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2185932 | Prove [imath]\lim_{x\to a }{ f(x)\over g(x)}={ L\over M}[/imath]
Suppose lim x→a f(x) = L and lim x→a g(x) = M, prove that lim x→a [f(x)/g(x)]= L/M provided M ≠ 0 and g(x) ≠ 0, for x sufficiently close to a (but not equal to a). What does mean by for x sufficiently close to a (but not equal to a) and how can I prove that? thanks | 1929572 | Proving the quotient rule for limits
In real analysis, we have been asked to finish a proof of the quotient rule for limits (Given that [imath]f(x)[/imath] approaches [imath]L[/imath] and [imath]g(x)[/imath] approaches M as [imath]x[/imath] approaches a, prove that [imath]\frac{f(x)}{g(x)}[/imath] approaches [imath]\frac{L}{M}[/imath]. I know that I could rewrite the quotient as multiplication and prove it that way but that is not the way the proof we are completing starts off. Here is how the proof in the book starts: We showed in a previous exercise that [imath]|g(x)| > \frac{|M|}{2}[/imath] for all [imath]x[/imath] belonging to [imath]D[/imath] near [imath]a[/imath]. In particular, [imath]g(x)[/imath] does not equal [imath]0[/imath] for all [imath]x[/imath] belonging to [imath]D[/imath] near [imath]a[/imath]. Consequently, [imath]\frac{f(x)}{g(x)}[/imath] makes sense near [imath]a[/imath]. Let [imath]\epsilon > 0[/imath] be given. The proof is based on the triangle inequality: [imath]\begin{align*} \left|\frac{f(x)}{g(x)} - \frac{L}{M}\right| &= \left|\frac{f(x)}{g(x)} - \frac{L}{g(x)} + \frac{L}{g(x)} - \frac{L}{M}\right|\\ & = \left|\frac{f(x)}{g(x)} - \frac{L}{g(x)} + L \frac{M - g(x)}{Mg(x)}\right| \\ &\leq \left|\frac{1}{g(x)}\right| |f(x) - L| + \left|\frac{L}{M}\right| \left|\frac{1}{g(x)}\right| |M - g(x)|. \end{align*}[/imath] We are expected to pick up the proof at this point. I was thinking that since I want everything above [imath]< \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon[/imath], that I would construct [imath]\epsilon_f[/imath] and [imath]\epsilon_g[/imath] based on [imath]\epsilon[/imath]. So, I said we want [imath]|\frac{1}{g(x)}| |f(x) - L| < \epsilon/2[/imath] and because the limit approaches [imath]L[/imath] end up with [imath]\epsilon_f[/imath] equals [imath]\frac{\epsilon|M|}{4}[/imath]. I intended to do the same for the g part; however, when I go to define [imath]\epsilon_g[/imath] it involves [imath]M[/imath]. Am I allowed to use [imath]M[/imath] in that definition for [imath]\epsilon_g[/imath]? And if not, any ideas on where to go from here? Thank you!! |
2190091 | Null Vector spaces proof
I have a work sheet on proofs but this is the first question and it would really help if I could be shown this one so I can attempt the last ones by myself. Thank you Let [imath]U, V, W[/imath] be finite-dimensional vector spaces and let [imath]P : U \to V[/imath], and [imath]Q : V \to W[/imath] be linear maps. Prove that [imath]null(Q \circ P) \le null(Q) + null(P)[/imath]. Hint: it may help to consider the restriction of [imath]P[/imath] to [imath]Ker(Q\circ P)[/imath]. | 202710 | Kernel of composition of linear transformations
Let [imath]f : U \to V[/imath] and [imath]g : V \to W[/imath] be linear transformations on the vector spaces [imath]U[/imath], [imath]V[/imath], and [imath]W[/imath]. Supposedly, [imath] \dim(\ker(g \circ f)) = \dim(\ker(f)) + \dim(\ker(g) \cap \operatorname{im}(f)). [/imath] How might I go about proving that? (Attempt:) The [imath]\dim(\ker(g)\cap\operatorname{im}(f))[/imath] term suggests to me that I should define a vector space [imath]V' = \ker(g) + \operatorname{im}(f)[/imath] to invoke the theorem that [imath] \dim(V') = \dim(\ker(g)) + \dim(\operatorname{im}(f)) - \dim(\ker(g) \cap \operatorname{im}(f)), [/imath] but I don't really see where to go from there. |
2190523 | Absolute value of real numbers and inequality
Let [imath]a[/imath] and [imath]b[/imath] be real numbers. Show that [imath]\vert a-b\vert < \epsilon \Rightarrow \vert a\vert<\vert b\vert+\epsilon[/imath] for [imath]\epsilon>0[/imath]. Looks quite easy but I'm not getting it. I tried to use triangular inequality in many forms but it doensn't come. | 1898422 | How to prove that [imath]|a-b|<\epsilon[/imath] implies [imath]|b|-\epsilon<|a|<|b|+\epsilon[/imath]?
Given [imath] a, b, \varepsilon \in \mathbb{R} [/imath] prove that [imath]|a-b|<\varepsilon \implies |b| - \varepsilon < |a| < |b|+\varepsilon. [/imath] Hi, I need help for proof this expression, which could be used arguments or results. I would appreciate any suggestions. |
2190755 | Does there exist a non-zero matrix [imath]B[/imath] such that [imath]AB = 0[/imath], given [imath]A[/imath] to be non-singular?
I am quite new to topics in linear algebra and try to verify my claim that there does not exist a non-zero matrix [imath]B[/imath] such that [imath]AB=0[/imath] given that [imath]A[/imath] is non-singular. I browsed the forum for related question but could not find a convincing answer so far. The forum only investigated cases when [imath]A[/imath] it self has not full rank and then found that there indeed exists such a matrix [imath]B[/imath] that is non zero. Can anybody help me to construct such a proof or provide me counter example that my claim is false? | 243160 | Finding matrix [imath]B[/imath] is not zero matrix where [imath]AB= 0[/imath]
[imath]A[/imath] is defined as an [imath]m\times m[/imath] matrix which is not invertible. How can i show that there is an [imath]m\times m[/imath] matrix [imath]B[/imath] where [imath]AB = 0[/imath] but [imath]B[/imath] is not equal to [imath]0[/imath]? For the solution of this question I think giving an example is not enough because it is too easy to solve this by giving an example, so how can I show that [imath]B[/imath] is not the [imath]0[/imath] matrix? |
2191083 | Calculating Basic Limit
How can I calculate the following limit, [imath]\lim_{x\to0}\frac{e^{-1/x^2}}x=0[/imath] If I plug [imath]x=0[/imath] I get the following, [imath]f'(0)=\lim_{x\to0}\frac{e^{-1/0^2}}0=\infty[/imath] | 2190330 | L'hopitals rule with limits
Given the following limit, [imath]\begin{align} \lim_{x\to 0}\frac{e^{-1/x^2}-0}{x-0}\\\\ & \end{align}[/imath] How do I calculate it? when pluggin in 0 I would get [imath]\frac{0}{0}[/imath] |
2191044 | Show that [imath]\limsup_{n\rightarrow\infty} a_n^\frac{1}{n} \leq \limsup_{n\rightarrow\infty} (a_{n+1} / a_n).[/imath]
I've been set the following question as part of my Real Analysis homework and I'm struggling with it: Let [imath](a_n)_n\in\mathbb{N}[/imath] be a sequence of positive numbers. Show that: [imath] \limsup_{n\rightarrow\infty} a_n^\frac{1}{n} \leq \limsup_{n\rightarrow\infty} (\frac{a_{n+1}}{a_n}) [/imath] I have reasoned that if the RHS takes value [imath]\infty[/imath] then the inequality is automatically satisfied so we can assume RHS [imath]= \lambda < +\infty[/imath] We got a hint in the lecture that went something like this: For [imath]m \geq n[/imath], [imath]a_m \leq a_n \lambda^{m-n} \Rightarrow a_m^{\frac{1}{n}} \leq (a_n \lambda^{m-n})^\frac{1}{n}[/imath] And then taking supremums a and limsups but I'm confused at this point and I'm not really sure where to take t from here. Any help would be greatly appreciated. | 2178272 | Finding limit using inequalities: [imath]\liminf \frac{a_{n+1}}{a_n} \le \liminf (a_n)^ {1/n}\le\limsup (a_n)^ {1/n}\le \limsup \frac{a_{n+1}}{a_n}[/imath]
The purpose of this exercise is to prove that [imath]\lim \frac{n}{(n!)^{1/n}}=e[/imath] when [imath]n[/imath] goes to infinity. In order to find the limit, the following inequality is used when [imath]n[/imath] goes to infinity with [imath]{a_n}[/imath] a sequence of positive terms: [imath] \liminf \frac{a_{n+1}}{a_n} \le \liminf (a_n)^ {1/n}\le\limsup (a_n)^ {1/n}\le \limsup \frac{a_{n+1}}{a_n}[/imath] What is the proof of this inequality? |
2189287 | Prove: if [imath]a \mid c[/imath] and [imath]b \mid d[/imath] then [imath]ab \mid cd[/imath]
Can this be proven? Or does it need a counter example? For any [imath]a, b, c, d\in\mathbb{N}[/imath], if [imath]a \mid c[/imath] and [imath]b \mid d[/imath] then [imath]ab \mid cd[/imath] Need help. | 2181325 | If [imath]a|c[/imath] and [imath]b|d[/imath], is it true that [imath]ab|cd[/imath]?
Determine if this is true or not: For any natural numbers [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], and [imath]d[/imath], if [imath]a | c[/imath] and [imath]b | d[/imath] then [imath]ab|cd[/imath]. My answer was True, I went: [imath]a = 2[/imath], [imath]b = 4[/imath], [imath]c = 6[/imath], and [imath]d = 8[/imath]. If [imath]2|6[/imath] and [imath]4|8[/imath], then [imath](2 * 4) | (6 * 8)[/imath]. Is there another way to do it using, for example, a contradiction proof? |
292738 | Show that [imath]({\mathbb{Q}},+)[/imath] is not finitely generated using the Fundamental Theorem of Finitely Generated Abelian Groups.
Can anyone please help me out on how to use the fundamental theorem of finitely generated abelian groups to prove that [imath]({\mathbb{Q}},+)[/imath] is not finitely generated? | 2702807 | Is [imath]\mathbb Q[/imath] finitely generated as a [imath]\mathbb Z[/imath]-module?
Is [imath]\mathbb Q[/imath] finitely generated as a [imath]\mathbb Z[/imath]-module? If I were to take a guess, I would say no but only because I am not sure how to generate all of [imath]\mathbb Q[/imath] from a finite set. But is this possible? |
2186533 | Does there exist a local diffeomorphism of [imath]\mathbb{R}^2[/imath] onto [imath]\mathbb{S}^2[/imath]?
Suppose there exists such a map [imath]f:\mathbb{R}^2\to\mathbb{S}^2[/imath]. Basic (topological) cover space theory tells us that this map cannot be a covering map. If we abandon the condition that [imath]f[/imath] must be a local diffeomorphism, then clearly such a map exists – we can simply use the map [imath]f(\theta,\phi) = (\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi).[/imath] However, I suspect (but cannot prove) that if we impose the condition of being a local diffeomorphism, there does not exist any such map. Algebraic topology seems to be of no direct use here (since I don't see how to use smoothness), and I don't know of any theorems in differential geometry that would be very useful in proving the nonexistence of such a map. | 100884 | Local diffeomorphism from [imath]\mathbb R^2[/imath] onto [imath]S^2[/imath]
Is there any local diffeomorphism from [imath]\mathbb R^2[/imath] onto [imath]S^2[/imath]? |
2189708 | Finding expected value of Bernoulli random variables
Consider a sequence of [imath]n[/imath] Bernoulli trials with [imath]P(\text{success})=p[/imath]. Let [imath]X_i[/imath] and [imath]X_j[/imath] be indicator variables of the number of "success" in [imath]i[/imath]th and [imath]j[/imath]th runs with [imath]i<j[/imath]. In other words [imath]X_i=1[/imath] if the [imath]ith[/imath] trial is success and [imath]0[/imath] otherwise.. Given the total number of success was [imath]m[/imath]. I am asked to compute the correlation coefficient for [imath]X_i[/imath] and [imath]X_j[/imath]. To compute the co-variance I will need to find [imath]E(X_i)[/imath] and [imath]E(X_j)[/imath] first. By definition [imath]E(X_i)=p[/imath] =[imath]E(X_j)[/imath]. But how should I find E([imath]X^2_i[/imath])? | 2189320 | Expected value of Bernoulli random variables
Consider a sequence of [imath]n[/imath] Bernoulli trials with [imath]P(\text{success})=p[/imath]. Let [imath]X_i[/imath] and [imath]X_j[/imath] be indicator variables of the number of "success" in [imath]i[/imath]th and [imath]j[/imath]th runs. Given the total number of success was [imath]m[/imath], [imath]m<n[/imath]. I am asked to compute the correlation coefficient for [imath]X_i[/imath] and [imath]X_j[/imath]. Now I know the formula for correlation coefficient, but to compute the co-variance I will need to find [imath]E(X_i)[/imath] and [imath]E(X_j)[/imath] first. By definition does [imath]E(X_i)=p_i[/imath] and [imath]E(X_j)=p_j[/imath]? I feel this seems a bit too easy but I can't see whats wrong with it. |
2191047 | inverse element in a set of polynomials
Let [imath]f\in\mathbb{F}[x][/imath] be a monic, irreducible polynomial of degree [imath]d[/imath]. We define [imath]\mathbb{K}=\{g\in\mathbb{F}[x] \ | \ deg(g)<deg(f)\}[/imath]. Let [imath]g\in\mathbb{K}[/imath], assume that [imath]g\neq 0_\mathbb{F}[/imath], prove that there exist a polynomial [imath]q\in\mathbb{K}[/imath] such that [imath]g\cdot q=s\cdot f+1_\mathbb{F}[/imath] for some polynomial [imath]s\in\mathbb{F}[x][/imath]. | 631057 | Let [imath]K[/imath] be a field and [imath]f(x)\in K[x][/imath]. Prove that [imath]K[x]/(f(x))[/imath] is a field if and only if [imath]f(x)[/imath] is irreducible in [imath]K[x][/imath].
Let [imath]K[/imath] be a field and [imath]f(x)\in K[x][/imath]. Prove that [imath]K[x]/(f(x))[/imath] is a field if and only if [imath]f(x)[/imath] is irreducible in [imath]K[x][/imath]. How to prove? I really have no idea... Thank you a lot. |
2191231 | Continuous at a point
[imath]F(x,y) =x^2 y /x^2+y^2 , (0,0) [/imath]. The answer is that F is not continues at [imath](0,0)[/imath] because [imath]\lim (x,y)>(0,0) ~ f(x,y)[/imath] does not exist [imath](0,0) \in D_f[/imath] where [imath]D_f[/imath] is the domain of [imath]f[/imath]. I try a method where you let [imath]y=g_1(x)=x[/imath] And [imath]y=g_2(x)=x^2[/imath] which is it to curves that pass the point [imath](0,0)[/imath] etc And the limit is equal which means that the function is exist ! And why does point [imath](0,0)[/imath] belong to [imath]D_f[/imath] ? | 155623 | Proving that the function [imath]f(x,y)=\frac{x^2y}{x^2 + y^2}[/imath] with [imath]f(0,0)=0[/imath] is continuous at [imath](0,0)[/imath].
How would you prove or disprove that the function given by [imath]f(x,y) = \begin{cases} \dfrac{x^2y}{x^2 + y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}[/imath] is continuous at [imath](0,0[/imath])? |
2191686 | notation in definition of range and domain
In my set theory book (A book of set theory, Charles,C,Pinter) domain and range of graph [imath]G[/imath] is defined by [imath]dom G=\{x:\exists y \ni (x,y)\in G\}[/imath] [imath]ran G=\{y:\exists x \ni (x,y)\in G\}[/imath] But I don't know what [imath]\exists x \ni (x,y)[/imath] exactly means. Ordered pair is defined with [imath](x,y)=\{\{x\},\{x,y\}\}[/imath] so [imath]x\in \{x\}\in \{\{x\},\{x,y\}\}=(x,y)[/imath]. I think we should write [imath]x \in (x,y)[/imath] instead of [imath]x \ni (x,y)[/imath] What makes that definition possible? | 15455 | Backwards epsilon
What does the [imath]\ni[/imath] (backwards element of) symbol mean? It doesn't appear in the Wikipedia list of mathematical symbols, and a Google search for "backwards element of" or "backwards epsilon" turns up contradictory (or unreliable) information. It seems it can mean both "such that", or "contains as an element". Is this correct, and if so, which is the more common usage? |
2191137 | Integration closed contours
[imath]\int_{-\infty}^{\infty} \frac{x \sin(x)}{x^2 +4} dx[/imath] I do not understand how to evaluate integrals by using a closed contour and passing to a limit, can someone show me using this example | 2172184 | Compute [imath]\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx[/imath]
[imath]\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx[/imath] Some rearranging eventually gives [imath]\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx = \frac{-i}{2}\int_{-\infty}^{\infty} \frac{xe^{2ix}}{9+x^{2}}[/imath] Consider [imath]f(z) = \frac{ze^{2iz}}{9+z^{2}}[/imath] and the contour [imath]\gamma[/imath] of the semicircle laying in the upper half of the plane: Let [imath]\gamma_{R}[/imath] denote the circular part with radius [imath]R[/imath] and [imath]\gamma_{L}[/imath] denote the part lying on the real axis with length [imath]2R[/imath]. Computing the residue at the only pole, [imath]z = 3i[/imath], we have that [imath]\oint_{\gamma}f(z) \, dz = \frac{i\pi}{e^{6}} [/imath] On the other hand, \begin{align*} \oint_{\gamma}f(z) \, dz &= \oint_{\gamma_{R}}f(z) \, dz + \oint_{\gamma_{L}} f(z) \, dz \\ &= \oint_{\gamma_{R}} f(z) \, dz + \int_{-R}^{R} \frac{xe^{2ix}}{9+x^{2}} \, dx \end{align*} We may evaluate the first integral in the usual way by parameterizing the contour and taking [imath]z = Re^{i\theta}[/imath]. \begin{align*} \oint_{\gamma_{R}} f(z) \, dz = \int_{0}^{\pi} \, \frac{R^{2}e^{2i\theta}e^{2i\cos\theta}e^{-2R\sin\theta}}{(9+R^{2}e^{2i\theta})} d\theta \end{align*} I'm aiming to show that this integral goes to [imath]0[/imath] as [imath]R[/imath] goes to infinity. This gives the desired result as the factor of [imath]\frac{-i}{2}[/imath] is all that is missing according to WolframAlpha. I'm not sure how to finish it though. |
2189551 | How to prove the continuity of the function of several at [imath](0,0)[/imath]
How to verify the continuity of the following function at [imath](0,0)[/imath] : \begin{cases}f(x,y)=\cfrac{x^3}{x^2+y^2}& \text{if } (x,y)\neq (0,0)\\ f(0,0)=0\end{cases} | 492288 | Is there limit [imath] \lim_{(x,y) \to (0,0)} \frac{x^3}{x^2 + y^2}[/imath]?
How to show if the limit [imath] \lim_{(x,y) \to (0,0)} \frac{x^3}{x^2 + y^2}[/imath] exists? I suspect that there is, as I can't find any path that would show that limit doesn't exist, and WolframAlpha also suggests that the limit is (0,0). In general, can you recommend any tips how to learn to approach similar limit problems (fractions and polynomials like this)? edit That is, excluding the polar coordinate conversion method? |
2192411 | Prove that if [imath]\{a_n\}[/imath] converges to [imath]a[/imath], that [imath]\{(a_n)^2\}[/imath] converges to [imath]a^2[/imath] by definition
Prove that if [imath]\{a_n\}[/imath] converges to [imath]a[/imath], that [imath]\{(a_n)^2\}[/imath] converges to [imath]a^2[/imath] by definition. Pf. Suppose [imath]\{a_n\}[/imath] converges to [imath]a[/imath]. This means: [imath]\forall \epsilon >0, \exists N>0, \text{s.t, if} \text{ }n>N, |a_n-a|< \epsilon[/imath] I want to show that: [imath]\forall \epsilon >0, \exists N>0, \text{s.t, if} \text{ }n>N, |a_n^2-a^2|< \epsilon[/imath] How do I use the given info? I don't know how I can manipulate it to get [imath]a_n^2[/imath] | 1019453 | Suppose that [imath](s_n)[/imath] converges to s. Prove that [imath](s_n^2)[/imath] converges to [imath]s^2[/imath]
I am really struggling with my proofs class, I don't really understand how to prove a statement like this, or what the epsilon is standing for.. Suppose that [imath](s_n)[/imath] converges to s. Prove that [imath](s_n^2)[/imath] converges to [imath]s^2[/imath] directly without using the fact that [imath]lim(s_nt_n)=st[/imath] Suppose that [imath](s_n)[/imath] converges to s. Then, since [imath](s_n)[/imath] is convergent, there exists an [imath]M_1[/imath] such that [imath]|S_n|<M_1[/imath], for all [imath]n\in \!\,\mathbb{N} \!\,[/imath]. Let [imath]M=M_1+|S|[/imath] (I don't completely understand this step...). Then given [imath]ε>0[/imath], there exists N such that n>N implies that [imath]|s_n-s|<ε/M[/imath]. Thus for n>N, we have [imath]|s_n^2-s^2|=|s_n-s|*|s_n+s|<(ε/M)(M)=ε[/imath]. Hence, [imath]s_n^2[/imath] converges to [imath]s^2[/imath]. |
2192780 | A subgroup that contains the powers of 2 of all elements contains the commutator
The question is as follow: [imath]G[/imath] is a group. [imath]N[/imath] is the smallest subgroup of G the contains {[imath] g^2 | g \in G[/imath]}. Prove that [imath]N \triangleleft G[/imath] and [imath]G/N[/imath] is abelian. So my plan is to use the theorem that states that: [imath]G'\leq N \iff N\triangleleft G , G/N [/imath] is abelian so I have problems to show now that [imath]G'\leq N[/imath] any help is appreciated!. | 2192244 | Commutator subgroup and subgroup generated by square.
While reading Cours d'algèbre by D. Perrin, I found the following claim: Proposition. If [imath]G[/imath] is a group, then [imath]D(G)\subseteq G^2[/imath], where [imath]G^2[/imath] is the subgroup generated by the squares of [imath]G[/imath]. I understand that it suffices to prove that for all [imath]x,y\in G[/imath], [imath][x,y]:=xyx^{-1}y^{-1}[/imath] is a product of squares. Equivalently, since [imath]G^2[/imath] is characteristic, hence normal, it suffices to establish that [imath]G/G^2[/imath] is abelian. For now, none of my attempts have been fruitful. Any enlightenment will be greatly appreciated! |
2048395 | Let p be a prime. If range(f) has an element of order p, then G has an element of order p.
I'm working through A Book Of Abstract Algebra, and in the chapter on homomorphisms, it has this question: Let [imath]f:G\to H[/imath] be a group homomorphism. Let [imath]p[/imath] be a prime. If [imath]\mathrm{range}(f)[/imath] has an element of order [imath]p[/imath], then [imath]G[/imath] has an element of order [imath]p[/imath]. I actually can't see how this could be true. Take G = the set of all integers, H = the parity group (odd, even). The homomorphism maps every integer to odd or even. H has one element of order 2: odd. 2 is prime. But there's no element with order 2 in G -- every element has infinite order. | 1284765 | Abstract algebra: homomorphism and an element of a prime order
Let me ask Prob.F6 in Pinter's algebra book. 'Let [imath]p[/imath] be a prime and [imath]f[/imath] be a homomorphism from [imath]G[/imath] to [imath]H[/imath] where [imath]G[/imath] and [imath]H[/imath] are groups. If range of [imath]f[/imath] has an element of order [imath]p[/imath], then [imath]G[/imath] has an element of order [imath]p[/imath].' If [imath]f[/imath] is injective, life gets easier. But without the injectivity of the homomorphism, i feel lost... Would somebody help please? |
976536 | Number of possible permutations of n1 1's, n2 2's, n3 3's, n4 4's such that no two adjacent elements are same?
Given [imath]n_1 [/imath] number of [imath]1 [/imath]'s, [imath]n_2 [/imath] number of [imath]2 [/imath]'s, [imath]n_3 [/imath] number of [imath]3 [/imath]'s, [imath]n_4 [/imath] number of [imath]4 [/imath]'s. form a sequence using all these numbers such that two adjacent numbers should not be same. I have tries lot of things but nothing worked. can somebody tell me how to solve this problem? | 2889232 | Arranging finite balls of different color such that no balls of the same color are adjacent
The problem involves arranging balls of different color in a line such that no two balls of the same color are adjacent to each other. Eventually I would like to find a generalization of having [imath]m[/imath] different colors of balls and [imath]n[/imath] balls of each color, where each ball besides color are indistinguishable. But for now I will start with [imath]3[/imath] colors, red blue and yellow, and having [imath]3[/imath] of each color. If we had an infinite number of each ball, then i know the answer would be [imath]3(2^8)[/imath] but this is including arrangements where we might only have red and blue balls alternating and not have any yellows. One thought was to consider "grouping" the colors in its [imath]3![/imath] permutations and finding the number of arrangements of the groups. But this wouldn't consider arrangements such as [imath]RYRYBYBRB[/imath] I feel like recurrence may be needed for a generalization but without being able to find this simple case I don't know if I can use it. Searching other questions of this sort doesn't help me either as it isn't this type of restriction. Brute forcing it is one option but it seems like a mess even for this smaller case, so it would only be good for smaller cases. Any help would be greatly appreciated |
2192449 | Euclidean domain [imath]\mathbb{Z}[\sqrt{d}][/imath]
I am trying to generalized, for which integral values of [imath]d[/imath], [imath]\mathbb{Z}[\sqrt{d}] = \{ a + b\sqrt{d} \vert a,b\in\mathbb{Z}\}[/imath] is an Euclidean domain? I am interested specially in positive integral values of [imath]d[/imath]. | 58561 | Norm-Euclidean rings?
For which integer [imath]d[/imath] is the ring [imath]\mathbb{Z}[\sqrt{d}][/imath] norm-Euclidean? Here I'm referring to [imath]\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a,b \in \mathbb{Z}\}[/imath], not the ring of integers of [imath]\mathbb{Q}[\sqrt{d}][/imath]. For [imath]d < 0[/imath], it is easy to show that only [imath]d = -1, -2[/imath] suffice; but what about [imath]d>0[/imath]? Thanks. |
2192614 | Prove or disprove C∖(A∪B)=(C∖A)∩(C∖B)
So I've tried a bunch of sets and I believe the statement to be true. However I am not quite sure how to formulate a correct proof. I know so far that for an element x in set [imath]C[/imath], it cannot be in set [imath]A[/imath] or [imath]B[/imath]. And on the right side, we are taking the union of the sets where for an element [imath]x[/imath], [imath]C\setminus A[/imath] and [imath]C\setminus B[/imath] is [imath]x[/imath] that is in [imath]C[/imath] but not [imath]A[/imath] and [imath]B[/imath]. | 2006136 | How to prove a set equality?
Statement: Let [imath]A[/imath], [imath]B[/imath], and [imath]C[/imath] be sets. Then [imath](A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)[/imath]. Proof: Let [imath](x,y) \in ( A \setminus B) \cap (A \setminus C)[/imath], where [imath] x \in A[/imath] and [imath]y \in (B \cap C[/imath] by definition of intersection. So, [imath](x,y) \in (A \setminus B \cap C)[/imath] which show that [imath](A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)[/imath]. \ Now let [imath](x,y) \in A \setminus(B \cup C)[/imath], where [imath] x \in A [/imath]. So, [imath] (x,y) \in (A \setminus B) \cap (A \setminus C)[/imath] by definition of intersection, which shows that [imath]A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)[/imath]. Therefore, by the Axiom Set of Equality we know that [imath](A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)[/imath] . Am I on the right track to proving that this is a true statement or is there a better way to show this? |
2149434 | If[imath]f_n[/imath] a sequence of integrable functions on a set of finite measure which converge to f uniformly, then the limit and integral can be swapped.
Hello I came across this problem and tried to prove it, I would just like to confirm that my proof is correct please as I think there is something wrong with it. The problem says the following: Suppose [imath](f_n)[/imath] is a sequence of integrable functions on a set A of finite measure. Show that if the sequence is uniformly convergent on A, then [imath]\lim_{n\to \infty} \int_{A} f_n dm= \int_{A} \lim_{n\to \infty} f_n dm[/imath]. My attenpt: Let [imath]\epsilon>0[/imath] be fixed but arbitrary. Then by uniform convergence,there exists an N s.t. [imath]|f_n-f|<\epsilon[/imath] for all n>N. Then [imath]0\leq \int_A |f_n-f| \leq \epsilon m(A)[/imath] for all n>N. Now since [imath]\epsilon>0[/imath] arbitrary and [imath]m(A)<\infty[/imath] then [imath]\int_A |f_n-f| \to 0[/imath] which implies [imath]0<|\int_{A} (f_n-f)| \leq \int_{A} |f_n-f|[/imath] and now taking the limit gives the desired result. Could someone verify my proof? I am not sure when steps like since epsilon is arbitrary work.Also where did I use integrability of the [imath]f_n[/imath] ? Thanks. | 24171 | lebesgue integral uniform convergence
Let [imath]f_n, f : [a,b] \to R.[/imath] If [imath]f_n \to f[/imath] uniformly then show that the lebesgue integrals are equal. ie. [imath]\int f = lim \int f_n[/imath] This is clearly true for continuous functions, but how do I handle the case of non-continuous functions? |
2192728 | prove a group of certain order is never simple
There is no simple group of order [imath]400[/imath]. [imath]400=2^{4}\times 5^{2}[/imath],so by sylow theorem, there is a 2-sylow group [imath]A[/imath] or 5-sylow group [imath]B[/imath],then by group action(left multiplication) we have a group homorphism from [imath]G[/imath] to symmtry groups [imath]S_{5^{2}}[/imath] or [imath]S_{2^{4}}[/imath].Kernel of these two morphisms are all normal subgroup of [imath]G[/imath] if they are nontrivial.It's enough for us to prove that one of these two homorphisms have nontrivial kernel.But I don't know how to prove it. Since there is an answer in mathstackexchange by using sylow theorem not group action, I think this should not be duplicated. | 79632 | No group of order 400 is simple
I've been given the question of showing that no group of order [imath]400[/imath] is simple. I've tried to attack it via the Sylow theorems for about a week now, but all the tricks and methods I know seem to be failing horribly. Things I've tried: Trying to produce a contradiction by giving a map into [imath]S_n[/imath] by elements acting by conjugation on Sylow 5-subgroups doesn't work, since there are 16 such Sylow 5-subgroups, and 400 divides [imath]16![/imath], so it might very well be an injection and therefore we can't obviously find a nontrivial kernel. Trying element counting is messy and I can't get it to come out the way I want- for instance, we can show that each of the Sylow 5-subgroups is isomorphic to [imath]\mathbb{Z}_5\times \mathbb{Z}_5[/imath], so there should be at least 125 elements of order divisible by only 5, but I can't see this producing a contradiction with any of the things I can find out about Sylow 2-subgroups. Anyways, I'm probably missing something fairly obvious, and I would appreciate any hints, solutions, or other help that you could give. |
2193345 | Measurability of a 2 variable function
Let [imath]f[/imath] be a real valued measurable function on [imath]\mathbb R^2[/imath] such that each section [imath]f_x[/imath] is Borel measurable and each section [imath]f^y[/imath] is continuous, where [imath]f_x(y)=f(x,y)[/imath] and [imath]f^y(x)=f(x,y)[/imath]. We need to show that [imath]f[/imath] is Borel measurable. I am not getting how to approach for it. Any suggestion would be appreciated. Thanks in advance. | 2066145 | [imath]f_x[/imath] is Borel measurable and [imath]f^y[/imath] is continuous then [imath]f[/imath] is Borel measurable
I have to prove the following: Let [imath]f: \mathbb{R^2}\to \mathbb{R}[/imath] such that [imath]f_x:y\to f(x,y)[/imath] is Borel measurable for all [imath]x\in\mathbb{R}[/imath] and that [imath]f^y:x\to f(x,y)[/imath] is continuous for all [imath]y\in\mathbb{R}[/imath]. Prove that [imath]f[/imath] is Borel measurable. What I have tried to do is to find a sequence of functions [imath]f_n(x,y)[/imath] s.t for a fixed [imath]y[/imath] [imath]f_n(.,y)[/imath] is a linear approximation of [imath]f(.,y)[/imath].. |
2192153 | Consider the following model for the populations of prey x ≥ 0 and predators y ≥ 0,
[imath]\dot{x} = x(1-\mu x-y), \quad \quad \dot{y} = y(-1+x-y)[/imath] where [imath]\mu > 0[/imath] is a constant. The number and type of equilibria depend on the value of μ; in fact there are three essentially different cases. Find the three corresponding ranges of μ and the number, type and stability of the equilibria in each case. (Ignore a borderline case with a zero eigenvalue, but look carefully at any other borderline cases.) I've already worked out the following: [imath](x_0, y_0) = (0,0), \quad (0,-1), \quad (\frac{1}{\mu},0), \quad (\frac{2}{\mu +1},\frac{1-\mu}{\mu+1})[/imath] [imath]\therefore[/imath] If [imath]\mu = -1[/imath] or [imath]\mu = 0[/imath] then there exist only 3 equilibrium points. However, if [imath][\mu \in \mathbb{R} : \mu \ne 0, -1][/imath] then 4 equilibria exists. Thus from here, assume [imath]\mu \ne 0,-1[/imath] Jacobian: [imath]J = \begin{bmatrix} 1-2\mu x -y & -x\\y & -1+x-2y\end{bmatrix}[/imath] I've also worked out the Jacobian for [imath](\frac{1}{\mu},0)[/imath] and found the polynomial for the EVal's to be [imath]\frac{-b \pm \sqrt{b^2 - 4b}}{2}[/imath] where [imath]b = \frac{\mu -1}{\mu}[/imath]. Now where do I go from here to complete the question? Also please do keep in mind that I'm studying Maths not Physics. Many thanks. | 2191385 | Solving Predator-Prey equations
Consider the following model of predator-prey dynamics : [imath]\dot x = x(\lambda − x − y),\ \\ \dot y = y(−1 + x − y)[/imath] The number and type of equilibria of the system depend on the parameter [imath]\lambda[/imath], and there are essentially three different cases corresponding to three different ranges of [imath]\lambda[/imath]. Find these three ranges for [imath]\lambda[/imath] and the number, type and stability of the equilibria in each case. There are two values of [imath]\lambda[/imath] that give rise to borderline cases; one of these corresponds to a zero eigenvalue and may be ignored, but for the other borderline case you should make a careful analysis of the stability type. I Seem to be quite a dunce at dynamics, I would appreciate some help on how to go about this problem. |
2193342 | Continuous map of differentiable manifolds is differentiable if differentiable functions pull back to differentiable functions
This is Exercise 3.1.A. in Vakil's notes Suppose that [imath]\pi: X\rightarrow Y[/imath] is a continuous map of differentiable manifolds. Show that [imath]\pi[/imath] is differentiable if differentialble functions pull back to differentiable functions, i.e., if pullback by [imath]\pi[/imath] gives mpa [imath]\mathcal{O}_Y\rightarrow \pi_*\mathcal{O}_X[/imath]. Let [imath]f: V\subseteq Y\rightarrow \mathbb{R}[/imath] be a differentiable function on an open subset of [imath]Y[/imath]. Let [imath](U,\phi)[/imath] be a chart of [imath]X[/imath], [imath](V,\psi)[/imath] be a chart of [imath]Y[/imath]. Then [imath]f\circ\pi(\phi^{-1})[/imath] is differentiable and [imath]f\circ\psi^{-1}[/imath] is differentiable. We need to show that [imath]\psi\circ\pi\circ\phi^{-1}[/imath] is differentiable. I don't know how to prove this. Can we say that since [imath](U,\psi\circ\pi)[/imath] constructs a chart of [imath]X[/imath], by the compatibility, [imath]\psi\circ\pi\circ\phi^{-1}[/imath] is differentiable? This seems true but this does not use the fact that [imath]f[/imath] is differentiable. Sorry I have no background of Differentiable Manifolds. Any help would be appreciated. | 2193343 | Continuous map of differentiable manifolds is differentiable if differentiable functions pull back to differentiable functions
This is Exercise 3.1.A. in Vakil's notes Suppose that [imath]\pi: X\rightarrow Y[/imath] is a continuous map of differentiable manifolds. Show that [imath]\pi[/imath] is differentiable if differentialble functions pull back to differentiable functions, i.e., if pullback by [imath]\pi[/imath] gives mpa [imath]\mathcal{O}_Y\rightarrow \pi_*\mathcal{O}_X[/imath]. Let [imath]f: V\subseteq Y\rightarrow \mathbb{R}[/imath] be a differentiable function on an open subset of [imath]Y[/imath]. Let [imath](U,\phi)[/imath] be a chart of [imath]X[/imath], [imath](V,\psi)[/imath] be a chart of [imath]Y[/imath]. Then [imath]f\circ\pi(\phi^{-1})[/imath] is differentiable and [imath]f\circ\psi^{-1}[/imath] is differentiable. We need to show that [imath]\psi\circ\pi\circ\phi^{-1}[/imath] is differentiable. I don't know how to prove this. Can we say that since [imath](U,\psi\circ\pi)[/imath] constructs a chart of [imath]X[/imath], by the compatibility, [imath]\psi\circ\pi\circ\phi^{-1}[/imath] is differentiable? This seems true but this does not use the fact that [imath]f[/imath] is differentiable. Sorry I have no background of Differentiable Manifolds. Any help would be appreciated. I also saw this post: pullback of continuous maps of manifolds, but I don't understand how [imath]\gamma[/imath] is smooth in the answer. [imath]\gamma=\beta\circ\rho[/imath], where [imath]\rho[/imath] is defined to be smooth, but [imath]\beta[/imath] is not. |
2193854 | How can I prove this series converges?
Wolfram Alpha says the following series converges, but I can't figure out how to prove it. [imath]\sum_{n=1}^{\infty}\frac{\sin(n)+\sqrt{n}}{n^2+5}[/imath] Can I use a comparison test with a simple harmonic p series, or is there a better way? | 2193155 | Determining whether the series [imath]\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}[/imath] is convergent or divergent by comparison test
I am given the series: [imath]\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}[/imath] and I am asked to determine whether it is convergent or not. I know I need to use the comparison test to determine this. I can make a comparison with a harmonic p series ([imath]a_n=\frac{1}{n^p}[/imath] where p > 1, series converges). I argue that as the denominator grows more rapidly than the numerator, I need only look at the denominators: [imath]\frac{1}{n^2+5}\le\frac{1}{n^2}[/imath] [imath]\frac{1}{n^2}[/imath] is a harmonic p series where [imath]p>1[/imath] which converges. As [imath]\frac{\sqrt{n}+\sin(n)}{n^2+5}[/imath] is less than that, by the comparison test, [imath]\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}[/imath] is convergent. Is this a valid argument for this question? |
2193031 | Banach space is direct sum of closed subspaces. Do these subspaces become continuous embeddings?
I am working on the following question: [imath]X[/imath] is Banach. [imath]A, B[/imath] are closed subspaces such that [imath]x \in X[/imath] can be uniquely represented as the sum [imath]x = a + b, a \in A , b \in B[/imath]. Show that [imath]\exists k \in \mathbb{R}: \| a\| \leq k \|x\| \\ \| b\| \leq k \|x\| \hspace{2cm} \forall x \in X[/imath]. I am completely stuck. I can't figure out how to use the closedness other than to conclude completeness of the space. Or perhaps to conclude that the projection maps will be bounded. But I can't proceed. Any nudge would be helpful. | 2192898 | If every [imath]x\in X[/imath] is uniquely [imath]x=y+z[/imath] then [imath]\|z\|+\|y\|\leq C\|x\|[/imath]
Problem Statement Let [imath](X,\|\cdot \|)[/imath] be a Banach space and [imath]Y[/imath], [imath]Z[/imath] closed subspaces of [imath]X[/imath]. If every [imath]x \in X[/imath] can be uniquely represented as [imath]x=y+z[/imath] for [imath]y \in Y[/imath] and [imath]z \in Z[/imath] then show that there exists [imath]c[/imath] such that [imath]\|y\|\leq C\|x\|[/imath] and [imath]\|z\| \leq C\|x\|[/imath] Attempt If [imath]L:X \rightarrow Y[/imath] given by [imath]L(x)=L(y+z)=y,[/imath] then we will be done if we can show that [imath]L[/imath] is a bounded linear operator. Alternatively, by adding the two inequalities, we can show that the norm [imath]\|\cdot\|[/imath] is equivalent to [imath]\| \cdot \|_{Y\times Z}[/imath], where [imath]\|(y,z)\|=\|y\|+\|z\|[/imath]. I attempted the first approach via the closed graph theorem but I am unable to make any headway. |
2194384 | reasonable but erroneous claim
I came across the following problem in Abbott's Understanding Analysis: Consider the reasonable but erroneous claim that [imath]\lim_{x\to10} \frac{1}{[[x]]} = \frac{1}{10},[/imath] where [imath][[x]][/imath] refers to the greatest integer value of [imath]x[/imath]. Find the largest [imath]\delta[/imath] that represents a proper response to the challenge of [imath]\epsilon = \frac{1}{2}[/imath] and find the largest [imath]\epsilon[/imath] challenge for which there is no suitable [imath]\delta[/imath] response possible. A student came to me with the solution to the first part as [imath]\delta = 5[/imath]. The reasoning for this was the following: [imath]|\frac{1}{[[x]]}-\frac{1}{10}| = |\frac{10-[[x]]}{10[[x]]}| = |\frac{[[x]]-10}{10[[x]]}| < \frac{|x-10|}{10}[/imath] and so we choose [imath]\delta = 10\epsilon[/imath], which means that for [imath]\epsilon = \frac{1}{2}[/imath], we have [imath]\delta = 5[/imath]. Now to my question on this: If we consider [imath]|\frac{1}{10}-\frac{1}{4}|=\frac{3}{20}[/imath] and [imath]|\frac{1}{10}-\frac{1}{16}|=\frac{3}{80},[/imath] then would this suggest that there is a larger [imath]\delta[/imath] (namely 6) that is a proper response to the challenge of [imath]\epsilon = \frac{1}{2}[/imath]? I may have some fundamental misinterpretation here and so I welcome any help. The proposed solution to the second claim is [imath]\epsilon = \frac{1}{10}[/imath]. The reasoning is that, as before: [imath]|\frac{1}{[[x]]}-\frac{1}{10}|<\frac{|x-10|}{10}.[/imath] So, picking [imath]\epsilon = \frac{1}{10}[/imath] will do the trick. I don't feel right about this answer either since [imath]\delta = 1[/imath] will yield [imath]|\frac{1}{[[x]]}-\frac{1}{10}|[/imath] as either [imath]\frac{1}{90}[/imath] or [imath]0[/imath] since [imath]|\frac{1}{10}-\frac{1}{10}| = 0[/imath] and [imath]|\frac{1}{10}-\frac{1}{9}| = \frac{1}{90}[/imath]. Again, I might have some sort of fundamental misunderstanding here. I appreciate any help that can be offered. Thanks in advance! | 1377558 | Challenging [imath]\lim_{x \rightarrow 10} \frac{1}{\lfloor x \rfloor} = \frac{1}{10}[/imath] for [imath]\epsilon=\frac{1}{2}[/imath].
Consider the (incorrect) claim that [imath]\lim_{x \rightarrow 10} \frac{1}{\lfloor x \rfloor} = \frac{1}{10}.[/imath] How might I find the largest [imath]\delta[/imath] such that I can challenge [imath]\epsilon = 1/2[/imath]? Clearly I need to use the [imath]\epsilon[/imath]-[imath]\delta[/imath] definition of a limit, but my problem is that I don't know how to find the largest delta. Here is what I know: [imath]|f(x) - (1/10)| <1/2[/imath] [imath]0<|x-10|< \delta[/imath] |
2194241 | find the principle part of Laurent expansion
Find the principle part of the Laurent expansion of the function [imath]f(z) = \frac{1}{(e^z - 1)^2}[/imath] about [imath]z = 0[/imath]. Hint: First show that [imath]f[/imath] has a pole of order [imath]2[/imath] at [imath]z = 0[/imath]. I already found the pole of order [imath]2[/imath] using Laurent expansion. But I am not able to proceed further with finding the principle part. Thank you | 2191572 | Principal part of Laurent expansion.
Trying to find the principal part of the Laurent Expansion of [imath](\exp(z)-1)^{-2}[/imath] about [imath]z = 0[/imath]. I know that because there is a pole of order two at [imath]z=0[/imath] I need to only find the constants for [imath]n=-1,-2[/imath]. |
1899511 | If H ≤ Z(G) ≤ G, where G is a finite group,Z(G) is its center, and (G:H) = p for some prime p, then G is abelian.
If [imath]H \leq Z(G) \leq G[/imath], where G is a finite group, [imath]Z(G)[/imath] is its center, and [imath](G:H) = p[/imath] for some prime [imath]p[/imath], then G is abelian. Well because H < Z(G) we have that H commutes with all of G. Also |G/H|=p. I was thinking of connecting this result to the fact that if [imath]G/Z(G)[/imath] is cyclic, then [imath]G[/imath] is abelian, but was unable to. Then I thought about using the result that [imath](G/H)/(Z(G)/H) = G/Z(G),[/imath] but that leads to nowhere. Any suggestions? | 2194383 | Prove that G is abelian if (G:H) is prime, where H is in Z(G)
I'm attempting to prove this, but am quite frankly stuck and haven't made any progress. I know that [imath]\frac{|G|}{|H|}=p[/imath], and I know by Cauchy's thm, that [imath]\exists x \in G[/imath] such that [imath]x^p[/imath] = e, and that any group of prime order is cyclic which implies Abelian, but that's as many relevant details I can think of. Any hints or tips would be appreciated. Thanks! |
2193724 | Is there a convex function smaller than a given function.
If [imath]g(x)[/imath] is positive on [imath](0,\infty)[/imath] and [imath]g(x)\to \infty[/imath] as [imath]x\to \infty[/imath]. Is there a convex function [imath]h(x)[/imath] on [imath](0,\infty)[/imath] such that [imath]h(x)\le g(x), h(x)\to \infty[/imath] as [imath]x\to \infty[/imath]. I didn't know how to solve this one. But I find a solution says that [imath]h(x)[/imath] exists iff [imath]\liminf_{x\to \infty} \frac{\log(g(x))}{x} = \alpha \gt 0[/imath]. I don't know how to do it in any direction. Can somebody give me a hint on what I should do? Thanks. There is a similar question in Rudin Real and Complex analysis on the interval [imath](0,1)[/imath]. But I think they are not same question. At least for that one the answer is affirmative but this answer suggests otherwise. | 50549 | Given a real function [imath]g[/imath] satisfying certain conditions, can we construct a convex [imath]h[/imath] with [imath]h \le g[/imath]?
The following is Exercise 8 from Chapter 3 of Rudin's Real and Complex Analysis (not a homework problem, just for fun). Let [imath]g[/imath] be a positive function on [imath](0, 1)[/imath] such that [imath]g(x) \to \infty[/imath] as [imath]x \to 0[/imath]. Does there exist a convex function [imath]h[/imath] on [imath](0, 1)[/imath] such that [imath]h \le g[/imath] and [imath]h(x) \to \infty[/imath] as [imath]x \to 0[/imath]? This seems true, but I can't show it. All I've been able to do is to reduce the problem to [imath]g[/imath] being a monotone step function or a strictly monotone piecewise-linear function but this doesn't seem to get me anywhere. I'm not sure how to use the property of [imath]g[/imath] to construct the explicit [imath]h[/imath]. |
2194490 | Use method of direct proof to prove the statement
If [imath]n[/imath] is a natural number such that [imath] n \geq 2[/imath], then the numbers [imath]n! + 2, n! + 3, n! + 4... n! + n[/imath] are all composite. (Thus, for any n greater than or equal to 2, one can find n consecutive composite numbers) I started with just plugging in numbers to see if they were composite and they were. But I don't know how to prove it for all natural numbers. | 1945871 | Prove that [imath]n! + k[/imath] is a composite number
Recall that if [imath]n[/imath] is a positive integer, [imath]n! = n(n-1)\cdots 3\cdot2\cdot1[/imath]. a) Let [imath]n[/imath], [imath]k[/imath] be positive integers with [imath]1< k \le n[/imath]. Prove that [imath]n! + k[/imath] is composite. b) Using part a, find [imath]100[/imath] consecutive integers all of which are composite. c) In part b you gave a sequence of [imath]100[/imath] consecutive integers all of which are composite. You should be able to use the same idea to find [imath]\ell[/imath] consecutive integers which are composite for any positive integer [imath]\ell[/imath]. In other words, you can find sequences of any length which consist of consecutive integers and contain no primes. Does this contradict the fact that there are an infinite number of prime numbers? |
2137581 | Why cofinality is a cardinal.
In the proof of the lemma For every limit ordinal [imath]\alpha[/imath], [imath]\textrm{cf } \alpha[/imath] is a regular cardinal Jech says that It is easy to see that if [imath]\alpha[/imath] is not a cardinal, then using a mapping of [imath]|\alpha|[/imath] onto [imath]\alpha[/imath] one can construct a cofinal sequence in [imath]\alpha[/imath] of length [imath]\leq |\alpha|[/imath], and therefore [imath]\textrm{cf } \alpha < \alpha[/imath]. I've spent a while trying to construct this cofinal sequence and was wondering if I was on the right path: From a bijection [imath]f[/imath] between [imath]\kappa = |\alpha|[/imath] and [imath]\alpha[/imath] take [imath]S = \{\beta \in \kappa : \ f(\gamma) < f(\beta)[/imath] for all [imath]\gamma < \beta\}[/imath] then if [imath]\xi[/imath] is the order type of [imath]S[/imath] and [imath]g[/imath] is the isomorphism from [imath]\xi[/imath] to [imath]S[/imath] the function [imath]f\circ g[/imath] is a cofinal sequence in [imath]\alpha[/imath] and [imath]\xi < \alpha[/imath]. Is this on the right track, or is there an easier way of doing it? | 3053049 | If [imath]\alpha[/imath] is a limit ordinal, then [imath]\operatorname{cf}(\alpha)[/imath] is a limit ordinal
In the textbook Introduction to Set Theory by Hrbacek and Jech, Section 9.2, the authors first introduce the definition of increasing sequence of ordinals: Then they introduce cofinality: My question: how do we prove that [imath]\operatorname{cf}(\alpha)[/imath] is a limit ordinal? If I take a sequence [imath]\langle \alpha_\nu \mid \nu<1 \rangle[/imath] where [imath]\alpha_0=\omega[/imath]. It is clear that the limit of this sequence is [imath]\omega[/imath] and thus [imath]\operatorname{cf}(\omega)=1[/imath], which is a successor ordinal. I don't know what's wrong with my reasoning. |
1151595 | Complete Residue System of an odd modulo m
Show that if [imath]m[/imath] is odd then [imath]\frac{(-m+1)}{2}, \frac{(-m+3)}{2}, \cdots, \frac{(m-3)}{2}, \frac{(m-1)}{2}[/imath] is a complete residue system modulo m. I understand the definition of a complete residue system but is having a hard time incorporating the odd part of this statement. | 3004067 | Prime Numbers: Proving that a "minimal residue" exists
In a class of mine the lecturer introduced the term of "minimal residue" as follows: Definition: If [imath]p[/imath] is an odd prime there is just one residue of [imath]n \text{ mod } p[/imath] between [imath]-\frac{1}{2}p[/imath] and [imath]\frac{1}{2}p[/imath]. It is called minimal residue. I do not see why such a minimal residue should necessarily exist or why it should be unique. I tried to google the term but I could not find anything about it. Could you tell me why this definition makes sense? |
2195185 | How to solve this in an efficient way (without calculators)
Solve for [imath]x[/imath], if [imath](x+4)(x+7)(x+8)(x+11)+20=0.[/imath] Is there an easier way to solve this than trying to multiply all the values together? I've tried multiplying all of them together, and I get an equation of the fourth degree which I find very hard to factorise. I'm hoping there is an easier way to solve this question. Any help is appreciated.Thanks :). The equation I get after multiplying is [imath]x^4+30x^3+325x^2+1500x+2484=0,[/imath] and its roots are [imath]-6, \quad -9, \quad \frac{-15\pm\sqrt{41}}2 .[/imath] | 1008509 | Fast way to come up with solutions to [imath]x(x-1)(x-2)(x-3)=1[/imath]?
I can solve this equation [imath]x(x-1)(x-2)(x-3)=1[/imath] using the usual method but I am looking for a fast analytical method to solve this. Any hints ? |
2194730 | Fibonacci Induction Proof in terms of Phi
I am trying to prove this equation with an induction proof: [imath]\ F(n) = \frac{(1 + \phi)^n - (-\phi)^n}{\sqrt{5}}[/imath] where [imath]\ \phi = \frac{\sqrt{5} - 1}{2} [/imath] I have started off by proving two base cases: [imath]\ F(1)[/imath] and [imath]\ F(2):[/imath] [imath]\ F(1) = \frac{(1 + \phi)^1 - (-\phi)^1}{\sqrt{5}} = \frac{(1 + \phi + \phi)}{\sqrt{5}} = \frac{1 + 2\phi}{\sqrt{5}} = \frac{1 + \sqrt{5} - 1}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5}} = 1 [/imath] [imath]\ F(2) = \frac{(1 + \phi)^2 - (-\phi)^2}{\sqrt{5}} = \frac{(1 + 2\phi + \phi^2 - \phi^2)}{\sqrt{5}} = \frac{(1 + 2\phi)}{\sqrt{5}} = \frac{(1 + \sqrt{5} - 1)}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5}} = 1[/imath] However, I do not know how to go about proving the inductive step: [imath]\ F(n + 1) [/imath]. Can someone show me how to proceed from here? | 1712429 | Proof a formula of the Fibonacci sequence with induction
It turns out that the Fibonacci sequence satisfies the following explicit formula: For all integers [imath]F_{n} ≥ 0[/imath], [imath]F_{n} = \frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1} - (\frac{1-\sqrt{5}}{2})^{n+1}][/imath] Verify that the sequence defined by this formula satisfies the recurrence relation [imath]F_{k} = F_{k-1} + F_{k-2}[/imath] for all integers [imath]k ≥ 2[/imath]. |
2195718 | Sequence and series convergence
Question: Let [imath]\{a_n\}[/imath] be a sequence defined by [imath]a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}, \text{ if } n \geq 1[/imath] Does [imath]\{a_n\}[/imath] converge or diverge? My solution: I just wrote down some terms. [imath]a_1 = 2, a_2 = 1, a_3 = \frac{1}{2}, a_4 = 5/13, a_5 = 13/34[/imath]. Basically eventually [imath]a_n[/imath] will be zero, so I say this converges to [imath]\frac{1}{3}[/imath]. What is the most simple but a good solution to prove this? | 2188284 | Let [imath]\{a_n\}[/imath] be defined as follows: [imath]a_1 = 2[/imath], [imath]a_{n+1}=\frac{1}{3-a_n}[/imath], if [imath]n \geq 1[/imath]. Does [imath]\{a_n\}[/imath] converge?
Let [imath]\{a_n\}[/imath] be defined as follows: [imath]a_1 = 2, a_{n+1}=\dfrac{1}{3-a_n}[/imath] [imath]n \geq 1[/imath]. Does [imath]\{a_n\}[/imath] converge? Using the monotone convergence thereom, if the sequence is both bounded below and decreasing, [imath]\forall n\in\mathbb{N}, n\geq 1[/imath], then we can say the sequence converges. I'm having trouble proving that the sequence is bounded below by induction. Claim: The sequence is bounded below by [imath]0[/imath]. Base Case: [imath]n=1[/imath] [imath]a_1 = 2 \geq 0[/imath], so this holds. Induction Step Suppose that [imath]a_k \geq 0[/imath], for some [imath]k \in\mathbb{N}[/imath]. (IH) Prove [imath]a_k \geq 0\to a_{k+1} \geq 0[/imath] I have some difficulty here. [imath]a_{k}=\dfrac{1}{3-a_{k-1}} \geq 0[/imath] [imath]a_{k+1}=\dfrac{1}{3-a_{k}}[/imath], but we know [imath]a_k \geq 0[/imath], but we don't know how much bigger. It could be that [imath]a_k = 4[/imath], and then I have [imath]a_{k+1}=\dfrac{1}{3-4}[/imath], which is negative, and does not confirm with my claim of being bounded below by [imath]0[/imath]. Have I gone somewhere wrong in my induction? |
2195328 | Divisor of a homogeneous polynomial
Let [imath]f \in \Bbb C[x_1, ..., x_n][/imath] be a homogeneous polynomial of degree [imath]d[/imath], and assume that [imath]f=gh[/imath], where [imath]g,h \in \Bbb C[x_1, ..., x_n][/imath]. Is it true that [imath]g[/imath] has to be a homogeneous polynomial of degree [imath]k \leq d[/imath]. I tried to write [imath]f(\lambda x_1, ..., \lambda x_n) = \lambda^d f(x_1, \dots, x_n) = g(\lambda x_1, ..., \lambda x_n)h(\lambda x_1, ..., \lambda x_n) = \lambda^d g(x_1, \dots, x_n) h(x_1, \dots, x_n)[/imath] for all [imath]\lambda, x_i \in \Bbb C, \lambda \neq 0[/imath]. What can I do next? | 1379475 | Decomposition of a homogeneous polynomial
Let [imath]k[/imath] be a field. Suppose I have a homogeneous polynomial [imath]f[/imath] in [imath]k[x,y,z][/imath]. If [imath]f[/imath] is reducible, does it always decompose as a product of homogeneous polynomials? Thanks! |
2195175 | Wilson's Theorem Problem
How can we proof that if [imath]p[/imath] is prime and [imath]k[/imath] is integer number that [imath]1<k<p-1[/imath] then [imath](p-k)!(k-1)! \equiv (-1)^k\pmod p[/imath] | 461331 | Prove [imath](p-k)!(k-1)!\equiv (-1)^k \text{ mod p }[/imath]
Here is a question from my number theory class. Prove [imath](p-k)!(k-1)!\equiv (-1)^k \text{ mod p} [/imath] Help please! |
2194523 | How to show [imath]A\times (B\cup C) = (A\times B)\cup (A\times C)[/imath]
How to show [imath]A \times (B \cup C) = (A \times B) \cup(A \times C)[/imath], where [imath]A[/imath], [imath]B[/imath], and [imath]C[/imath] are sets. i.e. If [imath]A=\{1,2\}[/imath], [imath]B=\{2,3\}[/imath] then [imath]A \times B = \{(1,2), (1,3), (2,2), (2,3)\}[/imath]. I started with the proposition, suppose [imath]x[/imath] is an element of [imath]LHS[/imath], i.e. [imath]A \times (B \cup C)[/imath]. Help is appreciated! Thanks. | 550782 | Union of Cartesian products: [imath]X \times (Y \cup Z)= (X \times Y) \cup(X\times Z)[/imath]
How do I prove or disprove [imath]X \times (Y \cup Z)= (X \times Y) \cup(X\times Z)[/imath] for all sets [imath]X[/imath], [imath]Y[/imath], and [imath]Z[/imath]? I'm lost on the steps here. |
2196084 | Is there some trick to evaluating the following integral
I have been staring at this integral for a while and I don't quite know how to take the first step. Is there some trick that I have to use to manipulate the function first? Integral in question: [imath]\int^{2\pi}_0 \frac{dx}{a + \cos x}[/imath] Any help/hints/insights is deeply appreciated. | 2177754 | Integral of [imath] 1 \, / \, (1 + a \, \cos(x) )[/imath]
Let [imath]a<1[/imath] be a positive constant. How can I compute the following integral? [imath] \int_{0}^{2\pi} \frac{1}{1 + a \, \, \cos(x)} dx [/imath] |
2194627 | Describing Bijections
I have a question that asks: "Describe bijections [imath]f[/imath] : [imath](A\times B) \times C \rightarrow (A\times B)\times C, A\times B \rightarrow B\times A[/imath] I have no idea how to start, or precisely what the question wants: do I find some function explicitly, or state something else? | 2196048 | How to find a explicit bijection for some [imath]f:A\times B \rightarrow B \times A[/imath]
The question I have states: "Describe a bijection [imath]f:A\times B\rightarrow B\times A[/imath]" The simplicity and lack of a domain or codomain that I can look at has caused this question to throw me off. Is describing [imath]f(a,b)=(b,a)[/imath] where [imath]a\in A, b\in B[/imath] enough to do this? To what extent do I have to prove that this is a bijection? I can say that [imath](x,y)=(x',y')[/imath] if and only if [imath](x=x')\ \cap \ (y=y')[/imath] and prove injectivity and surjectivity and so on, but I am a bit confused.. |
2196250 | Are [imath]T1[/imath] spaces precisely the topological spaces where singletons are closed?
A [imath]T1[/imath] space is a topological space where, for any 2 points, each has a neighborhood not containing the other. I wonder if every singleton being closed is an equivalent condition, and what the proof for that is. I also wonder if the following condition is equivalent: for any 2 points [imath]x[/imath] and [imath]y[/imath], every neighborhood of [imath]x[/imath] is disjoint from every neighborhood of [imath]y[/imath]. | 1830929 | In [imath]T1[/imath] space, all singleton sets are closed?
The definition of [imath]T1[/imath]-Space is: A topological space [imath]X[/imath] is said to be [imath]T1[/imath] if for each pair of distinct points [imath]a,b, [/imath] [imath]\exists[/imath] open sets [imath]U,V[/imath] s.t [imath]a\in U, b\notin U, a\notin V, b\in V[/imath]. What I'm confused about is in a [imath]T1[/imath] space, all singleton subsets of [imath]X[/imath] are closed. Let [imath]t,v \in X[/imath]. Then I think the singleton sets {[imath]t[/imath]} ,{[imath]v[/imath]} satisfy the definition of [imath]T1[/imath] in [imath]U[/imath] and [imath]V[/imath] what I wrote above. (i.e $t[imath]\in${$t$},$v\notin${$t$}, $t[/imath]\notin${[imath]v[/imath]}, [imath]v[/imath][imath]\in[/imath]{[imath]v[/imath]}.) I learned the theorem showing this result and I can understand the proof of it, but I'm still confused as to why this is not a counterexample. |
2195444 | Closest points between two line segments. If infinite, choose "closest to centers".
Let [imath]x, y, z, a, b, c, d[/imath] be vectors, and [imath]t,s[/imath] be scalars. Let line segments be [imath]y = a + bt, \ t \in [0,1]; \ \ z = c + ds, \ s \in [0,1][/imath]. The distance [imath]d(z(s), y(t)) = \sqrt{(z(s)- y(t))\bullet (z(s) - y(t))}[/imath]. How do I minimize the distance and output two points one for each segment. If there are an infinite number of solutions, make a "nice-looking" line between the segments, such that if they are paralell and form the sides of a rectangle, then the line is just the line that splits the rectangle. Then choose the output points to be the ends of this beautiful line. Algorithm? How do I solve this? Calculus? | 846054 | Closest points on two line segments
I am looking for a general formulation to find the closest points on two line segments. What I was thinking about is to define our lines as: [imath] P1 + s (P2-P1)[/imath] [imath] Q1 + t (Q2-Q1)[/imath] Where [imath]P1 , P2, Q1[/imath] and [imath]Q2[/imath] are the beginning and the end points on each segment. Now we should go through an optimization problem as: [imath]\min f(s,t)[/imath] such that [imath]0<s<1[/imath] and [imath]0<t<1[/imath]. Where [imath]f(s,t)[/imath] is the point-to-point distance function. Is there any straight forward solution? |
2195916 | If [imath]A[/imath] is an empty set, how should I understand [imath]\forall x\in A[/imath]?
It might look quite stupid, but I had become little confused when understanding empty functions. Anyway, my question is, If there is a statement [imath]P(x)[/imath] starting with "for [imath]\forall x\in A[/imath],..." and [imath]A[/imath] is an empty set, should I understand this as because the assumption is false, the conclusion is absolutely true? If not, how should I? Well, the place where I got stuck was this: For every set X, there exists a unique empty function [imath]f : \emptyset \rightarrow X[/imath]. To prove this I should set two empty functions [imath]f_1, f_2[/imath], and show that [imath]\forall x\in \emptyset[/imath], [imath]f_1(x)=f_2(x)[/imath]. When thinking as I stated above, since the assumption is false, the conclusion is true. But instead if we think about a statement [imath]\forall x\in \emptyset[/imath], [imath]f_1(x)\neq f_2(x)[/imath], this may be also true....(?) | 50492 | True, false, or meaningless?
Are the following two assertions always true, always false or meaningless? [imath]\exists i \in \emptyset[/imath] [imath]\forall i \in \emptyset[/imath] Because it seems that one encounters expressions of this kind fairly similar in mathematics, especially if we are dealing with degenerate cases of definitions. Let me give an example (in graph theory) of such a case. There, one can formalize the idea the two vertices are connected in the following way: Let [imath]G=(V,E)[/imath] be a graph and let [imath]v,w\in V [/imath]. We define [imath]v[/imath] and [imath]w[/imath] to be "connected" if: [imath]\exists n \in \mathbb{N}, \ \exists \alpha: \left\{ 1,\ldots,n \right\} \rightarrow V, \ \alpha_0=v \ \& \ \alpha_n=w \ \ \forall i\in \mathbb{N}, 0\leqslant i \ \& \ i<n:\ \left\{ \alpha_i, \alpha_{i+1} \right\} \in E[/imath] Now, if we would ask, if every node is connected to itself (a fact which intuitively we would want to be true), we would have to exhibit an [imath]n[/imath] and a sequence [imath]\alpha[/imath], such that [bla bla...]. The obvious choice for [imath]n[/imath] is 0. But for this choice of [imath]n[/imath], no mattter which sequence [imath]\alpha[/imath] we would consider, the set of the [imath]i[/imath]'s would be empty, because the set of the [imath]i[/imath]'s is actually [imath]\left\{ i\in \mathbb{N} | 0\leqslant j \ \& \ j<n \right\}[/imath]. But for [imath]n=0[/imath] this set is the empty set. Thus, because we quantify [imath]\forall i \in \left\{ i\in \mathbb{N} | 0\leqslant j \ \& \ j<0 \right\}=\emptyset[/imath], should the statement be true/false by definition? EDIT: Sorry, for my unclear formulation and to everybody who on my fault interpreted it wrongly. The way Carl Mummert or Listing interpreted it, was what I meant. |
2196604 | Set of all roots of unity
We know that in the set of nth roots of unity there are exactly [imath]n[/imath] distinct complex numbers: [imath]C_n = \{z \in \mathbb{C} : z^{n} = 1 \} = \{e^{i2k\pi/n}, \; k = 0, 1, ...,n-1 \} [/imath] and they lie on the unit circle. Now collect all roots of unity for every [imath]n \in \mathbb{N}[/imath] and consider the set [imath]C = \{C_n, \; n \in \mathbb{N}\}[/imath]. Is the set [imath]C[/imath] the unit circle, that is [imath]\{C_n, \; n \in \mathbb{N}\} = \{z \in \mathbb{C} : |z| = 1 \}[/imath] ? Obviously [imath]\{C_n, \; n \in \mathbb{N}\} \subseteq \{z \in \mathbb{C} : |z| = 1 \}[/imath], but I didn't manage to prove the other inclusion (if it is true). | 184666 | Help proving the primitive roots of unity are dense in the unit circle.
I'm having difficulty understanding how to prove that the primitive roots of unity are in fact dense on the unit circle. I have the following so far: The unit circle can be written [imath]D=\{x\in\mathbb{C}:|x|=1\}[/imath]. The set of primitive [imath]m[/imath]-th roots of unity is [imath]A_m=\{\zeta_k:\zeta_k^m=1,\zeta_k\text{ is primitive}\}[/imath]. Hence, the set of all primitive roots [imath]A[/imath] is given by the union of [imath]A_m[/imath] over [imath]m=1,2,3,\ldots[/imath]. But I can't seem to get started on how to prove that [imath]A[/imath] is dense in [imath]D[/imath]. |
2196514 | condition for equality of two matrics
Given two matrices [imath]A,B[/imath] and a base [imath]\{v_i\}[/imath], I want to show that [imath]A=B[/imath]. Is it sufficient to show [imath]Av_i=Bv_i[/imath] for all [imath]i[/imath]? | 1254204 | Av = Bv for all v implies A = B?
If [imath]A[/imath] and [imath]B[/imath] are two [imath]4\times3[/imath] matrices such that [imath]A\mathbb{v}=B\mathbb{v}[/imath] for all [imath]\mathbb{v}\in\mathbb{R}^3[/imath], then [imath]A=B[/imath]? If it's not true, can you give me an example of it? Thank you. |
2195887 | Induced right exact sequence when we apply hom
I am trying to prove the following proposition rigorously given at Michael Atiyah commutative algebra book. There is one step which I am currently stuck in. Suppose that [imath]N,N^{\prime}, \ and \ N^{\prime \prime}[/imath] are A-modules. The following exact sequence [imath]0 \xrightarrow \ N^{\prime} \xrightarrow{u} N \xrightarrow{v} N^{\prime \prime}[/imath] is exact iff for all A-modules M we have that the following sequence is exact [imath]0 \xrightarrow \ Hom(M,N^{\prime}) \xrightarrow{\bar{u}} Hom(M,N) \xrightarrow{\bar{v}} Hom(M,N^{\prime \prime}).[/imath] I proved the reverse direction, however in the forward direction I have one step I am currently stuck in. Here is what I have so far. So we want to show that [imath]ker(\bar{v}) = im(\bar{u})[/imath]. I showed the inclusion [imath]im(\bar{u}) \subset ker(\bar{v})[/imath]. Suppose that [imath]\psi \in ker(\bar{v})[/imath]. Then, we have [imath]\bar{v}(\psi) = v \circ \psi = 0[/imath]. Then [imath]im(\psi) \subset ker(v) = im(u) \cong N^{\prime}[/imath]. Then, we can reinterpret [imath]\psi : M \rightarrow N[/imath] as a morphism [imath]\phi : M \rightarrow N^{\prime}[/imath]. Denote this new map by [imath]\phi[/imath]. Now I am having troubles showing that [imath]\bar{u}(\phi) = \psi[/imath]. Please maybe provide a hint as I would like to discover this by myself. In my question I am not asking for a proof. I am asking to check certain aspect of my argument and improving on that. Thus, this post isn't a duplicate. | 47401 | Hom is a left-exact functor
If [imath]0 \to A \to B\to C[/imath] is a left exact sequence of [imath]R[/imath]-module, then for any [imath]R[/imath]-module [imath]M[/imath], [imath]0 \to Hom_R(M,A)\to Hom_R(M,B)\to Hom_R(M,C)[/imath] is left exact. I proved the above, and highlighted what I'm a little unfamiliar with: Let [imath]0 \to A\ \xrightarrow{i}\ B\ \xrightarrow{f}\ C[/imath] and [imath]0 \to Hom(M,A)\ \xrightarrow{Hom(M,i)}\ Hom(M,B)\ \xrightarrow{Hom(M,f)}\ Hom(M,C)[/imath]. We need to show that [imath]\ker(Hom(M,f))=Hom(M,i)(Hom(M,A))[/imath]. Let [imath]i \circ \varphi \in RHS[/imath]. Then [imath]f \circ i \circ \varphi : M \to C[/imath] is [imath]0[/imath] since [imath]f \circ i \circ \varphi(M) \subseteq f( i(A)) = f(ker(f))=0[/imath]. Conversely, let [imath]\psi \in LHS[/imath]. Then [imath]f \circ \psi = 0[/imath] so that [imath] f(\psi(M))=0[/imath]. Hence [imath]\psi(M) \subseteq ker(f)=i(A)[/imath]. Since the image of [imath]\psi[/imath] is contained in the image of [imath]i[/imath], we may factor [imath]\psi[/imath] as [imath]\psi=i \varphi[/imath] with [imath]\varphi : M \to A[/imath]. (Here is my trial, but I'm not fully understanding this: Since [imath]i[/imath] is injective, [imath]i(A)[/imath] is isomorphic with [imath]A[/imath]. So [imath]i^{-1}(\psi (M)) \subseteq A[/imath] and if we let [imath]\varphi=i^{-1} \psi[/imath], then [imath]\psi = i \varphi[/imath].) And I have one more question: The above looks very messy, especially the notation. Is there a better proof/understanding about it? |
2196815 | If [imath]f:\mathbb{R}\to\mathbb{R}[/imath] then [imath]|f(\mathbb{N})|\leq\aleph_0[/imath]
I am stuck at this problem for long time: Prove that if [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is some real-valued function, Then [imath]|f(\mathbb{N})|\leq\aleph_0[/imath]. In other words, prove that there exists a one-to-one function from [imath]f(\mathbb{N})=\{f(n)|n\in\mathbb{N}\}[/imath] to [imath]\mathbb{N}[/imath]. If [imath]f(\mathbb{N})[/imath] is finite then it is clear that [imath]|f(\mathbb{N})|\leq\aleph_0[/imath], But if [imath]f(\mathbb{N})[/imath] is infinite then I got stuck. Thanks for any hint/help. | 611293 | Image of the set of natural numbers under any function is denumerable.
Hi everyone I'd like to know if the following reasoning is correct, any suggestion would be great. Thanks. Proposition: Let [imath]Y[/imath] be a set and let [imath]f: \mathbb{N}\rightarrow Y[/imath] be a function. Then [imath]f[\mathbb{N}][/imath] is at most countable. Proof: If [imath]Y[/imath] is finite the result follows, this is because[imath]f[\mathbb{N}]\subset Y[/imath] and any subset of a finite set is finite and has at most the size of the set in which is contained. We may assume that the set [imath]Y[/imath] is infinite. We set [imath]A:= \{\,n\in\mathbb{N}:f(i)\not=f(n) \,\,\text{ for any }i< n\, \}[/imath]. And we define the function [imath]g:A \rightarrow f[\mathbb{N}][/imath] as the restriction of [imath]f[/imath] in [imath]A[/imath]. We claim that the function [imath]g[/imath] is a bijective map. From the definition of [imath]A[/imath] we already know that the map is [imath]1:1[/imath]. We will show that also is onto. Let [imath]y\in f[\mathbb{N}][/imath]. Suppose for the sake of contradiction that there is no [imath]n\in A[/imath] such that [imath]g(n)=y[/imath]. Since [imath]y[/imath] lies in the image of [imath]f[/imath], then there is some [imath]j\in \mathbb{N}[/imath] for which [imath]y=f(j)[/imath]. Then either [imath]f(j)\not= f(i)[/imath] for any [imath]i<j[/imath] or there exists some [imath]i<j[/imath] such that [imath]f(j)= f(i)[/imath]. For the former, [imath]j[/imath] lies in [imath]A[/imath] (by its definition). For the latter, the set [imath]\{\, n\in \mathbb{N}: f(j)=f(n)\, \text{and }\,n\not=j \}[/imath] is non empty, and hence has a minimum element. Let [imath]m[/imath] be its minimum element. Then [imath]f(m)\not=f(n)[/imath] for any [imath]n<m[/imath] and thus [imath]\,m\in A[/imath] but [imath]g(m)=f(m)= f(j)=y[/imath]. Thus any case leads to a contradiction. The result follows by reductio ad absurdum. Hence [imath]f[/imath] is surjective and injective, i.e., a bijective map. Since there is a bijection between [imath]A[/imath] and the images of the natural numbers under [imath]f[/imath], both have the same cardinality. Also we already know that [imath]A\subset \mathbb{N}[/imath] and any set of the natural numbers is denumerable, thus [imath]f[\mathbb{N}][/imath] is countable. [imath]\Box[/imath] |
2189902 | Example of function between monoids that preserves operations but does not preserve identity
A function [imath]f[/imath] between monoids is a homomorphism if [imath]f[/imath] preserves operations and the identity. Then, there is some function between monoids that preserves operations but does not preserve the identity. I've tried to get it thinking about [imath](\mathbb N,+)[/imath], [imath](\mathbb N\backslash\{0\} ,\times)[/imath] and other elementary monoids, but without success. Could someone help me with some hint? Thank you! | 2133269 | Exercise 1 pg 33 from "Algebra - T. W. Hungerford" - example for a semigroup-hom but not monoid-hom..
I have from Exercise 1 pg 33 - "Algebra - T. W. Hungerford" following: Question to proof: Show by example that the first conclusione may be false if [imath]G,\, H[/imath] are monoids that are not group. I thinked: Proof: let be for example two monoids [imath](G:=\{\emptyset\}, \bot)[/imath] and [imath](H:=\{\emptyset, \{\emptyset\}\}, \top)[/imath] with: [imath]\begin{array}{c|c} \bot & \emptyset \\ \hline \emptyset & \emptyset \end{array} \, \, \, \,\, \,\, \,\, \,\, \,\, \,\, \,\, \,\, \, \begin{array}{c|cc} \top & \emptyset & \{\emptyset\}\\ \hline \emptyset & \emptyset & \emptyset \\ \{\emptyset\} & \emptyset & \{\emptyset\} \end{array}[/imath] [imath]G[/imath] with [imath]e_G:=\emptyset[/imath] like neutral element and [imath]H[/imath] with [imath]e_H:=\{\emptyset\}[/imath] like neutral element. Let be [imath]f[/imath] a function from [imath]G[/imath] to [imath]H[/imath] with [imath]\emptyset \mapsto \emptyset[/imath] (namely [imath]x \mapsto x[/imath]), the function is Homomorphismus in fact [imath] \begin{align} {\color{Red}f}{\color{Red}(}{\color{Red}x}{ \color{Red}\bot }{\color{Red}y}{\color{Red})}=&f( \emptyset \bot \emptyset)= {\color{Green}f}{\color{Green}(}{\color{Green}\emptyset}{\color{Green})}= \emptyset \\ {\color{Red}f}{\color{Red}(}{\color{Red}x}{\color{Red})} {\color{Red}\top} {\color{Red}f}{\color{Red}(}{\color{Red}y}{\color{Red})}=&f(\emptyset) \top f(\emptyset)= \emptyset \top \emptyset= \emptyset \\ &\text{with }x,y \in G \end{align}[/imath] but it means that [imath] \begin{align} f(e_G)&\neq e_H \\f(e_G)={\color{Green}f}{\color{Green}(}{\color{Green}\emptyset}{\color{Green})}=\emptyset &\neq \{\emptyset\}= e_H \end{align}[/imath] Therefore [imath]f[/imath] ist Semigroup-Hom but not Monoid-Hom... Is it correct? |
2195596 | In a ring [imath]a \cdot a=a[/imath] then [imath]a+a=0[/imath]
I am stuck in a question for Algebraic Structures: I need to prove the following: Let [imath]R[/imath] be a ring. If [imath]a^2=a[/imath] for any [imath]a\in R[/imath], then [imath]a+a=0[/imath] for any [imath]a\in R[/imath]. (of course [imath]+[/imath], [imath]\cdot[/imath] and [imath]0[/imath] regarding [imath]R[/imath]) P.S. This is similar to a different question I asked (if [imath]a \cdot a = 0[/imath] then [imath]a + a = 0[/imath]) but I wasn't able to apply the ideas from there. In that question we assumed that [imath]R[/imath] is a unity ring. | 2192674 | In a ring [imath]a*a=0[/imath] then [imath]a+a=0[/imath]
I am stuck in a question for Algebraic Structures: I need to prove: If given any ring [imath]R[/imath], for any [imath]a[/imath] in [imath]R[/imath] [imath]a\cdot a=0[/imath] then [imath]a+a=0[/imath]. (of course [imath]+[/imath], [imath]*[/imath] and [imath]0[/imath] regarding [imath]R[/imath]) I didn't find any question regarding this question. |
2196259 | Every bounded sequence converges
Q) Prove or disprove : Every bounded sequence converges. (Make sure to fully justify your answer. That is provide a proof if true otherwise provide a counterexample and justify why your counterexample satisfy the desired criteria. Solution My counter example is [imath]a_n = (-1)^n [/imath], I know I could just leave it as this but I want to prove that this diverges but I've never worked with a proof like that so could someone check it? Firstly: {-1,1,-1,...} Therefore the sequence is bounded above by 1 and bounded below by -1. Therefore this sequence is bounded. Proof that this sequence diverges: Assuming the contradiction that [imath]\{a_n\}[/imath] converges. WTS: [imath]\exists L \in \mathbb R, \forall \epsilon > 0, \exists N > 0[/imath], such that for all [imath]n \in \mathbb N[/imath], if [imath]n > N[/imath], then [imath]|(-1)^n - L| < \epsilon[/imath] Let [imath]\epsilon = 1[/imath] n is odd: if [imath]n > N[/imath], then [imath]|L+1| < 1[/imath] [imath] \Leftrightarrow -2 < L < 0[/imath] n is even: if [imath]n > N[/imath], then [imath]|L-1| < 1[/imath] [imath]\Leftrightarrow 0 < L < 2[/imath] Thus [imath]L \in (-2,0)[/imath] and [imath]L \in (0,2)[/imath]. Therefore this is a contradiction and [imath](-1)^n[/imath] diverges. Therefore not all bounded sequences converge. If this is correct I dont get why this actually proves it diverges, mainly this part: "[imath]L \in (-2,0)[/imath] and [imath]L \in (0,2)[/imath]." Could someone explain that? I was looking at a similar proof when writing this. EDIT: Not a duplicate because I'm asking my additional proof is correct. | 2194778 | Prove or disprove : Every bounded sequence converges.
I know that all convergent sequences are bounded. I also know that the converse is not true in general and a counter example would be [imath]\{1,-1,1,-1, \ldots\}[/imath] How do I properly disprove this? |
2196882 | Ordinals and topological space.
Let [imath]\alpha[/imath] be an ordinal , and define [imath]X_{\alpha} = \alpha \times [0,1)[/imath] to be a topological space with the lexicographic order. Denote by [imath]\omega[/imath] the first countable ordinal and [imath]\Omega[/imath] the first ordinal that is not countable. I need to prove the following : if [imath]\alpha[/imath] is countable then [imath]X_{\alpha}[/imath] is isomorphic to [imath][0,1)[/imath] as ordered set and to conclude that [imath]X_{\alpha}[/imath] is homeomorphic to [imath][0,1)[/imath]. and, for each point in [imath]X_{\Omega}[/imath] except the min. element there is a neighborhood that homeomorphic to [imath]R[/imath]. Im a bit stuck on this question, for the first part i thought sending [imath](n,x)[/imath] to [imath]x[/imath] but this does not work. i tried also sending [imath](n,x)[/imath] to anywhere in the interval [imath](1/(n+1) , 1/n)[/imath] - this also does not work. I would really appreciate a detailed answer , as im not familiar that much with ordinals. Thanks for helping. | 2187586 | Prove that for every countable ordinal [imath]\alpha[/imath], [imath]\alpha\times[0,1)[/imath] is order isomoprhic to [imath][0,1)[/imath]
I need to prove that for every countable ordinal [imath]\alpha[/imath], [imath]X_{\alpha} = \alpha\times[0,1)[/imath] is order isomorphic to [imath][0,1)[/imath], where the order on [imath]X_{\alpha}[/imath] is the lexicographical order. I wanted to go by induction on the countable ordinals. I was able to manually construct a function [imath]f:X_{\omega}\to[0,1)[/imath] that is an order isomorphism, so the base case I have. So, suppose by induction that for some countable ordinal [imath]\alpha[/imath], for all [imath]\beta < \alpha[/imath], [imath]X_{\alpha}\cong [0,1)[/imath]. I need to show that [imath]X_{\alpha+1}\cong[0,1)[/imath]. So my attempt was to use the fact that [imath]X_{\alpha + 1} = \left(\alpha\cup\{\alpha\}\right)\times[0,1)[/imath] and to use the induction hypothesis to get an order isomorphism [imath]f'[/imath] between [imath]\alpha\times[0,1)[/imath] and [imath][0,1)[/imath], and to use the trivial isomorphism [imath]f''[/imath] between [imath]\{\alpha\}\times[0,1)[/imath] and [imath][0,1)[/imath], and then take the union of these functions [imath]f = f'\cup f''[/imath]. But then a problem arises when trying to compare [imath]f(\alpha,r)[/imath] and [imath]f(\gamma,q)[/imath] where [imath]\gamma < \alpha[/imath], since nothing promises me that [imath]f'(\gamma,q) < f''(\alpha,r)[/imath]. I assume after I figure out for successor ordinal, the generalization to a limit ordinal will not be difficult? |
2196474 | Probability that a Wiener process greater than zero at two given moments
What is [imath]P(W_t>0,W_s>0)[/imath]? Since increments are independent it may be helpful to split it into two processes as [imath]W_t=W_{t-s}+W_s[/imath]. Each of it has Gaussian probability density function and I think that pdf of sum of two independent processes may be evaluated as in case of independent random variables: [imath]f_{W_s+W_{t-s}} (t) = \int\limits_{-\infty}^{+\infty} f_{W_s}(s) f_{W_{t-s}}(t-s) ds = \frac{1}{\sqrt{4\pi \sigma^2}}e^{\frac{-(-2\mu+t)^2}{4\sigma^2}} \sim N(2\mu,2\sigma)[/imath]. What should I do next? If I integrate it over the area [imath]0<t<+\infty[/imath] than I will find probability that [imath]W_s+W_{t-s}>0[/imath], but I can't proceed forward. | 1637459 | Probability Brownian motion is positive at two points
Let [imath]0<s<t[/imath] and [imath](B_r)_r[/imath] is Brownian motion. Does anybody know what [imath]P(B_s>0,B_t>0)[/imath] is? I think I remember it was some [imath]arctan[/imath]-law but I don't know the exact form. So I do not need a proof, stating the result would be completely sufficient, but I could not find it using google. Thank you for your help. |
1958762 | Did I derive a new form of the gamma function?
I wish to extend the factorial to non-integer arguments in a unique way, given the following conditions: [imath]n!=n(n-1)![/imath] [imath]1!=1[/imath] To anyone interested in viewing the final form before reading the whole post: [imath]x!=\exp\left[\int_0^x\left(-\gamma+\int_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right][/imath] [imath]f(x):=\ln(x!)[/imath] [imath]f(x)=\ln(x!)=\ln(x)+\ln((x-1)!)=\ln(x)+f(x-1)[/imath] [imath]f(x)=f(x-1)+\ln(x)[/imath] [imath]\frac d{dx}f(x)=\frac d{dx}f(x-1)+\ln(x)[/imath] [imath]f'(x)=f'(x-1)+\frac1x\tag1[/imath] [imath]f'(x)=f'(x-2)+\frac1{x-1}+\frac1x[/imath] [imath]=f'(0)+1+\frac12+\frac13+\dots+\frac1x[/imath] for [imath]x\in\mathbb N[/imath]: [imath]f'(x)=f'(0)+\sum_{n=1}^x\frac1n\tag2[/imath] Euler has a nice extension of the harmonic numbers to non-integer arguments, [imath]f'(x)=f'(0)+\int_0^1\frac{1-t^x}{1-t}dt\tag{2.1}[/imath] from the FTOC we have [imath]\ln(x!)=\int_0^x\left(f'(0)+\int_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi[/imath] [imath]x!=\exp\left[\int_0^x\left(f'(0)+\int_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right]\tag3[/imath] And with [imath]f'(0)=-\gamma[/imath], the Euler mascheroni constant, we should get the gamma function. Or we may just let it sit as an unknown parameter. My questions are if this captures all possible extensions of the factorial with the given conditions, since, if it did, it'd be a pretty good general extension to the factorial? Given a few more assumptions, it is easy enough to set bounds to what [imath]f'(0)[/imath] might be as well. Notably, this representation fails when considering [imath]\Re(x)\le-1[/imath], but coupled with the first condition, it is extendable to all [imath]x[/imath], except of course the negative integers. robjohn♦ notes an extension to the harmonic numbers that converges for [imath]x\in\mathbb C[/imath], except the negative integers: [imath]\int_0^1\frac{1-t^\phi}{1-t}dt=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+\phi}\right)[/imath] Any suggestions on things I could've improved and flaws in this would be nice. Edit: Using the second condition and [imath]x=1[/imath], we may have [imath]1=\exp\left[\int_0^1\left(f'(0)+\int_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right][/imath] [imath]\implies f'(0)=-\int_0^1\int_0^1\frac{1-t^\phi}{1-t}dt\ d\phi[/imath] [imath]f'(0)=-\gamma[/imath] where [imath]\gamma[/imath] is the Euler-mascheroni constant. Using this we get a new form of the gamma function(?): [imath]\boxed{x!=\exp\left[\int_0^x\left(-\gamma+\int_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right]}\tag4[/imath] [imath]=\exp\left[\int_0^x\left(-\gamma+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+\phi}\right)\right)d\phi\right][/imath] I'm not sure how to deal with trivial manipulations of this expression, as surely someone is gonna say "hey, just multiply everything by [imath](1+\sin(2\pi x))[/imath] and it will still satisfy the conditions, right?" But regardless, I think this is a pretty cool new gamma function? Also, references to this if it's not new. If someone could make a graph of this to look at, you would be great. | 2310018 | Why does [imath]\Psi(n)=H_{n-1}-\gamma[/imath]
According to wikipedia: The gamma function obeys the equation [imath]\Gamma(z+1)=z\Gamma(z)[/imath]. Taking the derivative with respect to [imath]z[/imath] gives: [imath]\Gamma'(z+1)=z\Gamma'(z)+\Gamma(z)[/imath] Dividing by [imath]\Gamma(z+1)[/imath] or the equivalent [imath]z\Gamma(z)[/imath] gives: [imath]\frac{\Gamma'(z+1)}{\Gamma(z+1)}=\frac{\Gamma'(z)}{\Gamma(z)}+\frac{1}{z}[/imath] or: [imath]\psi(z+1)=\psi(z)+\frac{1}{z}[/imath] Since the harmonic numbers are defined as [imath]H_n=\sum_{k=1}^n\frac{1}{k}\tag{1}[/imath] the digamma function is related to it by: [imath]\psi(n)=H_{n-1}-\gamma[/imath] where [imath]H_n[/imath] is the nth harmonic number, and [imath]\gamma[/imath] is the Euler–Mascheroni constant. This doesn't make sense to me. The two values subtracted appear to be equal: How does [imath]\psi(z+1)=\psi(z)+\frac{1}{z}\tag{2}[/imath] differ from [imath]\sum_1^n \frac{1}{z}=H_n[/imath] If the sum in (2) doesn't terminate at 1, the definition of the sum is undefined for gamma at negative integer values. If wikipedia is saying non integer values of [imath]\Psi[/imath] of a finite sum i.e. [imath]\frac{1}{1.5}+\frac{1}{2.5}+\frac{1}{3.5}\dots[/imath] differ from [imath]H_n[/imath] by [imath]\gamma[/imath] I'm equally doubtful as [imath]\gamma[/imath] is only equal to infinite sums and I don't see how this value could equal the infinite difference between the integral and sum [imath]\frac{1}{n}[/imath]. What am I not understanding. Please help. Edit: as noted in the comments, could someone explain/link/prove why [imath]\frac{\Gamma'(0)}{\Gamma(0)}=\gamma[/imath] |
2197261 | Determine the equation of the tangent to a curve at a point.
I have the following problem. Determine the equation of the tangent to the curve defined by [imath]f(x)=x^2-6x+14[/imath] at the point [imath](1,9)[/imath], and then sketch it. I'm having trouble with this question as I have no lessons whatsoever on it, the only example is telling me to complete the square to sketch it, and its not working out when I try that. Any help would be appreciated. | 516007 | Find the equation of the tangent line to the curve at the given point. [imath]y = 1+2x-x^3[/imath] at [imath](1,2)[/imath]
I have the equation [imath]y = 1+2x-x^3[/imath] and the point [imath](1,2)[/imath]. When I work it out I come up with the derivative of [imath]2-3x^2[/imath]. When I apply the point I come up with a slope of [imath]-1[/imath] and a tangent line of [imath]y=4-x[/imath]. Can someone work it out and confirm my answer or show me where I am going wrong? |
2197241 | Example of ring satisfying nilpotent condition
Let [imath]R[/imath] be a commutative ring with identity, let [imath]N = \{x|x \in R, x^k=0[/imath] for some natural number k[imath]\}[/imath], and let [imath]L = \{x|x \in R, 1+xr[/imath] invertible for all [imath] r \in R\}[/imath]. I am trying to find an example of a ring R such that [imath]N[/imath] is strictly smaller than [imath]L[/imath]. It is not hard to show that [imath]N \subseteq L[/imath] and so I have reduced this to finding an element in R that is in [imath]L[/imath] but not in [imath]N[/imath]. It seems like [imath]R=[/imath] the real numbers might work, since for example [imath]a \in R[/imath] is in [imath]L[/imath] since [imath](1+ar)(\frac{1}{1+ar})=1[/imath] but [imath]a^k \neq 0[/imath] for any natural number k if [imath]a \neq 0[/imath]. I'd appreciate any help with this since I'm not sure whether or not I'm on the right track here and most of the other rings such as [imath]Z/nZ[/imath] I've considered don't seem to have this property. Thanks! | 634157 | Nilradical strictly smaller than Jacobson radical.
In a preparation question for an exam, I am asked to give an example of a ring [imath]A[/imath] such that the nilradical [imath]\operatorname{Nil}(A)[/imath] is strictly smaller that the Jacobson radical [imath]J(A)[/imath]. Here's how I solved the problem: It is enough to find some ring [imath]A[/imath] with a prime ideal [imath]p[/imath] which is strictly contained in some maximal ideal [imath]m[/imath]. Then the localization [imath]A_m[/imath] is a local ring with maximal ideal [imath]m[/imath], and by the correspondence between ideals of [imath]A_m[/imath] and ideals of [imath]A[/imath] contained in [imath]m[/imath] we have that [imath]J(A)=m\supsetneq p\supset\operatorname{Nil}(A)[/imath]. An example is [imath]A=\mathbb{Z}[x][/imath], [imath]p=(x^2+1)[/imath] and [imath]m=(x^2+1,2)[/imath]. Now I was wondering: are there also any "elementary" examples satisfying the condition above? By this I mean basically examples where I don't have to localize the ring. |
2196438 | Geometric meaning of the [imath]C^1[/imath] norm
Can someone explain the geometric meaning of the [imath]C^1[/imath] norm on [imath]C^1[X][/imath], where [imath]X[/imath] is compact. Thanks in advance. | 148349 | Open and closed balls in [imath]C[a,b][/imath]
Let [imath]X[/imath] be a non empty set and let [imath]C[a,b][/imath] denote the set of all real or complex valued continuous functions on [imath]X[/imath] with a metric induced by the supremum norm. How to find open and closed balls in [imath]C[a,b][/imath]? Can we see them geometrically? For example what is an open ball [imath]B(x_0;1)[/imath] i.e. ball centered at [imath]x_0[/imath] with radius [imath]1[/imath] in [imath]C[a,b][/imath]. I can visualize them in [imath]\mathbb R^n[/imath] but when it comes to functional spaces I have no clue how to identify them? Thanks for helping me. |
2196571 | How can I prove this limit?
This is the problem I have to solve, I tried many times using the definition, but the result is not the one required. Let [imath](X_n)_{n\in \Bbb N}[/imath] be a sequence of independent and identically distributed random variables with [imath] \Bbb P(X_i=0)=p[/imath] and [imath] \Bbb P(X_i=1)=1-p[/imath] for all [imath] i \in \Bbb N [/imath]. Show that for [imath] p \in [0,1) [/imath] [imath] \lim_{n\to \infty} {\sum_{i=1}^n X_i \over \sum_{i=1}^n X^2_i } = 1 [/imath] almost surely. Thanks in advance | 2192863 | Show [imath]\lim_{n \to \infty} \sum_{i=1}^n Y_i/\sum_{i=1}^n Y_i^2 = 1[/imath] for Bernoulli distributed random variables [imath]Y_i[/imath]
I want to show that: [imath]\lim_{n \rightarrow \infty} \frac{\sum_{i=1}^{n} Y_i}{\sum_{i=1}^{n} Y_i^2} = 1[/imath] almost surely given that [imath](Y_n)[/imath] is a sequence of independent and identically distributed random variables, [imath]P(Y_i = 0) = a, P(Y_i = 1) = 1-a[/imath]. I am confused about this 'almost surely'. What is a sequence of random variables? I guess somehow I can maybe convert the sum into a product and then use the independence...Any hints? Edit: found out that 'almost 'surely' means that the probability of it is 1. Also, in our case we actually just have an IID. |
2196459 | Proving the following equality
I have to prove the following equality for a nonnegative integer-valued random variable X and I do not know where to start. I tried few times but everytime it makes no sense. [imath]\sum_{i=0}^\infty i \Bbb P (X>i)= \frac 12 (\mathbf E(X^2) - \mathbf E(X))[/imath] Thanks in advance | 2196142 | Proof that [imath]\sum\limits^{\infty}_{i=0}{iP(X > i)} = \frac12(E(X²)-E(X))[/imath] for [imath]X[/imath] nonnegative and integer valued
The following is a homework example: For a nonnegative integer-valued random variable [imath]X[/imath], show that [imath]\sum^{\infty}_{i=0}{iP(X > i)} = 1/2(E(X²)-E(X)).[/imath] The only way I could think of doing this proof is to rewrite the left side starting with: [imath](1).[/imath] [imath]\sum^{\infty}_{i=0}{i} \sum^{\infty}_{j=i+1}{P(X=j}) = 1/2(E(X²)-E(X)).[/imath] and continue from there to get to an expression similar to: [imath](2).[/imath] [imath]1/2 * [\sum^{\infty}_{i=0}{i²*P(X=i)} - \sum^{\infty}_{i=0}{i*P(X=i))}] = 1/2(E(X²)-E(X)).[/imath] However I don't know how to continue to transform [imath](1)[/imath] into [imath](2)[/imath]. Could you give me a hint there? Is the approach even correct? |
2197529 | Testing end points of the interval of convergence of a series.
Find the interval of convergence including endpoints of [imath]\sum_{n=0}^{\infty} \frac{n(x+3)^n}{2^n(n^2+1)}.[/imath] I can find the interval of convergence but I don't know how to test endpoints [imath]x=-5[/imath] and [imath]x=-1[/imath]. I don't know whether the series convergences at [imath]x[/imath] = [imath]-5[/imath] and [imath]x[/imath]=[imath]-1[/imath] and need help. I don't think the series converges at [imath]x= -5[/imath] and [imath]x= -1[/imath]. No this is not a duplicate question because the starting index of the series is different. | 2197398 | Test whether the endpoints of the interval of convergence should be included in the interval of convergence
Find the interval of convergence including endpoints of [imath]\sum_{n=1}^{\infty} \frac{n(x+3)^n}{2^n(n^2+1)}.[/imath] I can find the interval of convergence but I don't know how to test endpoints [imath]x=-5[/imath] and [imath]x=-1[/imath]. I don't know whether the series convergences at [imath]x[/imath] = [imath]-5[/imath] and [imath]x[/imath]=[imath]-1[/imath] and need help. I don't think the series converges at [imath]x= -5[/imath] and [imath]x= -1[/imath]. |
2197791 | Intersection of neighborhood of identity is closure of identity.
Let [imath](G, .)[/imath] be a group. Further assume that [imath]G[/imath] is endowed with some topology such that the map [imath]G × G → G[/imath] given by [imath](x, y) → xy[/imath] and the map [imath]G → G[/imath] given by [imath]x → x ^{−1}[/imath] are continuous. In this case [imath](G, .)[/imath] is said to be a topological group wrt the topology on it. Let [imath]e[/imath] be the identity element of [imath]G[/imath]. Let [imath]H_0[/imath] denote intersection of all neighbourhoods of identity. Prove that it is closure of [imath]{e}[/imath]. Attempt : [imath]e \in H_0[/imath] If [imath]x \in H_0=(\cap U_\alpha[/imath] where [imath]U_\alpha[/imath] are neighbourhoods of identity), then [imath]x \in U_\alpha, \forall \alpha[/imath] , so [imath]x \in \bar e[/imath] , [imath]\bar e =[/imath] closure of identity [imath]{e}[/imath], so [imath]H_0 \subset \bar e[/imath]. Take [imath]y \in \bar e \setminus H_0[/imath]. Then [imath]y \in \bar e[/imath] but [imath]\exists U_\beta[/imath] such that [imath]y \notin U_\beta[/imath]. I'm done if I contradict this isn't it? Now , since [imath]y \in \bar e[/imath], any nbhd of [imath]y[/imath] say [imath]U_y \cap \{e\} \neq \emptyset[/imath], so, [imath]\{e\} \in U_y \cap U_\beta[/imath]. So [imath]\exists[/imath] [imath]V \subset U_y \cap U_\beta[/imath] open which is a nbhd of [imath]e[/imath]. I don't know if I am doing it correct and how to proceed. | 2039907 | Closure of identity element in a topological group is the intersection of its neighborhood system
Consider a topological group [imath]G[/imath]. I want to prove that if [imath]\cal{U}[/imath] is the set of all neighborhoods of [imath]e[/imath] (the identity element), then [imath]\overline{\{e\}}=\cap_{U\in \cal{U}}U[/imath]. What I thought was: i) If [imath]x\in \overline{\{e\}}[/imath], then every neighborhood [imath]V[/imath] of [imath]x[/imath] intersects [imath]\{e\}[/imath], hence every [imath]V[/imath] is a neighborhood of [imath]e[/imath] as well, i.e. [imath]V\in \cal{U}[/imath]. ii) On the other hand, if [imath]x\in \cap_{U\in \cal{U}}U[/imath], then [imath]x\in U[/imath] for all [imath]U \in \cal{U}[/imath], meaning every [imath]U[/imath] is a neighborhood of [imath]x[/imath] intersecting [imath]\{e\}[/imath]. Now, I think that the missing part is an argument justifying that there is no neighborhood of [imath]e[/imath] that is not a neighborhood of [imath]x[/imath] (part i)), and that there is no neighborhood of [imath]x[/imath] that is not a neighborhood of [imath]e[/imath] (part ii)). My idea for this is to argue that there is a homeomorphism taking [imath]e[/imath] to [imath]x[/imath], therefore the neighborhood system of one cannot have more elements that the neighborhood system of the other. Is this correct? If not, is there a way I can fix it? |
2198280 | A function that is differentiable on irrationals but discontinuous on rationals
Can I construct a function [imath]f:(0,1)\to\mathbb R[/imath] that is discontinuous on [imath]\mathbb Q\cap(0,1)[/imath], but [imath]\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{|h|^k}=0[/imath] holds for all [imath]x\in [0,1]\backslash\mathbb Q[/imath] and [imath]k>0[/imath] I try [imath]f=\sum_{n=1}^\infty\frac1{2^n}\chi_{(0,q_n]}[/imath] where [imath]\{q_n\}_{n=1}^\infty[/imath] is the sequence of rationals on [imath](0,1)[/imath]. But it doesn't seem to be differential, even when the converge speed is faster than [imath]\frac1{2^n}[/imath]. | 2115 | Discontinuous at rationals and differentiable at irrationals?
We know that there exist real functions which are continuous at each irrational and dis- continuous at each rational number. But does there exist a function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] that is differentiable at every irrational and discontinuous at every rational? |
2198626 | Proof involving homomorphism between [imath]\Bbb Z^n[/imath] and an abelian group G
Let [imath](e_1, ..., e_n)[/imath] be the standard [imath]\Bbb Z[/imath]-basis of [imath]\Bbb Z^n[/imath]. Let [imath]x_1,..., x_n[/imath] be elements of an abelian group [imath]G[/imath]. I want to prove that there exists a homomorphism [imath]f[/imath] [imath]:[/imath] [imath]\Bbb Z^n[/imath] [imath]\rightarrow[/imath] [imath]G[/imath] such that [imath]f[/imath][imath](e_i)[/imath] [imath]=[/imath] [imath]x_i[/imath] for all [imath]i[/imath]. Please can anyone lend a hand here? | 2193980 | Question on group homomorphisms involving the standard Z-basis
Let [imath](e_1, ..., e_n)[/imath] be the standard [imath]\Bbb Z[/imath]-basis of [imath]\Bbb Z^n[/imath]. Let [imath]x_1,..., x_n[/imath] be elements of an abelian group [imath]G[/imath]. Prove that there exists a homomorphism [imath]f[/imath] [imath]:[/imath] [imath]\Bbb Z^n[/imath] [imath]\rightarrow[/imath] [imath]G[/imath] such that [imath]f[/imath][imath](e_i)[/imath] [imath]=[/imath] [imath]x_i[/imath] for all [imath]i[/imath]. Can anyone please help me out here? |
1184445 | Torus diffeomorphic to [imath]S^1\times S^1[/imath].
This is an exercise from Guillemin/Pollack's Differential Topology. In a previous exercise, I'm asked to give a complete set of parametrizations of [imath]S^1\times S^1[/imath], which I've succeeded in (I think) by the likes of [imath]\begin{align}\varphi_1(u,v) &= (u,\sqrt{1-u^2},v,\sqrt{1-v^2}) \\ \varphi_2(u,v) &= (u,-\sqrt{1-u^2},v,\sqrt{1-v^2}) \end{align} \\ \vdots\\ \text{etc.}[/imath] (defined over [imath][0,1]^2).[/imath] Then, the torus [imath]\Bbb T_{a,b}[/imath] (my notation) is defined as the set of points in [imath]\Bbb R^3[/imath] that are at distance [imath]b < a[/imath] from the circumference in the plane [imath]\{z = 0\}[/imath] of radius [imath]a[/imath] centered at the origin. The exercise is to prove that [imath]\Bbb T_{a,b}[/imath] is diffeomorphic to [imath]S^1\times S^1[/imath]. I'm very stuck because I can't seem to characterize neighborhoods of points in [imath]\Bbb T_{a,b}[/imath], can someone help? Answers or just hints are welcome (but I might ask follow up questions in the latter case :) ). | 414679 | Prove that a standard torus is diffeomorphic to [imath] \mathbb S^1\times \mathbb S^1[/imath]
I was asked to prove that a standard torus(which means we don't consider those pathological cases where it intersects with itself, e.g horn torus) is diffeomorphic to [imath] \mathbb S^1\times \mathbb S^1[/imath]. I was thinking if we could prove it this way: Since every point on the torus can be uniquely defined with a pair of angles [imath](\theta_1, \theta_2)[/imath]. Then we construct a diffeomorphism [imath]\phi(\theta_1, \theta_2)=(\tilde{\theta}_1 ,\tilde{\theta}_2)[/imath] which maps every point on the torus to every point on [imath]\mathbb S^1 \times \mathbb S^1[/imath]. Since the map is definitely bijective and smooth with a smooth inverse. We're basically done... THERE MUST BE SOMETHING WRONG I THINK. Thanks a lot for everyone's help! |
2199704 | The mathematical notation for a subset with given conditions
Assuming I have a set [imath]A[/imath] of numbers, I'd like to mathematically notate a subset [imath]B[/imath], such that the cardinality of this subset is [imath]x[/imath] and its elements are the greatest [imath]x[/imath] elements of [imath]A[/imath]. So, how can I mathematically notate this? Thank you very much for any help. | 463502 | Mathematical expression for [imath]n[/imath] largest values in a set
I would like to know how to express mathematically [imath]n[/imath] largest values of a given set [imath]A[/imath]. For example: I have an unsorted set [imath]A[/imath] with [imath]100[/imath] values [imath]\in \mathbb{N}[/imath] and want to get a new set [imath]B[/imath] with the [imath]5[/imath] largest values. Thank you in advance! |
2198807 | Solve: [imath]\overline{z}=z^n[/imath]
[imath]\overline{z}=z^n[/imath] where [imath]n\in \mathbb{N}[/imath] So I have started with moving the polar representation as the expression is in the n-th power [imath]rcis(-\theta)=r^ncis(n\theta)[/imath] I can not multiply both sides in [imath]r^{(-n)}[/imath] as [imath]n\in \mathbb{N}[/imath] how should I continue? | 1547451 | How would I prove that if an integer [imath]n>2[/imath], then [imath]\bar{z}=z^{n-1}[/imath] has [imath]n+1[/imath] solutions
If [imath]n[/imath] is an integer and [imath]n>2[/imath], then [imath]\bar{z}=z^{n-1}[/imath] has [imath]n+1[/imath] solutions. Does this have something to do with the rational root theorem? |
2199743 | Is the extension [imath] \mathbb{Q}(\sqrt{2}, \sqrt{3}) [/imath] simple?
Intuitively, it is not. Can anyone give a proof? | 1311598 | Show [imath]\mathbb{Q}(\sqrt{2},\sqrt{3})[/imath] is a simple extension field of [imath]\mathbb{Q}[/imath].
Show [imath]\mathbb{Q}(\sqrt{2},\sqrt{3})[/imath] is a simple extension field of [imath]\mathbb{Q}[/imath]. Since [imath]\sqrt{2},\sqrt{3}\in \mathbb{Q}(\sqrt{2},\sqrt{3})[/imath], and [imath]\sqrt{2}\sqrt{3}=\sqrt{6}\in \mathbb{Q}(\sqrt{2},\sqrt{3})[/imath], so [imath]\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{6})[/imath], hence [imath]\mathbb{Q}(\sqrt{2},\sqrt{3})[/imath] is a simple extension field of [imath]\mathbb{Q}[/imath]. Does my argument right? The only information I have is : A field extension [imath]E[/imath] of [imath]F[/imath] is called a simple extension if [imath]E = F(\alpha)[/imath] for some [imath]\alpha \in E[/imath]. |
2198414 | Inequalities involving Absolute Values
[imath]\forall \ a, b \in \mathbb{R}, prove \hspace{5mm}\frac{|a+b|}{1 + |a+b|} \leqslant \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}[/imath] I decided to take different cases to solve this problem. For [imath]a, b \geqslant 0[/imath] and for [imath]a, b < 0[/imath], the inequality is obvious as [imath]\frac{|a+b|}{1 + |a+b|} = \frac{a}{1+|a+b|} + \frac{b}{1+|a+b|}[/imath] when [imath]a, b \geqslant 0[/imath] and [imath]\frac{|a+b|}{1 + |a+b|} = \frac{-a}{1+|a+b|} + \frac{-b}{1+|a+b|}[/imath] when [imath]a, b < 0[/imath] For [imath]a \geqslant 0, b < 0[/imath] and [imath]|a| > |b|[/imath], I was able to prove the inequality. However, I'm unable to prove in the case where [imath]a \geqslant 0, b < 0[/imath] and [imath]|a| < |b|[/imath]. | 194314 | Prove [imath]\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}[/imath].
Prove that for every [imath]a, b\in R\setminus\{0\}[/imath] is correct this inequality: [imath]\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.[/imath] |
2200147 | Find an equation of the plane. New
The plane that passes through the point [imath](−1,1,2)[/imath] and contains the line of intersection of the planes [imath]x+y−z=5[/imath] and [imath]2x−y+3z=1[/imath]. | 137629 | Find equation of a plane that passes through point and contains the intersection line of 2 other planes
Find equation of a plane that passes through point P [imath](-1,4,2)[/imath] that contains the intersection line of the planes [imath]\begin{align*} 4x-y+z-2&=0\\ 2x+y-2z-3&=0 \end{align*}[/imath] Attempt: I found the the direction vector of the intersection line by taking the cross product of vectors normal to the known planes. I got [imath]\langle 1,10,6\rangle[/imath]. Now, I need to find a vector normal to the plane I am looking for. To do that I need one more point on that plane. So how do I proceed? |
2200973 | Two sets which contain each other as elements
In Zermelo-Fraenkel set theory do there exist sets [imath]a,b[/imath] such that [imath]a \in b[/imath] and [imath]b \in a[/imath]. I think that no such sets exist but I am not sure how to prove why this is the case. | 127297 | Cyclic containment of sets
Is it possible in ZFC that you have a cyclic containment of sets, e.g., a inclusion like [imath]A \in B[/imath] and [imath]B \in A[/imath]? I never took set theory classes, I am just curious. |
2201185 | What's the fastest way to determine all the subgroups of the additive group [imath]\mathbb{Z}_{24}[/imath]
Question is as in title. I know that all of the subgroups of [imath]\mathbb{Z}_{24}[/imath] (under addition) must be cyclic, and I could find them by finding the generating groups for each element of [imath]\mathbb{Z}_{24}[/imath] - but surely there is a quicker way? Would appreciate any help, Jack | 823875 | Subgroups of [imath](\mathbb Z_n,+)[/imath]
The problem is to define all subgroups of [imath](\mathbb Z_n,+), n \in \mathbb N[/imath]. My guess is if n is prime number, then there is only trivial subgroups. If n is not prime, then I can factorize it, and every prime divisor will generate it's own subgroup in [imath](\mathbb Z_n,+)[/imath]. That means that [imath](\mathbb Z_n,+) \cong (\mathbb Z_h,+) \times (\mathbb Z_k,+) \times \dots[/imath] , [imath]h,k \in \mathbb N[/imath] are the prime factors of [imath]n[/imath]. The problem is that I don't know how to prove it. It's pretty easy to show that, for example, in [imath](\mathbb Z_6,+)[/imath] [imath] [2]_6[/imath] and [imath][3]_6[/imath] generate their own subgroups and [imath][1]_6[/imath] generate entire [imath](\mathbb Z_6,+)[/imath], but I don't know how to show that [imath][5]_6[/imath] do the same, except by showing it all: [imath][5]^2_6 = [4]_6[/imath] and so on. |
1902184 | Which entire functions [imath]h(z)[/imath] can be written as [imath]h(z)=f(z+1)-f(z)[/imath] for some entire function [imath]f[/imath]?
Question: For which entire functions [imath]h(z)[/imath] does there exist an entire function [imath]f(z)[/imath] such that [imath]h(z)=f(z+1)-f(z)[/imath]? What I have tried: Suppose that [imath]f:\mathbb{C}\to\mathbb{C}[/imath] is an entire function, and let [imath]\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n[/imath] be its Taylor series expansion. Then [imath]\displaystyle f(z+1)=\sum_{n=0}^\infty a_n\sum_{i=0}^n\binom{n}{i}z^i=\sum_{n=0}^\infty z^n\left[\sum_{j=n}^\infty a_j\binom{j}{n}\right].[/imath] Therefore [imath]f(z+1)-f(z)=\sum_{n=0}^\infty z^n\left[\sum_{j=n+1}^\infty a_j\binom{j}{n}\right].[/imath] For [imath]\{a_n\}\in\mathbb{C}^\infty[/imath], define [imath]c_n=\displaystyle\sum_{j=n+1}^\infty a_j\binom{j}{n}[/imath], if this sequence converges. If [imath]\{a_n\}[/imath] is a sequence for which each [imath]c_n[/imath] converges, then define [imath]\Pi(\{a_n\})=\{c_n\}[/imath]. Lets let [imath]\mathscr{H}[/imath] denote the collection of all sequences of complex numbers [imath]\{a_n\}[/imath] such that [imath]\Pi(\{a_n\})[/imath] is well-defined. Let [imath]\mathscr{H}_e[/imath] denote the collection of sequences such that [imath]\displaystyle\sum_{n=0}^\infty a_nz^n[/imath] has infinite radius of convergence. It is not hard to see from the above work that [imath]\mathscr{H}_e\subset\mathscr{H}[/imath]. Let [imath]\mathscr{H}_0\subset\mathscr{H}_e[/imath] be the collection of finite sequences (ie those corresponding to polynomials). I know that [imath]\Pi:\mathscr{H}_0\to\mathscr{H}_0[/imath] is surjective. I want to know what [imath]\Pi(\mathscr{H}_e)[/imath] is (I now know that [imath]\Pi:\mathscr{H}_e\to\mathscr{H}_e[/imath] is not surjective as noted in my comment below). Bonus question: If we mod out [imath]\mathscr{H}[/imath] by the relation [imath]\{a_n\}\sim\{b_n\}[/imath] if and only if [imath]a_k=b_k[/imath] for each [imath]k>0[/imath], then [imath]\Pi:\mathscr{H}_0\to\mathscr{H}_0[/imath] is injective. Is [imath]\Pi:\mathscr{H}\to\mathscr{H}[/imath] injective when modded out similarly? | 1174799 | Existence of an holomorphic function
Is there a simple way to prove this fact : For all holomorphic functions [imath]f : \mathbb C \to \mathbb C[/imath], there is an holomorphic function [imath]\psi : \mathbb C \to \mathbb C[/imath] such that [imath]\psi(z+1) = \psi(z) + f(z) [/imath] The solution I know use Galois covering spaces and summand of automorphy. Thanks in advance. |
2201835 | A tensor of rank [imath]3[/imath] satisfies [imath]T_{ijk}=T_{jik}[/imath] and [imath]T_{ijk}=-T_{ikj}[/imath]. Show that [imath]T_{ijk}=0[/imath]
A tensor of rank [imath]3[/imath] satisfies [imath]T_{ijk}=T_{jik}[/imath] and [imath]T_{ijk}=-T_{ikj}[/imath]. Show that [imath]T_{ijk}=0[/imath] I really don't know what we can use to show this. All I am able to do is rearrange the dummy indices to get a few more equations but they're of no use. Any help is appreciated. Thanks | 1257470 | Tensors which are symmetric and antisymmetric in overlapping groups
Say I have the following tensor [imath]T_{abc}[/imath] such that [imath] T_{(a[b)c]} [/imath] Ergo, it is symmetric in indices [imath]a[/imath] and [imath]b[/imath] and antisymmetric in [imath]b[/imath] and [imath]c[/imath]. Keeping in mind the various properties that (anti-)symmetry entails such as switching and the like, can [imath]a[/imath] and [imath]c[/imath] actually be switched around? I am trying to show that tensors having such a property vanish, but I get stuck when I get [imath]T_{bac}[/imath]. |
262410 | Find all polynomials that fix [imath]\mathbb Q[/imath] and the irrationals
Problem: Describe all polynomials [imath]\mathbb{R}\rightarrow\mathbb{R}[/imath] with coefficients in [imath]\mathbb C[/imath] which send rational numbers to rational numbers and irrational numbers to irrational numbers. | 2202087 | [imath]f(x)=ax+b[/imath] for some [imath]a,b\in\mathbb{Q}[/imath] if [imath]f(\mathbb{Q})\subset\mathbb{Q}[/imath] and [imath]f(\mathbb{R-Q})\subseteq \mathbb{R}-\mathbb{Q}[/imath]
let [imath]f[/imath] be a polynomial function such that [imath]f(\mathbb{Q})\subset \mathbb{Q}[/imath] and [imath]f(\mathbb{R}-\mathbb{Q}) \subset \mathbb{R}- \mathbb{Q}[/imath] . Prove that [imath]f(x)=ax+b[/imath] for some [imath]a,b\in\mathbb{Q}[/imath] My approach : I started by defining [imath]g(x)=f(x)-f(0)[/imath] . Now [imath]g(x)[/imath] satisfies the properties [imath]g(\mathbb{Q})\subset \mathbb{Q}[/imath] , [imath]g(\mathbb{R}-\mathbb{Q})\subset\mathbb{R}-\mathbb{Q}[/imath] and [imath]g(0)=0[/imath] . [imath]g[/imath] is a polynomial and [imath]g(0)=0[/imath] . [imath]0[/imath] is root of [imath]g[/imath] and hence [imath]g(x)=x h(x)[/imath] for some polynomial [imath]h(x)[/imath] . It will be enough if one can prove that [imath]h(x)[/imath] is constant . I'm stuck here . Any form help will be highly appreciated. |
2201328 | If [imath]a=(12,5)[/imath] and [imath]b=(6,8)[/imath] give two orthogonal vectors [imath]U_1[/imath] and [imath]U_2[/imath] such that [imath]U_1[/imath] lies in [imath]a[/imath] and [imath]U_1+U_2=b[/imath]
If [imath]a=(12,5)[/imath] and [imath]b=(6,8)[/imath] give two orthogonal vectors [imath]U_1[/imath] and [imath]U_2[/imath] such that: [imath]1)[/imath] [imath]U_1[/imath] lies in [imath]a[/imath] [imath]2)[/imath] [imath]U_1+U_2=b[/imath] I am not really sure what the problem is even asking and I would like some assistance in understanding what the problem is asking about. I have a vague idea on the problem as it is supposed to require some sort of usage of projections but I do not know how projections apply in this problem. | 2175335 | If [imath]a=\langle12,5\rangle[/imath] and [imath]b=\langle6,8\rangle[/imath], give orthogonal vectors [imath]u_1[/imath] and [imath]u_2[/imath] that [imath]u_1[/imath] lies on a and [imath]u_1+u_2=b[/imath]
Question If [imath]a=\langle12,5\rangle[/imath] and [imath]b=\langle6,8\rangle[/imath], give orthogonal vectors [imath]u_1[/imath] and [imath]u_2[/imath] that [imath]u_1[/imath] lies on a and [imath]u_1+u_2=b[/imath] My steps: I am not sure whether or not [imath]u_1[/imath] has to lie in the same direction as vector a. but I know that it has to intersect it then I recognize that [imath]u_1[/imath] and [imath]u_2[/imath] has to lie perpendicular on vector b, but I do not know what to do from here a detailed explanation is much appreciated. |
2202646 | Countability of Gaussian Integers
I am attempting to show that Gaussian Integers are countable. Is it valid to map [imath]a + bi[/imath] to an ordered pair [imath](a, b)[/imath] and then map this to the set of rationals [imath]a/b[/imath] ? I am unsure if this works since [imath]a/b[/imath] is not defined at [imath]b = 0[/imath] and am unsure of a different way to go about this. Any hints welcome, thanks! | 2196023 | Proving Gaussian Integers are countable
I know that Gaussian Integers are a subset of complex numbers. They are numbers in the form [imath]$G = \{a + ib \,\vert\, a,b \in \mathbb{Z}\}$[/imath] So to prove that a set is countable, I need to find a function [imath]$G \rightarrow \mathbb{N}$[/imath] such that every [imath]$n$[/imath] has finitely many preimages (from tricki.org). How should I go about proving that it is countable? I don't really understand what preimages mean. |
2200776 | Given [imath]a_1,\dots ,a_n\in V[/imath], the set [imath]W[/imath] of all linear combinations of [imath]a_1,\dots ,a_n[/imath] is a subspace of [imath]V[/imath].
I have to prove the following: Given [imath]a_1,\dots ,a_n\in V[/imath], the set [imath]W[/imath] of all linear combinations of [imath]a_1,\dots ,a_n[/imath] is a subspace of [imath]V[/imath]. I'm having a little bit of trouble because the two notions seems to be the same, but I guess the solution is as follows: From the definition of vector space, for each [imath]a,b\in V[/imath], [imath]a+b\in V[/imath] and [imath]\alpha a\in V[/imath]. Then for all [imath]\alpha_1,\alpha_2, \dots \alpha_n [/imath] we have [imath]\alpha_1 a_1+ \alpha_2a_2+ \dots \alpha_na_n \in V [/imath] as a consequence of successive aplications of [imath]1[/imath] and we have [imath]\alpha_1 a_1+ \alpha_2a_2+ \dots \alpha_na_n\in W[/imath] by definition. From this, we obtain that [imath]V=W[/imath] and hence, as [imath]V[/imath] is a subspace of [imath]V[/imath], then [imath]V[/imath] is a subspace of [imath]W[/imath]. Is my proof correct? | 2162249 | Let [imath]W[/imath] be the set of all linear combinations of [imath]v_1,...,v_n[/imath]. Then [imath]W[/imath] is a subspace of [imath]V[/imath].
Let [imath]W[/imath] be the set of all linear combinations of [imath]v_1,...,v_n[/imath]. Then [imath]W[/imath] is a subspace of [imath]V[/imath]. Proof. Let [imath]y_1,...,y_n[/imath] be numbers. Then [imath](x_1 v_1+...+x_n v_n)+(y_1 v_1+...+y_n v_n)=(x_1+y_1)v_1+...+(x_n+y_n)v_n[/imath]. Thus the sum of two elements of [imath]W[/imath] is again an element of [imath]W[/imath]. Furthermore, if [imath]c[/imath] is a number, then [imath]c(x_1 v_1+...+x_n v_n)=cx_1 v_1+...+cx_n v_n[/imath] is a linear combination of [imath]v_1,...,v_n[/imath], and hence is an element of [imath]W[/imath]. Finally, [imath]\mathbb{0}=0v_1+...+0v_n[/imath] is an element of [imath]W[/imath]. This proves that [imath]W[/imath] is a subspace of [imath]V[/imath]. My question is: How could writer say that'the sum of two elements of [imath]W[/imath] is again an element of [imath]W[/imath]'? Also, How could writer say that [imath]v[/imath] is closed under scaalar multiplication. Can you explain clearly? |
2114991 | Algebraic Multiplicity of Eigen Value of Matrix and Its Conjugate
Suppose [imath]A∈M_n(\mathbb{C})[/imath] is a nonsingular matrix and [imath]\bar{A}[/imath] is conjugate of [imath]A[/imath], i.e. entries of [imath]\bar{A}[/imath] is obtained by replace entries of [imath]A[/imath] with their conjugate. Let [imath]\lambda[/imath] be the real negative eigenvalue of [imath]A\bar{A}[/imath]. Prove the algebraic multiplicity of [imath]\lambda[/imath] is even. I've tried it if every entries of [imath]A[/imath] are real, the algebraic multiplicity of [imath]\lambda[/imath] is [imath]0[/imath]. But I can't generalize it if some entries of [imath]A[/imath] are not pure real or pure imaginary. | 135954 | Multiplicity of eigenvalues
Suppose [imath]A[/imath] is an [imath]n\times n[/imath] complex matrix. How to show the following two properties If [imath]\lambda[/imath] is an eigenvalue of [imath]A\bar{A}[/imath], so is [imath]\bar{\lambda}[/imath]. Here [imath]\bar{A}[/imath] means the entrywise conjugate of [imath]A[/imath]. The algebraic multiplicity of negative eigenvalues (if any) of [imath]A\bar{A}[/imath] are even. |
2201467 | Help understanding proof. Show that [imath]\varphi(n)=\frac{1}{2}n[/imath] if and only if [imath]n=2^{k}[/imath] for some [imath]k \geq 1[/imath]
I have this proof - I can do the second direction (starting with "on the other hand") and I understand some of the logic in this first direction...but I can't figure out WHY he did the first step? I want to know mostly why "we can write [imath]n=2^{k}p^{m}[/imath]" at the beginning? Also, why does the non-existence of such [imath]p[/imath] imply the result? Proof: Firstly, let [imath]\varphi(n)=\frac{1}{2}n[/imath], in which case we can write [imath]n=2^{k}p^{m}[/imath] so that [imath]\varphi(2^{k}p^{m})(1=\frac{1}{2})(1-\frac{1}{p})=\frac{1}{2}2^{k}p^{m}[/imath]. But, when attempting to solve for [imath]p[/imath] in the factor [imath](1-\frac{1}{p})=1[/imath], we find that we want to find [imath]p[/imath] such that[imath]-\frac{1}{p}=0[/imath], for which there exists no number [imath]p[/imath]. So, [imath]n[/imath] must contain only [imath]2[/imath]s. Thus, [imath]n=2^{k}[/imath] for some [imath]k[/imath]. On the other hand, let [imath]n=2^{k}[/imath] for some integer [imath]k[/imath]. Then we can calculate [imath]\varphi(n)=2^{k}(1-\frac{1}{2})=2^{k}(\frac{1}{2})=\frac{1}{2}n [/imath] | 171769 | [imath]\phi(n)=\frac{n}{2}[/imath] if and only if [imath]n=2^k[/imath] for some positive integer k
Show that [imath]\phi(n)=\frac{n}{2}[/imath] if and only if [imath]n=2^k[/imath] for some positive integer k. I think I have it figured and would like to see if I am on the right track. Thank you. |
1534500 | Integral/Derivation of Modulo/Greatest Integer
RELEVANT DEFINITIONS We define the floor of a real number [imath]$x$[/imath] to be the unique integer [imath]$n = \lfloor x \rfloor$[/imath] such that [imath]$n \leq x < n$[/imath]. We define the modulo of a real number [imath]$x$[/imath] with respect to a real number [imath]$w$[/imath] to be the real number value [imath]$y = x \% w = x - w \lfloor \frac {x}{w} \rfloor$[/imath]. MATHEMATICAL CONTEXT Integration is a useful tool by which to analyze functions. The derivative is similarly a useful tool as well. Determining a means by which to integrate and differentiate functions that are trickier than most functions or requiring a different skill set than normally taught in a calculus course is therefore beneficial to others. POTENTIAL USAGE The modulo function or "saw tooth wave" (as some sites such as Wolfram Alpha appear to call it) can be used to represent discrete wave functions which are sometimes used in signal processing. I'm not entirely sure what the floor function might be useful for in the real world other than for approximating functions. For example, [imath]$x \lfloor x \rfloor$[/imath] approximates the general shape of [imath]$x^2$[/imath]. Therefore it stands to reason that there are cases for which clever replacement of [imath]$x$[/imath] with [imath]$\lfloor x \rfloor$[/imath] (or something similar) may be useful in coming up with a decent approximation of a function that could potentially be integrated more easily than KNOWLEDGE LEVEL At the time I made this post according to the date I was in a calculus 2 course that covered your standard integration methods and other single dimension function related things such as polar coordinate integrals. There was defintiely no interaction between that course or it's predecessor with the two titular functions. As for my knowledge level at this time... if I were asking this question now I wouldn't be asking it at all because I'd already be able to figure it out myself. REASON FOR ASKING Nothing ever talked about how to deal with floor or modulo when combined with integration/differentiation, so I wanted to know how. THE PROBLEM STATEMENT Let [imath]$f(x)$[/imath] and [imath]$g(x)$[/imath] be functions of [imath]$x$[/imath]. For simplicity's sake and because of the knowledge level being considered let us also assume they can be written as an expression of [imath]$x$[/imath], [imath]$+$[/imath], [imath]$\cdot$[/imath], [imath]$\^$[/imath], [imath]$\ln$[/imath], [imath]$\cos$[/imath], and [imath]$\sin$[/imath]. We don't want them to be anything real unusual, but an answer is allowed to treat them as continuous differentiable and integrable functions if that makes the answer simpler (for example if the reasoning generalizes in some fashion). Is there a reasonably simple identity that allows one to find the solutions to: [imath]$\int \lfloor f(x) \rfloor dx$[/imath] [imath]$\int f(x) \% f(y) dx$[/imath] [imath]$\frac {d}{dx} (f(x) \% f(y))$[/imath] [imath]$\frac {d}{dx} (f(x))$[/imath] If there isn't such a rule or identity or integration method, could someone explain why? Is it simply because they aren't useful enough to warrant messing with and therefore the knowledge base regarding these functions is limited at best, or is there something deeper here that makes the existence of such an identity impossible? MY ATTEMPT(S) I honestly had no idea how to solve this when I originally asked it. I recall that in the original post was some really horrible hand-waved reasoning about asymptotes and such that I think ultimately just boiled down to "I can do this if the number of discontinuities work out to being finite because then I just know the answer" - that's equivalent to saying I can find some integrals in very niche specific cases by flat out guessing. While that's a valid way to find integrals in the real world if you're dealing with a problem that you might be able to make an educated guess about, it is not real helpful here as an attempt. That's why I'm not rewriting it. As for an attempt now... I think the expiration date has already passed on that. Providing an attempt 4 years later (actually a solution) would be best done in an answer. | 1610068 | Conjectures Involving floor(x)/piecewise continuity with regards to Integration and Differential Equations
[This answer has been heavily edited in response to a long chain of comments on Eric Stucky's answer.] I came up with these few theorems and I am curious whether or not my hypothesis are true. I don't really have a method for proving these, mind you. I'm more considering these as things I've noticed to be true. I just don't know whether they are always true. Some of these are definitions to which I will append "def." to the end of the title. Think of these as a given. It's like how the integral is defined abstractly as area. I'm defining terms. Jump Series Theorem def. A function [imath]f[/imath] [consisting of floor] is said to have a jump series if there is a series [imath]\Sigma=\sum_{n=1}^\infty a_nH_n[/imath] of scaled step functions such that [imath]f+\Sigma[/imath] no longer has any discontinuity attributable to floor. The sequence of scaling factors [imath]a_n[/imath] is called the jump sequence for the function. Definition of Implied Derivative def. We define the evaluation of the implied derivative of a function [imath]y[/imath] at the point [imath]x[/imath] as the set of values [imath]y^{\to}(x) = \{y | y = \lim_{h \to 0^+} \frac {y(x+h) - f(h)}{h} \lor y = \lim_{h \to 0^-} \frac {y(x+h) - f(h)}{h}\}[/imath]. It is in this sense that the implied derivative has the potential of being a multi-valued operator. Jump Series Resolution Theorem [The original wording is preserved; I have a formalism but it assumes that having a jump function is equivalent to being piecewise continuous, which I'm not sure is true.] The integral of a floor-based function is floor treated as a constant minus the appropriate jump series of the integral. Integration Continuity Theorem The (ordinary) indefinite integral of any function [imath]f[/imath] having a jump series must be continuous (or fixably discontinuous) on the domain of [imath]f[/imath]. However, this can vary for the implied indefinite integral. Calculus Jump Theorem def. [Again, the original wording is preserved; the meaning of this phrase has not been properly illuminated] Any form of jumping within a function representable as a floor function can be considered as a portion of the constant of the implied integration for floor. Jump Location Theorem For any function [imath]a(x)[/imath] with a jump series, the function [imath]\lfloor a(x)\rfloor[/imath], has points of discontinuity precisely at those [imath]x[/imath] such that [imath]a(x) - [a(x)] = 0[/imath]. Composite Floor Function Jump Sequence Theorem One If [imath]a[/imath] is a function such that [imath]a(x) - \lfloor a(x) \rfloor = 0[/imath] when [imath]x - \lfloor x \rfloor = 0[/imath], then the jump sequence of the composite function [imath]f(x - \lfloor a\rfloor)[/imath] is a constant value [imath]f(0) - f(1)[/imath]. Implied Differential Equations Conjecture Any continuous solution to an equation constructed with implied derivatives of various orders is the same as the solution to the similarly constructed differential equation. |
2202808 | Combinatorics summation proof involving binomial coefficients
Prove that [imath]\sum_{k=0}^{n}\binom{n}{k}(-1)^kk^n = (-1)^nn![/imath] Please can anyone help me out here? | 465172 | Show this equality (The factorial as an alternate sum with binomial coefficients).
Why does the following equality hold? [imath]n!=\sum_{k=1}^n (-1)^{n-k} \binom{n}{k} k^n[/imath] |
2203525 | For all [imath]x,y\geq 0[/imath] show that [imath]\frac{x^2+y^2}{4}\leq e^{x+y-2}[/imath]
How do I show that for all [imath]x,y\geq 0[/imath], [imath]\frac{x^2+y^2}{4}\leq e^{x+y-2}[/imath]? Tried using the mean value theorem but couldn't figure out how to use it properly. | 286695 | Show that [imath]\forall (x,y)[/imath] in the first quadrant: [imath]\frac {x^2+y^2}{4}\leq e^{x+y-2}[/imath]
I have the folowing exercise (which I've been thinking quite a while and couldn't figure out): Show that [imath]\forall (x,y)[/imath] in the first quadrant: [imath]\frac {x^2+y^2}{4}\leq e^{x+y-2}[/imath] My idea was to work with maxima and minima, but I'm stuck... Any help will be much appreciated! |
2203386 | Probability of winning when rolling a standard fair die with two players
We have two players, lets call them A and B, a fair die is rolled and if either player gets a 5 or a 6 they win, but if they get a 2,3 or 4 the game continues (until someone gets a 5 or a 6). If two 1's come up in a row the game ends in a draw, if A roll's first find the probability that A wins ? I have considered the case where there are no draws ie getting a 1,2,3 or a 4 continues the game and this is simply an infinite geometric series. s.t [imath]P(A)=\frac{2}{6} + (\frac{4}{6})^2\times \frac{2}{6} +(\frac{4}{6})^4\times \frac{2}{6} +... [/imath] but I am not sure how to do this when we consider draws? | 2194626 | Discrete probability problem: what is the probability that you will win the game?
Suppose there is a fair six faced dice, and you roll it once at each round. The rule is as follows: 1). If you roll 1 or 2 at an odd-number round, you win the game and the game ends immediately. 2). If you roll 1 or 2 at an even-number round, you lose the game and it ends immediately. 3). If you roll 3 at any round, then it is a draw and the game ends immediately. If two consecutive [imath]3[/imath]'s appear then the game ends immediately in a draw. 4). Otherwise the game goes on. Now, what is the probability that you will win the game? I attempted to compute the probability of winning at the 1,3,5,...-th round separately and sum them up. However, due to rule 3) it is ever more complicated to compute this for large round numbers. Another attempt was to calculate the probability of stopping at each round and to get a closed form expression for this sequence, but this is also hard, because it depends backwards on the previous rounds in an endless manner. Simply put, if we only consider 1) and 2), or only consider 3), the problem isn't hard, but a combination of the three really messes things up. Can anyone help? Thanks in advance. EDIT: I'm very sorry for the mistake when I transcribed rule 3), I have corrected it. |
2203677 | The limit of [imath]{x^{x^x}}[/imath] as [imath]x\to 0^+[/imath]
I wanted to determine the following: \begin{align*} \lim_{x\to 0^+}{x^{x^x}} &= 0 \end{align*} There has been a previous question posted on it before but the arguments on there do not seem entirely formal: Limit of [imath]{x^{x^x}}[/imath] as [imath]x\to 0^+[/imath] I wanted to see if we could use the fact that we know that [imath]\lim_{x\to 0^+}{x^x} =1[/imath] [imath]\lim_{x\to 0^+}{x^{x^x}} = \lim_{x\to 0^+}e^{x^xlnx}=e^{{\lim_{x\to 0^+}x^xlnx}} [/imath] but I can't see this helping me as [imath]lnx[/imath] has an infinite limit here. Any thoughts as to how to get a nice solution to this? | 439410 | Limit of [imath]{x^{x^x}}[/imath] as [imath]x\to 0^+[/imath]
Can you please explain why \begin{align*} \lim_{x\to 0^+}{x^{x^x}} &= 0 \end{align*} |
2203649 | True or false statement: For all [imath]a \in \mathbb{Z}[/imath] and [imath]p[/imath] prime we have that [imath]a^p \equiv a \mod{p}[/imath]
True or false statement: For all [imath]a \in \mathbb{Z}[/imath] and [imath]p[/imath] prime we have that [imath]a^p \equiv a \pmod{p}[/imath] I formed it to [imath]a^p \bmod p = a \bmod p[/imath] and then inserted several different numbers and I always got the same result. Statement must be true but I don't really have a proof (not required because only a "yes" or "no" is needed). Anyway, I would be interested in knowing if I'm right and especially how this could be shown. | 1757211 | Fermat's little theorem's proof for a negative integer
I'm in the process of proving Fermat's little theorem. For a prime integers [imath]p[/imath] we have [imath]a^p \equiv a \mod{p}[/imath] I proved it for a non-negative [imath]a[/imath], not I need to generalize the case to an arbitrary [imath]a \in \mathbb{Z}[/imath]. That is, I need to prove that give a negative integer [imath]a[/imath] we have [imath]a^p \equiv a \mod{p}[/imath] using the fact that it is so for a non-negative [imath]a[/imath]. |
2202379 | Definition of the functional derivative
In this book, define a functional derivative of a functional [imath]F[f][/imath] as follows: \begin{equation} \frac{\delta f}{\delta f(x)} = \lim_{\epsilon \to 0} \frac{F[f(x^\prime) + \epsilon\delta(x-x^\prime)] - F[f(x^\prime)]}{\epsilon}. \end{equation} I don't understand why change of functional F is [imath]\epsilon\delta(x-x^\prime)[/imath]. Why not define \begin{equation} \frac{\delta f}{\delta f(x)} = \lim_{\epsilon \to 0} \frac{F[f(x^\prime) + \epsilon] - F[f(x^\prime)]}{\epsilon}~? \end{equation} What is the meaning of [imath]\epsilon\delta(x-x^\prime)[/imath]? | 2202616 | Definition of functional derivative
In this book, A derivative of a function is defined as follows: \begin{equation} \frac{df}{dx} = \lim_{\epsilon \to 0} \frac{f(x+\epsilon) - f(x)}{\epsilon}. \end{equation} And define a functional derivative of a functional [imath]F[f][/imath] as follows: \begin{equation} \frac{\delta f}{\delta f(x)} = \lim_{\epsilon \to 0} \frac{F[f(x^\prime) + \epsilon\delta(x-x^\prime)] - F[f(x^\prime)]}{\epsilon}. \end{equation} I don't understand why change of functional F is [imath]\epsilon\delta(x-x^\prime)[/imath]. Why not define \begin{equation} \frac{\delta f}{\delta f(x)} = \lim_{\epsilon \to 0} \frac{F[f(x^\prime) + \epsilon] - F[f(x^\prime)]}{\epsilon}~? \end{equation} What is the meaning of [imath]\epsilon\delta(x-x^\prime)[/imath]? |
2204011 | Show that if a transitive subgroup [imath]G\subset S_n[/imath] of the symmetric group [imath]S_n[/imath] contains a n-1 cycle and transposition,then [imath]G=S_n[/imath].
Show that if a transitive subgroup [imath]G\subset S_n[/imath] of the symmetric group [imath]S_n[/imath] contains a n-1 cycle and transposition,then [imath]G=S_n[/imath]. Firstly, without loss of the genralization, we can assume the n-1 cycle is [imath](2, 3,4,....,n-1)[/imath], and the transposition is (1,2), but how can we prove that (1,2) and (2,3,...,n-1) generates [imath]S_n[/imath] | 1335994 | Transitive subgroup of symmetric group [imath]S_n[/imath] containing an [imath](n-1)[/imath]-cycle and a transposition
Suppose [imath]G[/imath] is a transitive subgroup of [imath]S_n[/imath] such that it there exist [imath]\sigma, \tau \in G[/imath] such that [imath]\sigma[/imath] is an [imath](n-1)[/imath]-cycle and [imath]\tau[/imath] is a transposition. Prove that [imath]G = S_n[/imath]. I just don't understand how to mathematically use the transitive nature of the subgroup. |
1436242 | Show that [imath]A[x] \cap A[x^{-1}][/imath] is integral over [imath]A[/imath].
Let [imath]R[/imath] be a commutative ring, [imath]A[/imath] a subring of [imath]R[/imath], and [imath]x[/imath] a unit in [imath]R[/imath]. Show that every [imath]y \in A[x] \cap A[x^{-1}][/imath] is integral over [imath]A[/imath]. I'm supposed to use the fact that there exists an integer [imath]n[/imath] such that the A-module [imath]M = Ax +..... +Ax^{n}[/imath] is stable under multiplication by [imath]y[/imath]. How do I prove the existence of such an [imath]n[/imath] and then proceed using the claim? | 791689 | Prove that [imath]B[x] \cap B[x^{-1}][/imath] is integral over [imath]B[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be two commutative rings with a unit element, with [imath]B[/imath] subring of [imath]A[/imath]. Suppose [imath]x[/imath] is an invertible element in [imath]A[/imath]. Then prove that the intersection of the two rings [imath]B[x] \cap B[x^{-1}][/imath] is integral over [imath]B[/imath], i.e., prove that for any [imath]a \in B[x] \cap B[x^{-1}][/imath] there is a monic polynomial [imath]f[/imath] with coefficients in [imath]B[/imath] such that [imath]f(a)=0[/imath]. |
2202405 | Field of Fractions [imath]F[/imath] is a finitely generated [imath]R[/imath]-module iff [imath]F[/imath] is equal to its integral domain
Given [imath]R[/imath] an integral domain with field of fractions [imath]F[/imath], I want to prove [imath]F[/imath] is a finitely generated [imath]R[/imath]-module if and only if [imath]R = F[/imath]. For [imath](\Rightarrow)[/imath]: I tried to first define a generating set based on the definition i.e. finitely many [imath]m_1,\dots,m_n[/imath] such that [imath]\sum_i Rm_i = M[/imath]. Then somehow show every element in [imath]R[/imath] is invertible. From where [imath]D=F[/imath] but how... For [imath](\Leftarrow)[/imath]: If [imath]R = F[/imath], I am not sure, but I know by definition, the embedding of [imath]R[/imath] in [imath]F[/imath] maps each [imath]n \in R[/imath] to the fraction [imath]\frac{an}{a}[/imath] for nonzero [imath]a[/imath] in [imath]R[/imath]. In further details, I am a bit lost though. | 598089 | Commutative integral domain does not finitely generate its field of fractions
I want to prove that if we have a commutative integral domain [imath]D[/imath] with field of fractions [imath]F\neq D[/imath] then [imath]F[/imath] is not finitely generated as a [imath]D[/imath]-module. (In this question it may be the case that [imath]1\not\in D[/imath].) My original plan was to localize at a maximal ideal and then use Nakayama's lemma but as we don't have a [imath]1[/imath] our localization may not be local (in fact we need not have maximal ideals). So I'm a bit lost as to where to go? Thanks for any help. |
2204199 | How many sequences of [imath]n[/imath] tosses of a coin that do not contain two consecutive heads have [imath]H[/imath] as the first toss and [imath]T[/imath] as the second?
If you toss a coin [imath]n[/imath] times, there are [imath]2^n[/imath] posible sequences of heads and tails. Let [imath]E_n[/imath] be the set of sequences which do NOT contain two consecutive heads and [imath]e_n[/imath], the number of sequences in [imath]E_n[/imath]. Thus [imath]E_3[/imath] [imath]=[/imath] [imath]\{[/imath] [imath]TTT[/imath], [imath]TTH[/imath], [imath]THT[/imath], [imath]HTT[/imath], [imath]HTH[/imath] [imath]\}[/imath] and [imath]e_3[/imath] [imath]=[/imath] [imath]5[/imath]. How many elements of [imath]E_n[/imath] have [imath]H[/imath] as the first toss and [imath]T[/imath] as the second? | 2204139 | How many sequences of [imath]n[/imath] tosses of a coin that do not contain two consecutive heads have tails as the first toss?
If you toss a coin [imath]n[/imath] times, there are [imath]2^n[/imath] possible sequences of heads and tails. Let [imath]E_n[/imath] be the set of sequences which do NOT contain two consecutive heads and [imath]e_n[/imath], the number of sequences in [imath]E_n[/imath]. Thus [imath]E_3[/imath] [imath]=[/imath] [imath]\{[/imath] [imath]TTT[/imath], [imath]TTH[/imath], [imath]THT[/imath], [imath]HTT[/imath], [imath]HTH[/imath] [imath]\}[/imath] and [imath]e_3[/imath] [imath]=[/imath] [imath]3[/imath]. How many elements of [imath]E_n[/imath] have [imath]T[/imath] as the first toss? Please can anyone help me out here? |
2203924 | Algebraic Integers and Polynomial Roots
Given that [imath]k_1,...,k_n[/imath] are algebraic integers and the complex number [imath]c[/imath] is a root of the polynomial [imath]X^n+k_1X^{n-1}+..+k_n[/imath] it is an exercise to prove that [imath]c[/imath] is also an algebraic integer. I know that, by definition, an algebraic integer is a complex number that is a root of some monic polynomial with integer coefficients. My question then is I don't quite get what is there exactly to be proved here. Am I missing something? Edit: So, can I say for any polynomial [imath]f[/imath] with algebraic coefficients and root [imath]c[/imath], there is a polynomial [imath]g[/imath] with integer coefficients and with same root [imath]c[/imath]. Then perhaps multiplying some polynomials with some roots will give a polynomial with integer coefficients? But I am not sure of the details~ | 472085 | Roots of monic polynomial over a number ring
If [imath]R[/imath] is a number ring with number field [imath]K[/imath] and [imath]f[/imath] is a monic polynomial over [imath]R[/imath], then I want to show that any root of [imath]f[/imath] is an algebraic integer. |
2204504 | Let [imath]S[/imath]:[imath]V[/imath]→[imath]W[/imath] and [imath]T[/imath]:[imath]V[/imath]→[imath]W[/imath] be linear transformations...
Let [imath]S[/imath]:[imath]V[/imath]→[imath]W[/imath] and [imath]T[/imath]:[imath]V[/imath]→[imath]W[/imath] be linear transformations. Given a in ℝ, define functions [imath](S+T):V→W[/imath] and (a[imath]T[/imath]):[imath]V→W[/imath] by [imath](S+T)[/imath](v)=S(v)+T(v) and (a[imath]T[/imath])(v)=a[imath]T[/imath](v) for all v in [imath]V[/imath]. Show that [imath]S+T[/imath] and a[imath]T[/imath] are linear transformations. | 2204443 | Show that [imath]S + T[/imath] and [imath]aT[/imath] are linear
Let [imath]S: V \to W[/imath] and [imath]T: V \to W[/imath] be linear transformations. Given [imath]a[/imath] in [imath]\mathbb{R}[/imath], define functions [imath](S+T): V \to W[/imath] and [imath](aT): V \to W[/imath] by [imath](S+T)(v)=S(v) + T(v)[/imath] and [imath](aT)(v)= aT(v)[/imath] for all [imath]v[/imath] in [imath]V[/imath]. Show that [imath]S+T[/imath] and [imath]aT[/imath] are linear transformations. |
2205011 | Calculate [imath]E(max(A, B))[/imath]
If I got two independent random variables [imath]A \sim R(0,1)[/imath] and [imath]B \sim R(0,1)[/imath] and I have to calculate [imath]E(max(A, B))[/imath] wouldn't that then just be to pick one of them since they are equal in size? So it would be the same [imath]E(A)[/imath]? What would E(max{A,B}) be? | 197299 | Expected value of maximum of two random variables from uniform distribution
If I have two variables [imath]X[/imath] and [imath]Y[/imath] which randomly take on values uniformly from the range [imath][a,b][/imath] (all values equally probable), what is the expected value for [imath]\max(X,Y)[/imath]? |
2205739 | Isomorphism Types of Abelian Groups?
I'm stuck on the following problem: Find the smallest number [imath]n[/imath] for which the following statement is true: "The number of isomorphism types of abelian groups with [imath]n[/imath] elements is equal to 10". How many such [imath]n[/imath] are there? My thoughts: I think that the smallest number is [imath]n=144[/imath]. How would I prove this? Also, am I correct in saying that there are infinitely many such numbers [imath]n[/imath]? Any help is appreciated! | 786525 | Abelian groups of order n.
Is there a number [imath]$n$[/imath] such that there are exactly 1 million abelian groups of order [imath]$n$[/imath]? Can anyone please explain. I would yes because numbers are infinitive, and so any number n can be expressed as a direct product of cyclic groups of order n. Can anyone please help me understand. Thank you. |
2204397 | Let [imath]a,b,c[/imath] be positive numbers. Prove that [imath]a^2b+b^2c+c^2a\leq a^3+b^3+c^3[/imath]
I tried moving all the variables onto one side and attempting to factor it into squares and proving the inequality since squares are always greater than or equal to 0. The inequality is too complex to factor through completing the square (or "rectangle"). How do I solve the inequality? | 1025588 | Proof of Inequality using AM-GM
I just started doing AM-GM inequalities for the first time about two hours ago. In those two hours, I have completed exactly two problems. I am stuck on this third one! Here is the problem: If [imath]a, b, c \gt 0[/imath] prove that [imath] a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a.[/imath] I am going crazy over this! A hint or proof would be much appreciated. Also any general advice for proving AM-GM inequalities would bring me happiness to my heart. Thank you! |
2206182 | [imath]A \preceq B => B^{-1} \preceq A^{-1}[/imath]
[imath]A \preceq B[/imath] indicates for covariance matrices A and B, B-A is SPSD (symmetric positive semi definite). Hence, I want to show that [imath]B^{-1} \preceq A^{-1}[/imath] , which seems straight-forward but I can't come up with a way to solve this. | 882705 | If [imath]A \succeq B[/imath] is it true that [imath]B^{-1} \succeq A^{-1}[/imath]
If [imath]A[/imath] and [imath]B[/imath] are two positive definite matrices such that [imath]A - B[/imath] is nonnegative definite, is it true that [imath]B^{-1} - A^{-1}[/imath] is positive definite? The doubt came to me when working with confidence regions in multivariate statistics that are usually obtained as hyper ellipsoids. |
2206401 | Finding distribution of two random variables
Given two independent random variables [imath]X_1\in NB(p,r_1)[/imath] and [imath]X_2\in NB(p,r_2)[/imath], find the distribution of random variable [imath]X_1+X_2[/imath] I came up with an answer that is obviously wrong since it's different from what's on the textbook ( [imath]X_1+X_2 \in NB(p,r+s)[/imath]), how do I go about showing this? | 1054048 | Negative binomial distribution - sum of two random variables
Suppose [imath]X, Y[/imath] are independent random variables with [imath]X\sim NB(r,p)[/imath] and [imath]Y\sim NB(s,p)[/imath]. Then [imath]X + Y \sim NB(r+s,p)[/imath] How do I go about proving this? I'm not sure where to begin, I'd be glad for any hint. |
2205817 | Possible typo in problem of Braun's Differential Equations
In Martin Braun's book Differential Equations, problem 16 in section 1.10 says: Consider the initial-value problem [imath]y'= t^2+y^2[/imath], [imath]y(0)=0[/imath] ... (*) and Iet [imath]R[/imath] be the rectangle [imath]0\leq t \leq a, - b \leq y \leq b[/imath] (a) Show that the solution [imath]y(t)[/imath] of (*) exists for [imath]0 \leq t \leq min(a, b/(a^2+b^2))[/imath] (b) Show that the maximum value of [imath]b/(a^2 +b^2)[/imath], for a fixed [imath]a[/imath], is [imath]1/2a[/imath]. (c) Show that [imath]\alpha=min(a,\frac{1}{2}a)[/imath] is largest when [imath]a=\frac{1}{2^{1/2}}[/imath]. (d) Conclude that the solution [imath]y ( t)[/imath] of (*) exists for [imath]0 \leq t \leq \frac{1}{2^{1/2}} [/imath] The problem is, in the incise b) i proof by calculus of the maximum of the function [imath]f(b)=\frac{a^2}{a^2+b^2}[/imath], the maximun is reached when [imath]f(a)=\dfrac{1}{2a}[/imath], but in incise c) the book puts [imath]\alpha=min(a,\frac{1}{2}a)[/imath], my question is, that is an error? or i'm wrong? it should be [imath]\alpha=min(a,\frac{1}{2a})[/imath]? or not? and, how can i prove that? thanks and, i apologize if the question is clumsy, but i tried to solve this problem for hours and the book confused me, because the notation. | 2104192 | ODE [imath]\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.[/imath]
Consider the ODE [imath]\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.[/imath] I have to find the interval of unique solution by using Picard method. As in my local book it is solved as [imath]|f(x,y)|=|x^{2}+y^{2}|\leq a^{2}+b^{2}=M. [/imath] [imath]h=\min\{a,\frac{b}{M}\}[/imath] where [imath]M=a^{2}+b^{2}[/imath]. So [imath]h=\min\{a,\frac{b}{a^{2}+b^{2}}\}[/imath]. Then my main problem is the author do like [imath]a=\frac{b}{a^{2}+b^{2}}[/imath] which gives a quadratic in [imath]b[/imath] as [imath]ab^{2}-b+a^{3}=0[/imath] with discriminant as [imath]1-4a^{4}=0[/imath] gives [imath]a=\frac{1}{\sqrt{2}}[/imath] so [imath]|h|\leq\frac{1}{\sqrt{2}}.[/imath] Please suggest me how he find [imath]h=\min\{a,\frac{b}{a^{2}+b^{2}}\}[/imath]? Why he put [imath]a=\frac{b}{a^{2}+b^{2}}[/imath]? Thanks in advance. |
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