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2266162 | Can the canonical linear function of a vector space to his double dual space be surjective?
I am having problems to prove the following: Let V be a K-Vectorspace and [imath]dim(V)=\infty[/imath]. Prove that the canonical linear transformation [imath]e: V \rightarrow V^{**}[/imath] is not surjective. I think I get the idea of the proof, since if [imath](v_{i})_{i \in I}[/imath] is a basis of [imath]V[/imath] and [imath](v_{i}^{*})_{i \in I}[/imath] is the dual family in [imath]V^{*}[/imath] then it might - or even can't - span the whole space [imath]V^{*}[/imath]. But I am neither sure how I can show this, nor the actual proof. Help would be much appreciated. Thanks in advance! | 179367 | Canonical Isomorphism Between [imath]\mathbf{V}[/imath] and [imath](\mathbf{V}^*)^*[/imath]
For the finite-dimensional case, we have a canonical isomorphism between [imath]\mathbf{V}[/imath], a vector space with the usual addition and scalar multiplication, and [imath](\mathbf{V}^*)^*[/imath], the "dual of the dual of [imath]\mathbf{V}[/imath]." This canonical isomorphism means that the isomorphism is always the same, independent of additional choices. We can define a map [imath]I : \mathbf{V} \to (\mathbf{V}^*)^*[/imath] by [imath]x \mapsto I(x) \in (\mathbf{V}^*)^* \ \text{ where } \ I(x)(f) = f(x) \ \text{for any } \ f \in \mathbf{V}^*[/imath] My Question: what can go wrong in the infinite-dimensional case? The notes I am studying remark that if [imath]\mathbf{V}[/imath] is finite-dimensional, then [imath]I[/imath] is an isomorphism, but in the infinite-dimensional case we can go wrong? How? |
2265114 | [imath]p:(E,e_0) \to (B,b_0)[/imath] be a covering map ; is the induced homomorphism between the first singular homology groups injective?
Let [imath]p:(E,e_0) \to (B,b_0)[/imath] be a covering map ; then I know that the induced homomorphism between first fundamental groups [imath]\pi_1(p) :\pi_1(E,e_0) \to \pi_1(B,b_0)[/imath] is a monomorphism . My question is , is the induced homomorphism between first singular homology groups [imath]H_1(f) : H_1(E) \to H_1(B)[/imath] a monomorphism i.e. injective ? | 2229232 | Induced maps in homology are injective
In a topological course I have learned that a covering map [imath]p:\bar{X}\rightarrow X[/imath] induces a map on fundamental groups and that this induced map is a monomorphism. Now I wonder myself if this is also true for [imath]H_1(p)[/imath]. Via the Hurewicz-isomorphism [imath]H_1(p)[/imath] is isomorphic to the abelianization of the fundamental group. |
2265515 | Is the function [imath]f(x,y)=xy^{\frac{1}{3}}[/imath] differentiable at [imath](0,0)[/imath]?
Is the function [imath]f(x,y)=xy^{\frac{1}{3}}[/imath] differentiable at [imath](0,0)[/imath]? I'm trying to find the limit of [imath] \lim\limits_{(x,y)\rightarrow(0,0)}\frac{xy^{\frac{1}{3}}}{\sqrt{x^2+y^2}} \;, [/imath] but when I use the inequality [imath]x^2+y^2\geq 2|x||y|[/imath], I get stuck at [imath] 0 \leq \lim\limits_{(x,y)\rightarrow(0,0)}\frac{|x||y|^{\frac{1}{3}}}{\sqrt{x^2+y^2}} \leq \lim\limits_{(x,y)\rightarrow(0,0)}\frac{(x^2+y^2)|y|^{\frac{-2}{3}}}{2\sqrt{x^2+y^2}} = \lim\limits_{(x,y)\rightarrow(0,0)}\frac{1}{2}\sqrt{x^2+y^2}|y|^{\frac{-2}{3}} =\ ? [/imath] | 2052226 | Is [imath]f(x,y)=(xy)^{2/3}[/imath] differentiable at [imath](0,0)[/imath]?
I looked at the graph of [imath]f(x,y)=(xy)^{2/3}[/imath] at the origin, and no matter how far I zoom in it has four corners along [imath]x=\pm y[/imath], so it doesn't look like it is differentiable as [imath]z=0[/imath] doesn't look like a tangent plane. I also found that its partial derivatives are not continuous at [imath](0,0)[/imath], e.g. [imath] \lim_{(ky^2,y) \to (0,0)} \frac{\partial f}{\partial x} = \lim_{(ky^2,y) \to (0,0)} \frac{2}{3}x^{-\frac{1}{3}}y^{\frac{2}{3}} = \frac{2}{3}k^{-\frac{1}{3}} [/imath] (although [imath]\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0[/imath]) I also tried to prove/disprove its differentiability by definition, and for it to be differentiable I need [imath] \lim_{(x,y) \to (0,0)} \frac{f(x,y)-0}{\sqrt{x^2+y^2}} =0 [/imath] but then I got stuck. Am I even on the right track? [UPDATE] I need to do the same thing for [imath]f(x,y)=(xy)^{\frac{1}{3}}[/imath]... I think the answer is that [imath](xy)^{2/3}[/imath] is differentiable but [imath](xy)^{1/3}[/imath] is not, according to my professor but I'm not sure |
2267190 | If N is a subgroup normal of the group G such that [imath]N \cap G'=\{e_{G}\}[/imath], show that [imath]N \subset Z(G).[/imath]
If N is a subgroup normal of the group G such that [imath]N \cap G'=\{e_{G}\}[/imath], show that [imath]N \subset Z(G).[/imath] Where [imath]G'[/imath] is the subgroup commutator. I did like this: Let [imath]g \in G [/imath] and [imath] x \in N[/imath] arbitrary, as N is normal subgroup then [imath] x ^{ - 1} \in N [/imath] Now let us consider the element [imath]xgx^{-1}g^{-1}[/imath]. On the other hand, [imath] xgx^{-1}g^{-1}\in N [/imath]. In fact, because of the hypothesis of N being normal subgroup of G, we have that [imath] x (gx^{-1} g^{-1}) \in N [/imath]. Therefore \begin{eqnarray*} xgx^{- 1}g^{- 1} \in N \quad \text{and} \quad xgx^{-1} g^{- 1} \in G' & \Rightarrow & xgx^{-1}g^{- 1} \in N\cap G' \\ &\Rightarrow & xgx^{-1} g^{-1}= e_{G} \\ &\Rightarrow & xg=gx \Rightarrow x\in Z(G). \end{eqnarray*} | 2257021 | Prove that [imath]N \subset Z(G)[/imath]
Assume that [imath]N[/imath] is a normal subgroup of [imath]G[/imath] such that [imath]N \cap G'=\{e\}[/imath]. Prove that [imath]N \subset Z(G)[/imath]. Note 1: [imath]Z(G)[/imath] is the center of the group [imath]G[/imath], and [imath]G'[/imath] is the set of all commutators of [imath]G[/imath]. Note 2: Assume that [imath]n \in N[/imath]. We want to show that for all [imath]g \in G,[/imath] [imath]ng=gn[/imath]. But, I don't know how to use the fact that [imath]N \cap G'=\{e\}[/imath] to reach this. Any hints? suggestions? answers? |
2267368 | Irreducible polynomial over rational numbers.
Prove that polynomial [imath]h(x)=\prod_{i=1}^{n} (x-a_i)+1[/imath] is irreducible over [imath]\mathbb Q[/imath], where [imath]a_i \in \mathbb Z, a_i \neq a_j[/imath] for [imath]i \neq j[/imath]. Except for [imath]h(x)=(x-a-1)^{2}=(x-a)(x-a-2)+1[/imath], and [imath]h(x)=(x-a)(x-a-1)(x-a-2)(x-a-3)+1=((x-a-1)(x-a-2)-1)^{2}[/imath] Let's prove irreducibility over [imath]\mathbb Z[/imath]. Then we'll use Gauss's lemma to prove irreducibility over [imath]\mathbb Q[/imath]. Let [imath]h(x) = h_1(x) h_2(x)[/imath], where [imath]deg(h_1), deg(h_2)<deg(h)[/imath]. [imath]h(a_i) = h_1(a_i)*h_2(a_i)=1[/imath]. Then [imath]h_1(a_i) = 1[/imath] and [imath]h_2(a_i) = 1[/imath], or [imath]h_1(a_i) = -1[/imath] and [imath]h_2(a_i) = -1[/imath]. What can I do next? Does [imath]h_1-h_2[/imath] got [imath]n[/imath] roots? | 1304679 | Proving that a polynomial of the form [imath](x-a_1)\cdots(x-a_n) + 1[/imath] is irreducible over [imath]\mathbb{Q}[/imath]
I want to prove that for any set of distinct integers [imath]a_1,\ldots,a_n[/imath], the polynomial [imath]h = (x-a_1)\cdots(x-a_n) + 1[/imath] is irreducible over the field [imath]\mathbb{Q}[/imath], except for the following special cases which are reducible: [imath]\left.\begin{cases} a_1 = a\\ a_2 = a+2 \end{cases}\right\} \implies h = (x-a-1)^2[/imath] and [imath]\left.\begin{cases} a_1 = a\\ a_2 = a+1\\ a_3 = a+2\\ a_4 = a+3 \end{cases}\right\} \implies h = ((x-a-1)(x-a-2)-1)^2[/imath] |
2267670 | Fibonacci sequence problem
Question: Show that if [imath] [/imath] [imath] k |n \implies F_k |F_n[/imath] [imath] [/imath] Comments: I ran into this problem the other day and have not figured it out yet. I think it has something to do with the order of the period of the Fibonacci sequence modulo [imath]m[/imath], for some [imath] m \in \Bbb{N}[/imath] | 378116 | Proving that [imath]n|m\implies f_n|f_m[/imath]
Question: Let [imath]m,n\in\mathbb{N}[/imath], prove that if [imath]n|m[/imath], [imath]F_n|F_m[/imath]. I've tried to use induction, but I don't really know where to start since there's [imath]2[/imath] numbers: [imath]n[/imath] and [imath]m\ \dots[/imath] I did induction before with just [imath]1[/imath] number, like proving [imath]1+2+\cdots+n=\frac{n(n+1)}{2}[/imath], that's only dealing with one number [imath]n[/imath], but with this I have no clue how to do it. Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start? Thanks for the help! |
2264051 | order of [imath]\mathbb{Z}_3^*/(\mathbb{Z}_3^*)^3[/imath]
I would like to know what is the order of [imath]\mathbb{Z}_3^*/(\mathbb{Z}_3^*)^3[/imath] as a group, where [imath]\mathbb{Z}_3[/imath] represents the 3-adic integers. I know how to find the units of [imath]\mathbb{Z}_3[/imath], but I do not know how to compute its cube. Please help me with this. | 2263344 | What is the group structure of 3-adic group of the cubes of units?
From what I understand about the group of units of the 3-adic integers, it consists all [imath]a[/imath] that can be expressed as a = [imath]\sum_{i=0}^\infty a_i 3^i[/imath] where [imath]a_0 \neq 0[/imath]. How can we recognize which 3-adic integers are the cubes of units? I was also told that cubes of units of 3-adic integers form a group. What does this group look like? |
1937779 | Proving that [imath](T S^n) \times \mathbb R[/imath] is diffeomorphic to [imath]S^n\times \mathbb R^{n+1}[/imath]
I'm asked to show a diffeomorphism [imath]\psi[/imath] between [imath](T S^n) \times \mathbb R[/imath] and [imath]S^n\times \mathbb R^{n+1}[/imath]. I'm not sure on how to proceed here. I have written down [imath](T S^n) \times \mathbb R[/imath] and [imath]S^n\times \mathbb R^{n+1}[/imath] as [imath](T S^n) \times \mathbb R= \big\{ (x, v, r): x\in S^n, v\in x^\perp, r\in \mathbb R \big\}, [/imath] [imath]S^n\times \mathbb R^{n+1}= \big\{ (x, w): x\in S^n, w\in \mathbb R^{n+1} \big\} [/imath] Can you help me in finding such diffeomorphism? | 2266245 | T[imath]\mathbb{S}^{n} \times \mathbb{R}[/imath] is diffeomorphic to [imath]\mathbb{S}^{n}\times \mathbb{R}^{n+1}[/imath]
My strategy is find a diffeomorphism [imath]\psi: T\mathbb{S}^{n} \times \mathbb{R} \to \mathbb{S}^{n}\times \mathbb{R}^{n+1}[/imath] from other diffeomorphisms. First let [imath](\mathbb{R},Id_{\mathbb{R}})[/imath] the 1-dimensional euclidean space with canonical chart, which is a global chart. Then [imath]T\mathbb{R} \cong \mathbb{R}\times \mathbb{R}[/imath], thus [imath]T\mathbb{S}^{n} \times \mathbb{R} \times \mathbb{R} \xrightarrow{(Id)^{-1}} T\mathbb{S}^{n}\times T\mathbb{R}[/imath], Now, the product of tangent bundles are diffeomorphic to tangent bundle of T([imath]\mathbb{S}^{n}\times \mathbb{R})[/imath], thus [imath]T\mathbb{S}^{n} \times \mathbb{R} \times \mathbb{R} \xrightarrow{(Id)^{-1}} T\mathbb{S}^{n}\times T\mathbb{R} \xrightarrow{\phi} T(\mathbb{S^{n}\times \mathbb{R})}[/imath]. where [imath]\phi:T(\mathbb{S}^{n} \times \mathbb{R}) \to T\mathbb{S}^{n}\times T\ \mathbb{R}[/imath] is a diffeomorphism mentioned above. My question is there exists a global chart for [imath]\mathbb{S}^{n}\times \mathbb{R}[/imath]? If it is true this problem is over. Right? Thanks |
1538285 | Calculating probability that you are dealt a full house
What is the probability that you are dealt a "full house"? (Three cards of one rank and two cards of another rank.) why can't I calculate the probability in the following way? The number of ways I can select 2 suits from one rank = [imath]13 \choose 2[/imath] The number of ways I can select 2 suits from one rank = [imath]4 \choose 2[/imath] The number of ways I can select 3 suits from another rank = [imath]4 \choose 3[/imath] so p should be [imath]\frac{{13 \choose 2}{4 \choose 2}{4 \choose 3}}{52\choose5}[/imath] My question is how can I tweak my approach and reach the right answer. | 1861010 | Why am I under-counting when calculating the probability of a full house?
I was trying to answer this question. Find the probability of getting a full house from a [imath]52[/imath] card deck. That is, find the probability of picking a pair of cards with the same rank (face value), and a triple with equal rank (different from the rank of the pair of course). My idea was this: Take one rank. The number of ways you can get a pair from a rank is [imath]C(4, 2)[/imath]. Take another rank. The number of ways you can get a triple from that rank is [imath]C(4, 3)[/imath]. Therefore, the number of ways you can get a full house from just [imath]2[/imath] ranks is [imath]C(4, 2) \cdot C(4, 3)[/imath]. But there are [imath]13[/imath] different ranks, and the number of ways you can select [imath]2[/imath] ranks out of [imath]13[/imath] is [imath]C(13, 2)[/imath]. For each of these pairs of ranks, you can have [imath]C(4, 2) \cdot C(4, 3)[/imath] full houses. Therefore the number of possible full houses is [imath]C(4, 2) \cdot C(4, 3) \cdot C(13, 2)[/imath], and the probability of selecting a full house is [imath]\dfrac {C(4, 2) \cdot C(4, 3) \cdot C(13, 2)}{C(52, 5)} \approx 0.000720288[/imath] However, this is way lower than the actual answer of [imath].00144[/imath]. Where did I go wrong here? |
1126244 | Time taken to empty a hemispherical shaped tank
The tank has a radius of [imath]2[/imath]m when initially filled and has an outlet of cross section [imath]12[/imath] cm2 Outlet flow as I calculated goes according to the law [imath]V(t)=0.6\sqrt{2gh(t)}[/imath]. Having found out the general description of the tank and the volume flowing out in time [imath]dt Qdt=A v(t) ; Q=A v(t)[/imath] in time [imath]T[/imath] After attempting further calculations, I do not know how to proceed to find the time taken to actually empty the tank | 1988851 | Time for the tank to be empty
A hemispherical tank of radius 2 metres is initially full of water and has an outlet of 12 cm[imath]^2[/imath] cross-sectional are at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law [imath]v(t) = 0.6 \sqrt{2gh(t)}[/imath], where [imath]v(t)[/imath] and [imath]h(t)[/imath] are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time [imath]t[/imath], and [imath]g[/imath] is the acceleration due to gravity. Find the time it takes to empty the tank. I know to write a differential equation by relating the decrease of water level to the outflow in order to solve. But don't know how and as we are not not given the initial height how can we solve it. |
2268186 | Hamiltonian path implies Hamiltonian cycle?
If a graph [imath]G[/imath] is such that from every vertex [imath]v[/imath] there is a Hamiltonian path, does that imply that there is a Hamiltonian cycle? I've tried a constructive proof, but to no avail. No other ideas occur to me. I'd rather have just hints :-) No fun otherwise... | 502073 | If a graph has a Hamilton Path starting at every vertex, must it contain a Hamilton Circuit?
If a graph [imath]G[/imath] has at least three vertices, and has a Hamilton Path starting at every vertex, must it contain a Hamilton Circuit? I have been struggling with this problem. It seems that because every vertex both starts and does not start a Hamilton Path, every vertex is contained in a circuit. But I cannot extend this idea to show that there is a Hamilton Circuit. |
2266826 | In general, how to determine the cardinality of the set of expression over a language?
For a given language L, with its set of variables [imath]V[/imath], function symbols [imath]F[/imath], relation symbols [imath]R[/imath]. Once we are given the cardinality of [imath]V[/imath], [imath]F[/imath], [imath]R[/imath]. Is there any way to define the cardinality of the set of expressions over this language? I think maybe there can be a systematic way to find the cardinality (which may be very messy, but if it is such a way to deal with that, I really wish to learn about it also). And also I expect to learn (if there are) some useful tricks to determine it simply use the relation of the size of the set of expressions and the size of finite strings (which can be find much easily in this related question below). Related question: Logic: cardinality of the set of formulas The cardinality of a language over a set of variables [imath]V[/imath] with [imath]\#V= κ[/imath] Both of above seems talked about the set of finite strings So could someone give some way to determine the size of the set of expression? Any help will be appreciate. | 2264169 | The cardinality of a language over a set of variables [imath]V[/imath] with [imath]\#V= κ[/imath]
In predicate logic, to define a language, we consider the following [imath]3[/imath] sets: (1) [imath]V[/imath], the set of variables; (2) [imath]F[/imath], the sets of function symbols; (3) [imath]P[/imath], the set of predicate symbols with their arities. Now I am concidering about determine the cardinality of a language over a set with arbitary cardinality, [imath]κ[/imath](Note that [imath]κ[/imath] can be either finite, countablely infinite or uncountablely infinite ). May I please ask: If the cardinality of [imath]V[/imath], [imath]\#V = κ[/imath], what conditions on [imath]V[/imath] , [imath]F[/imath] and [imath]P[/imath] will give us a language over [imath]V[/imath], [imath]F[/imath] and [imath]P[/imath] with cardinality [imath]κ[/imath]? And may I please ask if it is the fact that if [imath]\kappa[/imath] is finite, then to get a language with cardinality [imath]\kappa[/imath] we need [imath](V+F+P)^{<\omega}=(\kappa+F+P)^{<\omega}=\kappa[/imath], so we can never have a language of cardinality [imath]\kappa[/imath]. I think it is something weird here. Am I wrong with something? I had googled it but I have get nothing from websites, now I am confused. Could someone please help? Thanks so much! EDIT: Latter I am told that in my course, the cardinality of a language refers to the set of expression instead of this of the set of strings. So now I need to consider further about when a string makes sense. EDIT':Now I am considering that if we can find out a general way to determine the cardinality of expression, which may related to combinatorics. See this question: In general, how to determine the cardinality of the set of expression over a language? Any help would be appreciate. |
1222093 | Laurent Series Expansion for [imath]\sin(\frac{1}{z})[/imath]
I am having trouble with Laurent series expansions. I am supposed to find the Laurent Series Expansion for [imath]\sin(\frac{1}{z})[/imath] around [imath]0<|z| < \infty[/imath]. I know the definitions of [imath]a_n[/imath] and [imath]b_n[/imath] for the Laurent expansion but that is all. Is there another way of arriving at the same, without using the definition? | 1564141 | Laurent Series of [imath]f(z) = \sin {\frac{1}{z}}[/imath] at [imath]z=0[/imath]
I am attempting to find the Laurent expansion at [imath]z=0[/imath] for [imath]f(z) = \sin {\frac{1}{z}}[/imath] where [imath]z[/imath] is a complex number. I rewrote the function as [imath]f(z)=\frac{ e^{iz^{-1}} - e^{-iz^{-1}}}{2i}[/imath] [imath]=\frac{1}{2i} \sum_{n=0}^{\infty}[i^n-(-i)^n]\frac{z^{-n}}{n!} [/imath] What do I do next? |
2268766 | Finding value of integration with infinite limit
Finding value of [imath]\displaystyle \int^\infty_{-\infty}\frac{e^{2x}-e^x}{x(e^{2x}+1)(e^x+1)} \, dx[/imath] Attempt: Let [imath]\displaystyle I = \int^\infty_{-\infty}\frac{e^{2x}-e^x}{x(e^{2x}+1)(e^x+1)} \, dx = 2\int^\infty_0 \frac{e^{2x}-e^x}{x(e^{2x}+1)(e^x+1)} \, dx[/imath] So [imath]I = 2\int^\infty_0 \bigg(\frac{1}{e^x+1}-\frac 1 {e^{2x}+1}\bigg)\frac 1 x \, dx = 2\int_0^\infty \frac 1 {(e^x+1)x} \, dx - 2\int_0^\infty \frac{1}{(e^{2x}+1)x} \, dx[/imath] Could some help me how to solve it, thanks. | 808871 | Prove that [imath]\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x=\log 2[/imath]
This integral popped up recently [imath]\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x = \log 2[/imath] A solution using both real and complex analysis is welcome. I tried rewriting it using symmetry, and then the series expansion of [imath]1/(1+e^{nx})[/imath], however this did not quite make it. \begin{align*} I & = 2\int_{0}^{\infty} \frac{e^{2x}-e^x}{x e^{3x}(1+e^{-2x})(1+e^{-x})}\mathrm{d}x \\ & = 2 \int_{0}^{\infty}\frac{e^{-2x} -e^{-x}}{x} \left(\sum_{n=0}^\infty (-1)^n e^{-n x}\right)\left(\sum_{m=0}^\infty (-1)^me^{-2mx}\right) \mathrm{d}x \end{align*} The first part reminds me of a Frullani integral (it evaluates to [imath]\log 2[/imath]). However I am unsure if this is the correct path, any help would be appreciated. =) |
2268996 | Suppose that [imath]\sum_{n=0}^{\infty} a_n^{2}x^{n}[/imath] converges exactly on [-6,6].
Suppose that [imath]\sum_{n=0}^{\infty} a_n^{2}x^{n}[/imath] converges exactly on [-6,6]. Find the largest set on which you can guarantee that [imath]\sum_{n=0}^{\infty} a_nx^{n}[/imath] converges, and prove your answer. I think the largest set is [imath](-\sqrt{6},\sqrt{6})[/imath]. I am not sure how to prove this answer. | 2261069 | Interval of Convergence examples (-6,6)
There was a book problem I recently solved going like this: [imath]\sum_{n=0}^\infty a_n^2x^n[/imath] converges on [imath][-6,6][/imath] What is the largest interval [imath]\sum_{n=0}^\infty a_nx^n[/imath] is guaranteed to converge on? I already understand this: [imath]\limsup_{n\rightarrow\infty}|a_n^2|^{1/n}=1/6[/imath] [imath]\Rightarrow\limsup_{n\rightarrow\infty}|a_n|^{1/n}=1/\sqrt6[/imath] [imath]\Rightarrow[/imath]Radius of Convergence is [imath](-\sqrt6,\sqrt6)[/imath] for [imath]\sum_{n=0}^\infty a_nx^n[/imath] What I was wondering however, is what are examples of what the values for [imath]a_n[/imath] could be where the endpoints dont converge for the series, but the series with [imath]a_n^2[/imath] converges on [-6,6]? |
1400554 | Existence a diffeomorphism on [imath]\mathbb R^n[/imath]
How may I show that for any [imath]p,q\in\mathbb R[/imath], there exist a diffeomorphism [imath]F:\mathbb R\rightarrow\mathbb R[/imath] such that [imath]F(p)=q[/imath] and [imath]F[/imath] to be an identity function outside of a some neighborhood of [imath]p[/imath] ? My attempts: By the bump function [imath]\varphi[/imath] such that [imath]\varphi_{\bigl|[-1,1]}=1[/imath] and [imath]\varphi_{\bigl|\mathbb R-(-2,2)}=0[/imath], I defined [imath]g(x):=a\varphi(x)+x[/imath] for a fixed parameter [imath]a[/imath]. If I can show [imath]g'(x)>0[/imath] then inverse function theorem and a change of variable solve it. 1- Why [imath]g'(x)>0[/imath]? 2- How can I prove it for [imath]\mathbb{R^n}[/imath] ? | 325277 | [imath]\mathbb{R}\to\mathbb{R}[/imath] Diffeomorphism with special property
[imath]\phi:\mathbb{R}\to\mathbb{R}[/imath] be a diffeomorphism with the following property [imath]a\in\mathbb{R},|a|<\frac{1}{10}[/imath] (i)[imath]\phi(a)=0[/imath] (ii) [imath]\phi(x)=x[/imath] when [imath]|x|>1[/imath] and how to generalize that in [imath]\mathbb{R}^n[/imath] please help. |
2269472 | Is the game with given conditions fair?
A gambler rolls two unbiased dice and stands to loss of [imath] \[/imath]2[imath] if he fails to throw a six, to win [/imath] \[imath]4[/imath] if he throws one six and to win [imath] \[/imath]10$ if he throws two sixes. Is the game fair? I can find probability for all three cases but to decide whether game is fair or not? Could someone help me with this? | 839628 | What is a fair game?
Suppose [imath]X_n[/imath] is the fortune of a gambler after [imath]n[/imath] th game. Then the game is called fair (Breiman 1968) if [imath]E[X_{n+1} \mid X_1, \dots, X_n] = X_n \forall n[/imath] My question is why a fair game is not defined as the following [imath]E[X_{n+1}] = E[X_n] \forall n[/imath] i.e. [imath]E[X_{n+1}- X_n]=0[/imath]. This should be the proper definition as a fair game is where avg. gain is zero. Nothing conditioning should be there. |
2268695 | Constructing diffeomorphism of [imath]R^n[/imath]
Let [imath]R > r > 0[/imath] and [imath]a,b\in B(0,r)\subset R^n[/imath] where [imath]n \geq 2[/imath]. Construct a diffeomorphism [imath]f[/imath] of [imath]R^n[/imath] satisfying [imath]f(a)=b, f(x)=x [/imath] for all [imath]||x|| > R[/imath] and [imath]det(Df_x) =1 [/imath] for all [imath]x\in R^n[/imath]. I was trying with [imath]f(x)=\rho(x) g(x) + x[/imath] where [imath]\rho[/imath] is a bump function and [imath]g[/imath] is smooth function with [imath]g(a)=b-a[/imath]. But problem is that this type of function may not be diffeomorhism and also jaccobian of [imath]f[/imath] is not necessarily 1. | 2267317 | [imath]R>r>0[/imath] , [imath]||a||,||b|| ; is there a diffeomorphism on \mathbb R^n switching a,b ;identity for ||x||>R ; pulls back volume form to itself?[/imath]
Let [imath]n>1[/imath] , consider [imath]\mathbb R^n[/imath] with standard volume form [imath]\omega _0 =dx_1 \wedge ...\wedge dx_n[/imath] ; let [imath]r,R[/imath] be two positive real numbers with [imath]R>r[/imath] . Then is it true that for every [imath]a,b \in B(0;r)[/imath] , [imath]\exists[/imath] a diffeomorphism [imath]f:\mathbb R^n \to \mathbb R^n [/imath] such that [imath]f(a)=b ; f(x)=x , \forall x \in \mathbb R^n [/imath] with [imath]||x||>R[/imath] and [imath]f^*\omega _0=\omega _0[/imath] ? As [imath]f^*(dx_1 \wedge ...\wedge dx_n)=(det f' )(dx_1 \wedge ...\wedge dx_n)[/imath] , so basically given [imath]R>r>0[/imath] and [imath]||a||,||b||<r[/imath] , we have to find a diffeomorphism [imath]f : \mathbb R^n \to \mathbb R^n [/imath] with [imath]f(a)=b[/imath] ; [imath]f|_{\mathbb R^n \setminus B[0;R]}=Id[/imath] and [imath]det f'(p)=1 , \forall p \in \mathbb R^n [/imath] . Definitely this cannot work out in [imath]\mathbb R[/imath] , because the only differentiable function satisfying [imath]det f'(p)=1 , \forall p[/imath] and [imath]f(a)=b[/imath] is [imath]f(x)=x+b-a[/imath] and this function never gives identity unless [imath]a =b[/imath] |
2270193 | Limit of 1/x as x approaches infinity
Why is [imath]\lim_{x\to\infty} \frac{1}{x}[/imath] equal to [imath]0[/imath]? | 1276646 | limit as [imath]x[/imath] approaches infinity of [imath]\frac{1}{x}[/imath]
How can I show [imath]\lim_{x \to \infty} \frac{1}{x} = 0[/imath] using epsilon delta proof. Its pretty obvious that the limit is zero, but I am still new at epsilon proofs. |
2209627 | Group Orbits of this space under O(3)
Consider the vector space: [imath] \mathbb{V} = \{ \ A \in M_{3 \times 3}(\mathbb{R}) \ | \ A^{T} = A \mathrm{\ and\ }\mathrm{Tr}(A) = 0 \} [/imath] As well as the Lie Group: [imath] O(3) = \{ \ A \in M_{3 \times 3}(\mathbb{R}) \ | \ A^{T}A = A A^{T} = \mathbb{I} \} [/imath] Define the group action [imath]\Phi : O(3) \times \mathbb{V} \to \mathbb{V}[/imath] as [imath]\Phi(A,X)= \Phi_{A}(X) = AXA^{T}[/imath]. I am to determine the group orbit[imath]s[/imath] of this action. The way I am understanding this, is that I am to examine the structure of the set [imath] S := \{ \ \Phi_{A}(X) \in M_{3\times 3}(\mathbb{R}) \ | \ A \in O(3), X \in \mathbb{V} \ \} [/imath] I know that symmetric matrices are orthogonally diagonalizable, which means that I can write [imath]X = P D P^{T}[/imath] for any [imath]X \in \mathbb{V}[/imath] (where [imath]P \in O(3)[/imath] and [imath]D[/imath] is diagonal and traceless). This tells me that [imath]\Phi_{A}(X) = AXA^{T} = APDP^{T}A^{T} = \Phi_{AP}(D)[/imath], where [imath]AP[/imath] is just another matrix in [imath]O(3)[/imath]. This tells me: [imath] S = \{ \ \Phi_{A}(D) \in M_{3\times 3}(\mathbb{R}) \ | \ A \in O(3), D = \mathrm{diag}(a,b,c) \mathrm{\ with\ }a+b+c=0 \ \} [/imath] So the orbits of the whole vector space are exactly the orbits of the diagonal matrices in this subspace. Can I narrow this down further? | 2178119 | Determining the Orbits/Orbit Space of [imath]O(3)[/imath] on Real 3 by 3 Traceless Symmetric Matrices
Let [imath]O(n)[/imath] be the set of [imath]3 \times 3[/imath] orthogonal matrices, and [imath]S[/imath] be the set of traceless real symmetric [imath]3 \times 3[/imath] matrices. Now [imath]O(n)[/imath] will act on [imath]S[/imath] by conjugation. Note that by the spectral theorem [imath] \forall A \in S[/imath] in the orbit of [imath]A[/imath] there is a diagonal matrix [imath]D=diag(\lambda_1, \lambda_2,\lambda_3)[/imath] with [imath]\lambda_1\le\lambda_2 \le\lambda_3 [/imath]. Determine the all the orbits of the action & the orbit space of the action. What I've done: Since all elements of [imath]S[/imath] are diagonlizable that the orbit of any element should be the orbit of its' diagonalized form because [imath]\forall A \in S, \exists P \in O(3)[/imath] st [imath]PAP^t=D [/imath] with [imath]D[/imath] a diagonal matrix and moreover for any other [imath]Q\in O(3)[/imath] we get [imath]QDQ^t=QPAP^tQ^t=(QP)A(QP)^t=RAR^t[/imath]. And so would that imply the orbit space is just the set of traceless diagonal matrices as [imath]tr(QDQ^t)=tr(DQ^tQ)=tr(Q)[/imath] (cyclic property of trace) and [imath]Q\in S[/imath]. Any help is appreciated. |
2271382 | For how many ones is [imath]N =11\dots1[/imath] prime?
Suppose [imath]N=111\dots1[/imath] ([imath]n[/imath] ones). For which numbers [imath]n[/imath] is [imath]N[/imath] prime? [imath]n \ne 1, 3k[/imath] is all progress I have made! | 300805 | When is the number 11111....1 a prime number?
For which [imath]n[/imath] is the sum: [imath]\sum_{k=0}^{n}10^k[/imath] a prime number? Are they finite? |
2267563 | Proving an inequality involving the logarithm function: [imath]\frac{1}{n+1} \leq \ln \left(1+\frac 1n\right) \leq \frac 1n[/imath]
The question is to prove the inequality [imath]\frac{1}{n+1} \leq \ln \left(1+\frac 1n\right) \leq \frac 1n\\\forall n \geq 1, n\in \mathbb N[/imath] I tried using Taylor expansion but couldn't figure out anything. Any ideas? Thanks. | 2605447 | Show that $\frac{1}{k}-\ln\left(\frac{k+1}{k}\right)$ is bounded by [imath]\frac{1}{k^2}[/imath]
I want to show that the following inequality holds for all [imath]k\in\mathbb{N},k\geq 1[/imath]: [imath]$$ \frac{1}{k}-\ln\left(\frac{k+1}{k}\right)\leq\frac{1}{k^2} $$[/imath] I already know that it is bounded. It also know that [imath]1+x\leq e^x[/imath] for all [imath]x\in\mathbb{R}[/imath]. and that [imath]e^x\leq\frac{1}{1-x}[/imath] for all [imath]x\in [0,1)[/imath]. |
2271400 | If [imath]M[/imath] is an [imath]R[/imath]-module and [imath]A,B,C[/imath] are submodules of [imath]M[/imath] and [imath]B[/imath] is a submodule of [imath]C[/imath] then [imath]A+ (B\cap C)=(A + B)\cap C[/imath]
How to show that [imath]A+ (B\cap C)=(A + B)\cap C[/imath] if [imath]M[/imath] is an [imath]R[/imath]-module and [imath]A,B,C[/imath] are submodules of [imath]M[/imath] and [imath]B[/imath] is a submodule of [imath]C[/imath]? | 138954 | A necessary and sufficient condition for [imath]A\cap(B+C)=(A\cap B)+C[/imath] to hold for submodules
How to prove that if A, B, C are submodules of M module then [imath] A \cap (B+C) = (A\cap B)+C \iff C \subseteq A [/imath] Any help would be appreciated! |
2271642 | Is there a totally disconnected topological space which is not Hausdorff?
Definition: A topological space [imath]X[/imath] is totally disconnected if for all [imath]x \in X[/imath] , the connected component of [imath]X[/imath] containing [imath]x[/imath] is [imath]\{x \}[/imath]. My question is: Is there a totally disconnected topological space so which is not Hausdorff? | 1507986 | Is totally disconnected space, Hausdorff?
Recall a space is totally disconnected if the only connected subsets are singletons (one-point subsets). Is a totally disconnected space, Hausdorff? I think it is true since if [imath]a [/imath] and [imath]b [/imath] are two distinct points, they can be separated two disjoint open sets, since the main space is totally disconnected (see Theorem at < http://www.emathzone.com/tutorials/general-topology/totally-disconnected-space.html >). Is this argument true? |
2271842 | Uniform Continuity and Continuity
Could someone please suggest a good way to check for continuity (using the epsilon-delta definition or maybe using some other method. An example is to show that [imath]f(x)=\sqrt x[/imath] i.e. [imath]f(x)=x^{1/2}[/imath] is uniformly continuous on the positive real numbers. Thank you! Your help will be greatly appreciated. | 2245647 | Prove [imath]f(x) = \sqrt x[/imath] is uniformly continuous on [imath][0,\infty)[/imath]
I have seen a proof in [imath]\sqrt x[/imath] is uniformly continuous Below shows an alternative proof. Please correct me if im wrong. Proof: For any given [imath]\varepsilon >0[/imath], Let [imath]\delta_1 = \frac{\varepsilon}{2}[/imath], [imath]\forall x,y \in [1,\infty)[/imath] with [imath]|x-y|<\delta_1[/imath] Since [imath]|\sqrt x + \sqrt y|\geq2[/imath] [imath]|\sqrt x - \sqrt y| = \frac{|x-y|}{|\sqrt x+\sqrt y|} < |x-y| < \delta_1 = \frac{\varepsilon}{2}[/imath] Hence, [imath]\sqrt x[/imath] is uniformly continuous on [imath][1,\infty)[/imath]. [imath]\sqrt x[/imath] is continuous on [0,1] , so [imath]\sqrt x[/imath] is uniformly continuous on [0,1]. So, there exist [imath]\delta_2 > 0[/imath] such that [imath]\forall x,y \in [0,1], |x-y|<\delta_2[/imath], [imath]|\sqrt x -\sqrt y| <\frac{\varepsilon}{2}[/imath] Let [imath]\delta = \min{(\delta_1,\delta_2)}[/imath] [imath]\forall x,y \in [0,\infty)[/imath] with [imath]|x-y|<\delta[/imath], Case 1: [imath]x,y \in [0,1][/imath] Proven above as [imath]|x-y| < \frac{\varepsilon}{2} < \varepsilon[/imath] Case 2: [imath]x,y \in [1,\infty)[/imath] Proven above as [imath]|x-y| < \frac{\varepsilon}{2} < \varepsilon[/imath] Case 3: [imath]x \in [0,1] , y \in [1,\infty][/imath] [imath]|\sqrt x-\sqrt y| = |\sqrt x -1+1-\sqrt y| \leq |\sqrt x-1| + |\sqrt y -1| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}= \varepsilon[/imath] by applying case 1 and case 2. |
2271149 | Rudin Functional Analysis Chapter 2 Problem 16
Suppose that [imath]X[/imath] and [imath]K[/imath] are metric spaces, and [imath]K[/imath] is compact, and the graph of [imath]f:X\rightarrow K[/imath] is a closed subset of [imath]X\times K[/imath]. Prove that [imath]f[/imath] is continuous. Show that compactness of [imath]K[/imath] cannot be omitted from the hypothesis, even when X is compact. My solution: i tried doing it by proving the continuity by sequential criterion and for this i start with a sequence [imath]\{x_n\}_{n=0}^{\infty}[/imath] converging to some [imath]x\in X[/imath]. Then [imath]S=\big(\bigcup\limits_{n=1}^{\infty} \{x_{n}\}\big)\cup \{x\}[/imath] is compact subset of X and hence [imath]S^*=\big(\bigcup\limits_{n=1}^{\infty} (x_{n},f(x_n))\big)\cup (x,f(x))[/imath] is closed in [imath]X\times K[/imath] and since [imath](S\cup \{x\})\times K[/imath] is compact hence [imath]S^*[/imath] is compact and hence have a convergent subsequence. i am not able to move forward. any type of help will be appreciated. Thanks in advance. | 45227 | The closed graph theorem in topology
Theorem A map [imath]\phi[/imath] maps a topological space [imath]X[/imath] to another [imath]Y[/imath], where [imath]X[/imath] is Hausdorff, [imath]Y[/imath] is compact and the graph of [imath]\phi[/imath] is closed. Then [imath]\phi[/imath] is continuous. Is it reallly necessary to include the condition that [imath]X[/imath] is Hausdorff? Since I see no reason, and I appear to have a proof without using the condition, I would like to know the answer. A proof: For any closed subspace [imath]C[/imath] of [imath]Y[/imath], the pre-image [imath]D[/imath] ought to be closed. For any element [imath]a[/imath] in the complement of [imath]D[/imath], we can use the compactness to show that there is a finite number of open sets in [imath]Y[/imath] such that the union of them covers the C, and then the corresponding open sets in [imath]X[/imath] is an open neighborhood of [imath]a[/imath] which has an empty intersection with [imath]D[/imath]; so [imath]D[/imath] is closed. P.S. in the proof, those open sets are obtained by the condition that the graph is closed and that [imath](a,c)[/imath] is not in the graph for any [imath]c[/imath] in [imath]C[/imath]. |
2269541 | A special solution for a functional inequality from [imath]\mathbb{R}\times \mathbb{R} [/imath] onto [imath]\mathbb{R}[/imath]
Is there a bijection solution [imath]\eta:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}[/imath] (with an explicit formula, if it is possible) for the inequality [imath]\min\{x,y\}\leq \eta(x,y)\leq \max\{x,y\}\;\; ; \;\; x\neq y?[/imath] | 2269494 | A special bijection between [imath]\mathbb{R}\times \mathbb{R}[/imath] and [imath]\mathbb{R}[/imath]
Is there any bijection [imath]\eta:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}[/imath] with the property [imath]\min\{x,y\}\leq \eta(x,y)\leq \max\{x,y\}[/imath]? |
2272334 | Function relations to determine
let [imath]f,g[/imath] be contiunous, bounded on [imath]\Bbb R[/imath]. Suppose that [imath]f(x+y)=f(x)f(y)-g(x)g(y),\quad g(x+y)=f(x)g(y)+f(y)g(x),\ \forall\ x,y\in\ \Bbb R.[/imath] [imath]f(0)=1[/imath], [imath]g(0)=0[/imath]. Show that for some [imath]a\in\Bbb R[/imath], [imath]f(x)=\cos ax, g(x)=\pm \sin a x[/imath]. | 641416 | Functional equations leading to sine and cosine
This question is a possibly harder version of: Find [imath]g'(x)[/imath] at [imath]x=0[/imath]. Question. Let [imath]f,g :\mathbb R\to\mathbb R[/imath], such that \begin{align} f(x-y)=f(x)\, g(y)-f(y)\, g(x), \tag{1}\\ g(x-y)=g(x)\, g(y)+f(x)\, f(y), \tag{2} \end{align} for all [imath]x,y \in \mathbb{R} [/imath]. If [imath]g[/imath] is continuous at [imath]x=0[/imath] and not identically zero, then there exists an [imath]\alpha\in\mathbb R[/imath], such [imath] f(x)=\sin \alpha x\quad\text{and}\quad g(x)=\cos \alpha x. [/imath] Is there a pair of discontinuous [imath]f[/imath] and [imath]g[/imath] satisfying [imath](1)[/imath] and [imath](2)[/imath]? Update. If [imath]\ell :\mathbb R\to\mathbb R[/imath] is a linear functional over [imath]\mathbb Q[/imath] (i.e., [imath]\ell(qx+ry)=q\ell(x)+r\ell(y)[/imath], for all [imath]x,y\in\mathbb R[/imath] and [imath]q,r\in\mathbb Q[/imath]), then [imath] \sin\big(\ell(x)\big), \quad \cos\big(\ell(x)\big), [/imath] satisfy [imath](1)[/imath] and [imath](2)[/imath]. Discontinuous such functionals do exist, and they are obtainable using Zorn's Lemma (equivalently the Axiom of Choice.) This takes care of the second question. |
2272447 | Compute the fresnel integral [imath]\int_0^x\cos(\frac{\pi u^2}{2})[/imath]du
Background: This is from Arfken et al, Mathematica methods. Probelm 12.6.1 part a) The question asks to give the asymptotic expansion by integrating by parts of [imath]C(x)=\int_0^x\cos\left (\frac{\pi u^2}{2}\right )\text{du}[/imath] Naïvely integrating du and differentiating [imath]\cos(\pi u^2/2)[/imath] I got: [imath]u\cos\left(\frac{\pi u^2}{2}\right)+\frac{\pi u^3}{3}\cos\left(\frac{\pi u^2}{2}\right)-\frac{\pi^2 u^5}{3\cdot5}\cos\left(\frac{\pi u^2}{2}\right)-\frac{\pi^3 u^7}{3\cdot5\cdot7}\cos\left(\frac{\pi u^2}{2}\right)+\frac{\pi^4 u^9}{3\cdot5\cdot7\cdot9}\cos\left(\frac{\pi u^2}{2}\right)\dots[/imath] [imath]=\sum_{n=0}^{\infty}\frac{u^{4n+1}\pi^{2n}}{(4n+1)!!}(-1)^n\cos\left(\frac{\pi u^2}{2}\right)+\sum_{n=0}^{\infty}\frac{u^{4n+3}\pi^{2n+1}}{(4n+3)!!}(-1)^n\sin\left(\frac{\pi u^2}{2}\right)[/imath] Could someone explain how to transform the integral into an asymptotic integral with 1/u in the integrand. The correct answer: I found the answer worked here, see Daniel Fischer's answer. | 991985 | Asymptotic expansion for Fresnel Integrals
If you take the fresnel integrals to be [imath]S(x) = \int_{0}^{x}\sin \left(\frac { \pi \cdot t^2}{2} \right) dt[/imath] How do you find the asymptotic expansion? I know it begins with a [imath]1/2[/imath] but how? |
2271933 | Bijection of complements
Let [imath]E[/imath] and [imath]F[/imath] be subsets of some set [imath]S[/imath], which may be finite or infinite. Suppose we are given a bijection [imath]g\colon E\to F[/imath]. Is there an explicit way to construct from [imath]g[/imath] a bijection [imath]h\colon S\setminus F\to S\setminus E[/imath]? By explicit, I mean that it is not enough to show that [imath]g[/imath] exists (this is clear, since the cardinalities of [imath]S\setminus F[/imath] and [imath]S\setminus E[/imath] are equal). | 140930 | Cardinal number subtraction
We all know that [imath]|\mathbb{N}| = \aleph_0[/imath]. Since [imath]|\{-1\} \cup \mathbb{N}| = \aleph_0[/imath] as well, I guess you could say that [imath]\aleph_0 + 1 = \aleph_0[/imath]. You can go on to derive that [imath]\aleph_0 + \aleph_0 = \aleph_0[/imath], and by induction [imath]n \times \aleph_0 = \aleph_0[/imath] (assuming that [imath]n[/imath] is finite, anyway). Now, here's a thing: What is [imath]\aleph_0 - \aleph_0[/imath]? Well, [imath]|\mathbb{P}| = \aleph_0[/imath], and [imath]|\mathbb{P} \backslash \mathbb{N}| = \aleph_0[/imath], so perhaps [imath]\aleph_0 - \aleph_0 = \aleph_0[/imath]? No, wait. Consider the set [imath]S = \{n \in \mathbb{N}: n > 5\}[/imath]. Now we have [imath]|S| = \aleph_0[/imath], and yet [imath]|\mathbb{N} \backslash S| = 5[/imath]. So maybe [imath]\aleph_0 - \aleph_0 = 5[/imath]? But that is absurd, since we can redefine [imath]S[/imath] to make the result any finite number we wish. Or we can define [imath]S[/imath] such that the result is countably infinite. Does this mean that the notion of [imath]\aleph_0 - \aleph_0[/imath] simply has no definite answer? Or am I just being too simplistic here? |
2271467 | Find the centroid of a curve
Problem: Find the centroid of the part of the large loop of the limacon [imath]r=1+2\cos(\theta)[/imath] that does not include the small loop. I know that in order to compute the centroid one needs to use following equations [imath]x=\frac{\iint_D xdydx}{m}[/imath] and [imath]y=\frac{\iint_D ydydx}{m},[/imath] where [imath]m=\iint_D dydx.[/imath] Note that we are assuming [imath]\rho=1,[/imath] which should work for this problem since I got the correct answer for a similar problem. The real issue that I am facing is that I am unable to set up the bounds on the integral because I do not what values of [imath]r,\theta[/imath] do I have to exclude in order to avoid the loop. Any help in this regard would be much appreciated. | 2271457 | Find the centroid of a curve.
Problem: Find the centroid of the part of the large loop of the limacon [imath]r=1+2\cos(\theta)[/imath] that does not include the small loop. I know that in order to compute the centroid one needs to use following equations [imath]x=\frac{\iint_D xdydx}{m}[/imath] and [imath]y=\frac{\iint_D ydydx}{m},[/imath] where [imath]m=\iint_D dydx.[/imath] Note that we are assuming [imath]\rho=1,[/imath] which should work for this problem since I got the correct answer for a similar problem. The real issue that I am facing is that I am unable to set up the bounds on the integral because I do not what values of [imath]r,\theta[/imath] do I have to exclude in order to avoid the loop. Any help in this regard would be much appreciated. |
2271521 | Deduce relations between sides of a triangle if [imath]\cos A + \cos C= 2 - 2\cos B[/imath]
If the relation between the cosines of the angles of a triangle is given by [imath]\cos A + \cos C= 2 - 2\cos B[/imath] then what relation do the sides of the triangle follow? That is, are they in Arithmetic, Geometric or Harmonic progression? I tried using trigonometric addition formulae and then deducing the results. I applied [imath]\ cosx + \ cosy [/imath] formula along with [imath]A + B + C = \pi[/imath] but I could not solve the question. Please Note: This question is intended to be a multiple choice question. | 2084979 | If, in a triangle, [imath]\cos(A) + \cos(B) + 2\cos(C) = 2[/imath] prove that the sides of the triangle are in AP
By using the formula : [imath] \cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) [/imath] I've managed to simplify it to : [imath] 2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)=\sin\left(\frac{C}{2}\right)[/imath] But I have no idea how to proceed. |
1007733 | Diophantine [imath]7^a+2=3^b[/imath]
I want to find the solutions [imath](a,b)\in\mathbb{Z}^+\times\mathbb{Z}^+[/imath] of [imath]7^a+2=3^b[/imath]. One such solution is [imath](a,b)=(1,2)[/imath]. Looking modulo [imath]4[/imath], we have [imath](-1)^a+2\equiv(-1)^b[/imath], so [imath]a[/imath] and [imath]b[/imath] are of different parity. Modulo [imath]3[/imath]: [imath]1+2\equiv 0[/imath], which is always true. Modulo [imath]7[/imath]: [imath]2\equiv 3^b[/imath], so [imath]b\equiv 2\pmod 6[/imath]. This means [imath]a[/imath] is odd. | 1149561 | Diophantine equation: [imath]7^x=3^y-2[/imath]
I've tried using mods but nothing is working on this one: solve in positive integers [imath]x,y[/imath] the diophantine equation [imath]7^x=3^y-2[/imath]. |
2273239 | Number of real solutions [imath]2\cos(\frac{x^2+x}{2})=2^x+2^{-x} [/imath]
The number of real solutions of [imath]2\cos(\frac{x^2+x}{2})=2^x+2^{-x} [/imath] is (1) 0 (2) 1 (3) 2 (4) infinitely many . My work : [imath] 1\geq \cos\left(\frac{x^2+x}{2}\right)=\frac{2^x+2^{-x} }{2}\geq 1 \qquad \text{by (AM-GM).} [/imath] So [imath]\frac{x^2+x}{2}=2n\pi[/imath] for all [imath]n\in \mathbb{Z}[/imath] . Now discriminant [imath]=1+2n\pi[/imath] is always positive for [imath]n\geq0[/imath] . But the equation is a quadratic so it has only two solution . Hence the answer must be 2 . PS: I'm aware that the this problem is already on the site but i posted as there were no complete solution . | 2272551 | No. Of real roots of an equation
Consider the function [imath] f(x) = 2\cos((x^2+x)/6) - 2^x - 2^{-x}. [/imath] Clearly [imath]x= 0[/imath] is a root of equation so number of real roots can not be 0 but how to figure out question how many ? From where does this observation start? |
2273412 | Necessary and sufficient condition for [imath]x \equiv u_1(mod \ m_1)[/imath], [imath]...[/imath], [imath]x \equiv u_n(mod \ m_n)[/imath] to have a solution
How can i find necessary and sufficient condition for this system to have a solution: [imath]x \equiv u_1(mod \ m_1)[/imath] [imath]...[/imath] [imath]x \equiv u_n(mod \ m_n)[/imath] | 1856879 | When a congruence system can be solved?
How to prove that a congruence system with [imath]n[/imath] equations can be solved if and only if all the equations can be solved two by two? \begin{cases} x \equiv a_1 \phantom ((mod\phantom mm_1) \\ x \equiv a_2 \phantom ((mod\phantom mm_2) \\ ... \\ x \equiv a_n \phantom ((mod\phantom mm_n) \end{cases} I know that if [imath]n=2[/imath], the system can be solved if and only if [imath]\gcd(m_1,m_2)\mid a_1-a_2[/imath]. The thesis is: the system is solvable [imath]\Longleftrightarrow[/imath] [imath]\gcd(m_i,m_j)|\mid a_i-a_j[/imath] for all [imath]i\neq j[/imath]. |
2272971 | property of brownian motion
I want to show this result: [imath]\mathbb{P} \left[ \lim_{t \rightarrow \infty} \frac{B_t}{t} = 0 \right] = 1[/imath] For [imath]B_t[/imath] a standard brownian motion. I have no idea how to do this. Naively I tried exchanging limit and probability but that doesn't work. I just need a hint to get started. EDIT: In general what is a good way to show almost sure convergence? | 1729071 | Why does the Borel-Cantelli lemma finish the job? - Law of Large Numbers Brownian Motion
The objective is to prove that \begin{align*} \text{[imath]\lim_{t \to \infty} \frac{B_t}{t} =0 \qquad[/imath] a.s.} \end{align*} By the strong Law of Large Numbers, we have that: \begin{align*} \text{[imath]\lim_{t \to \infty} \frac{B_t}{t} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n B_k - B_{k-1} = 0 \qquad[/imath] a.s.} \end{align*} In doing so, the idea is to show that the trajectory of [imath]B[/imath] cannot deviate too much from [imath]B_n[/imath] on [imath][n, n+1][/imath]. It can be shown that \begin{align*} \sum_{n=0}^\infty P\big(\big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) < \infty. \end{align*} Now, the question is, how does the Borel-Cantelli Lemma finish this prove? So, the Borel-Cantelli Lemma is saying that \begin{align*} &P\big( \limsup_{n \to \infty} \big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) = 0, \qquad \text{so} \\ &P\big( \bigcup_{n=1}^\infty \bigcap_{k\geq n} \big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) = 0 \qquad \text{and} \\ &\lim_{n \to \infty} P\big( \bigcap_{k\geq n} \big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) = 0. \end{align*} How to find again the fraction [imath]\frac{B_t}{t}[/imath]? |
2273604 | Number of solutions for [imath]x_1+ x_2+ x_3 + \cdots + x_k=n[/imath], where [imath]0\leq x_i\leq p[/imath] for all [imath]i[/imath]
I know that for [imath]1\leq x_i\leq p[/imath] the answer will be the coefficient of [imath]x^n[/imath] in [imath](x + x^2 + x^3 + ... + x^p)^k[/imath]. But what will be the answer for the constraint [imath]0 \leq x_i \leq p?[/imath] Also, how can I generate a definite formula or recurrence relation to program it? It will be difficult to calculate the answer by summing up the GP series and then calculating the coefficients using series expansion. Thank you! | 1429561 | Number of solutions of [imath]x_1+x_2+\dots+x_k=n[/imath] with [imath]x_i\le r[/imath]
Let [imath]n,k,r[/imath] be positive integers. The number of all nonnegative solutions of the Diophantine Equation [imath]x_1+x_2+\dots+x_k=n[/imath] is [imath]\binom{n+k-1}{n}[/imath]. Is there a general formula for the number of solutions of the equation [imath]x_1+x_2+\dots+x_k=n[/imath] with [imath]x_i\le r[/imath] for every [imath]i\in \{1,2,\dots,k\}[/imath]? If one defines [imath]A_i[/imath] to be the number of solutions with [imath]x_i>r[/imath] then the answer will be [imath]\binom{n+k-1}{n}-|A_1\cup\dots\cup A_k|[/imath]. I think it can give a complicated formula. What is the formula? |
2274639 | Explicit definition of sequence of partial sums of 1/n!
is there an explicit definition of the following sequence? [imath]s(n)=\sum_{i=0}^n \frac1{i!}[/imath] Sincerely | 1281543 | Finite Summation of Fractional Factorial Series
Is there a closed form solution for the following series? (Without Using Gamma Function): [imath] S=\sum _{i=1}^{n-1} \frac{1}{(i+1)!} [/imath] |
2275295 | smooth [imath]C^\infty[/imath] anywhere but analytical [imath]C^{\omega}[/imath] nowhere function
There is function which is smooth but not analytic in one point. Such as [imath]f(x)=\exp(-1/x^2)[/imath] Does there exist function that is smooth [imath]C^\infty[/imath] anywhere but analytical [imath]C^{\omega}[/imath] nowhere ? If exist, give an explicit example. If not, how to prove? | 620290 | Is it possible for a function to be smooth everywhere, analytic nowhere, yet Taylor series at any point converges in a nonzero radius?
It is well-known that the function [imath]f(x) = \begin{cases} e^{-1/x^2}, \mbox{if } x \ne 0 \\ 0, \mbox{if } x = 0\end{cases}[/imath] is smooth everywhere, yet not analytic at [imath]x = 0[/imath]. In particular, its Taylor series exists there, but it equals [imath]0 + 0x + 0x^2 + 0x^3 + ... = 0[/imath], so while it has radius of convergence [imath]\infty[/imath], it is not equal to [imath]f[/imath] even in a tiny neighborhood of [imath]0[/imath]. There is also a function [imath]f(x) = \sum_{n=0}^{\infty} e^{-\sqrt{2^n}} \cos(2^n x)[/imath] which is smooth everywhere (that is, [imath]C^{\infty}[/imath]) yet analytic nowhere. In particular, the Taylor series at every point has radius of convergence [imath]0[/imath]. In fact, "most" smooth functions are not analytic. But this gets me wondering. Could there exist some function which is smooth everywhere, analytic nowhere, yet its Taylor series at any point has nonzero radius of convergence, and so converges to something, but that something is not the function, not even in a tiny neighborhood about the point of expansion? If yes, what is an example of such a function? If no, what is the proof that such a thing is impossible? And also, if no, what sort of restrictions exist on the convergence of the T.s.? At how many/what distribution of points can it converge to something which is not the function? I note that if we multiply together the two functions just given above, we have another smooth-everywhere, analytic-nowhere function, but this time at [imath]0[/imath] we have a convergent Taylor series (the same zero series as before -- just use the generalized Leibniz rule) which doesn't converge to the function in even a tiny neighborhood of [imath]0[/imath]. EDIT (Dec 31, 2013): With some Googling I came across a post to mathoverflow: https://mathoverflow.net/a/81465 The Taylor series of the Fabius function at any dyadic rational actually has infinite radius of convergence (only finitely many terms are nonzero) but does not represent the function on any interval. So it seems it is possible to have a function whose Taylor series converges to "the wrong thing" at a dense set of expansion points. But it still doesn't answer the question of whether that is possible for all expansion points on the entire real line. |
2275229 | Analysing a finite harmonic series
If [imath]X=\frac{1}{1001}+\frac{1}{1002}+\dotsb +\frac{1}{3001},[/imath] then a) [imath]X<1[/imath] b) [imath]X>\frac{3}{2}[/imath] c) [imath]1<X<\frac{3}{2}[/imath] d) None of the above holds My approach- I was not able to analyse series the way question is expecting me to do, I could only do these basic observations as [imath]\frac{1}{1001} > \frac{1}{1002} > \frac{1}{1003}> \dotsb > \frac{1}{3001}[/imath] means [imath]X=\frac{1}{1001}+\frac{1}{1002}+\dotsb +\frac{1}{3001} > \frac{2001}{3001} \approx 0.66[/imath] and using same [imath]X=\frac{1}{1001}+ \frac{1}{1002} +\dotsb +\frac{1}{3001} < \frac{2001}{1001} \approx 2.[/imath] | 688432 | Prove that [imath]1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\ldots+\frac{1}{3001}<\frac43[/imath]
Prove that [imath]1<\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+\ldots+\dfrac{1}{3001}<\dfrac43 \, .[/imath] My work: [imath]\begin{eqnarray*} S&=&\bigg(\dfrac{1}{1001}+\dfrac{1}{3001}\bigg)+\bigg(\dfrac{1}{1002}+\dfrac{1}{3000}\bigg)+\ldots+\dfrac{1}{2001}\\ S&=&\dfrac{1}{4002}\bigg\{\bigg(\dfrac{1001+3001}{1001}+\dfrac{1001+3001}{3001}\bigg)+\ldots\bigg\}+\dfrac{1}{2001}\\ S&\ge& \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1 \end{eqnarray*}[/imath] I could derive the left hand inequality with a [imath]\ge[/imath] sign though but could not do anything about the right hand inequality. Please help. EDIT: I do not need to use the equality sign, I can rather use only strict inequality because there exist terms which are not equal so, the AM-HM inequality is actually a strict inequality here. [imath]S> \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1[/imath] Though I got the solution but I am looking for some non-calculus solutions too. |
2275386 | Value of an expression involving complex numbers
Suppose [imath]a[/imath] is a complex number such that [imath]a^2+a+\frac 1a +\frac 1{a^2} +1 =0[/imath] If [imath]m[/imath] is a positive integer find value of [imath]a^{2m}+a^m+\frac 1{a^m} + \frac 1{a^{2m}}[/imath] Setting [imath]a=|a|e^{i\theta}[/imath] we get [imath]2|a| \cos \theta + 2 |a|^2 \cos 2\theta+1=0[/imath] I could not get anything from this. Any ideas? Thanks. | 518060 | Complex Numbers....
Suppose a is a complex number such that: [imath]a^2+a+\frac{1}{a}+\frac{1}{a^2}+1=0[/imath] If m is a positive integer, find the value of: [imath](a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}[/imath] My Approach: After I could not solve it using the usual methods I tried a bit crazier approach. I thought that as the question suggests that the value of the expression does not depend upon the value of m, provided it is positive, hence the graph of the expression on Y-axis and m on X-axis would be parallel to X-axis and thus the slope be zero. So I differentiated it with respect to m and equated it to zero and after factorizing and solving I got [imath]a^m=-1[/imath] or [imath]a^m=1[/imath] or [imath]a^m=\left(\frac{-1}{4}+i\frac{\sqrt15}{4}\right)[/imath] or [imath]a^m=\left(\frac{-1}{4}-i\frac{\sqrt15}{4}\right)[/imath]. But if [imath]a^m=1[/imath] then [imath]a=1[/imath] which does not sattisfy the first equaion. Thus the value of [imath](a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=\frac{-9}{4}[/imath] OR [imath](a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=0[/imath] What other approach would you suggest? What are the flaws in my approach (if any)? |
1018757 | Weak convergence in [imath]l_p[/imath] implies pointwise convergence?
Could someone please share their thoughts on this one: Consider at [imath]l_p(Y)[/imath], for [imath]1<p<\infty[/imath] with the counting measure on [imath]Y[/imath]. Show that if a sequence weakly converges in [imath]l_p(Y)[/imath] then it would converge pointwise in Y. Show that the converse holds only when Y is finite. Thanks! | 1513111 | weak convergence in [imath]l^p[/imath] implies bounded and pointwise convergence
Let [imath]1<p<\infty[/imath], [imath](u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})[/imath] such that [imath]u_n[/imath] converges weakly to [imath]u\in l^p(\mathbb{N})[/imath]. Prove that (1) [imath](u_n)[/imath] is bounded and (2) for all fixed component [imath]j\in n\in\mathbb{N}[/imath] it is [imath]\lim_{n\to\infty} u_{n,j}=u_j[/imath] (in lecture we called it "pointwise convergence of sequences"). My questions are: (1) is "[imath](u_n)[/imath] bounded", i.e. [imath]\|u_n\|<\infty[/imath] for all [imath]n\in\mathbb{N}[/imath] already satisfied because [imath](u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})[/imath]? For (2) it is to show that [imath]|u_{n,j}-u_j|\to 0,\; n\to\infty[/imath] for fixed [imath]j\in\mathbb{N}[/imath]. But I don't know how to do this. |
2275458 | Prove that [imath]2^{3^n}+1[/imath] is divisible by [imath]3^{n+1}[/imath]
I'm stuck on the induction step where substitution is difficult. | 1591176 | [imath]3^{n+1}[/imath] divides [imath]2^{3^n}+1[/imath]
Describe all positive integers,n such that [imath]3^{n+1}[/imath]divides [imath]2^{3^n}+1[/imath]. I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,[imath]3^{n+1}[/imath] divides [imath]2^{3^n}+1[/imath]. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1. |
2272673 | How do you plot a catenary curve with two known points and a known length?
All the equations I've seen assume I know the sag, height, force, or the weight of the catenary I want to plot, but I don't. All I know is the total length ([imath]L[/imath]) of the wire and the position of the two points it's attached to ([imath]x_1[/imath], [imath]y_1[/imath], [imath]x_2[/imath], [imath]y_2[/imath], respectively). If I give it values so that [imath]L = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)[/imath], I'd expect to get a straight line, since [imath]L[/imath] is equal to the distance between the two points. If [imath]L[/imath] was larger, however, I'd expect it to sag like a catenary normally does. [imath]L[/imath] will never be less than the distance between the two points. How can I plot a curve knowing only this? Is it even possible? | 1000447 | Finding the catenary curve with given arclength through two given points
I need to find a general way of expressing a catenary, and I couldn't find anything online. Is it possible to explicitly find an explicit equation of a catenary which goes through [imath]P_1 = (x_1,y_1)[/imath] and [imath]P_2 = (x_2,y_2)[/imath] (assuming [imath]x_1 \neq x_2[/imath]) so that the length of the curve between the two points is [imath]l[/imath] (assuming [imath]l> d(P_1,P_2) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/imath])? |
2275708 | Applying analytic continuation
I'm reading the Stein-Shakarchi. In Chapter 3 there is a problem and I need to check my solution: Show that there is no function [imath]f[/imath], such that is holomorphic, defined on the open unit disk that extends continuously to its border, and is equal to the function [imath]g(z)=\frac{1}{z}[/imath] in the border. Here is my "solution": Extend the domain of [imath]f[/imath] in such a way that now is [imath]\mathbb{C}[/imath]. Now is [imath]\frac{1}{z}[/imath] if [imath]|z|\geq1[/imath], and is equal to [imath]f(z)[/imath] otherwise. By Morera's theorem every triangle has integral egual to 0 at the border of [imath]\mathbb{D}[/imath], so is holomorphic, in the other hand, f and the new g agrees on a set with accumulation point, so they must be equal. Finally, [imath]g[/imath] is not holomorphic in all [imath]\mathbb{C}[/imath], and [imath]f[/imath] is, so we get a contradiction. Is something wrong here? Do you have any other solution? | 749620 | Very quick question: function extends to 1/z on the boundary of unit disc
How can one show there is no holomorphic function [imath]f[/imath] on the open unit disc [imath]\mathbb{D}[/imath] such that it extends continuously to [imath]\frac{1}{z}[/imath] on [imath]\partial\mathbb{D}[/imath]? I mean [imath]f[/imath] takes value [imath]\frac{1}{z}[/imath] when [imath]|z|=1[/imath] and continuous here. Thank you. |
2275527 | Application of MVT for integrals
I have a question and I'm really struggling to grasp what I need to do. I have Rolle's theorem and Mean Value Theorem given to me, and the question follows; Let [imath]f(x)[/imath] be a continuous function on [imath][0,1][/imath] such that [imath]\int_{0}^{1} f(y) dy = 1[/imath] Prove that then [imath]f(x)=1[/imath] for some [imath]x\in[0,1][/imath]. I don't know if it's the way this is worded but I've never been so confused in my entire life! Thanks in advance. | 2218132 | How do I prove that then [imath]f(x) = 1[/imath] for some [imath]x ∈ [0, 1][/imath]?
We are asked the following: let [imath]f(x)[/imath] be a continuous function on [imath][0, 1][/imath] such that [imath]\int _ {0} ^ {1} \ f(y) dy = 1[/imath] prove that [imath]f(x) = 1[/imath] for some [imath]x ∈ [0, 1][/imath] The question asks to take into account Rolle's Theorem and Mean Value Theorem. I'm aware of the definitions, but I am really unsure how to answer this question and my tutors are not really responding till after easter. Could someone please help? Why is there a "y" involved? Thanks you |
2275713 | showing a sequence is Cauchy in a normed space
Suppose I have a sequence in a normed space. In this space, the Cauchy-ness and convergence of the sequence is determined completely by the specific norm defined on the space, correct? So when showing that a given sequence is Cauchy on this normed space, one would not be able to use any information about the sequence from an outside context (for example, that the sequence convergences under the standard absolute value norm in [imath]\mathbb{R}[/imath])? As you can see, I am a little confused about non-standard norms on vector spaces. If we have one, is that the ONLY norm we can have there? Or can we have many different norms, including the standard one, on the space simultaneously? I hope that makes sense. | 74057 | Cauchy sequence in a normed space
Let [imath]V[/imath] be a real vector space. Suppose that [imath]\Vert \cdot \Vert_1[/imath] and [imath]\Vert \cdot \Vert_2[/imath] are two norms on [imath]V[/imath] which are equivalent. I suspect the following to be true. Let [imath](x_n)_{n=0}^\infty[/imath] be a sequence in [imath]V[/imath]. Then [imath](x_n)_n[/imath] is a Cauchy sequence w.r.t. [imath]\Vert \cdot \Vert_1[/imath] if and only if [imath](x_n)_n[/imath] is a Cauchy sequence w.r.t. [imath]\Vert \cdot \Vert_2[/imath]. Is it true? I'm looking for a good way to explain why "equivalence" of norms is the right way of "comparing" norms without mentioning anything from topology besides "convergence". |
2276066 | Is [imath]\mathbb Z[/imath] a [imath]\mathbb Q[/imath]- module?
Could we define proper [imath]\mathbb Q[/imath] module structure on ring of integer [imath]\mathbb Z[/imath]? Any comments are welcome. | 151850 | Prove [imath]\mathbb{Z}[/imath] is not a vector space over a field
This is an exercise from Chapter 3 of Golan's linear algebra book. Problem: Show [imath]\mathbb{Z}[/imath] is not a vector space over a field. Solution attempt: Suppose there is a such a field and proceed by contradiction. I will write multiplication [imath]FV[/imath], where [imath]F[/imath] is in the field and [imath]V[/imath] is an element of [imath]\mathbb{Z}[/imath]. First we rule out the case where the field has characteristic 2. We would have [imath]0=(1_F+1_F)1=1_F1+1_F1=2[/imath] a contradiction. Now, consider the case where the field does not have characteristic 2. Then there is an element [imath]2^{-1}_F[/imath] in the field, and [imath]1=2_F(2^{-1}_F1)=2^{-1}_F1+2^{-1}_F1[/imath] Now [imath]2^{-1}_F1\in\mathbb{Z}[/imath] as it is an element of the vector space, but there is no element [imath]a\in\mathbb{Z}[/imath] with [imath]2a=1[/imath], so we have a contradiction. Is this correct? |
2275785 | Find a simpler description of the following set: [imath]\bigcup_{i=0}^{\infty}[i, i+2][/imath]?
I asked a similar question last night asking for an explanation of the statement, however I was unable to find how to prove such a statement, so I have a proof, however I think it is wrong, so I'm just asking for it to be checked and if it is, for it to be corrected, thanks! Question Describe the following set, and prove your answer correct. (Here brackets denote intervals on [imath]\mathbb{R}[/imath].) [imath]\bigcup_{i=0}^{\infty}[i, i+2][/imath] Working [imath][i,i+2]=[/imath]{[imath]x\in\mathbb{R}|i\leq x\leq i+2[/imath]} Thus, with the union we have {[imath]x\in\mathbb{R}|i\leq x\leq i+2[/imath]}[imath]=[0, \infty)[/imath] Now, note that [imath]\bigcup_{i=0}^{\infty}[i, i+2]=[0, 2]\cup [1, 3]\cup [2,4]\cup...\cup[k, k+2]\cup...[/imath] for some [imath]k\in\mathbb{R}[/imath] Consider [imath]x \notin [0, \infty)[/imath] then [imath]|x|>[0, \infty)[/imath], so [imath]\exists[/imath] some [imath]i\in\mathbb{N}[/imath] such that [imath]i<|x|[/imath]. Thus, [imath]x\notin (0, \infty][/imath] [imath]\therefore x[/imath] is not the union Thus, we have shown the union of [imath][i, i+2][/imath] is on the interval [imath][0, \infty)[/imath] so; [imath]\bigcup_{i=0}^{\infty}[i, i+2]=[0, \infty)[/imath] | 2274429 | What does [imath]\bigcup_{i=0}^{\infty}[i, i+2][/imath] mean?
I think I get it but I wanted to be sure before attempting the question: Question Describe the following set, and prove your answers correct. (Here brackets denote intervals on [imath]\mathbb{R}[/imath].) [imath]\bigcup_{i=0}^{\infty}[i, i+2][/imath] Does it mean: {[imath]0, 1, 2, 3, 4, 5...,\infty[/imath]} [imath]\cup[/imath] {[imath]2, 3, 4, 5, 6, 7,...,\infty[/imath]}? If so, what is the difference between: [imath]\bigcup_{i=0}^{\infty}[i, i+2][/imath] and [imath]\bigcup_{i=0}^{\infty}(i, i+2)[/imath] |
2276305 | Trying to prove iff for summation of sequence
Prove [imath]\sum_{n=1}^\infty a_n \in \mathbb R \mathbb \leftrightarrow \sum_{n=1}^\infty\frac{a_n}{1+a_n}\in \mathbb R[/imath] The hint says that for one direction argue that [imath]b_n = \frac{a_n}{\frac{a_n}{1+a_n}}[/imath] is bounded after some fixed cutoff point. I dont know where the [imath]b_n[/imath] is coming from or which test to use to prove this. | 956256 | Series in Real Analysis
If [imath]a_n \ge 0 [/imath] for all n, prove that [imath]\sum_{n=1}^\infty a_n[/imath] converges if and only if [imath]\sum_{n=1}^\infty {a_n\over 1+a_n}[/imath] converges. Here is my attempt! => Suppose that [imath]\epsilon \ge 0[/imath] is given and [imath]a_n[/imath] converges, then for all $N\le n \le m[imath]$\sum_{k=n+1}^m a_k \lt \epsilon.[/imath] let $b_n={a_n\over 1+a_n}[imath] and Notice that [/imath]b_n \lt a_n$ and Then, applying the comparison test [imath]|\sum_{k=n+1}^m b_k| \le \sum_{k=n+1}^m |b_k |\le\sum_{k=n+1}^m a_k\lt\epsilon[/imath]Hence,${a_n\over 1+a_n}$ converges. let me know if it is worng and I don't know how to start for the other direction. Any help is appreciated! |
2276866 | Why does [imath]a+b+c=3\;[/imath] imply [imath]\;a\sqrt{a+3} + b\sqrt{b+3} + c\sqrt{c+3} \ge 6[/imath]
Let [imath]a,b,c[/imath] be positive real numbers such that [imath]a+b+c=3[/imath]. Prove that [imath]a\sqrt{a+3} + b\sqrt{b+3} + c\sqrt{c+3} \geqslant 6[/imath] | 2262726 | [imath]a+b+c = 3[/imath], prove that :[imath]a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6[/imath]
[imath]a, b,c [/imath] are positive real numbers such that [imath]a+b+c = 3[/imath], prove that :[imath]a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6[/imath] Any ideas ? |
2277018 | Suppose [imath]x > 0[/imath]. Prove that [imath]\frac{\sqrt{x}}{x+1} \leq \frac{1}{2}[/imath].
Suppose [imath]x > 0[/imath]. Prove that [imath]\frac{\sqrt{x}}{x+1} \leq \frac{1}{2}[/imath] Hey everyone, here is a simple math prove question, but I had a hard to start this proving, please give me some ideas to deal with this. Thanks | 191070 | Prove that [imath]\sqrt{a}\leq\frac{1+a}{2}[/imath]
I have a math homework where it's being asked to prove that : [imath]\forall a \geq 0,\sqrt{a}\leq\frac{1+a}{2}[/imath] However, I don't have any idea how I should start this one... Any idea ? |
2276927 | Show that there exists [imath]\alpha[/imath] such that [imath]f(x) = \alpha x^2[/imath] for all [imath]x\in \mathbb{R}[/imath].
Problem: Let [imath]f[/imath] be a real valued continuous function on [imath]\mathbb{R}[/imath] . Suppose that [imath]f(x+y) + f(x-y) = 2[f(x)+f(y)][/imath] for any [imath]x,y\in \mathbb{R}[/imath]. Show that there exists [imath]\alpha[/imath] such that [imath]f(x) = \alpha x^2[/imath] for all [imath]x\in \mathbb{R}[/imath]. My attempt: Frankly, I'm not really sure where to begin. I can rewrite the problem as [imath]\frac{f(x+y) + f(x-y)}{f(x)+f(y)} = 2[/imath] which looks a little bit like the Lipschitz condition. Also, it seems like a clever substitution is going to happend with [imath]f(x-y)[/imath] and [imath]f(x+y)[/imath]. Beyond that I can't really get a hold on the problem. Any hints or suggestions? | 485718 | Functional Equation (no. of solutions): [imath]f(x+y) + f(x-y) = 2f(x) + 2f(y)[/imath]
Find all the functions [imath]f\colon\mathbb{Q} \to \mathbb{Q}[/imath] such that [imath]f(x+y) + f(x-y) = 2f(x) + 2f(y)[/imath], for all rationals [imath]x,y[/imath]. |
2275741 | How do you describe the following set: [imath]\bigcap_{i=0}^{\infty}(\frac{-1}{i}, \frac{1}{i})[/imath]
I've been stuck on this question since last night and the problem I'm having is I just don't know how to set out the answer or if I answered it correctly. Question Describe the following set, and prove your answer correct. (Here brackets denote intervals on [imath]\mathbb{R}[/imath].) [imath]\bigcap_{i=0}^{\infty}(\frac{-1}{i}, \frac{1}{i})[/imath] Working As I said I don't know how to set it out, so my answer is simply: [imath]() \cap (-1, 1) \cap (\frac{-1}{2}, \frac{1}{2}) \cap (\frac{-1}{3}, \frac{1}{3}) \cap ... \cap (\frac{-1}{k}, \frac{1}{k}) \cap ...[/imath] and as for the actual working, I just kind of used my head and it seems like there isn't any working, and it says to prove my answer correct so I don't know what to do for it. Thanks, any help is greatly appreciated! :) | 1304402 | What is the result of [imath]\bigcap_{n=1}^{\infty}{(-1/n; 1/n)}[/imath]
I would like to know the intersection of [imath](-1/n ; 1/n), \forall n \in N[/imath]. I am in trouble thinking it could be [imath]\{0\}[/imath] or [imath]\emptyset[/imath]. Can anyone help me? |
2277371 | How to solve [imath]\int_{-\infty}^{\infty}\frac{\sin^3 x}{x^3}dx[/imath]?
The hint is to use integration by parts, but I really don't see how to integrate by parts here. [imath]\int_{-\infty}^{\infty}\frac{\sin^3 x}{x^3}dx[/imath] This is a question from a complex analysis class, so I thought I would use something like semicircular contour. Any help will be appreciated. Thanks! | 34436 | Evaluating the contour integral [imath]\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx[/imath]
I am trying to show [imath]\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx = \frac{3\pi}{8}.[/imath] I believe the contour I should use is a semicircle in the upper half plane with a slight bump at the origin, so I miss the singularity. Lastly I have the hint to consider [imath]\int_{0}^{\infty}\frac{e^{3iz}-3e^{iz}+2}{z^{3}}\mathrm dz[/imath] around the contour I mentioned. Thanks for any help or hints! |
2277355 | Question about the minimum value of [imath]2^{\sin^2a}+2^{\cos^2a}[/imath].
How to find the minimum value of [imath]2^{\sin^2a}+2^{\cos^2a}[/imath] I don't understand how to start the sum. | 1957770 | Minimum value of [imath]2^{\sin^2x}+2^{\cos^2x}[/imath]
The question is what is the minimum value of [imath]2^{\sin^2x}+2^{\cos^2x}[/imath] I think if I put [imath]x=\frac\pi4[/imath] then I get a minimum of [imath]2\sqrt2[/imath]. But how do I prove this? |
2276118 | Given three polynomials [imath]a(x), b(x)[/imath], and [imath]c(x)[/imath] satisfying [imath]\gcd(a(x), b(x), c(x))=1[/imath]
Given three nonzero polynomials [imath]a(x), b(x)[/imath] and [imath]c(x)[/imath] satisfying [imath]\gcd(a(x), b(x), c(x))=1[/imath]. Please help me prove that there exists six polynomials [imath]f(x), g(x), h(x), u(x), v(x), w(x)[/imath] such that [imath]\begin{vmatrix} a(x) & b(x) & c(x) \\ f(x) & g(x) & h(x) \\ u(x) & v(x) & w(x) \end{vmatrix}=1[/imath] | 76645 | Can any set of [imath]n[/imath] relatively prime elements be extended to an invertible matrix?
Say you're given a ordered set of [imath]n[/imath] relatively prime elements, [imath]a_1,\dots,a_n[/imath] in a principal ideal domain [imath]D[/imath]. If I relabel these elements [imath]a_{11},\dots,a_{1n}[/imath] in the same order, is it possible to find some remaining [imath]a_{kj}[/imath] in [imath]D[/imath] such that [imath](a_{kj})[/imath] is an invertible [imath]n\times n[/imath] matrix over [imath]D[/imath]? |
2257828 | Why [imath]x^2+y^2 \leq 1[/imath] in sum of inverse sine functions
I am able to derive [imath]\sin^{-1}x+\sin^{-1}y=\sin^{-1}\left(x \sqrt{1-y^2}+y\sqrt{1-x^2}\right)[/imath] as follows. Let [imath]z=\sin^{-1}{x}+\sin^{-1}y.[/imath] Now if [imath]z \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/imath] taking [imath]\sin[/imath] on both sides we get [imath]\sin z=x \sqrt{1-y^2}+y\sqrt{1-x^2}.[/imath] Now taking [imath]\sin^{-1}[/imath] on both sides we get [imath]\sin^{-1}(\sin z)=\sin^{-1}\left(x \sqrt{1-y^2}+y\sqrt{1-x^2}\right).[/imath] But in [imath]\left[-\frac{\pi}{2},\frac{\pi}{2}\right][/imath] we have [imath]\sin^{-1}(\sin z)=z.[/imath] Hence [imath]z=\sin^{-1}\left(x \sqrt{1-y^2}+y\sqrt{1-x^2}\right)[/imath] but my book gave this is valid for [imath]x^2+y^2 \le 1[/imath] or [imath]x^2+y^2 \gt 1[/imath] when [imath]xy \gt 0[/imath]. Can I know why? | 672575 | Proof for the formula of sum of arcsine functions [imath] \arcsin x + \arcsin y [/imath]
It is known that the following holds good: [imath] \arcsin x + \arcsin y =\begin{cases} \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 \le 1 \;\text{ or }\; x^2+y^2 > 1, xy< 0\\ \pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, 0< x,y \le 1\\ -\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, -1< x,y \le 0 \end{cases} [/imath] But I couldn't find a proof for the above. I tried to prove this myself, but failed. I have no clue how to bring in [imath]x\sqrt{1-y^2} + y\sqrt{1-y^2}[/imath] from the conditions like [imath]x^2 + y^2 < 1[/imath]. Please understand that I don't have any problem in getting the 'crux' part of the RHS : [imath] \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) [/imath]. I face trouble only in checking the range of that 'crux' under the given conditions. |
2276158 | Can someone show me how to prove that [imath]f:X \rightarrow Y[/imath] is onto if and only if it has a right inverse?
[imath]f:X \rightarrow Y[/imath] is onto if and only if it has a right inverse: that is, a function [imath]g:Y \rightarrow X[/imath] such that [imath]f \circ g = 1_y[/imath] | 790274 | Prove if [imath]f[/imath] has a right inverse function: [imath]f\circ g=id_x[/imath] [imath]\iff[/imath] [imath]f[/imath] is onto [imath]Y[/imath]
We have the sets [imath]X,Y[/imath] and [imath]f:X\to Y, \ g: Y\to X [/imath] Prove if [imath]f[/imath] has a right inverse function: [imath]f\circ g=id_Y[/imath] [imath]\iff[/imath] [imath]f[/imath] is onto [imath]Y[/imath] (surjective). [imath]\Leftarrow \forall y\in Y : \exists x\in X[/imath] then [imath]f(y)=x[/imath] I need to show that [imath]f\circ g(x)=x[/imath] but I don't know how. I have no clue on to start in the other direction. BTW, I found below similar question but they're either not quite the same as this one or I didn't understand the answers: Left inverse iff injective; right inverse iff surjective |
2278145 | Let [imath]I[/imath] be an ideal of ring [imath]R[/imath] . If [imath]K[/imath] is ideal
Let [imath]I[/imath] be an ideal of ring [imath]R[/imath] . If [imath]K[/imath] is ideal in [imath]R/I[/imath]. Show that [imath]K=J/I[/imath] for some ideal [imath]J[/imath] which [imath]I\subset J[/imath] . I was thinking [imath]J=a+(-I)[/imath] which [imath]a\in K[/imath] but I stuck to prove that [imath]J/I\subset K[/imath] | 39370 | Ideal correspondence
I'm confusing the ideal correspondence theorem. Is the following right? Ideal correspondence: Let [imath]f:A \to B[/imath] be a ring homomorphism. Then there is a one-to-one order-preserving correspondence between ideals of [imath]f(A)[/imath] and ideals of [imath]A[/imath] which contain [imath]\ker(f)[/imath] by [imath]\alpha \mapsto f(\alpha)[/imath] (conversely [imath]\beta \mapsto f^{-1}(\beta)[/imath]), and prime ideals correspond to prime ideals, and maximal ideals correspond to maximal ideals. |
2277962 | Defining a measure on [imath][0,1][/imath] with a specified closed set F as support
If [imath]X[/imath] is a metric space and [imath]B[/imath] is the Borel [imath]\sigma[/imath]-algebra and [imath]\mu[/imath] is a measure on [imath](X,B)[/imath].Define support of [imath]\mu[/imath] as the smallest closed set [imath]F[/imath] such that [imath]\mu(F^{c})=0[/imath]. Let [imath]X=[0,1][/imath]. Show that if [imath]F[/imath] is a closed subset of [imath]X[/imath], then there exists a finite measure on [imath]X[/imath] whose support is [imath]F[/imath]. I thought that the obviuos choice to construct such a measure is as follows : [imath]\mu(A)=m(A\cap F)[/imath]. Where m is the lebesgue measure on [imath][0,1][/imath]. But this example doesn't seem to work. Thanks in advance for any help. | 824079 | If [imath]F[/imath] is closed, does there exist a finite measure on [imath][0,1][/imath] with support [imath]F[/imath]?
If [imath]X[/imath] is a metric space, [imath]\mathcal B[/imath] is the Borel [imath]\sigma[/imath]-algebra, and [imath]\mu[/imath] is a measure on [imath](X, \mathcal B)[/imath], then the [imath]support[/imath] of [imath]\mu[/imath] is the smallest closed set [imath]F[/imath] such that [imath]\mu(F^c) = 0[/imath]. Show that if [imath]F[/imath] is a closed subset of [imath][0,1][/imath], then there exists a finite measure on [imath][0,1][/imath] whose support is [imath]F[/imath]. My Solution: Let [imath]F[/imath] be a closed subset of [imath][0,1][/imath] and [imath]\mu_F[/imath] be defined by [imath]\mu_F(B) = m(B \cap F)[/imath] for any [imath]B \in \mathcal B_{[0,1]}[/imath]. Since [imath]F[/imath] is a closed subset of [imath][0,1][/imath], it is in [imath]\mathcal B_{[0,1]}[/imath] and [imath]\mu_F[/imath] is a measure. The measure is finite since [imath]\mu_F([0,1]) = m([0,1] \cap F) \leq m([0,1]) = 1[/imath] and since [imath]\mu_F(F^c) = m(F^c\cap F) = m(\emptyset) = 0[/imath], the support of [imath]\mu_F[/imath] is [imath]F[/imath]. My questions: Why do we need the condition that [imath]X[/imath] is a metric space to define the support? Is my answer correct? |
2278738 | Prove that [imath](x^2-y^2)(1/y-1/x) \geq 0[/imath] for [imath]x,y > 0[/imath]
Suppose [imath]x,y[/imath] are both positive real numbers. Prove that [imath](x^2 - y^2)(1/y - 1/x) \geq 0[/imath] I've tried several ways but still stuck on this one, please give me some ideas, thanks a lot. | 2278510 | Are positive real numbers [imath]x,y[/imath] allowed to be taken out during this proof?
Prove [imath]\left(x^2 - y^2\right)\left(\frac1y - \frac1x\right) \ge 0[/imath] where [imath]x[/imath] and [imath]y[/imath] are positive real numbers. Can I simplify [imath]\frac{(x^2 - y^2)(x-y)}{xy} \ge 0[/imath] and then [imath](x^2 - y^2)(x-y) \ge 0[/imath] cancelling out the [imath]xy[/imath]? Is this valid because then \begin{align} (x+y)(x-y)(x-y)&\ge 0\cdot(x-y)^2\ge 0 \end{align} which is always true and then prove backwards from here? |
2265683 | Closed convex subset in Hilbert space
I got stuck on the following question. Suppose that [imath]H[/imath] is a Hilbert space and [imath]V[/imath] is a nonempty convex closed subset of [imath]H[/imath]. Prove that there exists a unique [imath]u\in V[/imath] such that [imath]\|u\|\leq\|v\|[/imath] for all [imath]v\in V[/imath]. I know that by proposition, there is a unique [imath]u\in V[/imath] such that for [imath]h[/imath] in [imath]H[/imath] but not in [imath]V[/imath], [imath]\|h-u\|=\inf \{\|h-v\|:v\in V\}[/imath]. But how could I use this to prove this problem? Thanks. | 1425053 | Proof of Closest Point Theorem in Hilbert Space
The theorem: Let [imath]C[/imath] be a non-empty closed convex subset of a Hilbert space [imath]X[/imath], and let [imath]x \in X[/imath]. Then there exists a unique [imath]y_0 \in C[/imath] such that [imath]||x-y_0|| \le ||x - y||[/imath] all [imath]y \in C[/imath]. In other words there is a point in [imath]C[/imath] which is closest to [imath]x[/imath]. I'm reading a proof here (https://www0.maths.ox.ac.uk/system/files/coursematerial/2014/3075/48/15B4.2-webnotes-all.pdf, p.4) and have two questions about the initial steps in the prooof (I can follow the later steps.): "Let [imath]d = inf\{||x - y||: y \in C\}[/imath]" I think that the infimum should exist by the property (axiom) of completeness of the real numbers. [imath]\{||x - y||: y \in C\}[/imath] is a set of real numbers which is bounded below (all norms are [imath]\ge 0[/imath]) and therefore has a greatest lower bound - is this correct ? "Let [imath](y_n)[/imath] be a sequence of points in C such that [imath]||x - y_n|| \to d[/imath]". I think such a sequence should exist by the following reasoning: With [imath]d = inf\{||x - y||: y \in C\}[/imath] then for any [imath]\epsilon > 0[/imath] there must be a point [imath]y[/imath] in [imath]C[/imath] such that [imath]||x - y|| \lt d + \epsilon[/imath] (otherwise [imath]d[/imath] is not the infimum). If for this point [imath]||x - y|| = d [/imath] then [imath]y_0 = y[/imath] and we are finished: [imath](y_0)[/imath] is a one element convergent sequence. Otherwise [imath]||x - y|| = d + \epsilon`[/imath] where [imath]\epsilon` \le \epsilon[/imath] and I can repeat the process to find a different point where [imath]||x - y|| \lt d + \epsilon`/2[/imath] and so generate a convergent sequence. I'm not sure this is valid, and it occurs to me that unless [imath]C = \{y_0\}[/imath] there are probably an infinite number of points at each step and the Axiom of choice is then required to extract a convergent sequence. Any help would be appreciated. |
2278955 | Integral [imath] \int_0^\infty \int_0^\infty \frac{e^{-(x+y)}}{x + y}\,dx\,dy [/imath]
The integral [imath] \int_0^\infty \int_0^\infty \frac{e^{-(x+y)}}{x + y}\,dx\,dy [/imath] is (A) infinite (B) finite, but cannot be evaluated in closed form (C) [imath]1[/imath] (D) [imath]2[/imath] I have tried with integration by parts in succession and ending up with value [imath]1/2[/imath]. But there is no such option. The correct option is (C) [imath]1[/imath]. Where I am going wrong? | 2188103 | [imath] \int_{0}^{ \infty} \int_{0}^{ \infty} \frac { e^{-(x+y)}}{x+y} dx dy [/imath]
Question : The integral [imath] \int_{0}^{ \infty} \int_{0}^{ \infty} \frac { e^{-(x+y)}}{x+y} \mathop{dx}\mathop{dy} [/imath] is (a) infinite (b) finite but can not be evaluated in closed form (c) 1 (d) 2 . I tried substituting [imath]u=x+y[/imath] and [imath]v=y[/imath] that led me no where . I'm not even sure about convergence of integral .Any help would be greatly appreciated . |
2278973 | To prove [imath]\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n} \le \frac{1}{\sqrt{3n+1}}[/imath]
To prove [imath]P=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n}\le \frac{1}{\sqrt{3n+1}}[/imath] i have written [imath]P[/imath] as [imath]P=\frac{(2n)!}{2^{2n}(n!)^2}=\frac{(2n)!}{4^{n}(n!)^2}=\frac{\binom{2n}{n}}{4^n}[/imath] Now [imath]P=\frac{\binom{2n}{n}}{(1+3)^n} \lt \frac{\binom{2n}{n}}{1+3n}[/imath] since [imath](1+3)^n=1+3n+\binom{n}{2}3^2+\cdots[/imath] Any help here.. | 119773 | How does one prove that [imath]\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}?[/imath]
I would like to show that [imath]\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}[/imath] holds for all natural numbers. I got stuck here: [imath]\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.[/imath] I would appreciate your help. |
2279364 | Contour integration of logarithmic functions: [imath]\int_0^{\infty} \! \frac{\ln(x)}{x^2+a^2} \, \mathrm{d}x[/imath]
I am trying to solve using contour integration [imath]\int_0^{\infty} \! \frac{\ln(x)}{x^2+a^2} \, \mathrm{d}x = \frac{\pi \ln(a)}{2a}[/imath] Where [imath]a>0[/imath]. I am stuck in finding roots and residue for the given problem. Can somebody help? | 1375028 | [imath]\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}[/imath] using residues
I'm trying to find [imath]\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx[/imath] where [imath]\alpha>0[/imath] is real. My approach was to take an integral along the real line from [imath]1/R[/imath] to [imath]R[/imath], around the circle counterclockwise to [imath]-R[/imath], along the real line to [imath]-1/R[/imath], and then around the circle clockwise to [imath]1/R[/imath]. I have encountered 2 problems with this: This path encloses one pole, at [imath]z=\alpha i[/imath]. I found the residue at [imath]z=\alpha i[/imath] to be [imath]\frac{\ln(\alpha)+i\pi/2}{2\alpha i}[/imath]. However, this gives me that [imath]\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx=\frac{\pi(\ln(\alpha)+i\pi/2)}{2\alpha}[/imath]. Since I have a real function integrated over the real line, there cannot be an imaginary part. Where did I go wrong? (Also, doing a few examples, the correct answer seems to be [imath]\frac{\pi\ln(\alpha)}{2\alpha}[/imath], the same as I have but without the imaginary part.) At first chose my path so instead of going all the way around the upper semicircle, it only went 3/4 of the way around, as I wanted to avoid anything that might go wrong with the discontinuity of [imath]\log(x)[/imath] at the negative real axis. When I do this, though, I get a different answer than before (the denominator of the fraction is [imath]\alpha(1-e^{3\pi i/4})[/imath] instead of [imath]2\alpha[/imath]. What am I doing wrong that gives me different answers? |
2279608 | [imath]k[/imath] divides the product of [imath]k[/imath] consecutive integers (case [imath]k=2,3)[/imath]
How do I show that for any arbitrary integer [imath]a[/imath], [imath]2|(a+1)[/imath] and [imath]3a|(a+1)(a+2)[/imath]? [imath]2|(a+1)[/imath], if [imath]a[/imath] is even, [imath]2k=a[/imath] replacing [imath]a[/imath] , [imath]2k=a[/imath] for some [imath]k[/imath] in [imath]Z[/imath] \begin{align}a(a+1)&=2k(2k+1)\\ &=4k^2+2k\\ &=2(2k^2+k)\end{align} shows clearly that [imath]2[/imath] is divisible [imath](a+1)[/imath] but I can't prove the second part | 253359 | Proving by strong induction that [imath]\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) [/imath]
I'm trying to prove by induction the following statement without success: [imath]\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) [/imath] For the base case: [imath]n = 2[/imath], [imath]d = 2[/imath] [imath]2\mid 2(2+1)[/imath] which is true. Now, the confusion begins! I assume I would need to use the second induction principle to proof this because [imath]P(n)[/imath] and [imath]P(n+1)[/imath] are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me. I tried the following: - Trying to prove by simple induction. I did not go very far. - Trying to split my induction step in 2 parts: If [imath]d\mid n[/imath], I'm done. If [imath]d[/imath] does not divide [imath]n[/imath], then I would need to do a second proof and this is where I'm blocked. Anyone could tell me what's wrong in the approach I take to solve this problem? Any help would be appreciated! |
2279324 | Showing that [imath]\sum_{k=N+1}^\infty \frac {1}{k!} < \frac {1}{N!}[/imath]
I was constructing a proof through inequalities, but I am having a bit of problem showing the following step: [imath]\sum_{k=N+1}^\infty \frac {1}{k!} < \frac {1}{N!}[/imath] Is there any quick way to show this? | 1557954 | How to show [imath]\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}[/imath]
[imath]\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}[/imath] Can someone show why this estimate holds true? I tried quite a bit but couldn't really find a way to approach this. WolframAlpha says it is true but I don't know what the gamma function is. [imath] \sum_{k=n}^\infty{\frac{1}{k!}} = \frac{1}{n!} + \sum_{k = n+1}^\infty \frac{1}{k!}[/imath] So then I need to show that[imath] \sum_{k=n+1}^\infty{\frac{1}{k!}} \leq \frac{1}{(n+1)!} ~~\Big[\leq \frac{1}{n!}\Big][/imath] Is it possible to do this by induction? I don't really know how to approach this now. |
2221084 | [imath]c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}[/imath] so prove [imath]e - c_n \le \frac{1}{n! * n}[/imath]
[imath]c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}[/imath], so [imath]e - c_n \le \frac{1}{n! * n}[/imath] I absolutely have no idea how to solve it, could anyone tell me the approach? | 1856779 | Factorial sum estimate [imath]\sum_{n=m+1}^\infty \frac{1}{n!} \le \frac{1}{m\cdot m!}[/imath]
Prove that: [imath]\displaystyle \sum_{n=m+1}^\infty \dfrac{1}{n!} \le \dfrac{1}{m\cdot m!}[/imath] I have tried induction on [imath]m[/imath] but it does not work very well. Any suggestion? |
1024719 | Divergence [imath]\int_{-\pi}^{\pi} |D_n(x)|dx[/imath] for Dirichlet kernel as [imath]n\to\infty[/imath]
Let [imath]D_n(x)[/imath] be the Dirichlet kernel defined by [imath]D_n(x):=\frac{\sin\frac{(2n+1)x}{2}}{2\pi\sin\frac{x}{2}}[/imath]where [imath]D_n(0)[/imath] can be set to [imath]\frac{2n+1}{2\pi}[/imath] if we desire it to be continuous. Another expression for [imath]D_n[/imath] is [imath]D_n(x)=\frac{1}{\pi}\Bigg(\frac{1}{2}+\sum_{k=1}^n \cos kx \Bigg).[/imath]I read that [imath]\lim_{n\to\infty}\int_{-\pi}^{\pi}|D_n(x)|dx=+\infty[/imath], but I cannot prove it rigourously. How can it be done? Thank you so much!!! | 2974643 | How can one estimate [imath](1/2)\int_{-\pi}^{\pi}|D_N|dx[/imath],here DN is the Dirichlet kernel
The Dirichlet Kernel [imath]D_N =\frac{\sin(N+1/2)x}{\sin x/2}[/imath] How to show this [imath]\frac{1}{2\pi}\int_{\pi}^{\pi}|D_N|dx=\frac{4}{\pi^2}\log N+O(1)[/imath] |
2280118 | [imath]A[/imath] and [imath]B[/imath] are connected spaces and [imath]B[/imath] has limit points of [imath]A[/imath] so [imath]A \cup B[/imath] is also connected
I know that if [imath]A \cap B [/imath] is non empty then the union is connected, but how can I ensure that if the closure of [imath]A[/imath] and [imath]B[/imath] has commom points implies in [imath]A \cap B [/imath] is non empty? | 1431179 | Given [imath]A\cap \overline{B}\neq \emptyset[/imath], prove [imath]A\cup B[/imath] connected.
[imath]A,B[/imath] are connected subsets of a topological space [imath]X[/imath]. What I've tried: As [imath]B[/imath] is connected, so is its closure [imath]\overline{B}[/imath]. We have: [imath]\overline{B}\cap A\neq \emptyset[/imath] [imath]\overline{B} \cap B\neq \emptyset[/imath] hence, [imath]\overline{B}\cup A \cup B[/imath] is connected. However, this isn't what I want. I was thinking of trying to show that [imath]A\cap \overline{B} \neq \emptyset[/imath] implies [imath]A \cap B \neq \emptyset[/imath], though I don't think this would be true if [imath]A[/imath] is closed. Any hints? |
2278972 | exercice for limits problems
We consider the problem [imath] \begin{cases} x^2 y''+ x\, y'+\lambda\, y = 0 \\ y(1)=y(e)=0 \end{cases} [/imath] The question is to find the eigenvalues and eigenfunctions for this problem. I try to put [imath]y(x)=x^r[/imath] and plug it in the equation, then we obtain the characteristic equation [imath] r^2 + \lambda = 0 [/imath] In the case [imath]\lambda > 0[/imath] we put [imath]\lambda = \alpha^2[/imath] where [imath]\alpha \in \mathbb{R}^\star_+[/imath]. Then the characteristic equation is [imath] r^2 + \lambda = 0 [/imath] Case 1. If [imath]\lambda > 0[/imath] We put [imath]\lambda = \alpha^2[/imath] where [imath]\alpha \in \mathbb{R}^\star_+[/imath]. Then the general solution of the equation is [imath] y(x)=C_1 \cos(\alpha\, x)+ C_2 \sin(\alpha\, x). [/imath] We have [imath]y(1)=0 => C_1=0[/imath] and [imath]y(e)=0 => C_2 \sin(\alpha\, e)=0[/imath] [imath]\sin(\alpha\, e)=0 => \alpha\, e = \pi + 2\, k\, \pi, \ k \in \mathbb{Z}^\star[/imath] Then the eigenvalues are [imath]\lambda= \alpha^2[/imath] where [imath]\alpha= \dfrac{\pi}{e}+ \dfrac{2\,k\,\pi}{e}[/imath] My solution is correct? | 2278229 | limits problem-ordinary sencond order equation
we consider the problem [imath] \begin{cases} x^2 y''+xy'+\lambda y\\ y(1)=y(e)=0 \end{cases} [/imath] The question is to find the eigenvalues and eigenfunctions for this problem. I try to put [imath]y(x)=x^r[/imath] and plug it in the equation, then we obtain the caracteristic equation [imath]r^2+r+\lambda =0[/imath] In the case [imath]\lambda >0[/imath] we put [imath]\lambda= \alpha^2[/imath] where [imath]\alpha \in \mathbb{R}^\star_+[/imath]. Then the caracteristic equation is [imath] r^2+\lambda=0 [/imath] Case 1. If [imath]\lambda >0[/imath] We put [imath]\lambda = \alpha^2[/imath] where [imath]\alpha \in \mathbb{R}^\star_+[/imath]. Then the genral solution of the equation is [imath] y(x)=C_1 \cos(\alpha x)+ C_2 \sin(\alpha x). [/imath] We have [imath]y(1)=0 => C_1=0[/imath] and [imath]y(e)=0 => C_2 \sin(\alpha e)=0[/imath] [imath]Sin(\alpha e)=0 => \alpha e = \pi + 2 k \pi, \ k \in \mathbb{Z}^\star[/imath] My solution is correct? |
2281292 | Why am I continuously getting this answer wrong for method of undetermined coefficients?
I'm working on solving this inhomogeneous problem with the method of undetermined coefficients using tips from this website: Paul's online math notes. There is an equation halfway down that looks almost exactly the same as mine, so I referenced it for help, but I'm still not getting the right answer. The equation is this: [imath]y''-2y'+y=te^t[/imath] I started out with my "guess" equation as: [imath]Y_P(t)=e^t(At+B)[/imath] Then I took the first and second derivatives: [imath]Y_P'(t)=e^tAt+Be^t+Ae^t[/imath] [imath]Y_P''(t)=e^tAt+2Ae^t+Be^t[/imath] After finding that I plugged in the three equations to the LHS of the original equation and then separated like terms: [imath]e^tt(A-2A+A)+e^t(2A+B-2A-2B+B)=te^t[/imath] In each of the parentheses I end up with a value of [imath]0[/imath], which can't be right. What am I doing wrong? | 1983616 | [imath]y''-2y'+y=te^t[/imath] by method of undetermined coefficients
In order to solve [imath]y''-2y'+y=te^t[/imath] I did: [imath]y = (at+b)e^t\implies\\ y' = ate^t+ae^t+be^t\implies \\y''= ate^t+2ae^t+be^t[/imath] but when I put these in the differential equation I'm getting zeros eveywhere. Does somebody know what I'm doing wrong? |
2281256 | Convex in every argument but not convex
Suppose [imath]f(x_1, \dots, x_n): \mathbb{R}^n \to \mathbb{R}[/imath]. Is there an example of such [imath]f[/imath] that is convex in every argument and not convex? Are there any additional conditions, under which 'marginal' convexity would be sufficient to call [imath]f[/imath] convex? | 211531 | Proof that a coordinate-wise convex function is convex?
I feel like this should be straightforward, but does anyone have a proof of the following? Let [imath]f: \mathbb{R}^n \to \mathbb{R}[/imath] satisfy the following. For each coordinate [imath]i[/imath], for an arbitrary vector [imath]x_{i}[/imath], define [imath] f_i(y)[/imath] to be [imath]f[/imath] restricted to the [imath]i^{th}[/imath] parameter, fixing the others [imath] x_{i}[/imath]; then [imath] f_i(y)[/imath] is convex. Then [imath]f[/imath] is a convex function. Also, if there is a more standard math notation I should use to describe this problem, what would that be? (Edit: or a counterexample...) Thanks! I should add what I've tried. [imath]g(\lambda \vec{x} + (1-\lambda)\vec{y}) = g(\lambda x_1 + (1-\lambda)y_1,\dots,\lambda x_n + (1-\lambda)y_n)[/imath] [imath]\leq \lambda g(x_1,\lambda x_2 + (1-\lambda)y_2,\dots) + (1-\lambda) g(y_1,\lambda x_2 + (1-\lambda)y_2,\dots)[/imath] I don't see this going anywhere good, we end up with something nasty like [imath]\lambda^n g(\vec{x}) + \lambda^{n-1}(1-\lambda) \sum_i g(y_i,\vec{x}_{i}) + \dots + (1-\lambda)^n g(\vec{y})[/imath] I think. Maybe we can get it from there? |
2246842 | Let [imath]a>0[/imath] and [imath]x_1 > 0[/imath] and [imath]x_{n+1} = \sqrt{a + x_n}[/imath] for [imath]n \in \mathbb{N}[/imath]. Show that [imath]\{x_n\}_{n\ge 1}[/imath] converges
Let [imath]a>0[/imath] and [imath]x_1 > 0[/imath] and [imath]x_{n+1} = \sqrt{a+x_n}[/imath] for [imath]n \in \mathbb{N}[/imath]. Show that [imath]\{x_n\}_{n\ge 1}[/imath] converges Please help me with this problem. I am unable to prove it monotone and bounded. | 2280414 | Show that the following inductively defined sequence converges and find its limit.
Let a > 0 and let [imath]x_1 > 0[/imath]. Define [imath]x_{n+1} := \sqrt{a + x_n}[/imath] for n [imath]\in[/imath] N. Show that <[imath]x_n[/imath]> converges and find its limit. |
2281967 | Decomposing ordinal numbers
I believe every ordinal numbers [imath]\alpha[/imath] can be written as [imath]\alpha=\lambda+n[/imath] for a limit ordinal [imath]\lambda[/imath] and a natural number [imath]n[/imath]. Is this true? Is it easy to see? Is there some terminology for the [imath]\lambda[/imath] and the [imath]n[/imath], something like limit part and natural part? (in analogy to integer and fractional part of reals) | 34361 | Can any infinite ordinal be expressed as the sum of a limit ordinal and a finite ordinal?
I've been browsing through Jech's and Levy's texts on set theory, and the ideas of ordinals come up fairly quickly. The idea of a limit ordinal is introduced, which is an ordinal with no maximum element. My question is, can any infinite ordinal be written as the sum of a limit ordinal and a finite ordinal, possibly unique? My thinking was, if [imath]\alpha[/imath] is an infinite ordinal with no maximum, it is a limit ordinal, so [imath]\alpha=\alpha+0[/imath]. Otherwise, suppose [imath]\alpha[/imath] has some order type [imath]\{a_0,a_1,\ldots, b\}[/imath], so [imath]\alpha=\omega+1[/imath]. Similarly, if [imath]\alpha[/imath] has order type [imath]\{a_0,a_1,\ldots,b,c\}[/imath], we could write it as [imath]\omega+2[/imath]. (Thanks to Arturo Magidin, for pointing out that the following example I gave is not an ordinal.) But what about an order type like [imath]\{a_0,a_1,a_2,\ldots, b_2,b_1,b_0\}[/imath], this has order type [imath]\omega+\omega^*[/imath], would it still be possible to write is a sum of a limiting ordinal and a finite ordinal? Thanks. |
2275538 | Let [imath]A, B[/imath] be unitary rings and let [imath]f[/imath] be a surjective homomorphism of rings. Then [imath]f(Jac(A)) \subseteq Jac(B)[/imath]
Let [imath]A, B[/imath] be unitary rings, and let [imath]J(A)[/imath], [imath]J(B)[/imath] denote the Jacobson's radical of [imath]A[/imath] and [imath]B[/imath]. Let [imath]f[/imath] be a surjective homomorphism of rings. Then: [imath]f(J(A)) \subseteq J(B)[/imath] Attempt: It suffices to prove that the preimage of every maximal ideal of [imath]B[/imath] is a maximal ideal in [imath]A[/imath]. Let [imath]M \subseteq B[/imath] be an ideal of [imath]B[/imath], and let [imath]f^{⁻1}(M)[/imath] denote its preimage. Suppose [imath]J[/imath] is an ideal of [imath]A[/imath] such that [imath]f^{⁻1}(M) \subseteq J[/imath]. Since [imath]f[/imath] is a surjection, this implies [imath]M \subseteq f(J)[/imath]. Since [imath]M[/imath] is maximal, we have [imath]f(J)=M[/imath] or [imath]f(J)=B[/imath]. If [imath]f(J)=M[/imath], then [imath]J \subseteq f^{⁻1}(M)[/imath] and then [imath]J=f^{-1}(M)[/imath]. If [imath]f(J)=B[/imath], then I intended to prove that [imath]1_{A}[/imath] is in [imath]J[/imath], since any surjective homomorphism maps [imath]1_{A}[/imath] to [imath]1_{B}[/imath].The problem is that might be that exist other elements of [imath]A[/imath] mapped to [imath]1_{B}[/imath], and then we can't conclude that [imath]1_{A} \in J[/imath]. My questions: 1.Is the initial problem true? 2.Is the preimage of a maximal ideal, through an epimorphism, also a maximal ideal? 3.Are there any ring homomorphisms [imath]f:A\rightarrow B[/imath] that maps [imath]1_A[/imath] and another element to [imath]1_B[/imath]? | 2281554 | Properties of rad A in rings.
Let [imath]A[/imath] and [imath]B[/imath] rings, [imath]f: A \to B[/imath] is a surjective homomorphism. Then [imath]f(\operatorname{rad}A) \subset \operatorname{rad}B[/imath] and construct an example where the inclusion is strict. By [imath]\operatorname{rad}A[/imath] denote the Jacobson radical of [imath]A[/imath], i.e. the intersection of all maximal ideals of [imath]A[/imath]. But how can I prove this ? If [imath]x \in f(\operatorname{rad}A)[/imath] then there exist [imath]y \in \operatorname{rad}A[/imath] such that [imath]f(x)=y[/imath] and then how can I use the surjectivity of [imath]f[/imath] ? Could someone help me do this please. Thanks for your time and help. |
2282248 | The limit of the nth root of a to the n plus b to the n is the maximum of (a,b)
I've been asked to prove the following from Spivak's Calculus [imath]\lim_{n\to\infty}\sqrt[n]{a^n+b^n}=\max(a,b); a,b > 0[/imath] I understand that this is a proof by cases, and that our cases are [imath]a=b[/imath], [imath]a>b[/imath], and [imath]b>a[/imath]. I have done the [imath]a=b[/imath] case, but I am stuck on the [imath]a>b[/imath] and [imath]b>a[/imath] cases. Some hints would be appreciated- Thanks! | 1122067 | How to evaluate [imath] \lim_{x\rightarrow +\infty } \sqrt[x]{a^x+b^x} = ? [/imath]
If a>0 and b>0, [imath] \lim_{x\rightarrow +\infty } \sqrt[x]{a^x+b^x} = ? [/imath] What I was trying to do: Suppose a>b. Then, for sufficiently large values of x, [imath] a^x >> b^x [/imath]; so [imath]\sqrt[x]{a^x+b^x} \rightarrow \sqrt[x]{a^x} \rightarrow a [/imath] when [imath]x \rightarrow +\infty[/imath]. Is that idea correct? How can I formalize it? |
2282913 | Show that [imath]f[/imath] is zero homomorphism.
If [imath]n, m \in N^*[/imath] and a group homomorphism [imath]f: Z_n \to Z_m[/imath]. Show that [imath]f(Z_n) =0[/imath] if and only if [imath](n, m)=1[/imath]. | 1807358 | Number of homomorphisms between two cyclic groups.
Is it true that the number of homomorphisms between any two finite cyclic groups of order [imath]m\,\&\,n[/imath] is [imath]\gcd(m,n)[/imath]? I have posted an answer which I believe is true, just wanted to know different approaches to this problem. |
2283046 | What are the roots of the equation [imath]8x^3-6x+1=0[/imath]?
Let the roots be, [imath]a,b[/imath] and [imath]c[/imath] We have [imath]a+b+c=0\tag1[/imath] [imath]ab +bc +ca=-\frac68\tag2[/imath] [imath]abc = -\frac18\tag3[/imath] | 2157643 | How can I solve the equation [imath]x^3-x-1 = 0[/imath]?
Can someone give me a hint on how can I solve the equation [imath]x^3 - x - 1 =0?[/imath] Thank you! |
2283505 | Primitive element in [imath]L^H:=\{\alpha\in L:\thinspace \sigma\alpha=\alpha\thinspace[/imath] [imath]\forall\sigma\in H\}[/imath]
I'm studying Galois theory and I'm stuck in the following problem. Let [imath]L=K(x)[/imath] the field of rational expresions over a field [imath]K[/imath] with [imath]char(K)=p>0[/imath]. Let [imath]\sigma\in Aut(L/K)[/imath] the automorphism determinated by [imath]\sigma(x)=x+1[/imath]. Show that [imath]H:=\langle\sigma\rangle[/imath] is a finite subgroup of [imath]G=Aut(L/K)[/imath], and find [imath]\alpha\in L^H[/imath] such that [imath]L^G=K(\alpha)[/imath]. For the first part, I need to write explicitly what is H?. Any help will be appreciated ! | 346888 | Generator of rational functions unchanged under [imath]\sigma(X) = X + 1[/imath]
Let [imath]L = K(X)[/imath] be the field of rational functions over a field [imath]K[/imath] with characteristic [imath]p > 0[/imath], and let [imath]\sigma \in \operatorname{Aut}_K(L)[/imath] with [imath]\sigma(X) = X + 1[/imath]. Show that [imath]G = \left<\sigma\right>[/imath] is finite, and determine a generator of [imath]L^G = \{ f \in L: \rho(f) = f[/imath] for all [imath]\rho \in G \}[/imath] over [imath]K[/imath]. My solution so far: [imath]G = \left<\sigma\right>[/imath] is finite because char [imath](K) = p > 0[/imath] and therefore [imath]\sigma^p (X) = X + p = X [/imath]. A rational function unchanged under [imath]\sigma[/imath] will also be unchanged under any [imath]\sigma^n[/imath], so we don't have to worry about powers of [imath]\sigma[/imath] I think. Any ideas for finding a generator? Thanks. |
2283772 | The differences between [imath]\aleph[/imath] and [imath]\mathfrak{c}[/imath]
In elementary set theory, the cardinal of the real numbers is denoted by [imath]|\mathbb{R}| = \aleph[/imath]. After a few months here on MathExchange, I have seen quite a few times the notation [imath]\mathfrak{c}[/imath] for the exact same thing. Recently an answer of mine was edited such that every time I used [imath]\aleph[/imath] it was replaced with [imath]\mathfrak{c}[/imath] which made me wonder: Are there any actual differences between the two notation? Of course, they're defined to be the exact same thing, but maybe differences in the source of the notation, or where you should use each, or something like that... I appreciate your help in advance :) | 9475 | Symbol for the cardinality of the continuum
The usual symbol for the cardinality of the continuum (i.e. the real numbers) is Fraktur [imath]\mathfrak{c}[/imath]. However, I recall some sources also using [imath]\aleph[/imath] (with no subscript). This usage is not mentioned in Wikipedia or Mathworld, but I found some support for it over Google. Is the [imath]\aleph[/imath] notation standard? |
2284125 | Find all values of z for which [imath]z^6 = (z+1)^6[/imath].
I have [imath]\frac{z^6}{(z+1)^6} = 1 = e^{2k\pi i}[/imath]. | 2109609 | Find all [imath]x[/imath] such that [imath]x^6 = (x+1)^6[/imath].
Find all [imath]x[/imath] such that [imath]x^6=(x+1)^6.[/imath] So far, I have found the real solution [imath]x= -\frac{1}{2}[/imath], and the complex solution [imath]x = -\sqrt[3]{-1}[/imath]. Are there more, and if so what/how would be the most efficient find all the solutions to this problem? I am struggling to find the rest of the solutions. |
2284214 | why does math "break" when it comes to certain topics?
For example, [imath]1 + 2 + 3 + ... = -1/12[/imath] or [imath]2 + 4 + 8 + ... = -1[/imath] What caused it to break? Are there some properties of real numbers that do this? Sorry if it's too vague because I was looking at group theory and there were things that didn't hold. Another thing is why do some problems become ill-posed? For example, Wilkinson Polynomial, why do roots break by having too many multiplication of binomials? | 248798 | Assigning values to divergent series
I have been looking at divergent series on wikipedia and other sources and it seems people give finite "values" to specific ones. I understand that these values sometimes reflect the algebraic properties of the series in question, but do not actually represent what the series converges to, which is infinity. Why is it usefull to assign values to divergent series? The only theory I could come up with, is this: Say you have 2 divergent series, series' A and B, and you assign each a value, Series ([imath]A= \sum_{n=0}^\infty a_n[/imath]), which I assigned the value Q and series ([imath]B= \sum_{n=0}^\infty b_n[/imath] ), which I assigned the value P But it just so happens that series [imath]C=A-B= \sum_{n=0}^\infty (a_n-b_n)[/imath] converges. Could that imply that the actual value of series [imath]C[/imath] is the difference of the two assigned values to [imath]A[/imath] and [imath]B[/imath], that is [imath]\sum_{n=0}^\infty (a_n-b_n)=Q-P[/imath] ? If so, then that would make some sense to me, as to why people sometimes assign values to divergent series. |
2284003 | Finding the length of a complex vector
Let us say the vector [imath]u = (1 - i, 2 + 3i, 5)[/imath]. How would I find [imath]\|u\|[/imath]? I have tried solving this through: [imath]\sqrt{(1-i)^2 + (2+3i)^2 + 5^2}[/imath] But I ended up needing to find the square root of a complex number, which from what I understand is impossible. | 20050 | What are the rules for complex-component vectors and why?
I want to take the inverse of a dot product, where both vectors have complex components. In other words, if [imath]\textbf{A} \cdot \textbf{B} = d[/imath], and I know [imath]\textbf{A}[/imath] and [imath]d[/imath], I want to find a [imath]\textbf{B}[/imath]. I know that I cannot do so uniquely, which is fine; I have a procedure for creating a set of vectors that will satisfy [imath]\textbf{A} \cdot \textbf{B} = d[/imath]. But, it relies on finding vectors that are orthogonal to [imath]\textbf{A}[/imath] and each other. Normally this would not be a problem; just take [imath]\textbf{A}[/imath], zero all but two of its components, switch the last two, and negate one of them. Make sure it's a different pair every time. For more orthogonal vectors, you can take cross products. The difficulties arise when I consider vectors with complex components. I want to normalize each of the orthogonal vectors, which means divide by their magnitudes. I have read that you need to divide by [imath]\sqrt{|v_x|^2 + |v_y|^2 + |v_z|^2 ...}[/imath]. What is the justification for this? Since I am just normalizing vectors, do I absolutely have to take the magnitudes of the components? Also, My technique of finding the inverse dot product relies on the identity [imath]\textbf{A} \cdot \textbf{B} = d = |\textbf{A}||\textbf{B}|[/imath]. What modifications might I need to make? I can post more details if people want. Also, if anyone has links to stuff to read about, especially with regards to the reasons why, I would be most appreciative. |
2284105 | [imath]t[/imath]-continuity of Itô Integral
I'm reading Øksendal's SDEs book and in Theorem 3.2.5 he proves that the Itô integral [imath] \int_0^t f(s,\omega) dB_s(\omega)[/imath] has a version such that it is [imath]t[/imath]-continous. In order to do this he picks elementary functions [imath]\phi_n[/imath], such that [imath] E[\int_0^T (f-\phi_n)^2 dt]\rightarrow 0[/imath] as [imath]n[/imath] tends to infinity. Then he defines [imath]I_n(t,\omega)=\int_0^t \phi_n(s,\omega)dB_s(\omega)[/imath] and states that [imath]I(\cdot,\omega)[/imath] is continous for all [imath]n[/imath]. This is crucial to apply Doob's martingale inequality later and gives the statement without proof, but this doesn't seem trivial to me. My question is, why should be the Itô integral [imath]I_n(t,\omega)[/imath] be continous at all? Aren't there going to be jumps as we increase [imath]t[/imath] and more addends appear in the integral? | 111984 | Continuity of the Ito integral
Let [imath]B_t[/imath] be a 1-dimensional Brownian motion. I am following "Stochastic Differential Equations" by Bernt Øksendal. On the page 32 (it is displayed in the link I've put) there is a proof of existence of continuous version of the Ito integral. There is stated that for a function [imath] \phi_n(t,\omega) = \sum\limits_j e_j(\omega)\cdot 1_{[t_j,t_{j+1})}(t) [/imath] its integral [imath] I_n(t,\omega) = \int\limits_0^t\phi_n(s,\omega)dB_s(\omega) [/imath] is continuous in [imath]t[/imath]. From what I seen, I think that [imath] I_n(t,\omega) = \sum\limits_{t_j\leq t}e_j(\omega)(B_{t_{j+1}} - B_{t_j}) [/imath] which does have jumps. So I wonder, how can it be continuous in [imath]t[/imath]. |
2284505 | Induction.provide formula. prove it by induction.
I have done the formula but im not sure how to prove this by induction. Can anyone provide me a solution to this part? (solution was not provided) [imath]a_n=\frac{n}{4n+1}[/imath] is the explicit formula I got. | 2284350 | Guess the formula for [imath]\sum\frac 1{(4n-3)(4n+1)}[/imath] and prove by induction
For [imath]n \ge 1[/imath], let [imath]a_n = \dfrac1{1 \cdot 5} + \dfrac1{5 \cdot 9} + \cdots + \dfrac1{(4n-3)(4n+1)}.[/imath] Guess a simple explicit formula for [imath]a_n[/imath] and prove it by induction. So I have guessed the formula as : [imath]\frac{n}{4n+1}[/imath] But wasn't sure how to prove it by induction |
2284420 | Min and max of exponentialy distributed random variables
If [imath]X[/imath] and [imath]Y[/imath] are i.i.d. exponential as [imath]f_X(x)=\lambda e^{-\lambda x}[/imath] and [imath]f_Y(y)=\lambda e^{-\lambda y}[/imath], and [imath]Z=\min(X,Y)[/imath], then we may say that probability of event [imath]Z=X[/imath] or [imath]Z=Y[/imath] is [imath]\frac 1 2[/imath]. This may be valid for any kind of distribution. If [imath]X[/imath] and [imath]Y[/imath] are independent but not identical exponential as [imath]f_X(x)=\lambda_x e^{-\lambda_x x}[/imath] and [imath]f_Y(y)=\lambda_y e^{-\lambda_y y}[/imath], and [imath]Z=\min(X,Y)[/imath], then how I can find the probability of event [imath]Z=X[/imath] or [imath]Z=Y[/imath] ?. Most importantly, I need how to find this probability when [imath]X[/imath] and [imath]Y[/imath] have any kind of continuous distributions which are independent. | 1805587 | Prove that [imath]\mathbb P(X>Y) =\frac{b}{a + b}[/imath] if [imath]X, Y[/imath] are exponentially distributed with parameters [imath]a[/imath] and [imath]b[/imath].
Let [imath]X, Y[/imath] be an exponentially distributed random variables with parameters [imath]a, b[/imath]. Then [imath]X[/imath] has pdf: [imath]f_X(x) =\begin{cases} a e^{-a x},& x\geq 0\\ 0,& \text{otherwise}.\end{cases}[/imath] Suppose [imath]X[/imath] and [imath]Y[/imath] independent. Show that [imath]\mathbb P(X>Y) = \frac{b}{a+b}.[/imath] Now I thought the following: [imath]f(x,y) = f_X(x)\ f_Y(y) = abe^{-ax -by},\qquad\text{for } x,y > 0.[/imath] And then [imath]\mathbb P(X>Y) = \int_0^\infty \int_0^x a b e^{-ax -by}\,dydx[/imath] However, if I solve this (manually or using Wolframalpha), I can't seem to end up with [imath]\frac{b}{a+b}[/imath]. Any ideas? |
2284189 | Writing first [imath]5[/imath] terms of recurrence relation
How do I write the first [imath]5[/imath] terms for this recurrence relation? [imath]S_0=2\\ S_n=S_{n-1}^2 +S_{n-2}^2+\ldots+S_0^2[/imath] Since I can't substitute [imath]S_0[/imath] directly? | 2283092 | Confusion on how to solve this question about sequences.
Write down the first five values of each of the following recursive sequences: [imath] s_0 = 2, \quad s_n = s_{n−1}^2 + s_{n−2}^2 + \cdots + s_0^2 [/imath] for all integers [imath]n \geq 1[/imath]. Approach: Say for [imath]s_1= s_0^2 + s_1^2 + s_0^2[/imath] which means [imath]s_0^2 + s_0^2[/imath] since [imath]s_1[/imath] isn't greater or equal to [imath]1[/imath]. So is the answer [imath]8[/imath] or just [imath]s_0^2[/imath] which will be [imath]4[/imath]? I am confused by this part [imath][\ldots + s_0^2][/imath]. |
2284818 | The set of formal power series over a fieldis a local ring?
Let [imath]\mathbb{K}[/imath] be a field, and consider [imath]\mathbb{K}[[x]][/imath], the ring of formal power series with coefficients in [imath]\mathbb{K}[/imath], i.e. the set of expressions of the form [imath]\sum_{n=0}^{\infty}a_n x^n,\quad a_n\in\mathbb{K}[/imath] with the usual rules for addition and multiplication. How to show that [imath]\mathbb{K}[[x]][/imath] is local ring ? Thank you in advance | 989811 | If [imath]R[/imath] is a local ring, is [imath]R[[x]][/imath] (the ring of formal power series) also a local ring?
So, I was trying to find a counter-example that shows not every local ring's lattice of ideals is a chain. I think [imath]F[[x_1,\cdots,x_n]][/imath] is a good counter-example but I'm not able to show that [imath]F[[x_1,\cdots,x_n]][/imath] is a local ring. I read somewhere that [imath]F[[x]][/imath] is indeed a local ring. So here comes the question: If [imath]R[/imath] is a local ring, what can we say about [imath]R[[x]][/imath]? Is it a local ring too? I'm looking for a simple proof that doesn't use commutative algebra and localizations to show that. An elementary undergraduate level proof is appreciated. |
2284520 | maximal ideals of the Rad A
Let [imath]A[/imath] a semilocal ring and [imath]M_1,M_2,\dots,M_n[/imath] are all of its maximal ideals then the Jacobson radical [imath]rad(A) =M_1M_2\cdots M_n = M_1\cap M_2\cap \cdots \cap M_n[/imath] I tried to use the Chinese Remainder Theorem, How can I solve this? Thanks for your time and help. | 1057517 | Given distinct maximal ideals [imath]M_1,...,M_n[/imath], is [imath]M_1\cdots M_n[/imath] radical?
Let [imath]R[/imath] be a commutative ring with [imath]1[/imath] and [imath]M_1,...,M_n[/imath] be distinct maximal ideals in [imath]R[/imath]. What I want to show is [imath]M_1\cap\cdots \cap M_n=M_1\cdots M_n[/imath]. If I can show that [imath]M_1\cdots M_n[/imath] is radical, then the proof ends since [imath]M_1\cap\cdots\cap M_n=rad(M_1\cdots M_n)[/imath]. How can I prove this? |
2285195 | Find Groups Whose Direct Products are Isomorphic
Give an example of groups [imath]H_i[/imath] and [imath]K_i[/imath] such that [imath]H_1 \times H_2 \simeq K_1 \times K_2[/imath] and no [imath]H_i[/imath] is isomorphic to any [imath]K_i[/imath]. Would this work? Let [imath]H_1 = V_4[/imath], the Klein 4-group, [imath]H_2 = \{e\}[/imath], and [imath]K_1 = K_2 = Z_2[/imath]. Then [imath]Z_2 \times Z_2 \simeq V_4 \simeq V_4 \times \{e\}[/imath], and clearly neither [imath]V_4 \simeq Z_2[/imath] nor [imath]\{e\} \simeq Z_2[/imath]. | 2283143 | Non-trivial groups pairwise non-isomorphic with isomorphic direct products
Are there four non-trivial groups [imath]G_1,G_2,H_1,H_2[/imath] pairwise non-isomorphic but such that [imath]G_1 \times G_2[/imath] is isomorphic to [imath]H_1 \times H_2[/imath]? My idea was to consider the abelian groups and something related to their decomposition as direct product of cyclic groups, but it doesn't work. |
2285173 | Expression of expected value for counting process
In this document at page 20, one can read [imath]M(t)[/imath] as being defined as the expected value of [imath]N_t[/imath] where [imath]N_t[/imath] is the number of occurence of an event at time=t. As demonstration of the result, the author says: [imath]E[N_t] = \sum_{n=0}^{\infty}{P(N_t>n)}[/imath] I don't understand where it comes from. The expected value should be [imath]E[N_t] = \sum_{n=0}^{\infty}{n P(N_t=n)}[/imath] How do you obtain the author's expression? | 143097 | Formula similar to [imath]EX=\sum\limits_{i=1}^{\infty}P\left(X\geq i\right)[/imath] to compute [imath]E(X^n)[/imath]?
Is there a formula like [imath] EX=\sum_{i=1}^{\infty}P\left(X\geq i\right) [/imath] (which can be found on Wikipedia and holds for positive [imath]X[/imath]) for [imath]EX^{n}[/imath] ? And I don't mean this one, [imath] EX^{n}=\sum_{i=1}^{\infty}P\left(X\geq\sqrt[n]{i}\right), [/imath] which is immediate, if we take [imath]Y=X^{n}[/imath] and use the above formula for [imath]Y[/imath]. I mean a "more elegant" one - if there is one. |
2285470 | Intersection of powers of the maximal ideal
There certainly exist commutative rings [imath]R[/imath] with unity containing a maximal ideal [imath]m \subsetneq R[/imath] such that [imath]\bigcap_{k=1}^\infty m^k \neq \{0\}.[/imath] (Thanks to Dietrich Burde, rschwieb and Lord Shark the Unknown.) Question: What happens if we further assume that (i) [imath]R[/imath] is an integral domain, [imath]\rightarrow[/imath] (EDIT3:) No, see here: Can an ideal in a commutative integral domain be its own square?. or that (ii) the quotient [imath]R/m[/imath] is finite? [imath]\rightarrow[/imath] (EDIT2:) No, take [imath]R = \mathbb{C} \times \mathbb{F}_2[/imath] and [imath]m = \mathbb{C} \times \{ 0 \}[/imath]. Or more general, is there a way to describe those rings [imath]R[/imath] such that [imath]\bigcap_{k=1}^\infty m^k = \{0\}[/imath] holds? | 18835 | Intersection of powers of an ideal in a Noetherian ring
Given a Noetherian ring [imath]R[/imath] and a proper ideal [imath]I[/imath] of it. Is it true that [imath]\bigcap_{n\ge 1} I^n=0[/imath] as [imath]n[/imath] varies over all natural numbers? If not, is it true if [imath]I[/imath] is a maximal ideal? If not, is it true if [imath]I[/imath] is the maximal ideal of a local ring [imath]R[/imath]? If not, is it true under additional assumptions on [imath]R[/imath] (like [imath]R[/imath] is regular)? |
2286043 | How exactly do I explain why the proof is incorrect?
Explain in detail why the following “proof” is incorrect. Theorem [imath]1[/imath]. All positive integers are equal. Proof. We show that any two positive integers are equal, from which the result follows. We do this by induction on the maximum of the two numbers. Let [imath]P (n)[/imath] be the statement “if [imath]r[/imath] and [imath]s[/imath] are positive integers and max{[imath]r, s[/imath]} = [imath]n[/imath] then [imath]r = s[/imath]”. Clearly [imath]P (1)[/imath] is true. Suppose that [imath]P (n)[/imath] is true and let [imath]r, s[/imath] be positive integers whose maximum is [imath]n + 1[/imath]. Then max{[imath]r−1,s−1[/imath]} = [imath]n[/imath]. By the inductive hypothesis, [imath]r−1=s−1[/imath] and hence [imath]r=s[/imath]. Thus [imath]P(n+1)[/imath]is true. The result is now proved by mathematical induction. | 2281660 | Induction proof. Explain in detail why it’s incorrect
Can somebody give a clear explanation why this is incorrect? thank you Theorem 1: All positive integers are equal. Proof: We show that any two positive integers are equal, from which the result follows. We do this by induction on the maximum of the two numbers. Let [imath]P(n)[/imath] be the statement “if [imath]r[/imath] and [imath]s[/imath] are positive integers and [imath]\max \{ r, s \} = n[/imath] then [imath]r = s[/imath].” Clearly [imath]P(1)[/imath] is true. Suppose that [imath]P(n)[/imath] is true and let [imath]r[/imath] and [imath]s[/imath] be positive integers whose maximum is [imath]n + 1[/imath]. Then [imath]\max \{ r − 1, s − 1 \} = n[/imath]. By the inductive hypothesis, [imath]r − 1 = s − 1[/imath] and hence [imath]r = s[/imath]. Thus [imath]P(n + 1)[/imath] is true. The result is now proved by mathematical induction. |
2285774 | Proof of [imath]\Box (A\land B) \leftrightarrow \Box A \land \Box B[/imath] in propositional modal logics
I'm searching for an example proof of the above formula as distribution over and in basic proportional modal logic, I.e. in some of the standard systems K, S4 or S5. | 1318235 | Distribution Axiom of Modal Logic
Is it possible to prove the distribution axiom of modal logic? I have proven all the conclusions of propositional modal logic using this axiom, the definitions of the four standard modal operators, and the postulate that some contingent proposition is true. I wanted to know if it was possible to prove this distribution proposition (also called Kripke property): [imath] \square (p \rightarrow q) \rightarrow (\square p \rightarrow \square q)[/imath] It basically distributes the necessity operator over the components of every necessary conditional proposition. Most modal logic books accept this as an axiom in addition to others, but I want the fewest axioms possible as I have already reduced them to the distribution axiom, modal semantics and definitions, and (some contingent proposition is true). Is there any further reduction possible by proving the distribution axiom. And if so please provide a natural deduction proof of this distribution statement from previous principles in modal logic. |
2286148 | Determining if Galois group is cyclic
I found a short problem while reviewing some material. True/False: The splitting field over [imath]\mathbb Q[/imath] of the polynomial [imath]x^{17} − 5 \in \mathbb Q[x][/imath] has a cyclic Galois group. It's clear that this splitting field is a Galois extension, but I'm not sure how to solve the problem. Any help would be great. | 204552 | Computing the Galois group of polynomials [imath]x^n-a \in \mathbb{Q}[x][/imath]
I have some problems with this exercise. I don't know if it can be done. Consider the polynomial [imath] x^n - a \in \mathbb{Q} [/imath] Can I compute the Galois group of this over [imath]\mathbb{Q}[/imath]? Maybe having a nice "basis. The splitting field is given by [imath]\mathbb{Q}(\zeta_n,\alpha)[/imath] , where [imath]\zeta_n[/imath] is a primitive root of unity , and [imath]\alpha[/imath] is some number such that [imath]\alpha^n = a [/imath]. Well first of all, I want a [imath]\underline{"good basis"}[/imath] for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different (in this case the computation of the galois group is very simple). For example one easy case, it's when [imath]a>0[/imath], then [imath](a)^{\frac{1}{n}} \in \mathbb{R}[/imath] , so clearly the minimal polynomial of [imath](a)^{\frac{1}{n}} , \zeta_n[/imath] are distincs, and I'm done. If [imath]n[/imath] is odd then , it's also easy, since one root it's also real, for example [imath]x^3-3 [/imath], the real root is [imath] \root 3 \of { - 3} = - \root 3 \of 3 [/imath] , so I can consider the splitting field as [imath]\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )[/imath]. The difficult case is when [imath]n[/imath] is even and [imath]a<0[/imath] , for example [imath]x^8+20[/imath] or [imath]x^4+20[/imath] in some cases as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general. It can be done? [imath]\underline{Remark}[/imath] I'm searching a [imath]\underline{"good basis"}[/imath] for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different since in this case the computation of the galois group is very simple. |
1798428 | Show that [imath]\bar{A}=\{x \in M | d(x,a)=0\}[/imath]
Let [imath](M,d)[/imath] be a metric space. Let [imath]A[/imath] be an arbitary subset of [imath]M[/imath] and let [imath]x[/imath] be an arbitary point. Define [imath]d(x,A)=\inf \{d(x,y)\mid y \in A\}[/imath]. Show that [imath]\bar{A}= \{x \in M \mid d(x,A)=0\}[/imath] How would I approach this? How should I begin? | 1354748 | Suppose [imath]X[/imath] is a metric space and [imath]S \subseteq X.[/imath] Then, [imath]S^o=\{x \in X~|~dist(x,S^c)>0\}[/imath].
Suppose [imath]X[/imath] is a metric space and [imath]S \subseteq X.[/imath] Then, according to my textbook, [imath]S^o=\{x \in X~|~dist(x,S^c)>0\}[/imath]. (Notations Used: [imath]S^o[/imath] refers to interior of [imath]S[/imath] . If [imath]x \in X, dist(x,S) = \{\inf (d(x,s)~|~s \in S)\}[/imath] ) I think that [imath]S^o[/imath] should be instead equal to [imath]\{x \in S~|~dist(x,S^c)>0\}[/imath]. [imath]S^o[/imath] is defined as [imath]S - \partial S~~~~~..(1)[/imath] where [imath]\partial S = \{~x \in X ~|~dist(x,S)=0=dist(x,S^c)~\}~~~...(2)[/imath] So, to obtain [imath]S^o[/imath], remove all those points from [imath]S[/imath] which satisfy [imath]dist(x,S^c)=0[/imath]. This yields [imath]S^o=\{x \in S~|~dist(x,S^c)>0\}[/imath]. Is it possible that my textbook has a possible error with respect to this definition? Thank you very much for your help. |
2286465 | Sum of all entries of [imath]A^3[/imath]
Let [imath]A[/imath] be a [imath]5\times5[/imath] matrix with real entries such that the sum of the entries in each row of [imath]A[/imath] is [imath]1[/imath]. Then the sum of all entries in [imath]A^3[/imath] is (a) [imath]3[/imath] (b) [imath]15[/imath] (c) [imath]5[/imath] (d) [imath]125[/imath] [imath]1[/imath] is an eigenvalue of [imath]A[/imath], that's all I know. How could we go to the sum of all entries of [imath]A^3[/imath] ? | 1796331 | Sum of the entries in the matrix [imath]A^3[/imath]
Let [imath]A\neq I[/imath] be a [imath]5\times5[/imath] matrix with real entries such that the sum of the entries in each row of [imath]A[/imath] is [imath]1[/imath]. Then the sum of all the entries in [imath]A^3[/imath] is 1)[imath]\space 3[/imath] [imath]\qquad [/imath]2)[imath]\space 15[/imath] [imath]\qquad[/imath] 3)[imath]\space 5[/imath] [imath]\qquad[/imath] 4)[imath]\space 125[/imath] Solution: The answer is [imath]5[/imath] if [imath]A = I[/imath], but for [imath]A \neq I[/imath], I found that vectors [imath]\begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \end{pmatrix}[/imath] and [imath]\begin{pmatrix} -1\\ -1\\ -1\\ -1\\ -1\\ \end{pmatrix}[/imath] are eigen vectors for the matrix [imath]A[/imath] corresponding to eigen value [imath]1[/imath]. From here, I dont know how to proceed, please help me solve the question..and if there are mistakes in my argument, please tell me.. |
2286616 | Given that [imath]\frac{2^n +1}{n^2}[/imath] is a integer.Find [imath]n[/imath]
Determine all set of integers [imath]n\gt 1[/imath] such that [imath]\frac{2^n+1}{n^2}\;\;\text{ is a integer. }[/imath] I found that [imath]n[/imath] is an odd. Please help by providing me with a hint. | 203976 | Find all [imath]n>1[/imath] such that [imath]\dfrac{2^n+1}{n^2}[/imath] is an integer.
Find all [imath]n>1[/imath] such that [imath]\dfrac{2^n+1}{n^2}[/imath] is an integer. I know that [imath]n[/imath] must be odd, then I don't know how to carry on. Please help. Thank you. |
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