} M_i \Delta_i \\ &=& \sup_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \sup_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \sup_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &>& 0 \end{eqnarray*} In the same manner, \begin{eqnarray*} L(f,P) &=& \sum_{P([a,b])} m_i \Delta_i \\ &=& \sum_{P(B_\delta(x_0))} m_i \Delta_i \\ &=& \inf_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \inf_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \inf_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &\geq& 0 \end{eqnarray*} We can conclude from here that [imath]\int_{a}^{b} f(x) dx > 0[/imath]. Is this a valid proof? |
1242857 | The order of an element
The order of a unit [imath]a \pmod m [/imath] is the least [imath]n \geq 1[/imath] such that [imath]a^n \equiv 1 \pmod m[/imath]. my question is : Is true that number and its inverse have the same order? | 1008610 | An element of a group has the same order as its inverse
If [imath]a[/imath] is a group element, prove that [imath]a[/imath] and [imath]a^{-1}[/imath] have the same order. I tried doing this by contradiction. Assume [imath]|a|\neq|a^{-1}|[/imath] Let [imath]a^n=e[/imath] for some [imath]n\in \mathbb{Z}[/imath] and [imath](a^{-1})^m=e[/imath] for some [imath]m\in \mathbb{Z}[/imath], and we can assume that [imath]m < n[/imath]. Then [imath]e= e*e = (a^n)((a^{-1})^m) = a^{n-m}[/imath]. However, [imath]a^{n-m}=e[/imath] implies that [imath]n[/imath] is not the order of [imath]a[/imath], which is a contradiction and [imath]n=m[/imath]. But I realized this doesn't satisfy the condition if [imath]a[/imath] has infinite order. How do I prove that piece? |
1242253 | Section of a covering projection from a connected space
Let [imath]p:\overline{X}\rightarrow X[/imath] is a continuous mapping. A continuous map [imath]s:X\rightarrow \overline{X}[/imath] such that [imath]p\circ s =Id_X[/imath] is called a section of [imath]p[/imath]. Suppose [imath]\overline{X}[/imath] is connected also. Suppose that [imath]p[/imath] is a covering projection. Show that any section of [imath]p[/imath] is a homeomorphism onto [imath]\overline{X}[/imath]. As [imath]p\circ s=Id[/imath] we see that [imath]s[/imath] is injective. Suppose we can prove that [imath]s[/imath] is surjective. Then we have a bijective continuous function [imath]s[/imath] such that [imath]s^{-1}=p[/imath] which means that [imath]s[/imath] is homeomorphism. [imath]p\circ s =Id\Rightarrow p\circ s\circ s^{-1}=s^{-1}\Rightarrow p=s^{-1}.[/imath] So, enough to prove [imath]s[/imath] is surjective i.e., [imath]s(X)=\overline{X}[/imath]. As [imath]\overline{X}[/imath] is connected, showing [imath]s[/imath] is surjective is equivalent to prove [imath]s(X)[/imath] is both open and closed in [imath]\overline{X}[/imath]. Let [imath]\bar{x}=s(x)\in \overline{X}[/imath] for some [imath]x\in X[/imath]. As [imath](p\circ s )(x)=x[/imath] we have [imath]p(\bar{x})=x[/imath]. For this [imath]x\in X[/imath] there exists an open set [imath]U[/imath] in [imath]\overline{X}[/imath] containing [imath]\bar{x}[/imath] such that [imath]U\rightarrow p(U)[/imath] is a homeomorphism. As [imath]U[/imath] is open in [imath]\overline{X}[/imath] containing [imath]\bar{x}[/imath] and [imath]s[/imath] is continuous we have open set [imath]V[/imath] in [imath]X[/imath] with [imath]x\in V[/imath] and [imath]s(V)\subset U[/imath]. We have [imath](p\circ s)(V)=V\Rightarrow (p^{-1}\circ p\circ s)(V)=p^{-1}(V)\Rightarrow (p^{-1}\circ p)(s(V))=p^{-1}(V).[/imath] As [imath]s(V)\subset U[/imath] the function [imath]p^{-1}\circ p[/imath] is identity on [imath]s(V)[/imath] so we have [imath]s(V)=p^{-1}(V)[/imath] So, [imath]s(V)[/imath] is an open set. we have [imath]\bar{x}=s(x)\in s(V)\subset s(X)[/imath]. Thus [imath]s(X)[/imath] is open in [imath]\overline{X}[/imath]. I am not able to see why [imath]s(X)[/imath] is closed in [imath]\overline{X}[/imath]. This also says [imath]p[/imath] is also a homeomorphism. | 256951 | If a covering map has a section, is it a [imath]1[/imath]-fold cover?
If [imath]q: E\rightarrow X[/imath] is a covering map that has a section (i.e. [imath]f: X\rightarrow E, q\circ f=Id_X[/imath]) does that imply that [imath]E[/imath] is a [imath]1[/imath]-fold cover? |
1243089 | Basis of orthogonal complement subspace
Let [imath]A[/imath] be the matrix [imath] \begin{pmatrix} 1 & 1 & -1&-1 \\ 1 & 2 & -2 & 1 \\ \end{pmatrix} ,[/imath] let [imath]W[/imath] = ker [imath]A[/imath] and let [imath]W^\bot[/imath] be the orthogonal complement for W in [imath]\Bbb{R}^4[/imath]. (a) Find orthonormal bases for [imath]W[/imath] and [imath]W^\bot[/imath]. I found [imath]W=span \{(3,-2,-0,1),(0,1,1,0)\} \;[/imath] by evaluating [imath] \left[ \begin{array}{cccc|c} 1&1&-1&-1&0\\ 0 & 1 & -1 & 2&0 \end{array} \right] [/imath] but what I don't understand is why span of [imath]W^\bot[/imath] are the two rows of the row-echelon matrix i.e. [imath]W^\bot=span \{(1,1,-1,-1),(0,1,-1,2)\} \;[/imath]. I've spent hours but can't figure out the connection of [imath]W^\bot[/imath]and those two rows. Please help. | 1243036 | Spans of Orthogonal complements
Let [imath]A[/imath] be the matrix [imath] \begin{pmatrix} 1 & 1 & -1&-1 \\ 1 & 2 & -2 & 1 \\ \end{pmatrix} ,[/imath] let [imath]W[/imath] = ker [imath]A[/imath] and let [imath]W^\bot[/imath] be the orthogonal complement for W in [imath]\Bbb{R}^4[/imath]. (a) Find orthonormal bases for [imath]W[/imath] and [imath]W^\bot[/imath]. I found [imath]W=span \{(3,-2,-0,1),(0,1,1,0)\} \;[/imath] by evaluating [imath] \left[ \begin{array}{cccc|c} 1&1&-1&-1&0\\ 0 & 1 & -1 & 2&0 \end{array} \right] [/imath] but what I don't understand is why span of [imath]W^\bot[/imath] are the two rows of the row-echelon matrix i.e. [imath]W^\bot=span \{(1,1,-1,-1),(0,1,-1,2)\} \;[/imath]. I've spent hours but can't figure out the connection of [imath]W^\bot[/imath]and those two rows. Please help. |
1243190 | Find the max and min of [imath]f(x) = x^5 -x^4+x^2-x[/imath]
I get [imath]5x^4-4x^3+2x-1[/imath] for the derivative but I am not sure what to do after. The teacher told us that we would have to use Newton's method to solve the problem. | 1243144 | Finding local maximum and minimum
[imath]f(x) = x^5-x^4+x^2-x[/imath] on the interval [imath](-\infty,\infty)[/imath] The teacher told us that we would have to use Newton's method to solve this problem but I am not sure what to do after taking the derivative. |
1243637 | Simplify [imath](x_1+x_2+\dots+x_m)^p[/imath]?
Is there a way to simplify [imath](x_1+x_2+\dots+x_m)^p[/imath]? Thank you! | 151611 | simplify [imath](a_1 + a_2 +a_3+... +a_n)^m[/imath]
How to simplify this best [imath](a_1 + a_2 +a_3+... +a_n)^m[/imath] for [imath]m=n, m<n, m>n[/imath] I could only get [imath]\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ... [/imath] |
1243485 | Immersion of punctured torus into Euclidean
(a) Show there is an immersion of the punctured torus [imath]S^1\times S^1[/imath] - {a point} into [imath]R^2[/imath]. (b) generalized it to [imath]T^n[/imath] - {a point} into [imath]R^n[/imath] can you give concrete proof for these problem? Intuitively i can guess above statement is true by drawing some pictures of two wedge product of [imath]S^1[/imath] and deforming fattened. (I read this statement in the lecture-notes on class-material but it left rigorous proof as an exercise.) I found out this is an exercise in ch1-section3, exercise 3 in Hirsch, Differential topology, pdf version can be found [here].(http://www.maths.ed.ac.uk/~aar/papers/hirschdx.pdf) | 337237 | Immersing punctured torus
I am looking for a proof (as elementary as possible) of the fact that punctured n-dimensional torus admits an immersion to [imath]\mathbb{R}^n[/imath]. The 2-dimensional case seems to be evident, but I haven't got the idea how to give an explicit construction in the n-dim case. |
1243503 | If [imath]R/I \times R/J[/imath] is isomorphic to [imath]R/(I\cap J)[/imath] as [imath] R [/imath]-modules, then [imath]I + J = R[/imath].
If [imath]R[/imath] is a commutative ring with identity and [imath]I[/imath] and [imath]J[/imath] are ideals of [imath]R[/imath] such that [imath]R/I \times R/J[/imath] is isomorphic to [imath]R/(I\cap J)[/imath] as [imath]R[/imath]-modules, then [imath]I + J = R[/imath]. I know this is the converse of CRT and there are many posts on this. However, I can't find a post that provides a complete answer to this. This is the solution I found. However, I find it to be very complicated. What does [imath]\otimes_R[/imath] mean? What does tensor product mean? I have not learn tensor product. I prefer a solution without it and using only concepts from elementary ring theory course. Is there simpler answer to this? | 1152991 | Converse to Chinese Remainder Theorem
So as seen on this question Converse of the Chinese Remainder Theorem, we know that if [imath](n,m) \neq 1[/imath], then [imath]\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}[/imath], because the right hand side does not have an element of order [imath]nm[/imath]. But take a more general setting. Let [imath]R[/imath] be a commutative ring, and let [imath]A,B[/imath] be ideals in [imath]R[/imath]. [imath]A[/imath] and [imath]B[/imath] are said to be comaximal if [imath]A+B=R[/imath]. If [imath]A, B[/imath] are comaximal, then we have: [imath]R/AB \cong R/A \times R/B[/imath]. In this setup, is the converse true? If we have [imath]A, B[/imath] ideals in [imath]R[/imath] such that [imath]R/AB \cong R/A \times R/B[/imath], do we always have that [imath]A[/imath] and [imath]B[/imath] are comaximal? |
1244063 | Why [imath]\int _c^df^{-1}\left(y\right)\:dy+\int _a^b\:f\left(x\right)dx=b\cdot d-a\cdot c[/imath]?
Why [imath]\int _c^df^{-1}\left(y\right)\:dy+\int _a^b\:f\left(x\right)dx=b\cdot d-a\cdot c[/imath] ? where f is an bijective function and [imath]f(a)=b,f(c)=d,[/imath] I don't understand graph... I can't see on graph this equality, so have somebody patience to explain me on graph this equality? I think it is not duplicate, because he wants a rigorous proof, not a draw, and I want to understand from graph, I think is not duplicate... I don't know why they think is duplicate | 1115222 | Show rigorously that the sum of integrals of [imath]f[/imath] and of its inverse is [imath]bf(b)-af(a)[/imath]
Suppose [imath]f[/imath] is a continuous, strictly increasing function defined on a closed interval [imath][a,b][/imath] such that [imath]f^{-1}[/imath] is the inverse function of [imath]f[/imath]. Prove that, [imath]\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)[/imath] A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate. However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with [imath]f[/imath] continuous, I cannot apply change of variables technique. Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification. The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given [imath]f[/imath] is differentiable. However, this seems to hold for non-differentiable functions also. I would really want some help. Pictorial proofs and/or area arguments are invalid. |
1243013 | Binomial Theorem of Differentiation?
I noticed that [imath]\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)[/imath] and it's had me scratching for a little bit. It's easy to see how the cross terms add up but can anyone draw a direct comparison between the behavior being exhibited here and the binomial expansion? It seems to work for the case equivalent to the multinomial theorem as well ([imath]\frac{d^{m}}{dx^{m}} \prod_{k=1}^{n} f_k(x)[/imath]). Can someone produce a proof for this more general case (preferably by induction)? Any and all insights are welcome! | 1194948 | Why General Leibniz rule and Newton's Binomial are so similar?
The binomial expansion: [imath](x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}[/imath] The General Leibniz rule (used as a generalization of the product rule for derivatives): [imath](fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}[/imath] Both formulas can be obtained simply by induction; Newton's binomial also has a combinatorial proof (here's the relevant wikipedia page). It's striking how these formulas are similar; is there a possible connection between them? I was thinking that maybe the General Leibniz rule could be obtained using a combinatorial argument as well (hence the binomial coefficients)... |
1244613 | Divisors of Fermat numbers
For [imath]n > 1[/imath] let [imath]F_n = 2^{2^n} + 1[/imath] be a Fermat number and [imath]a = 2^{2^{n - 2}}(2^{2^{n - 1}} - 1)[/imath]. Prove that any divisor of the Fermat number [imath]F_n = 2^{2^{n}} + 1[/imath] is of the form [imath]k\,2^{n + 2} + 1[/imath]. I don't know where to begin to proof this theorem. Notice that this uses the stronger result of the divisors of fermat numbers. | 83938 | A theorem about prime divisors of generalized Fermat numbers?
A theorem of Édouard Lucas related to the Fermat numbers states that : Any prime divisor [imath]p[/imath] of [imath]F_n=2^{2^n}+1[/imath] is of the form [imath]p=k\cdot 2^{n+2}+1[/imath] whenever [imath]n[/imath] is greater than one. Does anyone know is there some similar theorem for generalized Fermat numbers: [imath]F_n(a)=a^{2^n}+1[/imath] ? I've been searching the internet but I couldn't find any similar theorem . |
879192 | What is this sequence of all permutations with gaps permissible
Let there be a sequence [imath]a_1, a_2, a_3,...,a_n[/imath] that represent some actions that you know are required to solve a problem. However, you do not know what order these actions need to be taken to solve that problem, but you know that all of them are required at some point. What is the shortest such sequence [imath]S[/imath] of actions with length [imath]m[/imath] in terms of [imath]n[/imath] such that all permutations of the actions are contained within [imath]S[/imath] with gaps and overlaps permissible? Example: Let's say you know that 3 steps are required to fix your computer, but you don't know in what order: Restarting a program. Installing a different program. Run antivirus. So one sequence containing all permutations [imath](123,132,213,231,312,321)[/imath] guaranteed would be [imath]S=123\ 123\ 123[/imath] (meaning [imath]m[/imath] has a guaranteed upper bound of [imath]n^2[/imath] if we extrapolate) but a shorter sequence would be [imath]S=1231231[/imath] (which I am not sure is of minimum length). How can we generate this minimum sequence? | 48808 | Shortest sequence containing all permutations
Given an integer [imath]n[/imath], define [imath]s(n)[/imath] to be the length of the shortest sequence [imath]S = (a_1, \cdots a_{s(n)})[/imath] such that every permutation of [imath]\{1,\cdots,n\}[/imath] is a subsequence of [imath]S[/imath]. If [imath]n=1[/imath], then [imath]S = (1)[/imath] is the shortest sequence containing all permutations of [imath]\{1\}[/imath], so s(1) = 1. If [imath]n=2[/imath], then [imath]S = (1, 2, 1)[/imath] contains all permutations of [imath]\{1,2\}[/imath] as a subsequence, so [imath]s(2)=3[/imath]. Is there a general formula for [imath]s(n)[/imath]? |
408538 | Principal [imath]G[/imath]-bundles as pull back bundles.
Let [imath]G[/imath] be a compact Lie group and consider a [imath]G[/imath]-universal bundle [imath]\pi: EG \to BG [/imath] where [imath]BG[/imath] is the classifying space for the group [imath]G[/imath] and the bundle [imath]\pi: EG \to BG [/imath] is defined as the principal [imath]G[/imath]-bundle such that [imath]EG[/imath] is a contractible space equipped with a free [imath]G[/imath]-action and [imath]BG:=EG / G[/imath]. Now let [imath]X[/imath] be a topological space. How can I prove that every principal [imath]G[/imath]-bundle [imath]\xi=(E,p,X)[/imath] with base-space [imath]X[/imath] can be realized as the pull-back of the universal bundle [imath]\pi: EG \to BG [/imath]? | 402980 | Universal property of universal bundles.
A classifying space for a group [imath]G[/imath] is a topological space [imath]BG[/imath] with a principle [imath]G[/imath]-bundle [imath]p : EG \to BG[/imath] where [imath]EG[/imath] is contractile, so that [imath]BG = EG/G[/imath]. A classifying space is universal in the sense that if [imath]q : E \to B[/imath] is a principle [imath]G[/imath]-bundle, then there is a continuous map [imath]B \rightarrow BG[/imath], unique up to homotopy, such that [imath]E[/imath] is the fiber product [imath]f^*EG[/imath]. Why this definition implies tha there is a one-to-one corrispondence between the isomorphism classes of [imath]G[/imath]-bundle on a space [imath]X[/imath] (that I denote with [imath]\mathcal{P}_G(X)[/imath]) and the homotopy classes of maps [imath]X \to BG[/imath] ([X,BG])? In practice, how can I show that [imath] [\mathcal{P}_G(X)] \simeq [X,BG] [/imath] |
1246027 | What is a function which differs when differentiated with respect to x and then with y to function differentiated with respect to y and then x?
What is a function which differs when differentiated with respect to [imath]x[/imath] and then with [imath]y[/imath] to function differentiated with respect to [imath]y[/imath] and then [imath]x[/imath]? Even if it differs at one particular point. | 725578 | When is partial differentiation commutitive
Consider a function: [imath]f:\mathbb R ^2\to\mathbb R [/imath] when does [imath]\dfrac {\partial f(x,t)}{\partial t\partial x}=\dfrac {\partial f(x,t)}{\partial x\partial t}[/imath] Thinking about it in terms of the limit definition of derivative, and thinking about it as taking slices of surfaces and measuring the slope on the edge, seems to be giving me the feeling that answer is always. However I have some memories of there being requirements like piecewise continuous, and smooth. What is the full set of conditions? |
1246101 | Why is e used for polar form of complex numbers?
This is a real basic question. Why is [imath]e[/imath] the base for polar form of complex numbers? In high school maths I learned that e is useful in derivatives etc. And it's conventional to use it for exponential growth/decay functions, although really we could use any base number for them if we tweak the exponent appropriately (eg. [imath]Ae^t = Ax^{kt})[/imath] for appropriate values of x and k). So could any number be used as the base for polar form of complex numbers? Is [imath]e[/imath] just convention or is there some special relationship to [imath]i[/imath] that makes this work? | 1233283 | Why did Euler use e to represent complex numbers?
From Euler we've learned that [imath]z=re^{i\theta}[/imath]. And it's easy to see that [imath]|z|^2=r^2[/imath], since [imath]re^{i\theta}\times re^{-i\theta}=r^2[/imath]. Why must we use e to represent these numbers correctly? It seems that I could arbitrarily choose a different exponent [imath]z=r\pi^{i\theta}[/imath] and get the same size for [imath]z[/imath] as I did before: [imath]|z|^2=r\pi^{i\theta}\times r\pi^{-i\theta}=r^2[/imath] What did I miss? |
1246269 | Stochastic Integral martingale if no [imath]dt[/imath] term?
There is a proposition in my book that For a process [imath]M_t[/imath] to be a martingale, it is necessary that its stochastic differential [imath]dM_t[/imath] has no [imath]dt[/imath] term. Why is this exactly? My guess is that it implies there is no change with respect to time [imath]t[/imath]. This is speculation, and is the story told with the 2nd property given regarding conditional expectation. A martingale only requires that: [imath]1. \space \forall t, E \|M_t \| < \infty[/imath] and [imath]2. \space \forall t,s>0, \space E[M_{t+s}|\mathcal{F}_t] = M_t[/imath] | 1035461 | show that the solution is a local martingale iff it has zero drift
Most financial maths textbook state the following: Given an [imath]n[/imath]-dimensional Ito-process defined by \begin{equation} X_t = X_0 + \int_0^{t} \alpha_s \,d W_s + \int_0^{t} \beta_s \,d s, \end{equation} where [imath](\alpha_t)_{t \geq0}[/imath] is a predictable process that is valued in the space of [imath]n \times d[/imath] matrices and [imath](W_t)_{t \geq 0}[/imath] is a [imath]d[/imath]-dimensional Brownian motion, \begin{equation} (X_t) \text{ is a local martingale } \quad \Longleftrightarrow \text{ It has zero drift.} \end{equation} Can anyone show me a reference for the proof of this statement or at least give a hint of how to construct this proof? (I know how to prove the ([imath]\Leftarrow[/imath]) direction, but I am not so sure about the other one.) |
1246322 | [imath]\mathbb{Z}/p^n[/imath] is an injective module over itself.
For any prime power [imath]p^n[/imath], prove that [imath]\mathbb{Z}/p^n[/imath] is an injective module over itself. | 1232291 | [imath]R/Ra[/imath] is an injective module over itself
Let [imath]R[/imath] be a PID, [imath]a\in R[/imath] be a nonzero nonunit in [imath]R[/imath]. Prove that [imath]R/Ra[/imath] is an injective module over itself. If [imath]R[/imath] is a PID, every [imath]R[/imath]- divisible module is injective, but the question concerns with [imath]R/Ra[/imath]-module, so I have no idea to solve this problem. Help me. |
1246482 | Is it possible to solve this system of equations?
Consider a system of equations given below: [imath] p_1 + p_2 + p_3 + p_4 + p_5 = 1 [/imath] [imath] x_1*p_1 + x_2*p_2 + x_3*p_3 +x_4*p_4 =0[/imath] [imath] x_1^2*p_1 + x_2^2*p_2 + x_3^2*p_3 +x_4^2*p_4 =1[/imath] [imath] x_1^3*p_1 + x_2^3*p_2 + x_3^3*p_3 +x_4^3*p_4 =v_1[/imath] [imath] x_1^4*p_1 + x_2^4*p_2 + x_3^4*p_3 +x_4^4*p_4 =v_2[/imath] [imath] x_1^5*p_1 + x_2^5*p_2 + x_3^5*p_3 +x_4^5*p_4 =v_3[/imath] [imath] x_1^6*p_1 + x_2^6*p_2 + x_3^6*p_3 +x_4^6*p_4 =v_4[/imath] [imath] x_1^7*p_1 + x_2^7*p_2 + x_3^7*p_3 +x_4^7*p_4 =v_5[/imath] [imath] x_1^8*p_1 + x_2^8*p_2 + x_3^8*p_3 +x_4^8*p_4 =v_6[/imath] Is there a method to solve [imath]x_i[/imath] and [imath]p_i[/imath] in terms of [imath]v_i[/imath]? Thank You! | 1238939 | [imath]10[/imath] Equations in [imath]10[/imath] variables
Define, [imath]F_k = xp^k + yq^k + zr^k + us^k + vt^k[/imath] Let, [imath]F_0 = 2[/imath] [imath]F_1 = 3[/imath] [imath]F_2 = 16[/imath] [imath]F_3 = 31[/imath] [imath]F_4 = 103[/imath] [imath]F_5 = 235[/imath] [imath]F_6 = 674[/imath] [imath]F_7 = \color{blue}{1667}[/imath] [imath]F_8 = 4526[/imath] [imath]F_9 = 11595[/imath] Solve for [imath]x, y, z, u, v, p, q, r, s, t[/imath] Can anyone post a solution if you have solved it? Thanks. [imath]\color{blue}{Edit}[/imath]: It seems there is a typo. Compare to OEIS A072684: [imath]2, 3, 16, 31, 103, 235, 674, \color{blue}{1669}, 4526, 11595,\dots[/imath] |
1246460 | The sum of n independent poisson([imath]\lambda_i[/imath]) random variables is poisson([imath]\sum_1^n \lambda_i[/imath]) random variable.
How can I show that the sum of n independet poisson random variables is a poisson random variable? I need to conclude that the sum of i.r.v with distribution [imath]Poisson(\lambda_i)[/imath], i=1,...,n, is [imath]Poisson(\sum_1^n \lambda_i)[/imath]. | 162839 | Induction for sum of Poisson distributed random variables
Given the identically distributed and independant random variables [imath]X_1,X_2,\ldots\sim\operatorname{Po}(\lambda)[/imath] and [imath]S_n=X_1+\ldots+X_n[/imath] show with induction that [imath]\Pr[S_n=k]=\frac{(n\lambda)^k}{k!}e^{-n\lambda}.[/imath] So far for [imath]n=1[/imath] via the definition of the density function for Poisson:[imath]\Pr[S_1=k]=\Pr[X_1=k]=\frac{e^{-\lambda}\lambda^k}{k!}[/imath] With [imath]n=2[/imath] and independency: [imath]\Pr[X_1+X_2=k]=\sum\limits_{n=0}^k\Pr[X_1=n]\Pr[X_2=k-n]=\sum\limits_{n=0}^k\frac{\lambda^n}{n!}e^{-\lambda}\frac{\lambda^{k-n}}{(k-n)!}e^{-\lambda}[/imath][imath]=\frac{1}{k!}e^{-2\lambda}\sum\limits_{n=0}^k\binom{k}{n}\lambda^n\lambda^{k-n}=\frac{(2\lambda)^k}{k!}e^{-2\lambda}[/imath] I assume now that with [imath]S_n\sim\operatorname{Po}(\lambda)[/imath] every [imath]X_i\sim\operatorname{Po}(\lambda/n)[/imath] and every Poisson variable can obviously split up, but how can I prove this? |
1246426 | How can I compute [imath]\sum_{n=0}^{\infty} 0.6^n[/imath]?
I am a computing teacher and just helping out some students with a math question. They have been asked to calculate the following: [imath]\sum_{n=0}^{\infty} 0.6^n[/imath] I am intrigued as to how one gets to the answer - i.e. the logic and process behind it if possible. | 370662 | Infinite Geometric Series Formula Derivation
We know that the formula for computing a geometric series is:[imath]\sum_{i=1}^{\infty}{a_0r^{n-1}} = \frac{a_0}{1-r}[/imath] Out of curiosity, I would like ask: Is there any ways the formula can be derived other than the following two ways? Method 1 (The way I found on my own): [imath]\sum_{i=1}^{\infty}{a_0r^{n-1}} \equiv S[/imath] [imath]S = a_0r^0+a_0r^1+a_0r^2+\cdots[/imath] [imath]S = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots\right)[/imath] [imath]S = r\left(a_0r^{-1} + S\right)[/imath] [imath]S = a_0 + rS[/imath] [imath](1-r)S = a_0[/imath] [imath]S = \frac{a_0}{(1-r)}[/imath] Note that for this to work, you must first confirm this: [imath]\lim_{n\to\infty} a_n = 0[/imath] Method 2 (The way I found on the web): [imath]\sum_{i=1}^{n}{a_0r^{n-1}} \equiv S_n[/imath] [imath]S_n = a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}[/imath] [imath]rS_n = r\left(a_0r^0+a_0r^1+a_0r^2+\cdots + a_0 r^{n-2} + a_0 r^{n-1}\right)[/imath] [imath]rS_n = a_0r^1 + a_0r^2 + a_0r^3 + \cdots + a_0 r^{n-1} + a_0 r^{n}[/imath] [imath]S_n-rS_n = a_0r^0 - a_0r^n[/imath] [imath](1-r)S_n = a_0 - a_0 r^n[/imath] [imath]S_n = \frac{a_0(1 - r^n)}{1-r}[/imath] Given:[imath]\left|r\right| < 1,[/imath] [imath]\lim_{n\to \infty} S_n = \lim_{n\to \infty}\frac{a_0(1 - r^n)}{1-r} = \frac{a_0}{1-r}[/imath] I personally prefer Method 1 because it is faster and more intuitive, as we don't have to multiply by [imath]r[/imath]. Method 1 for formula of partial sums: [imath]\sum_{i=1}^{n}{a_0r^{n-1}} \equiv S_n[/imath] [imath]S_n = a_0r^0+a_0r^1+a_0r^2+\cdots+a_0r^{n-2}+a_0r^{n-1}[/imath] [imath]S_n = r\left(a_0r^{-1} + a_0r^{0} + a_0r^1+\cdots+a_0r^{n-3}+a_0r^{n-2}\right)[/imath] [imath]S_n = r\left(a_0r^{-1} + S_n - a_0r^{n-1}\right)[/imath] [imath]S_n = a_0 + rS_n - a_0r^{n}[/imath] [imath](1-r)S_n = a_0 - a_0r^n[/imath] [imath]S_n = \frac{a_0(1 - r^n)}{(1-r)}[/imath] |
1185259 | How to explain the formula for the sum of a geometric series without calculus?
How to explain to a middle-school student the notion of a geometric series without any calculus (i.e. limits)? For example I want to convince my student that [imath]1 + \frac{1}{4} + \frac{1}{4^2} + \ldots + \frac{1}{4^n} = \frac{1 - (\frac{1}{4})^{n+1} }{ 1 - \frac{1}{4}}[/imath] at [imath]n \to \infty[/imath] gives 4/3? | 1434715 | Partial sums for [imath]\sum_{j=1}^\infty\frac{1}{3^j}[/imath] and similar.
I'm trying to take the limit of a series involving [imath]\sum_{j=1}^\infty\frac{1}{3^j}[/imath] and thinking that this might have a partial sum representation. Here it says that [imath]\sum_{i=1}^n \frac{1}{3^{i-1}} = \frac{3}{2}(1-\frac{1}{3^n})[/imath] is the partial sum representation for [imath]\sum_{i=1}^n \frac{1}{3^{i-1}}[/imath]: http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx But how is this derived? |
1247136 | Show that if [imath]r[/imath] is nilpotent in a ring with identity, then [imath]1-r[/imath] is a unit in [imath]R[/imath]
Let [imath]R[/imath] be a ring. An element [imath]r \in R[/imath] is called nilpotent if [imath]r^n=0[/imath] for some integer [imath]n \ge 1[/imath]. Show that if [imath]r[/imath] is nilpotent in a ring with identity, then [imath]1-r[/imath] is a unit in [imath]R[/imath]. Proof. Recall that in any ring we have [imath](−a)(−b) = (ab)[/imath] Thus in any ring with 1 (commutative or not) we have the following identities: [imath]1 −(−1)^na^n = (1 + a)(1 − a + a^2 − · · · + (−1)^{−1}a^{n−1})[/imath] If [imath]a^n = 0[/imath] then we obtain an inverse [imath]1 − a[/imath] Now, I'm not so familiar with series and am not sure if this suffices. I also assume this can be shown with matrices, integers....as long as it is a ring | 68191 | Let [imath]R[/imath] be a commutative ring with 1 then why does [imath]a\in N(R) \Rightarrow 1+a\in U(R)[/imath]?
Let [imath]R[/imath] be a commutative ring with 1, we define [imath]N(R):=\{ a\in R \mid \exists k\in \mathbb{N}:a^k=0\}[/imath] and [imath]U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.[/imath] Could anyone help me prove that if [imath]a\in N(R) \Rightarrow 1+a\in U(R)[/imath]? I've been trying to construst a [imath]b[/imath] such that [imath]ab=1[/imath] rather than doing it by contradiction, as I don't see how you could go about doing that. |
1247155 | Let [imath]X_1[/imath] and [imath]X_2[/imath] be connected topological spaces. I want to show that the product [imath]X_1 \times X_2[/imath] is connected.
Let [imath]X_1[/imath] and [imath]X_2[/imath] be connected topological spaces. I want to show that the product [imath]X_1 \times X_2[/imath] is connected. By definition the base from which the topology [imath]\mathcal J_{X_1 \times X_2} = \{\cup_{\alpha \in I} \mathcal B_{\alpha}, \mathcal B_{\alpha} \in \mathcal B \}[/imath] is induced is given as [imath]\mathcal B = \{ U_1 \times U_2 \mid U_i \in \mathcal J_i \}[/imath]. I've tried proving connectedness by assuming [imath]X_1 \times X_2 = U_1 \cup U_2[/imath] with [imath]U_1 \cap U_2 = \emptyset[/imath], [imath]U_1, U_2 \in \mathcal B_{X_1 \times X_2}[/imath], and then use the fact that [imath]X_i[/imath] is connected. If [imath]U_i = \emptyset[/imath] we are done, so one can assume [imath]U_i \neq \emptyset, X_1 \times X_2[/imath]. Can someone help me to show that [imath]X_1 \times X_2[/imath] is connected ? | 338027 | Product of connected spaces
You have two connected topological spaces [imath](A,B)[/imath]. Prove that [imath]A\times B[/imath] is also connected. I understand that I have to prove that there is a point in [imath]B[/imath] (call it [imath]b[/imath]), that makes [imath]A\times\{b\}[/imath] homeomorphic to [imath]A[/imath] making it connected to [imath]A\times B[/imath]. Then prove that [imath]\{a\}\times B[/imath] is connected in [imath]A\times B[/imath]. But I don't really know where to being with this. If you could help that would be appreciated. |
1187430 | Finding equation using hyperbolic transcendental functions.
I have tried and tried but cannot for the life of me see how one equation follows onto the other... can anybody help?? [imath]\Omega(\theta)=-b \coth(\operatorname{arsinh}(e^{a\theta} \sinh c_0 ))[/imath] [imath]\implies \Omega(\theta)=\sqrt{c_1 e^{-2a\theta} + b^2}[/imath] note that: [imath]c_0=-\operatorname{arcoth}\left(\frac{\Omega_0}{b}\right)[/imath] Eichberger quotes (at the bottom of page 5) "By inserting (11) into (8), using identities of the hyperbolic transcendental functions and carefully observing [imath]±[/imath] signs, we obtain [imath]Ω[/imath] as a function of [imath]θ[/imath]." http://www.roulette.gmxhome.de/roulette%5B1%5D.pdf | 1193127 | Rearranging equations using hyperbolic transcendental functions
I have tried and tried but cannot for the life of me see how one equation follows onto the other... can anybody help?? [imath]\Omega(\theta)=-b.\coth(\operatorname{arsinh}(\exp a\theta . \sinh(c_0)))[/imath] [imath]\implies \Omega(\theta)=\sqrt(c_1.\exp -2a\theta + b^2)[/imath] Note that: [imath]C_0=-\operatorname{arcoth}\left(\frac{\Omega_0}{b}\right)[/imath] Eichberger quotes (at the bottom of page 5) "By inserting (11) into (8), using identities of the hyperbolic transcendental functions and carefully observing ± signs, we obtain Ω as a function of θ." http://www.roulette.gmxhome.de/roulette%5B1%5D.pdf |
1247490 | Using the Mean Value theorem to find an inequality(s)
Can anyone help me with the following question? Use the Mean Value Theorem to show for all [imath]x>0[/imath] [imath]\frac{x}{1+x}\lt\ln{(1+{x})}\lt{x}[/imath] I can easily see that the first term is less than the last term but can't see my way past that point! | 1245994 | Using mean value theorem for multiple inequalities
Use the Mean Value Theorem to prove that [imath]\frac{(x-1)}{x} < \ln x < x-1[/imath] for [imath]x > 1[/imath]. I was thinking of breaking up the inequality into \frac{(x-1)}{x} < \ln x[imath], and [/imath]\ln x < x-1[imath] and just proving each individual inequality using the mean value theorem approach:[/imath] So, \ln x - \frac{x-1}{x} > 0[imath]. So, [/imath]\frac{x\ln x - x - 1}{x} > 0[imath], so let [/imath]g(x) = x\ln x - x - 1 > 0[imath]. Thus, [/imath]g'(x)$ |
1243455 | How to randomly select a point from the surface of a unit sphere ?
Construct in [imath]\Bbb R^k[/imath] a random variable [imath]X[/imath] that is uniformly distributed over the surface of the unit sphere in the sense that [imath]|X|=1[/imath] and [imath]UX[/imath] has the same distribution as [imath]X[/imath] for orthogonal transformations [imath]U[/imath]. Hint Let [imath]Z[/imath] be uniformly distributed in the unit ball in [imath]\Bbb R^k[/imath], define [imath]\phi(x) = \frac{x}{\|x\|}[/imath] and take [imath]X=\phi(Z)[/imath]. I understand the intuition behind the solution i.e. orthogonal transform does not change the angles between points therefore the area of the [imath]\phi[/imath] preimage of the orthogonal transformed set on the surface remains unchanged. But, how do I formalize it ? | 87230 | Picking random points in the volume of sphere with uniform probability
I have a sphere of radius [imath]R_{s}[/imath], and I would like to pick random points in its volume with uniform probability. How can I do so while preventing any sort of clustering around poles or the center of the sphere? Since I'm unable to answer my own question, here's another solution: Using the strategy suggested by Wolfram MathWorld for picking points on the surface of a sphere: Let [imath]\theta[/imath] be randomly distributed real numbers over the interval [imath][0,2\pi][/imath], let [imath]\phi=\arccos(2v−1)[/imath] where [imath]v[/imath] is a random real number over the interval [imath][0,1][/imath], and let [imath]r=R_s (\mathrm{rand}(0,1))^\frac13[/imath]. Converting from spherical coordinates, a random point in [imath](x,y,z)[/imath] inside the sphere would therefore be: [imath]((r\cos(\theta)\sin(\phi)),(r\sin(\theta)\sin(\phi)),(r\cos(\phi)))[/imath]. A quick test with a few thousand points in the unit sphere appears to show no clustering. However, I'd appreciate any feedback if someone sees a problem with this approach. |
1248790 | Does the improper integral [imath]\int_{0}^{\infty}\sin(x^2)\;\mathrm dx[/imath] converge?
Does the improper integral [imath]\int_{0}^{\infty}\sin(x^2)\;\mathrm dx[/imath] converge? So if it converges then [imath]\lim_{b \to\infty}\int_{0}^{b}\sin(x^2)\;\mathrm dx[/imath] exists and our integral converges to this limit. I have tried the substitution [imath]y=x^2[/imath] but this does not make the calculation easier. Another way I know is to use the integral test but it also seems to be useless now. | 378012 | Trig Fresnel Integral
[imath]\int_{0}^{\infty }\sin(x^{2})dx[/imath] I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you. I have not done complex analysis (only real analysis as I am a high school student) so how can I evaluate it using elementary functions (without complex analysis)? |
306681 | Examples of modules such that [imath]M\oplus N_1\simeq M\oplus N_2[/imath], but [imath]N_1\not\simeq N_2[/imath]?
Are there any examples of modules such that [imath]M\oplus N_1\simeq M\oplus N_2[/imath], but [imath]N_1\not\simeq N_2[/imath] as modules? I thought of taking [imath]\bigoplus_{i\geq 0}\mathbb{Z}\oplus\mathbb{Z}\simeq\bigoplus_{i\geq 0}\mathbb{Z}\oplus\{0\}[/imath] as [imath]\mathbb{Z}[/imath]-modules, but [imath]\mathbb{Z}\not\simeq\{0\}[/imath]. Is this correct? Even if it is, I felt it's kind of a cheap example, and was hoping to see something more interesting. | 1248681 | Is there a counterexample for the claim: if [imath]A \oplus B\cong A\oplus C[/imath] then [imath]B\cong C[/imath]?
Here [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are [imath]R[/imath]-modules. Is there a counterexample for the claim: if [imath]A \oplus B\cong A\oplus C[/imath] then [imath]B\cong C[/imath]? And what if [imath]B[/imath] and [imath]C[/imath] are finitely generated? |
1249332 | [imath] \forall \epsilon \in \mathbb{R}^{*}_{+} , | x -y | < \epsilon \iff x = y [/imath]
Prove that for all [imath] \epsilon > 0, \epsilon \in \mathbb{R} [/imath] for every [imath] x, y \in \mathbb{R} [/imath] if [imath] | x - y | < \epsilon \iff x = y[/imath] (this question has similar ones in, but this one has the full proof, not only the hint). | 1166255 | Prove : [imath]|x - y| \leqslant \epsilon[/imath] [imath]\forall \epsilon > 0[/imath] iff [imath]x = y[/imath]
I have to do this exercise for my math study, and I'm having trouble with doing the second part of it. Let [imath]x, y, \epsilon \in \mathbb{R}[/imath] and [imath]\epsilon > 0[/imath]. Prove: [imath]|x - y| \leqslant \epsilon[/imath] [imath]\forall \epsilon > 0 \Leftrightarrow x = y[/imath] I think I have the left implication [imath]\Leftarrow[/imath]: Assume [imath]x = y \Rightarrow x - y = 0 \Rightarrow|x - y| = 0 = |0| \Rightarrow|x - y| = 0 < \epsilon \Rightarrow |x - y| \leqslant \epsilon[/imath] [imath]\forall \epsilon > 0 \in \mathbb{R}[/imath] Is this argument correct? For the right implication, I do only have an idea of how to prove it. I think I have to assume first that [imath]x > y[/imath] and then [imath]x < y[/imath] and get a contradiction form both. Could you please explain me the right implication and tell me if my left implication is correct? Thanks in advance! |
945353 | Finding minimum value of a trigonometric expression
If [imath]A+B+C=\pi[/imath] then what is the minimal value of [imath]\tan^{2}A/2+\tan^{2}B/2+\tan^{2}C/2[/imath] | 842881 | Trigonometric Triangle Equality
[imath]A, B, C[/imath] are the angles of a triangle then [imath]tan^2(A/2)+tan^2(B/2)+tan^2(C/2)[/imath] is always greater than what integral value. |
1247279 | Showing that an indefinite integral of a function with a jump discontinuity is not differentiable at the jump discontinuity
I was trying to help this person with his question. Integrable but not differentiable function Showing continuity was relatively easy, but I'm having trouble with showing that the function is not differentiable at [imath]c[/imath]. I tried using the hint and using one-sided derivatives, but all I got was that [imath]F'(x) = f(x)[/imath] for all [imath]x\neq c[/imath], and all I managed to show from that was that [imath]\lim_\limits{x \to c} F'(x)[/imath] does not exist. But all that showed was that the derivative of [imath]F[/imath] wasn't continuous at [imath]c[/imath], not that [imath]F[/imath] wasn't differentiable at [imath]c[/imath]. So I tried going down to Riemann sums. But I got stuck when showing that the last bit was a contradiction, since the absolute value of an integral is at most the integral of the absolute value of the integrand, not the other way round. I was wondering if it's salvageable, or if there are other errors I didn't see. Thanks! | 1245984 | Integrable but not differentiable function
Suppose [imath]f(x)[/imath] is continuous on [imath][a, b][/imath] except at a point [imath]c[/imath] in [imath](a, b)[/imath] at which [imath]f(x)[/imath] has a jump discontinuity. For [imath]x[/imath] in [imath](a,b)[/imath], set [imath]F(x)=\int_a^xf(t) \, dt[/imath]. Show that [imath]F(x)[/imath] is continuous but not differentiable at [imath]x =c[/imath]. Hint: first explain why [imath]f(x)[/imath] is integrable on [imath][a,b][/imath], then consider one-sided derivatives of [imath]F(x)[/imath] at [imath]x =c[/imath]. I don't know how to step up because it seems conflict to me. [imath]f(x)[/imath] is not continuous because it has a jump but I have to prove it continuous, how? Can someone guide me, please. |
1249489 | Solving for [imath]x[/imath] after simplifying, please check my work!
I was presented with the question: [imath](\frac{x}{a})^{3.2} + (\frac{y}{b})^{3.2} = 1[/imath] While: [imath]\frac{a}{b} = \frac{174.1}{86}=c[/imath] I began to simplify and reached: [imath]x = c(b^{3.2}-y^{3.2})^{1/3.2}[/imath] [imath]y = \frac{(a^{3.2} - x^{3.2})^{1/3.2}}{c}[/imath] My goal is to solve for [imath]x[/imath]. Thus, I substituted [imath]y[/imath]: [imath]x = c(b^{3.2}-(b^{3.2}-((\frac{x^{3.2}}{c^{3.2}})^{1/3.2})^{3.2})^{1/3.2}[/imath] I then simplify: [imath]x = c(b^{3.2}-(b^{3.2}-(\frac{x^{3.2}}{c^{3.2}}))^{1/3.2}[/imath] How should I proceed? I am looking for [imath]x =[/imath] | 1249149 | Please check my simplification for x?
I'm hoping someone could check my work. I'm trying to simplify the equation for [imath]x[/imath]. [imath](\frac{x}{a})^{3.2} + (\frac{y}{b})^{3.2} = 1[/imath] [imath]\frac{a}{b} = \frac{174.1}{86}[/imath] Thus I began by substituting [imath]a[/imath] with: [imath]a = \frac{174.1}{86}b[/imath] [imath]((\frac{86}{174.1})(\frac{x}{b}))^{3.2} + (\frac{y}{b})^{3.2} = 1[/imath] [imath]((\frac{86}{174.1})(\frac{x}{b}))^{3.2} = 1 - (\frac{y}{b})^{3.2}[/imath] [imath](\frac{x}{b})^{3.2} = \frac{(1 - (\frac{y}{b})^{3.2})(174.1)^{3.2}}{(86)^{3.2}}[/imath] [imath]x^{3.2} = \frac{(1 - (\frac{y}{b})^{3.2})(174.1)^{3.2}(b)^{3.2}}{(86)^{3.2}}[/imath] [imath]x^{3.2} = \frac{(1 - (\frac{y}{b})^{3.2})((174.1)(b))^{3.2}}{(86)^{3.2}}[/imath] [imath]x^{\frac{16}{5}} = \frac{(1 - (\frac{y}{b})^{3.2})((174.1)(b))^{3.2}}{(86)^{3.2}}[/imath] [imath]x = [\frac{(1 - (\frac{y}{b})^{3.2})((174.1)(b))^{3.2}}{(86)^{3.2}}]^{\frac{5}{16}}[/imath] Is my work correct, or am I completely off the board? |
1249391 | Complicated Planar Geometry
A regular polygon [imath]\mathcal{P}[/imath] is inscribed in a circle [imath]\Gamma[/imath]. Let [imath]A[/imath], [imath]B[/imath], and [imath]C[/imath] be three consecutive vertices of the polygon [imath]\mathcal{P}[/imath], and let [imath]M[/imath] be a point on the arc [imath]AC[/imath] of [imath]\Gamma[/imath] that does not contain [imath]B[/imath]. Prove that [imath]MA \cdot MC = MB^2 - AB^2.[/imath] I have tried inscribing the polygon in a unit circle with no luck. Any help is greatly appreciated! | 1165665 | A Polygon is inscribed in a circle [imath]\Gamma[/imath]
A regular polygon P is inscribed in a circle [imath]\Gamma[/imath]. Let A, B, and C, be three consecutive vertices on the polygon P, and let M be a point on the arc AC of [imath]\Gamma[/imath] that does not contain B. Prove that [imath]MA\cdot MC=MB^2-AB^2[/imath] I inscribed the polygon P in the unit circle and let B=1. Past this, I'm not really sure how to proceed. It might be helpful to have M at 1 as well but given that the polygon would have the n-th roots of unity in it B seemed like a good choice. What would be a good way to proceed from here? |
1249680 | How to prove that every countable compact Hausdorff space is homeomorphic to a well-ordered set with its order topology?
I encountered this problem in a textbook: Let [imath]X[/imath] be a countable compact Hausdorff topological space, I was asked to prove that [imath]X[/imath] is always homeomorphic to a (necessarily countable) topological space [imath]Y[/imath] equipped with the order topology of a certain well-order [imath]\leqslant[/imath] on [imath]Y[/imath] (a well-ordering on [imath]Y[/imath] is a total order [imath]\leqslant[/imath] on [imath]Y[/imath] such that every subset of [imath](Y,\leqslant)[/imath] admits a minimal element, or equivalently, an ordinal). But I had no ideal where to start (I guessed that firstly I had to prove [imath]X[/imath] is first countable, but I failed). Can anyone help me about it? Any useful hint will be appreciated. | 38247 | Countable compact spaces as ordinals
I heard at some point (without seeing a proof) that every countable, compact space [imath]X[/imath] is homeomorphic to a countable successor ordinal with the usual order topology. Is this true? Perhaps someone can offer a sketch of the proof or suggest a topology/ordinal text which treats countable spaces in depth. I also wonder if these spaces are metrizable or if they can be embedded into [imath]\mathbb{R}[/imath]. By the way, I'm assuming [imath]X[/imath] is Tychonoff but perhaps this can be reduced to a weaker separation axiom. Edit: Just to clarify and avoid erroneous future editing,"countable compact space" means "compact space, whose underlying set is countable." This is different from "countably compact space." |
1250190 | projective plane and topology
Gluing the Mobius strip with a disk altogether along their boundaries gives [imath]\mathbb RP^2[/imath]. Please give me some hint or explanation how to solve it, why it is true. And give a topology on this space. | 77569 | [imath]\Bbb RP^2[/imath] as the union of a Möbius band and a disc
I spent some time trying to understand that the projective plane, [imath]\Bbb RP^2[/imath], can be viewed as the union of a Möbius band and a disc. I consider this using Homogeneous coordinates. But I still hope to really 'see' this fact in some way. |
1250302 | Infiniteness of set of primes such [imath]f[/imath] have root [imath]\mod p[/imath]
Let [imath]f \in \mathbb{Z}[x][/imath] be non constant. How to prove that exists infinitely many primes such [imath]f[/imath] have root in [imath]\mathbb{Z/_{(p)}}[/imath]. I spent much time, but with no benefits. | 1019538 | Primes dividing a polynomial
Let [imath]g(x)\in \mathbb{Z}[x][/imath], a nonconstant polynomial. Show that the set of primes [imath]p[/imath] such that [imath]p\mid g(n)[/imath] for some [imath]n\in \mathbb{Z}[/imath] is infinite. I don't know how to start. I have tried asuming that the set is finite but i haven't reached any contradiction. |
1250363 | Show that if [imath]R[/imath] and [imath]S[/imath] are ideal of a ring [imath]A[/imath] then the product [imath]R\cdot S[/imath] is a ideal of [imath]A[/imath].
How to prove that if [imath]R[/imath] and [imath]S[/imath] are ideal of a ring [imath]A[/imath] then the product [imath]R\cdot S[/imath] is a ideal. I can't show only that if [imath]x, y\in R\cdot S[/imath] then [imath]x-y\in R\cdot S[/imath]. The other axioms of ideal I showed | 329079 | Let [imath]R[/imath] be a commutative ring and let [imath]I[/imath] and [imath]J[/imath] be ideals of [imath]R[/imath]. Show [imath]IJ[/imath] is an ideal of [imath]R[/imath].
Let [imath]R[/imath] be a commutative ring. For ideals [imath]I[/imath], [imath]J \in R[/imath] define [imath]IJ[/imath] to be the set [imath]\{a_1b_1 +\ldots+a_nb_n : n\in\mathbb N[/imath]; [imath]a_i\in I[/imath]; [imath]b_j \in J\}.[/imath] Prove that [imath]IJ[/imath] is an ideal in [imath]R[/imath]. |
1250655 | If [imath]a=b[/imath] then [imath]a+c=b+c[/imath]?
A friend of mine just asked me how to prove that if [imath]a=b[/imath] then [imath]a+c=b+c[/imath], where [imath]a,b[/imath] and [imath]c[/imath] are real numbers, I'm not sure what I should answer. I have a book called introduction to logic and to the theory of the deductive sciences by Alfred Tarski, which is about propositional logic, and I remember reading that two things [imath]a[/imath] and [imath]b[/imath] are equal if any proposition that is true about [imath]a[/imath] is also true about [imath]b[/imath] and vice-versa. However I think this isn't very formal. I haven't taken any set theory course, I think that another way to justify it is to say that sum is a function and since the ordered pairs [imath](a,c)[/imath] and [imath](b,c)[/imath] are equal then [imath]+(a,c)=+(b,c)[/imath]. But I'm not too convinced. If we use the standard axioms how would we justify [imath]a+c=b+c[/imath] using the mainstream axioms of today. I think there is something Zermelo-Frankl with choice. Would these be enough, what properties of the real numbers do we need? Can we prove it using the usual construction of the real numbers and Zermelo-Frankl? As you can probably see I am not very knowledgeable about these topics, so I would like a delicate explanation. Many thanks and regards. | 1100564 | Is there a law that you can add or multiply to both sides of an equation?
It seems that given a statement [imath]a = b[/imath], that [imath]a + c = b + c[/imath] is assumed also to be true. Why isn't this an axiom of arithmetic, like the commutative law or associative law? Or is it a consequence of some other axiom of arithmetic? Thanks! Edit: I understand the intuitive meaning of equality. Answers that stated that [imath]a = b[/imath] means they are the same number or object make sense but what I'm asking is if there is an explicit law of replacement that allows us to make this intuitive truth a valid mathematical deduction. For example is there an axiom of Peano's Axioms or some other axiomatic system that allows for adding or multiplying both sides of an equation by the same number? In all the texts I've come across I've never seen an axiom that states if [imath]a = b[/imath] then [imath]a + c = b + c[/imath]. I have however seen if [imath]a < b[/imath] then [imath]a + c < b + c[/imath]. In my view [imath]<[/imath] and [imath]=[/imath] are similar so the absence of a definition for equality is strange. |
1211286 | Residue theorem application
How could we use the Residue theorem to calculate the following integral: [imath]\int_0^{2\pi} \frac{1}{1-2p\cos{x} + p^2} dx[/imath] where [imath]p[/imath] is a real constant, such that [imath]p\in ]0,1[[/imath] Thank you! | 211058 | Evaluating [imath]\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta[/imath]
I need solve this integral, and I tried various methods of solving and did not get it. The integral is: [imath]\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta,[/imath] where [imath]t[/imath] is a positive integer. |
1251000 | Evaluating [imath]\lim_{n\to\infty}\int_0^1x^nf(x)\,dx[/imath].
Let [imath]f[/imath] be a continuous function on [0,1]. Evaluate [imath]\lim_{n\to \infty} \int_0^1 x^nf(x)dx[/imath] My approach : Consider [imath]\int x^nf(x)dx = \frac{f(x)x^{n+1}}{n+1} - \frac{1}{n+1}\int x^{n+1}f(x)dx[/imath] by using integration by parts. then applying limit with [imath]n\to \infty[/imath] we get [imath]\int x^nf(x)dx = 0[/imath], regardless of the limits of integral. Is this approach correct? | 1248342 | A problem that I'm not sure whether to use Weierstrass Approximation Theorem
[imath]\text{Let }f:[0,1]\to \Bbb R\text{ be a continuous function}[/imath] Evaluate the function [imath]\lim_{n\to \infty}\int_{0}^{1}x^nf(x)\,dx[/imath] Here is my work: Given [imath]\epsilon \gt 0[/imath],since [imath]f(x) \in C^0[0,1][/imath], by Weierstrass Approximation Theorem, there is a polynomial [imath]p(x)[/imath] s.t. [imath]|f(x)-p(x)|\lt \varepsilon[/imath] So we have [imath] \begin{align} \int_{0}^{1}x^n|f(x)-p(x)|\,dx & \lt \int_{0}^{1}x^n\varepsilon\, dx\\ & = \left[\varepsilon\,\frac{x^{n+1}}{n+1} \right]_{0}^{1}\\ & = \frac{\varepsilon}{n+1} \end{align} [/imath] Since [imath]p(x)[/imath] is a polynomial, suppose it has degree [imath]m[/imath], and has the form [imath]p(x)=a_0+a_1x+a_2x^2+\cdots+a_mx^m[/imath] [imath]a_i\in \Bbb R\text{ for each }i[/imath]. Integrate it we get [imath]\int_{0}^{1}x^n\,p(x)\,dx=\frac{a_0}{n+1}+\frac{a_1}{n+2}+\frac{a_2}{n+3}+\cdots+\frac{a_m}{n+m+1}[/imath] By triangle inequality, [imath]0\le |f(x)|\le |f(x)-p(x)|+|p(x)|[/imath], so [imath] \begin{align} 0\le \int_{0}^{1}x^n\,|f(x)|\,dx & \le \int_{0}^{1}x^n|f(x)-p(x)|\,dx+\int_{0}^{1}x^n|p(x)|\,dx\\ & \le \int_{0}^{1}x^n|f(x)-p(x)|\,dx+\int_{0}^{1}x^n\sum_{i=0}^{m}\lvert a_ix^i\rvert\,dx \end{align} [/imath] But we know that both [imath]\int_{0}^{1}x^n|f(x)-p(x)|\,dx\text{ and }\int_{0}^{1}x^n\sum_{i=0}^{m}\lvert a_ix^i\rvert\,dx[/imath] goes to [imath]0 \text{ as }n\to \infty[/imath] Then by pinching, [imath]\int_{0}^{1}x^n\,|f(x)|\,dx\to 0\text{ as }n\to \infty[/imath] then [imath]\int_{0}^{1}x^n\,f(x)\,dx\to 0\text{ as }n\to \infty[/imath] I don't I'm do the right thing? because I just come up with this idea, and seems strange. |
1250355 | Product of any two arbitrary positive definite matrices is positive definite or NOT?
Suppose that , [imath]A[/imath] and [imath]B[/imath] are [imath]n\times n[/imath] positive definite matrices and > [imath]I[/imath] be [imath]n\times n[/imath] identity matrix. Then which of the followings are positive definite ? (i) [imath]A+B[/imath] (ii) [imath]ABA[/imath] (iii) [imath]A^2+I[/imath] (iv) [imath]AB[/imath] I know the definition of positive definite as : [imath]\color{red}{A_{n\times n}}[/imath] [imath]\color{red}{\text{is positive definite if it's quadratic form}} [/imath] [imath]\color{red}{x^TAx>0}[/imath] Since [imath]A[/imath] and [imath]B[/imath] are positive definite so, [imath]x^TAx>0[/imath] and [imath]x^TBx>0[/imath]. Then, [imath]x^T(A+B)x=x^TAx+x^TBx>0.[/imath] So [imath]A+B[/imath] is positive definite. I am confused about the product.. I saw a lot of questions in this site about the product of positive definiteness. But the answer in those questions it is assume that the matrices are symmetric. For example see the answer of this question. I want to know whether the product of any two arbitrary positive definite matrices is positive definite or NOT with a valid proof or counter example.... | 1010887 | Positive definiteness of the matrix [imath]A+B[/imath]
Let, [imath]A[/imath] & [imath]B[/imath] are [imath]n\times n[/imath] positive definite matrices & [imath]I[/imath] be the [imath]n\times n[/imath] identity matrix. Then which of the followings are positive definite? (a) [imath]A+B[/imath] (b) [imath]ABA[/imath] (c) [imath]A^{2}+I[/imath] (d) [imath]AB[/imath] I know that, [imath]A^{2}+I[/imath] is positive definite, as if [imath]\lambda[/imath] is an eigen value of [imath]A[/imath] then [imath](1+\lambda^2)[/imath] is an eigen value of [imath]A^2+I[/imath]. I think (d) is true.Suppose, [imath]\lambda_{1}[/imath] & [imath]\lambda_{2}[/imath] be two eigen values of [imath]A_{2\times 2}[/imath] matrix & [imath]\beta_{1}[/imath], [imath]\beta_{2}[/imath] be two eigen values of [imath]B_{2\times 2}[/imath] matrix. Now, [imath]det(AB)=det(A)det(B)=\lambda_{1}.\lambda_{2}.\beta_{1}.\beta_{2}.[/imath] As, [imath]\lambda_{1},\lambda_{2},\beta_{1},\beta_{2}[/imath] are all positive so eigen values of [imath]AB[/imath] are all positive, so [imath]AB[/imath] is positive definite. Similarly, [imath]ABA[/imath] is positive definite.But I am not sure & I have no idea about [imath]A+B[/imath]. |
1251077 | Proving that a linear functional is matrix trace
Let [imath]W=\operatorname{M}_{n\times n}(\mathbb{F})[/imath] (square matrices [imath]n\times n[/imath] over [imath]\mathbb{F}[/imath]), and [imath]f\in W^*[/imath]. If [imath]f(AB)=f(BA)[/imath] for every [imath]A,B\in W[/imath] and [imath]f(I)=n[/imath] prove that [imath]f(A)=\operatorname{tr}A[/imath] for every [imath]A\in W[/imath]. I can understand why it's true, the only part in [imath]AB[/imath] that is the same in [imath]BA[/imath] is the diagonal, and together with [imath]f(I)=n[/imath] it is pretty obvious. But how can I prove it? I tried writing the base for [imath]W[/imath] and then writing [imath]I[/imath] as a linear combination of that base, but I got stuck. Thank you. | 104854 | Characterization of the trace function
We know that the trace of a matrix is a linear map for all square matrices and that [imath]\operatorname{tr}(AB)=\operatorname{tr}(BA)[/imath] when the multiplication makes sense. On the Wikipedia page for trace, under properties, it says that these properties characterize the trace completely in the following sense: If [imath]f[/imath] is a linear function on the space of square matrices satisfying [imath]f(xy)=f(yx)[/imath], then [imath]f[/imath] and [imath]\operatorname{tr}[/imath] are proportional. A note on the bottom of the page gives the justification, but I do not understand the logic of it. Thanks |
1251060 | How prove that [imath]\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)[/imath]
How prove that sum [imath]\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)[/imath] | 544228 | Finite Sum [imath]\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}[/imath]
Question : Is the following true for any [imath]m\in\mathbb N[/imath]? [imath]\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}[/imath] Motivation : I reached [imath](\star)[/imath] by using computer. It seems true, but I can't prove it. Can anyone help? By the way, I've been able to prove [imath]\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}[/imath] by using [imath](\star)[/imath]. Proof : Let [imath]f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.[/imath] We know that [imath]f(x)\gt0[/imath] if [imath]0\lt x\le {\pi}/{2}[/imath], and that [imath]\lim_{x\to 0}f(x)=1/3[/imath]. Hence, letting [imath]f(0)=1/3[/imath], we know that [imath]f(x)[/imath] is continuous and positive at [imath]x=0[/imath]. Hence, since [imath]f(x)\ (0\le x\le {\pi}/2)[/imath] is bounded, there exists a constant [imath]C[/imath] such that [imath]0\lt f(x)\lt C[/imath]. Hence, substituting [imath]x={(k\pi)}/{(2n+1)}[/imath] for this, we get [imath]0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.[/imath] Then, the sum of these from [imath]1[/imath] to [imath]n[/imath] satisfies [imath]0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.[/imath] Here, we used [imath](\star)[/imath]. Then, considering [imath]n\to\infty[/imath] leads what we desired. |
1251544 | Proof of [imath]\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}[/imath]
Prove that for [imath]n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} [/imath] I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality | 1220203 | Proving [imath] 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}[/imath] for all [imath]n\geq 2[/imath] by induction
Question: Let [imath]P(n)[/imath] be the statement that [imath]1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} <2- \dfrac{1}{n}[/imath]. Prove by mathematical induction. Use [imath]P(2)[/imath] for base case. Attempt at solution: So I plugged in [imath]P(2)[/imath] for the base case, providing me with [imath]\dfrac{1}{4} < \dfrac{3}{2}[/imath] , which is true. I assume [imath]P(n)[/imath] is true, so I need to prove [imath]P(k) \implies P(k+1)[/imath]. So [imath]\dfrac{1}{(k+1)^2} < 2 - \dfrac{1}{k+1}[/imath]. I don't know where to go from here, do I assume that by the Inductive hypothesis that it's true? |
1055923 | Showing the group of Inner Automorphism are a normal subgroup of the Automorphisms
I am trying to show that the group of Inner Automorphisms are a normal subgroup of the Automorphisms of a group G. So far I have this. Let [imath]\phi_x:G\rightarrow G[/imath] be the inner automorphism defined as [imath]\phi_x(g)=xgx^{-1} \forall g \in G[/imath]. It is clear that [imath]Inn(G)\neq \emptyset[/imath] since [imath]e\in G[/imath] such that [imath]\phi_x(e)=xex^{-1}=e.[/imath] We will show closure and inverses. Observe that [imath]\phi_x\circ \phi_y(g)=\phi_x(ygy^{-1})=xygy^{-1}x^{-1}[/imath]. My question here is for composition of functions, what are you trying to show? Is it just that [imath]\phi_x\circ \phi_y(g)=\phi_{xy}(g)?[/imath]. Now for inverses. (I am trying to follow along with this question here Set of all Inner Automorphisms is a subgroup of the set of all Automorphisms of a group [imath]G[/imath]). So in my case [imath]\phi_e[/imath] is the identity automorphism. Thus [imath](\phi_x)^{-1}=\phi_{x^{-1}}[/imath]. Thus [imath](\phi_x)^{-1}\phi_h=\phi_{x^{-1}}\phi_h=\phi_{hx^{-1}}=\phi_e\in Inn(G)[/imath]. Thus the Inner automorphisms are a subgroup. Now to show normality. We need to show that [imath]\gamma\circ\phi_x\circ\gamma^{-1}\in Inn(G)[/imath]. Since [imath]\gamma\in Aut(G), \gamma[/imath] is a homomorphism. Thus For all [imath]g\in G[/imath] [imath]\gamma\circ\phi_x\circ\gamma^{-1}(g)=(\gamma\circ\phi_x)\gamma^{-1}(g)=\gamma(x\gamma^{-1}(g)x^{-1})=\gamma(x)\gamma(\gamma^{-1}(g))\gamma(x^{-1}=\gamma(x)g\gamma(x^{-1}).[/imath] But this is in the group Inn(G). Thus Inn(G) is a normal subgroup of the Aut(G). | 824062 | Inner automorphisms form a normal subgroup of [imath]\operatorname{Aut}(G)[/imath]
For an arbitrary group [imath](G,\cdot)[/imath] let [imath]\operatorname{Aut}(G) = \{f: G \to G \mid f \text{ is an isomorphism}\}[/imath] be the set of all automorphisms of the group [imath]G[/imath]. We assume that [imath](\operatorname{Aut}(G),\circ)[/imath] where [imath]\circ[/imath] is composition of mappings is a group. 1) Prove that for arbitrary [imath]a \in G[/imath] there is an automorphism [imath]p_a: G \to G;\quad p_a(x) = a^{-1}xa[/imath]. 2) Prove, that [imath]\operatorname{Inn}(G) = \{p_a \mid a \in G\}[/imath] is a normal subgroup of [imath](\operatorname{Aut}(G),\circ)[/imath]. |
1251253 | The annihilator induces a module
Let [imath]R[/imath] be a ring, and [imath]M[/imath] an [imath]R[/imath]-leftmodule. Let [imath]\operatorname{Ann}_R(M)[/imath] be the annihilator of M, meaning that [imath]r m = 0 \space\space\space\space \forall r \in \operatorname{Ann}_R(M), m \in M[/imath]. Let [imath]I \subseteq \operatorname{Ann}_R(M)[/imath] be a two-sided ideal. Show that M is naturally an [imath]R/I[/imath]-module. Thanks in advance! I'm not that used to annihilators, so any help would be appreciated. | 1224888 | Natural $(R/I)$-module structure for an [imath]R[/imath]-module [imath]M[/imath] annihilated by [imath]I[/imath]
Suppose that [imath]I[/imath] is a two-sided ideal in the ring [imath]R[/imath], and that [imath]M[/imath] is a module over the quotient ring [imath]R/I[/imath]. Why can we naturally regard [imath]M[/imath] as a [imath]R[/imath]-module that is annihilated by [imath]I[/imath]? Conversely, suppose that [imath]M[/imath] is a [imath]R[/imath]-module annihilated by [imath]I[/imath]. Why is [imath]M[/imath] naturally a module over [imath]$R/I$[/imath]? |
1251276 | A bizarre property
Studing some fact about p-adic numbers I read a bizarre property. A metric space S is called ultrametric when [imath]d(x, y) \le\max\{d(x, z), d(z, y)\} \forall(x,y,z) \in S^3[/imath]. Prove that all ball of S admits all any of its points as its center | 1089656 | Every point of an open ball is a centre for the open ball.
Suppose [imath]X[/imath] is a nonempty set and [imath]d[/imath] is an ultrametric on [imath]X[/imath] i.e.,[imath]d(x,y)\le\max\{d(x,z),d(z,y)\}[/imath] for all [imath]x,y \in X[/imath]. Suppose B is an open ball of [imath](X,d)[/imath]. Show that every point of B is a centre for B. I can't understand the statement I have to prove as how can an open ball have more than one centre . Please Help!! I found the problem from Searcoid's Book on Metric Space! Thank You!! |
1118866 | Let [imath]t_n[/imath] denote the [imath]n[/imath]th triangular number. For what values of [imath]n[/imath] does [imath]t_n[/imath] divide [imath]t_1^2+t_2^2+ \cdots +t_n^2[/imath]
Let [imath]t_n[/imath] denote the [imath]n[/imath]th triangular number. For what values of [imath]n[/imath] does [imath]t_n[/imath] divide [imath]t_1^2+t_2^2+ \cdots +t_n^2[/imath]. The hint says that because [imath]t_1^2+t_2^2+ \cdots +t_n^2 = t_n(3n^3 + 12n^2 + 13n + 2)/30[/imath], it suffices to determine those [imath]n[/imath] satisfying [imath]3n^3+12n^2+13n+2 \equiv 0[/imath] (mod 2*3*5). However, I don't see how to get [imath]t_1^2+t_2^2+ \cdots +t_n^2 = t_n(3n^3 + 12n^2 + 13n + 2)/30[/imath] this equiation in the first place. Also, I'm trying to solve the congruence [imath]3n^3+12n^2+13n+2 \equiv 0[/imath] for each mod 2, 3, 5 respectively but I have been unable so far to find the [imath]n[/imath] that satisfies these congruences. I would greatly appreciate any help. | 475526 | Congruence Equation [imath]3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}[/imath]
How to solve the following congruence equation? [imath]3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}[/imath] If [imath]t_n[/imath] be the [imath]n[/imath]th triangular number, then [imath]t_1^2+t_2^2+...+t_n^2=\frac{t_n(3n^3+12n^2+13n+2)}{30}[/imath] so if we solve this congruence equation we find the values of [imath]n[/imath] that [imath]t_n[/imath] divides [imath]t_1^2+t_2^2+...+t_n^2[/imath]. |
1252073 | How to prove a cube minus a cube is never a cube (in whole numbers)
How to prove [imath]x^3-y^3\neq z^3[/imath] where [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath] are whole numbers (integers greater than zero)? | 237556 | Reference request: Clean proof of Fermat's last theorem for [imath]n=3[/imath].
I have seen a proof for FLT, [imath]n=3[/imath] using factorisation in the ring of Eisenstein integers, but it's quite long and convoluted; I am wondering if there is a more 'advanced' proof which avoids infinite descent/messy repeated calculations. What's your favorite proof of Fermat's last theorem for [imath]n=3[/imath], and where can it be found written down? |
1252071 | What is the Taylor series for the function [imath]f(x)=\cos(x)[/imath] centered at [imath]a=(-\pi/4[/imath])?
The title is the extent of the problem. It is a problem from my Calculus II practice test that I am having trouble solving. | 330890 | General term of Taylor Series of [imath]\sin x[/imath] centered at [imath]\pi/4[/imath]
What is the general term for a Taylor series of [imath]\sin(x)[/imath] centered at [imath]\pi/4[/imath]? It should be [imath](-1)^{[??]} \times \sqrt{2}/2 \times \frac{(x-\pi/4)^n}{n!}[/imath] What power is [imath](-1)[/imath] supposed to be raised to? |
1252741 | Solving [imath]x^3 = -1[/imath] for complex numbers
How can I solve for the complex solutions of [imath] x^3 = -1 [/imath] | 192742 | How to solve [imath]x^3=-1[/imath]?
How to solve [imath]x^3=-1[/imath]? I got following: [imath]x^3=-1[/imath] [imath]x=(-1)^{\frac{1}{3}}[/imath] [imath]x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}[/imath]... |
1253129 | As the limit of [imath]n[/imath] goes to infinity, prove that [imath]x^n = 0[/imath] if [imath]\operatorname{abs}(x)<1[/imath].
As the limit of [imath]n[/imath] goes to infinity, prove that [imath]x^n = 0[/imath] if [imath]\operatorname{abs}(x)<1[/imath]. So I want to prove it this by observing that [imath]\operatorname{abs}(x) < 1[/imath] which means [imath]1/(\operatorname{abs}(x)) > 1[/imath]. How would I proceed from here? | 126042 | Proving that [imath]x^n[/imath] converges to [imath]0[/imath] whenever [imath]|x| < 1[/imath]
I've already proved that for any [imath]p > 0[/imath] and for any [imath]\alpha \in \mathbb{R}[/imath], the sequence [imath]\frac{n^\alpha}{(1 + p)^n}[/imath] converges to [imath]0[/imath]. Now, I want to prove that [imath]\lim_{n \to \infty} x^n = 0[/imath] as long as [imath]|x| < 1[/imath]. I've split it into two possible cases: (1) [imath]0 < x < 1[/imath]. Let [imath]p = \frac{1}{x} - 1[/imath], so that [imath]x = \frac{1}{1 + p}[/imath] and [imath]p > 0[/imath] (since [imath]x < 1[/imath]). Then we let [imath]\alpha = 0[/imath] to see that [imath]\lim_{n \to \infty} x^n = \lim_{n \to \infty} \frac{n^0}{(1 + p)^n} = 0[/imath] (2) [imath]-1 < x < 0[/imath]. Here we cannot use the sequence [imath]\frac{n^\alpha}{(1 + p)^n}[/imath] again, since if we let [imath]p = \frac{1}{x} - 1[/imath], then [imath]p[/imath] isn't necessarily positive (for example, if [imath]x = - \frac{1}{2}[/imath], then [imath]p = -3 < 0[/imath]). How can I prove Case (2)? |
1253665 | Prove [imath]\int_a^bf(x)dx = \int_{a+c}^{b+c}f(x-c)dx[/imath]
Here's the problem: Let [imath]f:[a,b] \to \Bbb R[/imath] and let [imath]c \in \Bbb R[/imath]. Prove that if [imath]\int_a^bf(x)dx[/imath] exists then so does [imath]\int_{a+c}^{b+c}f(x-c)dx[/imath] and these two integrals are equal. I've been working on it for a while now and it is really turning out to be less straight forward that it appears. Could someone show me how to do this? | 1247690 | [imath]\int_a^{b} f(x) dx[/imath] exists then so does [imath]\int_{a+c}^{b+c} f(x-c)dx[/imath]
I am with a friend doing some late night math at math club and we came across this problem: [imath]f : [a,b] \rightarrow \mathbb R[/imath] and let [imath]c \in \mathbb R[/imath]. I am trying to show that [imath]\int_a^{b} f(x) dx[/imath] exists then so does [imath]\int_{a+c}^{b+c} f(x-c)dx[/imath] and these two integrals are equal. This seems almost trivial to us as we quickly computed some integrals and this was obvious, however we are having some trouble proving this. Is there some rule where you can subtract a constant in the integral and add it in the integral bounds. I have never used this before. Help would be great! Just want to understand how this is actually proved. Thanks! |
1253673 | Is [imath]\sum (\sin n)^n[/imath] diverges?
I saw this series incidentally: [imath]\sum_{n=1}^\infty (\sin n)^n [/imath] Result from WolframAlpha seems to say the series diverges but I don't know how to prove it. Thanks for any help! | 811717 | Does the sequence [imath]\{\sin^n(n)\}[/imath] converge?
Does the sequence [imath]\{\sin^n(n)\}[/imath] converge? Does the series [imath]\sum\limits_{n=1}^\infty \sin^n(n)[/imath] converge? |
1253563 | Is this statement about manifold true?
Suppose [imath]M[/imath] is a closed [imath]k-[/imath]manifold in [imath]\mathbb R^n[/imath] without boundary, can we always find a smooth function [imath]f:\mathbb R^n\to\mathbb R^{n-k}[/imath] such that [imath]M[/imath] is the level set where [imath]f=0[/imath]? | 23764 | Is every embedded submanifold globally a level set?
It's a well-known theorem (Corollary 8.10 in Lee Smooth) that given a smooth map of manifolds [imath]\phi:M\rightarrow N[/imath] and a regular value [imath]p\in N[/imath] of [imath]\phi[/imath], the level set [imath]\phi^{-1}(p)\subset M[/imath] is a closed embedded submanifold. Is the converse true? That is, given an embedded submanifold [imath]S\subset M[/imath], is there necessarily a manifold [imath]N[/imath], smooth map [imath]\phi:M\rightarrow N[/imath], and regular value [imath]p\in N[/imath] of [imath]\phi[/imath] such that [imath]S=\phi^{-1}(p)[/imath]? Prop. 8.12 in Lee Smooth shows that this is true locally; specifically, Let [imath]S[/imath] be a subset of a smooth [imath]n[/imath]-manifold [imath]M[/imath]. Then [imath]S[/imath] is an embedded [imath]k[/imath]-submanifold of [imath]M[/imath] if and only if every point [imath]p\in S[/imath] has a neighborhood [imath]U\subset M[/imath] such that [imath]U\cap S[/imath] is a level set of a submersion [imath]\phi:U\rightarrow\mathbb{R}^{n-k}[/imath]. (and any level set of a submersion is of course the level set of a regular value). I feel like this is the kind of question where, if there is a counterexample, it probably is very simple, but I wasn't able to come up with one. |
1252066 | Probability of asymmetric random walk returning to the origin
Consider the random walk [imath]S_n[/imath] given by [imath]$ S_{n+1} = \left\{ \begin{array}{l} S_n+2 & \text{with probability $p$}\\ S_n - 1 & \text{with probability $1-p$} \end{array} \right. \ $[/imath] Assume that [imath]S_0 = n >0 [/imath]. What is the probability of eventually reaching the the point [imath]0[/imath]? Seems that the probability for this must be less than one; since it is not symmetric, it is not guaranteed to reach the origin if [imath]p>0.5[/imath]. It also seems like the positive starting value not being at the origin matters too, but I'm not sure how to compute the probability of eventually reaching the origin. | 1257934 | Biased Asymmetric Random Walk
Consider the random walk [imath]S_n[/imath] given by: [imath] S_{n+1}= \left\{ \begin{array}{ll} S_n + 2 & \mbox{w.p } p\\ S_n -1 & \mbox{w.p } 1-p \end{array} \right. [/imath] Assume that [imath]S_0=n > 0[/imath] with certainty. What is the probability of eventually reaching the origin (the point 0) ? Now I know the derivation for the same problem but when it's symmetric and unbiased namely the additions are [imath]1,-1[/imath] with probability [imath]\frac{1}{2}[/imath] for each. Can I use this result or that I need to make the derivation from the start? |
1253747 | Determining whether [imath]f(z)=\ln r + i\theta[/imath] (with domain [imath]\{z:r\gt , 0\lt \theta \lt 2\pi\}[/imath]) is analytic
Define [imath]f(z)=\ln r + i\theta[/imath] on the domain [imath]\{z:r\gt , 0\lt \theta \lt 2\pi\}[/imath]. This domain is just a punctured disk of radius [imath]\ln r[/imath], correct? How does one determine whether this is analytic, I can't see how I would take the CREs [imath]u(r,\theta) = \ln r[/imath] [imath]v(r,\theta) = \theta.[/imath] Should I convert this back to [imath]x+iy[/imath] form and proceed? How can I do such a thing with what appears to be a punctured open neighborhood? How do I show that the function is analytic and find its derivatives? | 1245754 | Cauchy-Riemann equations in polar form.
Show that in polar coordinates, the Cauchy-Riemann equations take the form [imath]\dfrac{\partial u}{\partial r} = \dfrac{1}r \dfrac{\partial v}{\partial \theta}[/imath] and [imath]\dfrac{1}r \dfrac{\partial u}{\partial \theta} = −\dfrac{\partial v}{\partial r}[/imath]. Use these equations to show that the logarithm function defined by [imath]\log z = \log r + i\theta[/imath] where [imath]z=re^{i\theta}[/imath] with [imath]-\pi<\theta<\pi[/imath] is holomorphic in the region [imath]r > 0[/imath] and [imath]-\pi<\theta<\pi[/imath]. What I have so far: Cauchy-Riemann Equations: Let [imath]f(z)[/imath] = [imath]u(x, y)[/imath] +[imath]iv(x, y)[/imath] be a function on an open domain with continuous partial derivatives in the underlying real variables. Then f is differentiable at [imath]z = x+iy[/imath] if and only if [imath]\frac{∂u}{∂ x}(x, y)[/imath] = [imath]\frac{∂ v}{∂ y}(x, y)[/imath] and [imath]\frac{∂u}{∂ y}(x, y)[/imath] = −[imath]\frac{∂ v}{∂ x}(x, y)[/imath]. So we have [imath]f'(z)= \frac{∂u}{∂ x}(z) +i \frac{∂ v}{∂ x}(z)[/imath]. Let [imath]f(z)[/imath] = [imath]f(re^{iθ})[/imath]= [imath]u(r,θ)[/imath] +[imath]iv(r,θ)[/imath] be a function on an open domain that does not contain zero and with continuous partial derivatives in the underlying real variables. Then f is differentiable at [imath]z[/imath] = [imath]re^{iθ}[/imath] if and only if [imath]r \frac{∂u}{∂r}=\frac{∂ v}{∂θ}[/imath] and [imath]\frac{∂u}{∂θ}[/imath] = [imath]−r \frac{∂v}{∂ r}[/imath]. Sorry, if this is not very good. I just decided to start learning complex analysis today... |
1253738 | Trigonometric integral of [imath]f(x)=(x^2)(\sin(x^2))[/imath].
I've tried with the chain rule and [imath]u[/imath]-subtitution ([imath]u=\sqrt{x}[/imath]) but I get nothing. Can you help me please? [imath]\int (sqrt{x})(\sin(x)) \ dx[/imath] | 239366 | How to integrate [imath]\int x^2 \sin^2(x)dx[/imath]
I don't know how to integrate [imath]\int x^2\sin^2(x)\,\mathrm dx[/imath]. I know that I should do it by parts, where I have [imath] u=x^2\quad v'=\sin^2x \\ u'=2x \quad v={-\sin x\cos x+x\over 2}[/imath] and now I have [imath] \int x^2\sin^2(x)\,\mathrm dx = {-\sin x\cos x+x\over 2} x^2 - \int 2x{-\sin x\cos x+x\over 2}\,\mathrm dx\\ ={-\sin x\cos x+x\over 2} x^2 - \int x(-\sin x\cos x+x)\,\mathrm dx\\ [/imath] so I have to calculate [imath] \int x(-\sin x\cos x+x)\,\mathrm dx=-\int x\sin x\cos x\,\mathrm dx+\int x^2\,\mathrm dx[/imath] I know that [imath]\int x^2\,\mathrm dx = {1 \over 3}x^3+C[/imath] but I don't know what should I do with [imath]\int x\sin x\cos x \,\mathrm dx[/imath] Should i use parts again or what? Please help. |
1253949 | Suppose [imath]a \in \mathbb{C}[/imath], [imath]|a| < 1[/imath], and [imath]f(z) = \dfrac{z - a}{1 - \overline{a}z}[/imath]. How to prove dependence of [imath]|f(z)|[/imath] on [imath]|z|[/imath]?
Let [imath]a \in \mathbb{C}[/imath], [imath]|a| < 1[/imath]. Also let [imath]f(z) = \dfrac{z - a}{1 - \overline{a}z}[/imath]. I am asked to prove that [imath]|f(z)| < 1[/imath] if [imath]|z| < 1[/imath] and that [imath]|f(z)| = 1[/imath] if [imath]|z| = 1[/imath]. What is a good way to proceed on this problem? | 921176 | Is there anything special with complex fraction [imath]\left|\frac{z-a}{1-\bar{a}{z}}\right|[/imath]?
Is there anything special with the form: [imath]\left|\frac{z-a}{1-\bar{a}{z}}\right|[/imath] ? With [imath]a[/imath] and [imath]z[/imath] are complex numbers. In fact, I saw it in a problem: If [imath]|z| = 1[/imath], prove that [imath]|\frac{z-a}{1-\bar{a}{z}}| = 1[/imath] If [imath]|z| < 1[/imath] and [imath]|a| < 1[/imath], prove that [imath]|\frac{z-a}{1-\bar{a}{z}}| < 1[/imath] I can easily prove the first one with expansion: [imath]z=\cos\theta + i\sin \theta \\ a = m +in[/imath] But it will be terrible to use in the second one. What's more, I found this form a little special so maybe there is some clever trick without using expansion? Thank you a lot! |
1252837 | Group Theory and Lagrange's Theorem: coprime subgroups.
Let [imath]G_1[/imath] and [imath]G_2[/imath] be finite groups, and let [imath]K≤G_1 \times G_2[/imath]. Let [imath]H_1 = \{ g \in G_1 : (g,e) \in K\}[/imath] and [imath]H_2 = \{g \in G_2 : (e,g) \in K\}[/imath] and suppose [imath]|G_1|[/imath] and [imath]|G_2|[/imath] are coprime. Then show that [imath]H_1 \times H_2 = K[/imath]. Show that if [imath]|G_1|[/imath] and [imath]|G_2|[/imath] are not coprime then this does not have to be the case. My working: I have proved that [imath]H_1 ≤ G_1[/imath], [imath]H_2 ≤ G_2[/imath] and [imath]H_1 \times H_2 ≤ K[/imath]. And since [imath]|G_1|, |G_2|[/imath] are coprime, we must have [imath]|H_1|, |H_2|[/imath] are also coprime. Then given that [imath]|G_1 \times G_2|[/imath] = [imath]|G_1||G_2|[/imath], I am trying to use Bézout's Lemma or some similar thing to show that [imath]H_1 \times H_2 = K[/imath]. As for the second part, I have no idea. | 764939 | [imath] |G_1 |[/imath] and [imath]|G_2 | [/imath] are coprime. Show that [imath]K = H_1 \times H_2[/imath]
Let [imath]G_1[/imath] and [imath]G_2[/imath] be finite groups and let [imath]K \le G_1 \times G_2[/imath] (i) Set [imath]H_1 = \{ g \in G_1 : (g,e) \in K \}[/imath] and [imath]H_2 = \{ g \in G_2 : (e,g) \in K \}[/imath]. Show that: [imath]H_1 \le G_1; \quad H_2 \le G_2; \quad H_1 \times H_2 \le K.[/imath] (ii) Suppose that [imath]|G_1|[/imath] and [imath]|G_2|[/imath] are coprime. Show that [imath]K = H_1 \times H_2[/imath]. (iii) Show that this result need not follow if [imath]|G_1|[/imath] and [imath]|G_2|[/imath] are not coprime. I have done part (i). I was doing part (ii) and got stuck: Since from above I showed that [imath]H_1 \times H_2 \subseteq K[/imath], now I only need to show that [imath]H_1 \times H_2 \supseteq K[/imath]. Let [imath](g_1,g_2) \in K[/imath] where [imath]g_1 \in G_1[/imath] and [imath]g_2 \in G_2[/imath]. Since [imath]K \leq G_1 \times G_2[/imath], we have [imath](g_1,e) \in K[/imath] and [imath](e,g_2) \in K[/imath]. Hence [imath]g_1 \in H_1[/imath] and [imath]g_2 \in H_2[/imath]. Finally we have [imath](g_1, g_2) \in H_1 \times H_2[/imath] As I haven't used the fact that order of G_1 and G_2 are coprime, there must be something fundamentally wrong with my proof. |
1254774 | Convolution of an integrable function an [imath]L^\infty[/imath] function
Let [imath]f[/imath] be an integrable function on [imath]\mathbb{R}[/imath], and [imath]g[/imath] be an [imath]L^\infty[/imath] function on [imath]\mathbb{R}[/imath]. Then, the convolution [imath]f*g[/imath] is said to be continuous and bounded on R. I managed to show that it is bounded, but can't show that it is continuous. The convolution is defined as an integral, so it seems intuitively clear that it is continuous. Could anyone show me how to rigorously prove this? | 570494 | Convolution is uniformly continuous and bounded
Suppose [imath]f\in L^\infty(\mathbb{R})[/imath] and [imath]K\in L^1(\mathbb{R})[/imath] with [imath]\int_\mathbb{R}K(x)dx=1[/imath]. Show that the convolution [imath]f\ast K[/imath] is a uniformly continuous and bounded function. The definition of the convolution is [imath](f\ast K)(x)=\int_\mathbb{R}f(x-y)K(y)dy[/imath]. There is the inequality [imath]\|f\ast K\|_\infty\leq\|f\|_\infty\|K\|_1[/imath], which yields that [imath]f\ast K[/imath] is bounded. But what about uniform continuity? |
1254700 | Prove that if [imath]A[/imath] is both open and closed, [imath]A=\mathbb R[/imath].
Suppose [imath]A[/imath] is a non-empty subset of [imath]\mathbb R[/imath]. Prove that if [imath]A[/imath] is both open and closed, [imath]A=\mathbb R[/imath]. I think I'm supposed to assume that [imath]A[/imath] is not equal to [imath]\mathbb R[/imath] and derive a contradiction. Would that mean [imath]A[/imath] complement is also both closed and open? I'm not sure if that would be the right approach to the proof. Any help is greatly appreciated! | 1249943 | Proving that if a set is both open and closed then it is equal to the real numbers
Prove that if [imath]A[/imath] is both open and closed then [imath]A = \mathbb{R}[/imath] also as one suggested let [imath]A \neq \emptyset[/imath] You may use what ever definition of open and closed you would like, just avoid going into metric spaces, haven't covered that topic yet. My question is well essentially how to prove this statement but considering I like to do things myself, I was wondering if anyone had any particular suggestions to help me solve this. attempted proof: Let [imath]a\in A[/imath] then since [imath]A[/imath] is open there exists an open interval [imath]N(a,\epsilon)[/imath] such that [imath]a\in N(a,\epsilon)\subset A[/imath] Then [imath]a[/imath] must be an interior point of [imath]A[/imath], but since [imath]A[/imath] is also closed then [imath]a[/imath] must also be an accumulation or limit point of [imath]A[/imath] as well. I am going to stop here because I am not sure if this is the right approach here, any suggestions would be greatly appreciated. |
1254843 | Area of surface
What is the area of the region that is bounded by the curve [imath]\vec{R}(t)=(\cos^3t, \sin^3t), 0\leq t<2\pi?[/imath] I have no idea how to start here or what i have to use. | 538554 | find area of the region [imath]x=a\cos^3\theta[/imath] [imath]y=a\sin^3\theta[/imath]
Find the area of the region enclosed by [imath]x=a\cos^3\theta[/imath] and [imath]y=a\sin^3\theta[/imath] What steps should I take in order to find the area? |
402841 | [imath]C([0, 1])[/imath] is not complete with respect to the norm [imath]\lVert f\rVert _1 = \int_0^1 \lvert f (x) \rvert \,dx[/imath]
Consider [imath]C([0, 1])[/imath], the linear space of continuous complex-valued functions on the interval [imath][0, 1][/imath], with the norm [imath]\displaystyle\lVert f\rVert_1 = \int_0^1 \lvert f(x)\rvert \,dx.[/imath] I have to show that [imath]C([0, 1])[/imath] is not complete with respect to this norm. I have found the following example from a book. Let [imath]f_n \in C[0,1][/imath] be given by [imath]f_n(x) := \begin{cases} 0 & \text{if $0 \le x \le \frac1{2}$}\\ n(x-\frac{1}{2}) & \text{if $\frac {1}{2} < x \le \frac {1}{2} + \frac {1}{n}$}\\ 1 & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $} \end{cases}[/imath] How to prove that [imath]f_n[/imath] is a Cauchy sequence with respect to [imath]\lVert \cdot\rVert_1[/imath]? If I use basic definition then I have to prove that [imath]\lVert f_n - f_m\rVert_1 < \epsilon[/imath] [imath]\forall n, m > N[/imath]. But I am finding it difficult to prove this. Please help me to understand how to prove that [imath]f_n[/imath] is Cauchy sequence in [imath]C([0, 1])[/imath]. Thanks | 820520 | Prove that [imath]C^1[0,1][/imath] is not complete with norms
I have to show that the [imath]C^1[0,1][/imath] is not complete with any of these norms: [imath]\|f\|_{\infty}=\sup_{x\in[0,1]}|f(x)|[/imath] [imath]\|f\|_{*}=|f(0)|+\int_0^1|f'(x)|dx[/imath] My attempt The right sequence for the first norm is [imath]f_n=\sqrt{x+\frac{1}{n}}[/imath]. Notice that [imath]\forall n\in\mathbb{N} : f_n\in C^1[0,1][/imath] Let [imath]f=\sqrt{x}[/imath] We see that [imath](f_n)[/imath] converges to [imath]f[/imath] in sup norm in [imath]C[0,1][/imath], thus it is Cauchy. [imath]C^1[0,1][/imath] is a subspace of [imath]C[0,1][/imath] and all terms of [imath](f_n)[/imath] are in [imath]C^1[0,1][/imath], so [imath](f_n)[/imath] is Cauchy in [imath]C^1[0,1][/imath] But [imath]f[/imath] is not in [imath]C^1[0,1][/imath]. So [imath]C^1[0,1][/imath] with sup norm is not complete. When it comes to the second norm, I think the same sequence will be also okay. Am I right? |
42930 | What is the expected value of the number of die rolls necessary to get a specific number?
Given a discrete random number generator, such as a six-sided die, what is the expected value of the number of rolls necessary to roll a specific number (e.g. a six)? I think the result should be given by E[imath]\langle[/imath]rolls[imath]\rangle[/imath] = [imath]\frac{1}{6}\sum_{n=0}^\infty{(\frac{5}{6})^n(n+1)}[/imath], but I don't know how to calculate the convergence of that sum. Also, how do I calculate the variance? | 2179670 | John keeps rolling a standard six-sided dice until it comes up as a 6. What is the expected number of dice rolls?
For this problem, I found a very complicated solution. Are there any easier methods? I am asking mainly because the final answer was what I intuitively guessed before solving the problem. It is also surprising that after so much work I get such an intuitive result. Problem: John keeps rolling a standard six-sided dice until it comes up as a 6. What is the expected number of dice rolls? My Solution: The probability that John rolls only once is [imath]\frac{1}{6}[/imath]. The probability that John rolls twice is [imath]\frac{5}{6}\cdot\frac{1}{6}[/imath]. The probability that John rolls three times is [imath]\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}[/imath]. Seeing the pattern, the probability that John rolls [imath]k[/imath] times is [imath]\frac{1}{6} \cdot \left(\frac{5}{6}\right)^{k-1}[/imath]. We verify this by adding all [imath]k[/imath] from [imath]1[/imath] onwards to [imath]\infty[/imath] and checking if that is [imath]1[/imath]. This works because positive integers are the only number of times that John can roll. [imath]\sum_{n=1}^{\infty} \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{i-1} = 1[/imath]What we want is the sum of the arithmetico-geometric sequence [imath]\sum_{n=1}^{\infty} \frac{i}{6} \cdot \left(\frac{5}{6}\right)^{i-1} = \hspace{1mm}?[/imath]To get a better idea of what this sequence is, we expand out a few terms. [imath]1 \cdot \frac{1}{6} + 2 \cdot \frac{5}{6}\cdot\frac{1}{6} + 3 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{2} + \cdots[/imath]. Now we factor the initial term, [imath]\frac{1}{6} \left(1 + 2 \cdot \frac{5}{6} + 3 \cdot \left(\frac{5}{6}\right)^{2} + \cdots \right)[/imath]. We let [imath]S = 1 + 2 \cdot \frac{5}{6} + 3\left(\frac{5}{6}\right)^{2} + \cdots[/imath]. We can now use the trick for any geometric sequence on this sequence. [imath]\frac{5}{6}S = \frac{5}{6} + 2\left( \frac{5}{6}\right)^2 + 3\left(\frac{5}{6}\right)^{3} + \cdots[/imath]. Subtracting the second equation from the first gives us [imath]\frac{1}{6}S = 1 + \frac{5}{6} + \left(\frac{5}{6}\right)^2 + \left(\frac{5}{6}\right)^3 + \cdots[/imath]. This leads to [imath]S = 36[/imath]. The expression we want is [imath]\frac{S}{6} = \boxed{6}[/imath] Edit: My question is different than the other one because I am asking for an explanation on why the answer is the intuitive one and if there are any better methods to solve while that question is only asking for the answer. |
1255116 | How to calculate the sum [imath]\sum_{i=0}^\infty i^nx^i[/imath]
Here I have [imath]x\in\mathbb{R}_+[/imath] and [imath]x < 1[/imath]. I would like to evaluate the following sum: [imath]\sum_{i=0}^\infty i^nx^i.[/imath] I know that [imath]\sum_{i=0}^\infty x^i=\dfrac{1}{1-x}.[/imath] So I started calculating the derivative of the above formula. I found something like: [imath]\sum_{i=0}^\infty i^nx^i=\dfrac{P_n(x)}{(1-x)^{n+1}},[/imath] where [imath]P_n(x)[/imath] is a polynomial of degree [imath]n[/imath]. [imath]P_0(x)=1,[/imath] [imath]P_1(x)=x,[/imath] [imath]P_2(x)=x+x^2,[/imath] [imath]P_3(x)=x+4x^2+x^3,[/imath] [imath]P_4(x)=x+11x^2+11x^3+x^4,[/imath] [imath]P_4(x)=x+26x^2+66x^3+26x^4+x^5,[/imath] [imath]\cdots.[/imath] Is there a general formula for [imath]P_n(x)[/imath]? | 1247336 | Compute the sum [imath]\sum_{k=1}^{\infty}k^mz^k[/imath] where [imath]|z|<1[/imath]
Do you know how to find the limit of [imath]\sum_{k=1}^{\infty}k^mz^k[/imath] where [imath]|z|<1[/imath] and m is a natural number? I've tried to google it in wiki but I do not understand the closed form (http://en.wikipedia.org/wiki/List_of_mathematical_series). |
1254565 | probability, random walk, Markov chain question
Let [imath]P[/imath] be a transition matrix for a regular Markov chain and let [imath]w[/imath] be it’s equilibrium vector. Show that [imath]w[/imath] has no zero entries. | 1257186 | Equilibrium Vector For Regular Markov Chain
Let [imath]P[/imath] be a transition matrix for a regular Markov Chain and let [imath]w[/imath] be it's equilibrium vector. Show that [imath]w[/imath] has no zero entries |
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