qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
2800189 | A question on Induction
Prove that: for any natural number [imath]n\ge10[/imath], [imath]2^n\ge n^3.[/imath] How can I prove by induction if [imath]n\ge10[/imath] and I must do that [imath]2^n\ge n^3[/imath]? I stop on this step: [imath]2^{n+1}\ge n^3+3n^2+3n+1.[/imath] Here's the question I'm trying to prove. I'm just not certain how I should approach the inductive / constructor step. I proved induction wrote [imath]n+1[/imath] instead of [imath]n[/imath]. I think I have everything right until the induction step. But I don’t know what I should do next step, because I had never solve inequality before I began ask questions on this site. | 1772026 | Use induction to prove that [imath]2^n \gt n^3[/imath] for every integer [imath]n \ge 10[/imath].
Use induction to prove that [imath]2^n \gt n^3[/imath] for every integer [imath]n \ge 10[/imath]. My method: If [imath]n = 10[/imath], [imath]2^n \gt n^3[/imath] where [imath]2^{10} \gt 10^3[/imath] which is equivalent to [imath]1024 \gt 1000[/imath], which holds for [imath]n = 10[/imath]. [imath]2^k \gt k^3[/imath]. [imath]2^{k + 1} \gt (k + 1)^3[/imath] [imath]2^{k + 1} \gt (k + 1)(k + 1)(k + 1)[/imath] [imath]2^{k + 1} \gt (k^2 + 2k + 1)(k + 1)[/imath] [imath]2^{k + 1} \gt (k^3 + 3k^2 + 3k + 1)[/imath] [imath](k + 1) \cdot 2^k \gt k^3 + 3k^2 + 3k + 1[/imath] Since [imath]2^{10} \gt 10^3[/imath], the inequality holds for [imath]n = 10[/imath]. Assume [imath]2^k \gt k^3[/imath] for [imath]k \ge 10[/imath]. We show that [imath]2^{k + 1} \gt (k + 1)^3[/imath]. Hence, [imath]2^{k + 1} = 2 \cdot 2^k \gt 2k^3[/imath] [imath]= k^3 + k^3 \ge k^3 + 10k^2[/imath] [imath]= k^3 + 10k^2 = k^3 + 4k^2 + 6k^2 \ge k^3 + 4k^2 + 6 \cdot 10 = k^3 + 4k^2 +60[/imath] [imath]k^3 + 4k^2 + 60 \gt k^3 + 3k^2 + 3k + 1[/imath] Therefore, [imath]2^{k + 1} \gt (k + 1)^3[/imath]. Hence, [imath]2^n \gt n^3[/imath] for every integer [imath]n \ge 10[/imath]. I was trying to fix this but I am not sure how to go about doing so. |
2800402 | Finding the probability that the matrix has full rank
There are only two entries, [imath]0[/imath] and [imath]1[/imath], over [imath]\mathbb{Z}_2[/imath]. Thus, only [imath]16[/imath] possible [imath]2\times 2[/imath] matrices over [imath]Z_2[/imath], and [imath]6[/imath] of them have full rank: [imath]\begin{pmatrix}0&1\\ 1&0\end{pmatrix} \quad \begin{pmatrix}1&1\\ > 1&0\end{pmatrix} \quad \begin{pmatrix}0&1\\ 1&0\end{pmatrix} \quad > \begin{pmatrix}0&1\\ 1&1\end{pmatrix} \quad \begin{pmatrix}1&1\\ > 0&1\end{pmatrix} \quad \begin{pmatrix}1&0\\ 0&1\end{pmatrix}[/imath] Randomly generate a [imath]n\times n[/imath] matrix over [imath]\mathbb {Z}_2[/imath] (where n is big, say, [imath]1000[/imath]). What's the probability that the matrix has full rank? I'd want to talk something regarding to this question. I found the probability is equal to: [imath]\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^n} \tag{1}[/imath] On other hand, what's difference between [imath]2^n[/imath] and [imath]2^{n^2}[/imath]? [imath]\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^{n^2}} \tag{2}[/imath] Can you tell whether or not I'm wrong? Regards! | 54246 | Probability that a random binary matrix is invertible?
What is the probability that a random [imath]\{0,1\}[/imath], [imath]n \times n[/imath] matrix is invertible? Assume the 0 and 1 are each present in an entry with probability [imath]\frac{1}{2}[/imath]. Is there an explicit formula as a function of [imath]n[/imath]? Does it tend to 1 as [imath]n[/imath] grows large? I'm sure this is all known... Thanks! |
2803807 | Existence of holomorphic function on the unit disc
Does there exist a holomorphic function [imath]f:D\to D[/imath] where [imath]D=\{z\in \mathbb{C} : |z|<1\}[/imath] such that [imath]f(1/2)=-1/2[/imath] and [imath]f'(1/4)=1[/imath]? I tried with Schwarz pick lemma, but for that I need the value of [imath]f(1/4)[/imath] which is not given. So kindly tell me ow to do this. Thanks. | 595404 | Holomorphic self map on the open unit disc
Does there exist a holomorphic function [imath]f:\mathbb{D}\longrightarrow\mathbb{D}[/imath] such that [imath]f(1/2)=-1/2[/imath] and [imath]f'(1/4)=1[/imath], where [imath]\mathbb{D}=\{z:|z|<1\}[/imath]? |
2804597 | Help with Proof: [imath]Aut_{Grp}(\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}) \cong S_3[/imath]
I need help improving my proof. It's a bit lacking in details, and I don't know how to phrase my reasoning. For any homeomorphism, identities must map to each other. Therefore, any automorphism of [imath]\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}[/imath] must map [imath](0,0) \mapsto (0,0)[/imath]. This leaves three other elements, label them [imath]1: (0,1), 2: (1,0), 3: (1,1)[/imath]. Since all of these elements are order two, and the group action of any two will give the third, they can be interchanged with each other freely. So, any bijection of these three elements will result in an automorphism. Hence, there is a function between the automorphisms of [imath]\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}[/imath] and the symmetric group [imath]S_3[/imath] To prove this is a homeomorphism, we show [imath]\mu(g\circ h) = \mu(g)\cdot\mu(h)[/imath]. This is true because [imath]g[/imath] and [imath]h[/imath] are automorphisms, and the bijection above supports this property, and this is circular reasoning that I can't seem to articulate more clearly. To prove this is a bijection, like before, the function maps the pairs to their labels. It's injective, because if two labels are equal, then their points are equal, and so it is with the functions inputs and outputs. It's also surjective because of labeling of inputs and outputs carrying over to functions. Is "labeling" enough to prove bijection? I don't want to just enumerate all the functions, there has to be a better way. The reasoning for homeomorphism and bijection are both really lame, but I don't know what sort of reasoning I can employ to make this more concise. This is in the "basic" Group Theory section of Paolo Aluffi's "Algebra: Chapter 0". I have not yet gotten to Rings, Fields, Modules, or any Linear Algebra. | 2392851 | Showing [imath]\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3[/imath]
So I believe I've finished the proof, but it all amounts to case-work and I really think (and hope) there's a better approach than this. Here's what I've done: Attempt: Within [imath]\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})[/imath] identities must be mapped to identities thus what defines each isomorphism is the bijective function between its last three elements: \begin{matrix} (0,0) & \mapsto & (0,0) \\ (0,1) & & (0,1) \\(1,0) & & (1,0) \\(1,1) & & (1,1) \\ \end{matrix} If every bijective function between these last three elements constitutes a homomorphism (including the first unchanged mapping of identities), then [imath]\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3[/imath] in that isomorphisms in [imath]\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})[/imath] can be identified with permutations of three elements, which is exactly what [imath]S_3[/imath] is. I then, essentially go case-by-case showing every one of the six different bijections is a homomorphism. For two of those bijections (which are just compositions of other bijections I've already checked) I apply the following lemma: If [imath]\phi :G \to H[/imath] and [imath]\psi : H \to K[/imath] are bijective homomorphisms, then [imath](\psi \circ \phi)[/imath] is a homomorphism. As proof, assume the premise. We then have [imath]\forall a,b \in G:[/imath] [imath]\psi (\phi (ab))= \psi(\phi(a) \phi(b))=\psi (\phi(a))\psi (\phi(b))[/imath] [imath]\implies (\psi \circ \phi)[/imath] is a homomorphism. So my questions are: Is the proof that I have sufficient? Is there a better proof for this problem? (A hint towards it rather than complete give-away would be much more appreciated) EDIT: I have indeed looked at this post already, but the approach carried out there uses vector spaces which I have not been introduced to or am supposed to use to solve this problem. |
2796711 | [imath]AB - BA = \lambda B[/imath] has a nontrivial solution, iff [imath]\lambda = \lambda_i - \lambda_j[/imath]?
I have the following question in hand. If [imath]\lambda_1,\cdots,\lambda_n[/imath] are the eigenvalues of a given matrix [imath]A \in M_n[/imath], then prove that the matrix equation [imath]AB - BA = \lambda B[/imath] has a nontrivial solution [imath]B \neq 0 \in M_n[/imath], if and only if [imath]\lambda = \lambda_i - \lambda_j[/imath] for some [imath]i,j[/imath]. | 738662 | If [imath]E[/imath] has eigenvalues [imath]\lambda_1, \ldots, \lambda_n[/imath], then [imath]B \mapsto [E,B] = EB - BE[/imath] has eigenvalues [imath]\lambda_i - \lambda_j[/imath].
Let [imath]E \in M_n(C)[/imath] be an [imath]n \times n[/imath] matrix with entries in a(n algebraically closed, characteristic 0) field [imath]C[/imath], with eigenvalues [imath]\lambda_1, \ldots, \lambda_n[/imath]. Show that the commutator map [imath]M_n(C) \to M_n(C)[/imath] given by [imath]B \mapsto [E,B] = EB - BE[/imath] has eigenvalues [imath]\lambda_i - \lambda_j[/imath]. This is an ingredient a proof of the fact that any regular singular equation [imath]y' = Ay[/imath] over [imath]C((z))[/imath] is equivalent to [imath]v' = Dz^{-1}v[/imath] with [imath]D[/imath] a constant matrix. See exercise (7) (a) (i) of Galois Theory of Differential Equations, Algebraic Groups and Lie Algebras by Marius van der Put (page 6 of the PDF). This is homework, so hints rather than full answers would be appreciated. Things I have found that may or may not be useful: [imath]EB - BE = (\lambda_i - \lambda_j)B[/imath] can be written [imath](E-\lambda_i)B = B(E- \lambda_j)[/imath]. [imath]EB[/imath] and [imath]BE[/imath] have the same eigenvalues. |
2805052 | What is Taylor's Inequality about?
I am confused by the Taylor's Inequality formula: [imath]|R_n(x)| \le \frac M{(n+1)!}|x−a|^{n+1}[/imath] What do [imath]R_n[/imath], [imath]M[/imath], [imath]n[/imath], [imath]x[/imath], [imath]a[/imath], stand for i.e. what to substitute inside each variable? What purpose does this formula serve/How to apply it in situations/problems? | 1174455 | What does the Taylor's Inequality mean?
Taylor's Ineqaulity If [imath]|f^{(n+1)}(x)|\leq M[/imath] for [imath]|x-a|\leq d[/imath], then the remainder [imath]R_n(x)[/imath] of the Taylor series satisfies the inequality [imath]|R_n(x)|\leq \dfrac{M}{(n+1)!}|x-a|^{n+1}\text{ for } |x-a|\leq d[/imath] I'm trying to understand the theorem. Can someone explain it to me intuitively what does it mean? |
2805444 | How can I solve the problems of Combination + indistinguishable objects + repetition allowed?
Can I solve the problems of Combination with indistinguishable objects plus repetition allowed using a formula? For instance, How many [imath]3[/imath]-letter combinations could be formed from the word [imath]BBA[/imath] if repetition of letters is allowed? We know the answer is [imath]4[/imath], as [imath]S=\{{AAA, BBB, ABB, AAB}\}[/imath]. Can these kinds of problems be solved using a formula, or, do I need to manually solve it every time? Note. The problem in this question is different from the problem posted in this problem which I posted earlier. Here we have [imath]2[/imath] letters and need to produce [imath]3[/imath]-element combinations (i.e. [imath]n<r[/imath]). On the other hand, in the previous problem, we had [imath]3[/imath] letters to choose from [imath]26[/imath] letters (i.e. [imath]n>r[/imath]). | 2805217 | What is the formula for Combination with repetition?
What is the number of ways we can choose three letters from [imath]\{A,B,...,Z\}[/imath] if repetition is allowed? This is not a permutation. Is there any formula to solve this kind of problems? |
2805359 | Without brute force, how can we show that 1729 is the smallest number that can be expressesed as the sum of two cubes in two different ways?
It's very known that 1729 is the smallest number that can be expressed as the sum of two cubes in two different ways, but is there a proof that not uses brute force? If we let [imath]n[/imath] be the smallest number such that [imath]n=a^3+b^3=c^3+d^3[/imath] with [imath]a\neq c[/imath], [imath]b \neq d[/imath] and [imath]a,b,c,d > 0[/imath], then: [imath]n = (a+b)(a^2-ab+b^2)[/imath] [imath]n=(c+d)(c^2-cd+d^2)[/imath] But I don't have any idea how to follow from here. If we take a look at 1729: 1729=13·19·7 It seems obvious that if a number is the product of only 2 primes then it can't be expressed as the sum of two cubes in two different ways, but I don't know how to prove either that the least number which such property must be product of 3 primes or more. Does anyone has any idea of how to approach this? | 487537 | Proof that [imath]1729[/imath] is the smallest taxicab number
For homework I have to produce the proof (algebraic or otherwise) to show that [imath]1729[/imath] HAS to be the smallest taxi cab number. A taxicab number means that it is the sum of two different cubes and can be made with [imath]2[/imath] sets of numbers. I have the list of the next ones and I was wondering if it was linked with the fact that it would have to be [imath]0[/imath] cubed if it got any lower which obviously wouldn't work. Any help appreciated, thanks in advance! |
2806002 | Visual representation of the [imath]V/\ker f[/imath] and first isomorphism theorem
I'm trying to gasp the visual representation of [imath]V/\ker f[/imath] and in continuation the first isomorphism theorem in linear algebra. My understanding is that, given the representation of a vector space [imath]V,[/imath] [imath]V/\ker f[/imath] is created by "erasing" all the dimensions [imath]f[/imath] maps to [imath]0.[/imath] That's where I get stuck. Any help would be appreciated P.s The way I have linear algebra in mind is how it is presented in 3blue1brown's channel, if this is of any importance https://www.youtube.com/watch?v=XkY2DOUCWMU&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=5 | 109106 | What's an intuitive way of looking at quotient spaces?
I understand the concept of [imath]\mathbb{Z}/n\mathbb{Z}[/imath], but I am having a really hard time understanding how this concept of quotients applies to vector spaces. Suppose [imath]V = \mathbb{F}[x][/imath] is a vector space and [imath]U \le V[/imath]. What exactly does [imath]V/U[/imath] represent? |
2804451 | Which one is bigger, [imath]e ^ {3 \pi}[/imath] or [imath] \space 3 ^ {e \pi}[/imath]?
I have this equation, and I've tried checking which is bigger, [imath] e ^ {3 \pi} \space ? \space 3 ^ {e \pi}[/imath] What I've tried: [imath] e ^ {3 \pi} \space ? \space 3 ^ {\pi e} / \sqrt[\pi]{} [/imath] [imath] e ^ {3} \space ? \space 3 ^ {e}[/imath] [imath] (e ^ {1\over e}) ^ {3e} \space ? \space (3 ^ {1\over 3}) ^{3e} / \sqrt[3e]{}[/imath] [imath] e ^ {1\over e} \space ? \space 3 ^ {1\over 3} / \ln()[/imath] [imath] {1\over e} \space ? \space {1\over 3} \ln(3) [/imath] and now because ln(3) > 1, I assumed the right side was bigger, apparentaly it wasn't. For future references. How do I go about checking which one is bigger? | 2126331 | [imath]a^b > b^a[/imath] or [imath]a^b < b^a[/imath] which is true if [imath]a>b[/imath] where [imath]a,b[/imath] are positive real numbers
I was trying to find a simple relationship between mutual exponentiation of two positive real values, if one of them is greater/smaller to another. |
2805256 | Calculating Dimension of Intersection of Kernels
Let T be sqaure matrix and regarded as a linear operator on a finite dimensional vector space V such that [imath]T^2 = 0[/imath]. [imath]\dim(\ker T \cap \ker T^t) =\dim \ker T+\dim\ker\ T^t−\dim(\ker T\dot+\ker T^t) [/imath] ([imath]\dot+ [/imath] denotes direct sum) Is the abvoe equation true? If so, why is it so? | 500511 | Dimension of the sum of two vector subspaces
[imath]\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2).[/imath] I want to make sure that my intuition is correct. Suppose we have two planes [imath]U_1,U_2[/imath] though the origin in [imath]\mathbb{R^3}[/imath]. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in [imath]\mathbb{R^3}[/imath]. To obtain the dimension of [imath]U_1[/imath] and [imath]U_2[/imath], we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3). *additional question(s): Can we generalize this notion to [imath]\mathbb{F^{n}}[/imath]? Suppose we have an additional case where [imath]U_1[/imath] and [imath]U_2[/imath] are planes in [imath]\mathbb{R^3}[/imath], but [imath]U_1 \subseteq U_2[/imath]. In this instance, [imath]dim(U_1 + U_2) < 3[/imath], because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces [imath]U_1,U_2[/imath] are two dimensional and [imath]U_1 \subseteq U_2[/imath], then their intersection is also two-dimensional, concluding [imath]dim(U_1+U_2)=2+2-2 = 2[/imath]. Is this proper intuition? |
2806340 | Does continuous imply measurable?
If [imath]f[/imath] is continuous, is it (Lebesgue) measurable ? Namely, [imath]\{x\ :\ f(x)<\alpha\}\in\mathcal M\ \forall\alpha[/imath] ? If not what are some counter examples? | 696124 | $f$ a real, continuous function, is it measurable?
Let [imath]f: \mathbb{R} \to \mathbb{R} [/imath] be a continuous function. I need to show that is a measurable function. I tried working with the definition: Let [imath]f: X \to \mathbb{R}[/imath] be a function. If [imath]f^{-1}(O)[/imath] is a measurable set for every open subset [imath]O[/imath] of [imath]\mathbb{R}[/imath], then [imath]$f$[/imath] is called a measurable function. Since [imath]f^{-1}(O)[/imath] also lies in [imath]\mathbb{R}[/imath], I think it is sufficient to show that every subset of [imath]\mathbb{R}[/imath] is measurable. But is this possible? So far I concluded that [imath]\mathbb{R}[/imath] itself is measurable, since [imath]$$\mu(A) = \mu(A \cap \mathbb{R}) + \mu(A \cap \mathbb{R}^c) = \mu(A) + \mu(\emptyset) = \mu(A).$$[/imath] How do I need to approach? |
2806413 | Coin flipping experiment
This was an exercise in an old exam in probability theory without a solution. We are flipping a coin repeatedly. The probability of tail is [imath]p[/imath] < [imath]\frac{1}{2}[/imath]. Let [imath]A_{k}[/imath] for [imath]k \geq 2[/imath] be the Event that under the throws from [imath]2^{k},2^{k}+1,...,2^{k+1}-1[/imath] at least [imath]k[/imath]-times in a row tails occurs. It is to Show that [imath] \mathbb{P}(A_{k}~~ \text{infinitely often}) = 0. [/imath] There is also the hint to consider [imath]A_{k} = \bigcup_{j=2^{k}}^{2^{k+1}-k}\{X_{j}=1,...,X_{j+k-1}=1\}[/imath] and to estimate [imath]\mathbb{P}(A_{k})[/imath]. Here [imath]X_{i}[/imath] is the outcome of the [imath]i[/imath]-th coin flipping, where [imath]X_{i} = 1[/imath] if the [imath]i[/imath]-th coin flippin gis tail. First I think Borell-Cantelli would help me here. But the hint confuses me a bit | 1821116 | Use inclusion-exclusion to find explicit formula for [imath]P(A_k)[/imath]. Where is independence used?
Probability with Martingales This was answered here using bounds and here using PGF. I would like to try inclusion-exclusion. Let [imath]H_1, H_2, ...[/imath] be independent events with [imath]P(H_k) = p[/imath] where [imath]H_k[/imath] means the kth toss is heads. It seems that ([imath]\cap[/imath] omitted) [imath]A_0 = H_1[/imath] [imath]A_1 = H_2 \cup H_3[/imath] [imath]A_2 = (H_4H_5) \cup (H_5H_6) \cup (H_6H_7) := \bigcup_{i=1}^{3} E_{2,i}[/imath] [imath]A_3 = (H_8H_9H_{10}) \cup ... \cup (H_{13}H_{14}H_{15}) := \bigcup_{i=1}^{6} E_{3,i}[/imath] [imath]\vdots[/imath] [imath]A_k = (H_{2^k}...H_{2^k + k-1}) \cup ... \cup (H_{2^{k+1} - k}...H_{2^{k+1} - 1}) := \bigcup_{i=1}^{2^k - k + 1} E_{k,i}[/imath] So by inclusion-exclusion, we have [imath]P(A_0) = p[/imath] [imath]P(A_1) = p + p - p^2[/imath] [imath]P(A_2) = 3p^2 - 2p^3[/imath] [imath]P(A_3) = \ldots[/imath] It gets tricky starting [imath]A_3[/imath] [imath]P(A_3) = P\left(\bigcup_{i=1}^{6} E_{3,i}\right)[/imath] Attacking that directly is going to lead to problems. What about this? [imath]P\left(\bigcup_{i=1}^{6} E_{3,i}\right)[/imath] [imath] = P\left(\bigcup_{i=1,4} E_{3,i} \cup \bigcup_{i=2,5} E_{3,i} \cup \bigcup_{i=3,6} E_{3,i} \right)[/imath] [imath] := P(F_{3,1} \cup F_{3,2} \cup F_{3,3})[/imath] Each [imath]F[/imath] is a union of disjoint sets. Also, I notice independence isn't being used. I guess the [imath]H_k[/imath]'s and the [imath]A_k[/imath]'s are independent. What about some subset of the the [imath]E[/imath]'s? The [imath]F[/imath]'s? |
2806380 | Compute [imath]\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx[/imath]
Compute [imath]I=\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx[/imath] My 2 attempts: First: We observe that [imath]\frac {1}{x^2+2x+2}=\frac {1}{(x+1)^2+1}[/imath] and [imath]\frac 1{(x+1)^2+1}=(\arctan{(x+1)})^{'}.[/imath] Then: [imath]\int_0^2\frac{\arctan{x}}{x^2+2x+2}dx=\int_0^2\arctan{x}(\arctan{(x+1)})^{'}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\frac{1\times\arctan{(x+1)}}{1+x^2}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\frac{(1+x^2-x^2)\times\arctan{(x+1)}}{1+x^2}dx=\left.\arctan{x}\arctan{(x+1)}\right|_0^2-\int_0^2\arctan{(x+1)}dx-\int_0^2\frac{x^2\arctan{(x+1)}}{1+x^2}dx[/imath] Where the second integral is pretty easy to solve using integration by parts but the second one is not very pleasant so... I thought I should try something else. Second attempt: Let [imath]u=\arctan(x+1)[/imath] then [imath]x=\tan u-1[/imath] then [imath]I=\int_{\arctan1}^{arctan3}\arctan{(\tan u -1)du}[/imath] then I could let [imath]w=\tan(u)-1[/imath] then [imath]dw=((w+1)^2+1)du[/imath]... but it's a lot of work and I really think you can do this more easily... Any hints? | 2798550 | Integral [imath]\int_0^2 \frac{\arctan x}{x^2+2x+2}dx[/imath]
I am tring to evaluate [imath]I=\int_0^2 \frac{\arctan x}{x^2+2x+2}dx[/imath] My first thing was to notice that [imath]\frac{1}{x^2+2x+2}=\frac{1}{(x+1)^2+1}=\frac{d}{dx}\arctan(x+1)[/imath] So I integrated by parts in order to get [imath]I=\arctan 2\arctan 3-\int_0^2\frac{\arctan(x+1)}{1+x^2}dx[/imath] I let [imath]x=u+1[/imath] but when I do that I get [imath]I=\arctan 2\arctan 3+\int_{-1}^1\frac{\arctan(u)}{1+(1+u)^2}du =\arctan 2\arctan 3[/imath] Now this is not close to the approximation given by wolfram. What have I done wrong and how to solve this? |
2806990 | How to prove that this recurrent sequence is Cauchy
Let [imath]f:(E,d)\to (E,d)[/imath] such that [imath]d(f(x),f(y))<d(x,y), \forall x,y\in E ,x\neq y[/imath] where [imath]E[/imath] is compact I want to prove that [imath]f[/imath] has a unique fixed point I consider the recurrent sequence [imath]\begin{cases} x_{n+1}=f^{n+1}(x_0)\\ x_0\in E\end{cases}[/imath] But how to prove that it is a Cauchy sequence? if [imath]d(f(x),f(y))<d(x,y)[/imath] and [imath]x_{n+1}=f^{n+1}(x_0)[/imath] if for example i want to to see [imath]\lim_{n\to+\infty} d(x_n,x_{n+1})[/imath] [imath]d(x_n,x_{n+1})= d(f(x_{n-1}),f(x_n))<d(x_{n-1},x_n)<...<d(x_0,x_1)[/imath] then the limit is not 0 how to prove that it is a Cauchy sequence ? | 2779756 | Edelstein Theorem
Let (M,d) be a compact metric space and [imath]d(f(x),f(y)) < d(x,y) [/imath] for all [imath] x\neq y[/imath] Prove that if [imath]f[/imath] is a continuous fuction then there is a unique [imath]x_0 \in M [/imath] such that [imath]f(x_0)=x_0[/imath] I know that since [imath]g(x) = d((f(x),f(y))[/imath] is continuous then a continuous function in a compact space reaches its maximum and minimum. Then by proving that the minimum of [imath]g(x)[/imath] = 0 can I conclude that [imath]x_0[/imath] is a fixed point? |
2802636 | Solve the IVP [imath]\vec{y'}=A\vec{y}\ [/imath] given [imath]\ (A-I)\vec{v}\ [/imath] and [imath]\ (A-I)^2\vec{v}[/imath]
Let [imath]A=\begin{pmatrix} 5 & -2 & -7 \\ -4 & 3 & 4 \\ 6 & -3 & -8 \end{pmatrix} [/imath] and [imath]\vec{v}=\left(\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right)[/imath]. The first part of the problem asked to calculate [imath](A-I)\vec{v}[/imath] and [imath](A-I)^2\vec{v}[/imath], which I calculated to be [imath]\left(\begin{array}{c} -1 \\ -2 \\ 0 \end{array}\right)[/imath] and [imath]\vec{0}[/imath] respectively. The question then asks to solve the initial value problem [imath]\vec{y'}=A\vec{y}[/imath] with [imath]\vec{y}(0)=\vec{v}[/imath]. I don't understand how to use what we calculated in the earlier part to help to solve this initial value problem. I determined from first principles that the eigenvalues are [imath]\lambda=1,1,-2[/imath] and I know that the solution of our initial value problem will take the form [imath]\vec{y}=e^{tA}\vec{v}\ [/imath], but I don't quite understand the significance of [imath](A-I)\vec{v}[/imath] and [imath](A-I)^2\vec{v}[/imath]. | 2804043 | IVP Differential Equation
I have stumbled across a very old exam question from my linear algebra course and the solutions are not available. I was wondering if my working/logic is correct and if any improvements can be made. Let [imath]\ B=\begin{pmatrix} 1 & -1 & 4 \\ 6 & -7 & 2 \\ -3 & 1 & -6 \end{pmatrix}\ [/imath] and [imath]\ \vec{v_1}=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}[/imath]. Part (i): Calculate [imath](B+5I)\vec{v_1}[/imath] and [imath](B+5I)^2\vec{v_1}[/imath]. Using simply matrix multiplication, I calculated [imath](B+5I)\vec{v_1}=\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}[/imath] and [imath](B+5I)^2\vec{v_1}=\vec{0}[/imath]. Part (ii): Find all eigenvalues and eigenvectors of B. From our calculations in part (i), we can see that [imath]\ \lambda=-5,-5[/imath] are two of the eigenvalues, with corresponding eigenvector [imath]\ \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}[/imath] and generalised eigenvector [imath]\ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}[/imath]. From the trace of B, the remaining eigenvalue is [imath]\lambda=-2[/imath] with corresponding eigenvector [imath]\begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}[/imath]. Part (iii): Giving reasons, write down a basis for the generalised eigenspace [imath]GE_{-5}[/imath] of B. I am unsure of this part of the question. If I had to guess, I would say a basis for [imath]GE_{-5}=\left\{\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\right\}[/imath]. Is this correct? Part (iv): Solve the initial value problem [imath]\vec{y'}=B\vec{y}[/imath] where [imath]\ B=\begin{pmatrix} 1 & -1 & 4 \\ 6 & -7 & 2 \\ -3 & 1 & -6 \end{pmatrix}\ [/imath] and [imath]\vec{y}(0)=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}[/imath]. Well, the solution to this IVP will take the form [imath]\vec{y}=e^{tB}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}[/imath]. I noticed that [imath]\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}[/imath] was a generalised eigenvector and hence we can write [imath]\vec{y}=e^{-5t}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}+e^{-5t}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}t[/imath]. Thus [imath]\vec{y}=e^{-5t}\begin{pmatrix} 1+t \\ 1+2t \\ -1-t \end{pmatrix}[/imath] |
2808556 | To find sum of infinite series.
How to find the sum of the following infinite series? [imath]\frac1{2 \cdot 3 \cdot 4}+\frac1{4 \cdot 5 \cdot 6}+\frac1{6 \cdot 7 \cdot 8}+\cdots[/imath] I can write its partial sum but am unable to proceed. | 2621718 | Sum of series [imath]\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots [/imath]
What is the limit of series [imath]\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots [/imath]? The [imath]n[/imath]th summand is [imath]\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}[/imath]. I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?) |
2808783 | A question about prime divisors
I am trying but still unsuccessfully solving the following question: For every positive integer [imath]n[/imath], show that any prime divisor of [imath]12n^2 + 1[/imath] is of the form [imath]6k + 1[/imath], where [imath]k[/imath] is an integer. I would like, if possible, to get a solution that does not refer to Legendre symbol or quadratic reciprocity. That is: I would like to get, if possible, a more elementary solution. I ask for help. | 2685813 | Prove that every prime factor of [imath]12x^2+1[/imath] has the form [imath]6n+1[/imath]
Let [imath]n[/imath] be an integer, prove that every prime factor of [imath]12x^2+1[/imath] has the form [imath]6m+1[/imath], where [imath]m[/imath] is an integer. I've gone as far as knowing that all primes [imath]p>3[/imath] are either [imath]6m+1[/imath] or [imath]6m-1[/imath], how do I use this, not sure whether to even use that. I know about the Legendre symbol and quadratic reciprocity |
2807641 | Uniformly convergence on the convergent circle for complex power series!
The power series [imath]f(z)=\sum_{n=0}^{\infty}a_nz^n[/imath] has convergent radius [imath]r>0[/imath], so [imath]f(z)[/imath] is analytic in [imath]|z|<r[/imath]. MOREOVER, we suppose [imath]f(z)=\sum_{n=0}^{\infty}a_nz^n[/imath] is continuous on the closed disk [imath]|z|\leq r[/imath]. My question is that: Does the series [imath]\sum_{n=0}^{\infty}a_nz^n[/imath] converge uniformly on the closed disk [imath]|z|\leq r[/imath]? In my opinion, I can give a example such as: the series [imath]f(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^2}[/imath] has radius [imath]r=1[/imath] and converges uniformly on the closed unit disk [imath]|z|\leq1[/imath]. I do not know whether my question is right or wrong. Any hints will welcome! | 390407 | Existence of a power series converging non-uniformly to a continuous function
I am wondering whether there exist a function [imath]f(z) = \sum_{n\geq0} a_n z^n[/imath] such that: [imath]f[/imath] converges and is continuous on the closed unit disk [imath]D[/imath] and the series [imath]\sum_n a_n z^n[/imath] does not converge uniformly on [imath]D[/imath]. I have tried to construct a counter-example, but with no success so far. |
2808465 | Integrate [imath]\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx[/imath]
Integrate [imath]\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx[/imath] [imath]\newcommand{\intd}[1]{\,\mathrm{d}#1}[/imath] My Attempt Put [imath]t=\tan x\implies\intd t=\sec^2x\intd x[/imath] \begin{align*} \int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx&=\int_0^{\pi/2}\frac{\tan x\cdot\sec^2x}{(1+m^2\tan^2 x)\sec^2x}\intd x\\ &=\int_0^{\infty}\frac{t}{(1+m^2t^2)(1+t^2)}\intd t \end{align*} Put [imath]t^2=y\implies 2t\intd t=\intd y[/imath] \begin{align*} \frac{1}{2}\int_0^\infty\frac{dy}{(1+m^2y)(1+y)}&=\frac{1}{2}\int_0^\infty\bigg[\frac{1}{1-m^2}\cdot\frac{1}{1+y}+\frac{m^2}{m^2-1}\cdot\frac{1}{1+m^2y}\bigg]dy\\ &=\frac{1}{2(1-m^2)}\int_0^\infty\frac{dy}{1+y}-\frac{m^2}{2(1-m^2)}\int_0^\infty\frac{dy}{1+m^2y}\\ &=\bigg[\frac{1}{2(1-m^2)}\log|1+y|-\frac{1}{2(1-m^2)}\log|1+m^2y|\bigg]_0^\infty\\ &=\bigg[\frac{1}{2(1-m^2)}\log|\frac{1+y}{1+m^2y}|\bigg]_0^\infty=\color{red}{\frac{1}{2(1-m^2)}\bigg[\frac{\infty}{\infty}-0\bigg]}\end{align*} I think I am getting stuck here as the substitutions does not seem to give the solution ? Doubt [imath] \lim_{y\to 0}\log|\frac{1+y}{1+m^2y}|=\log 1=0\\ \lim_{y\to \infty}\log|\frac{1+y}{1+m^2y}|=\lim_{y\to \infty}\log|\frac{\frac{1}{y}+1}{\frac{1}{y}+m^2}|=\log\frac{1}{m^2} [/imath] So what really we are doing with definite integrals ?. Does this mean that we are actually computing the upper and lower limits and taking the difference to find the definite integral ? Note: Expanding in terms of [imath]\sin x[/imath] and [imath]\cos x[/imath] gives the solution, no doubt about limit going to infinity, not looking for that. | 1005976 | Finding [imath]\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x[/imath]
How do we prove that [imath]I(m)=\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x=\frac{\log{m}}{m^2-1}[/imath] I see that [imath]I(m)=\frac{\partial}{\partial m} \int_{0}^{\pi/2} \arctan({m\tan x}) \ \mathrm{d}x[/imath] But I don't see how to use this fact. Can we? Please help me out, and if possible please post a solution using differentiation under the integral sign. |
2808984 | Separable metric space spanned by base
Suppose [imath]X[/imath] is a metric space . If [imath]X[/imath] happens to be separable ,then countably many open subsets are enough to form a base. How to prove this one? Any suggestion will be appreciated. | 1810581 | Every separable metric space has a countable base
A collection [imath]\{V_{\alpha}\}[/imath] of open subsets of [imath]X[/imath] is said to be a base for [imath]X[/imath] if the following is true: For every [imath]x\in X[/imath] and every open set [imath]G\subset X[/imath] such that [imath]x\in G[/imath], we have [imath]x \in V_\alpha \subset G[/imath] for some [imath]\alpha[/imath]. In other words, every open set in [imath]X[/imath] is the union of а subcollection of [imath]\{V_{\alpha}\}[/imath]. Prove that every separable metric space has a countable base. Proof: Let [imath](X,d)[/imath] be a metric space and [imath]D=\{d_j\}_{j\in \mathbb{N}}[/imath] be the countable and dense subset of [imath]X[/imath]. Let [imath]U[/imath] is an open set in [imath]X[/imath] then for [imath]x\in U[/imath] [imath]\exists \varepsilon_x>0[/imath] such that [imath]N_{\varepsilon_x}(x)\subset U[/imath]. Then [imath]\exists d_j\in D[/imath] such that [imath]d(x, d_j)<{\varepsilon_x}/{4}[/imath] [imath]\Rightarrow[/imath] [imath]d_j\in N_{{\varepsilon_x}/{4}}(x)\subset U[/imath]. It's easy to check that [imath]N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)[/imath]. Indeed, if [imath]z\in N_{{\varepsilon_x}/{4}}(d_j)[/imath] then [imath]d(z,d_j)<\varepsilon_x/4[/imath] and since [imath]d(x, d_j)<{\varepsilon_x}/{4}[/imath] then by triangle inequality: [imath]d(z,x)\leqslant d(z,d_j)+d(x, d_j)<{\varepsilon_x}/{4}+{\varepsilon_x}/{4}<{\varepsilon_x}.[/imath] Hence [imath]z\in N_{{\varepsilon_x}}(x)[/imath] and inclusion [imath]N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)[/imath] holds. By the way [imath]N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U[/imath] for [imath]r\in \mathbb{Q}[/imath] and [imath]r<{\varepsilon_x}/{4}[/imath]. Thus for every [imath]x\in U[/imath] we associate an open ball with center at some point [imath]d_j[/imath] of [imath]D[/imath] and rational radius [imath]r[/imath] such that [imath]x\in N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U[/imath]. Thus [imath]U=\bigcup\limits_{d_j\in U} N_{r_j}(d_j).[/imath] This set equality can be verified very easy. Inclusion [imath]\subseteq[/imath] I described in above paragraph. Converse inclusion is not difficult: if [imath]z\in \bigcup\limits_{d_j\in U} N_{r_j}(d_j)[/imath] then [imath]z\in N_{r_j}(d_j)[/imath] for some [imath]d_j\in U[/imath]. But these balls arranged so that [imath]N_{r_j}(d_j)\subset U[/imath] then [imath]z\in U[/imath]. Hence collection of balls [imath]\{N_{r}(d): r\in \mathbb{Q}, d\in D\}[/imath] is a countable base. Is my proof correct? Can anyone check it please? Sorry if this topic is repeated but I would like to know is my proof correct. |
2807443 | What does the equal sign mean in mathematics?
When we say [imath]a=a[/imath] means that are absolutely the same right? What does equal mean in [imath]3+2=7-2[/imath]? Does it mean that they are absolutely the same or that the value is just the same? | 1347569 | Explaining the meaning of equality
I've been tasked with explaining to a group of people what the notion of equality means in mathematics, I've come up with a working explanation, but would appreciate some input, suggestions etc. Mathematical equality [imath]X=Y[/imath] is a binary relation between two mathematical expressions [imath]X[/imath] and [imath]Y[/imath] which is essentially the statement that the two expressions represent the same mathematical object ([imath]X[/imath] and [imath]Y[/imath] define the same mathematical object) if the equality holds for any value of [imath]X[/imath] and any value of [imath]Y[/imath] (in other words, they share all and only the same properties). If the equality doesn't hold for all values of [imath]X[/imath] and [imath]Y[/imath] then the relation [imath]X=Y[/imath] is a conditional statement that holds for a certain set of values of [imath]X[/imath] and [imath]Y[/imath]. I was then hoping to move on to the notion of equivalence relations and how equality is the archetypal example. How does one define equality as an equivalence relation? Also, is the reason why the equivalence classes of equality (as an equivalence relation) contain only a single element (in each class) simply because the only element which satisfies all the conditions of equality with another element (i.e. shares all and only the same mathematical properties) is the element itself (i.e. each element is equal to itself and only itself). Hope this is clear enough, apologies if it isn't (I feel like I've confused myself a bit through trying to explain the concept). [imath]\underline{\textbf{Edit}}[/imath]: Would it be correct to define the notion of equality in terms of equality of sets, i.e. given two sets [imath]A[/imath] and [imath]B[/imath], then [imath]A=B[/imath] if and only if, for every [imath]x\in A[/imath] we have that [imath]x\in B[/imath], and for every [imath]x\in B[/imath] we have that [imath]x\in A[/imath]? If this is the case then I can see how equality of (natural) numbers follows, as if we define them as [imath]0=\emptyset,\;1=\lbrace\emptyset\rbrace,\;2=\lbrace\emptyset,\lbrace\emptyset\rbrace\rbrace,\ldots,n=\lbrace 0,\ldots,n-1\rbrace[/imath] then it is obvious that [imath]1=0[/imath] is not true because [imath]\emptyset =\lbrace\emptyset\rbrace[/imath] is not true (as [imath]\emptyset\in\lbrace\emptyset\rbrace[/imath], but [imath]\emptyset\notin\emptyset[/imath]). |
2809058 | Why does continuous linear functional [imath]f[/imath] in topological space [imath]V[/imath] have to have an inverse?
Why does continuous linear functional [imath]f[/imath] in topological space [imath]V[/imath] have to have an inverse? Like in the proof for kernel of continuous functional being closed. https://mathproblems123.wordpress.com/2011/01/27/functional-is-continuous-iff-kernel-is-closed/ If f is continuous then [imath]\ker f=f^{-1}(0)[/imath] ... But how does one know [imath]f^{-1}[/imath] exists? | 437829 | The preimage of continuous function on a closed set is closed.
My proof is very different from my reference, hence I am wondering is I got this right? Apparently, [imath]F[/imath] is continuous, and the identity matrix is closed. Now we want to show that the preimage of continuous function on closed set is closed. Let [imath]D[/imath] be a closed set, Consider a sequence [imath]x_n \to x_0[/imath] in which [imath]x_n \in f^{-1}(D)[/imath], and we will show that [imath]x_0 \in f^{-1}(D)[/imath]. Since [imath]f[/imath] is continuous, we have a convergent sequence [imath]\lim_{n\to \infty} f(x_n) = f(x_0) = y.[/imath] But we know [imath]y[/imath] is in the range, hence, [imath]x_0[/imath] is in the domain. So the preimage is also closed since it contains all the limit points. Thank you. |
770735 | Show [imath]E(\frac{1}{2\theta\sum_{i = 1}^n W_i}) = \frac{1}{2(n-1)}[/imath] given the pdf [imath]f(y\mid\theta) = \theta y^{\theta - 1}[/imath]
Let [imath]Y_1, \ldots, Y_n[/imath] be a random sample from the pdf [imath]f(y\mid\theta) = \theta y^{\theta - 1}, 0 < y < 1, \theta > 0 .[/imath] show that [imath]E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}[/imath] I have showed that [imath]\sum_{i = 1}^n -\ln(Y_i)[/imath] is sufficient for [imath]\theta[/imath], and if [imath]W_i = -\ln(Y_i)[/imath], I have showed that [imath]W_i[/imath] has an exponential distribution with mean [imath]1/\theta[/imath]. I have also showed that [imath]2\theta \sum_{i=1}^n W_i [/imath] has a [imath]\chi^2[/imath] distribution with [imath]2n[/imath] degrees of freedom. My first question can we treat [imath]2\theta\sum_{i=1}^n W_i[/imath] as just 1/y? If so, I got to the following step [imath]\int_0^\infty \frac{1}{y} f(y) dy = \frac{1}{\Gamma(n)} \frac{1}{2} \int _0^\infty u^{n-2} e^{-u} du[/imath]... I am wondering how to solve [imath]\int _0^\infty u^{n-2} e^{-u} du?[/imath] I know it is going to be [imath]\Gamma(n-1)[/imath] but what is the process for finding it? | 769563 | Given the pdf [imath]f(y\mid\theta) = \theta y^{\theta - 1}1_{0, show E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}[/imath]
Let [imath]Y_1, \ldots, Y_n[/imath] be a random sample from the pdf [imath]f(y\mid\theta) = \theta y^{\theta - 1}, 0 < y < 1, \theta > 0 .[/imath] show that [imath]E\left(\frac{1}{2\theta\sum_{i = 1}^n W_i}\right) = \frac{1}{2(n-1)}[/imath] I have showed that [imath]\sum_{i = 1}^n -\ln(Y_i)[/imath] is sufficient for [imath]\theta[/imath], and if [imath]W_i = -\ln(Y_i)[/imath], I have showed that [imath]W_i[/imath] has an exponential distribution with mean [imath]1/\theta[/imath]. I have also showed that [imath]2\theta \sum_{i=1}^n W_i [/imath] has a [imath]\chi^2[/imath] distribution with [imath]2n[/imath] degrees of freedom. I am wondering how to show it? |
2809198 | Does a converging sequence of measurable functions necessarily converge towards a measurable function?
Say [imath]f_n\geq0\ \forall n[/imath] are measurable in the sense that [imath]\{x\ :\ f(x)<\alpha\}\in\mathcal M\ \forall\alpha\in\Bbb R[/imath] where [imath]\mathcal M[/imath] is the set of Lebesgue measurable subsets of [imath]\Bbb R[/imath]. And [imath]f_n\rightarrow f[/imath] Is [imath]f :\Bbb R\mapsto\Bbb R[/imath] necessarily a measurable function in the same sense? | 1327081 | How to prove limit of measurable functions is measurable
I need help to prove the following theorem Suppose [imath]f[/imath] is the pointwise limit of a sequence of [imath]f_n[/imath], [imath]n = 1, 2, \cdots[/imath], where [imath]f_n[/imath] is a Borel measurable function on [imath]X[/imath]. Then [imath]f[/imath] is Borel measurable on [imath]X[/imath]. My idea is to use the standard definition like for every [imath]c[/imath],[imath]\{x:f(x)<c\}[/imath] is Borel measurable. But got stuck as how to do it for sequence of [imath]f_n[/imath]. |
2809458 | How do I use prove this function is bijective?
Suppose I have the following function [imath]e : \mathbb{N} \times \mathbb{N} \to \mathbb{N}_1[/imath], where [imath]\mathbb{N}_1[/imath] is the set of natural numbers starting from 1 instead of 0: [imath]e(m,n) = 2^m (2n+1)\ \ \ \forall m,n \in \mathbb{N}[/imath] I want to show that the map [imath]e[/imath] is bijective. So I have to show it is both injective and surjective. To show that it's injective, here are my thoughts: Show for all [imath]m,n[/imath], we have [imath]2^m[/imath] and [imath]2n+1[/imath] being strictly increasing, and so the function is also a strictly increasing one. Since it is strictly increase, there can only be a one-to-one input to output mapping, and so the function is injective trivially. Now surjection will be harder to prove. My thoughts are: This problem has actually 2 arguments instead of the standard 1 argument function, so I suppose I would need to prove individually for each [imath]m[/imath] and [imath]n[/imath]. Since they are natural numbers, induction come into mind. I should try to prove that for [imath]\forall m[/imath], starting from [imath]e(0,n)[/imath] and proving the inductive hypothesis, that [imath]2^m[/imath] is surjective. (is this true? A brief idea of Cantor's diagonalization argument makes me think this might be false). Then I try to prove for all [imath]2n+1[/imath] it is surjective. This can be done quite easily since if we keep [imath]m[/imath] constant, then the image [imath]2n+1[/imath] is essentially enumerable. In fact, can we use this argument to prove that [imath]e(m,n)[/imath] is enumerable for all [imath]m[/imath] we vary and [imath]n[/imath] is constant? Finally, having proven that for both cases where [imath]m[/imath] varies and [imath]n[/imath] is constant (and vice versa) they are injective, we can see that the resultant function is surjective as it is closed under multiplication. Is this the right way to define it? That is the most reasonable shot I can give so far but I've no idea whether this is correct. Note: the goal is to prove this without using the fundamental theorem of arithmetic. | 1575338 | Prove [imath]{2}^{n}\cdot(2m+1)-1 [/imath] is invertible
Prove that the function [imath]{2}^{n}\cdot(2m+1)-1[/imath] from [imath]\Bbb{N}\times\Bbb{N}\to\Bbb{N}[/imath] is invertible I know that a function to be invertible must be injective and surjective, I am not sure how to calculate this since in this case I need a pair [imath](x,y)[/imath] since the function comes from [imath]{\Bbb{N}}^{2}[/imath]. |
2809831 | Coefficient of [imath]x^2[/imath] in the polynomial
How to find the coefficient of [imath]x^2[/imath] in the polynomial [imath](1-x)(1+2x)(1-3x)....(1+14x)(1-15x)[/imath] Is there any particular way to look at such problems? | 1317174 | Coefficient Problem (polynomial expansion)
Let [imath]C[/imath] be the coefficient of [imath]x^2[/imath] in the expansion of the product [imath](1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).[/imath] Find [imath]|C|.[/imath] Just to begin, [imath](1-x)(1+2x) = -2x^2 + x + 1[/imath] [imath](1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1[/imath] But expanding on like this take too long. In the end the terms will be like: [imath](1 - 15x)(a_0 + a_1x^1 + a_2x^2 + ....)[/imath] Then just looking at the [imath]x^2[/imath], it will be: [imath](a_2x^2 - 15a_1x^2) = x^2(a_2 - 15a_1)[/imath] But it it still a very hard problem, any hints? |
2808966 | How to prove that [imath]N\setminus A[/imath] is finite?
[imath]A \subseteq \mathcal{R}(N)[/imath] and given that (by inductive definition): [imath]N ∈ S[/imath]. If [imath]a \in R[/imath], then [imath]R \setminus \{a\} \in A[/imath]. I need to prove that for each [imath]A \in S[/imath], [imath]N\setminus A[/imath] is finite. How can I prove this formally? | 2814443 | Set theory - induction - countable sets
need help with the following question: Let [imath]S \subseteq P(\mathbb N)[/imath] be defined inductively so that: [imath]\mathbb N \in S[/imath] for all[imath] ~ a \in \mathbb N \Longrightarrow \mathbb N\setminus\{a\}\in S[/imath] for all [imath]A,B \in S \Longrightarrow A\cap B \in S[/imath] Prove that if [imath]A\subseteq \mathbb N~ [/imath] and [imath]~ \mathbb N [/imath] \ [imath]A[/imath] is finite then [imath]A \in S[/imath] thanks! |
2810275 | Given a [imath]3 \times n[/imath] board, find the number of ways to fill it with [imath]2\times1[/imath] rectangular boxes.
We have been given n, find total number of ways to form a [imath]3\times n[/imath] board using [imath]2\times 1[/imath] boxes. | 1317885 | Dominos ([imath]2 \times 1[/imath] on [imath]2 \times n[/imath] and on [imath]3 \times 2n[/imath])
How many ways are there to tile dominos (with size [imath]2 × 1[/imath]) on a grid of [imath]2 × n[/imath]? How about on a grid of [imath]3 × 2n[/imath]? |
2810229 | Properties of the multiplication table of [imath]\mathbb{Z}_m[/imath]
For which [imath]m[/imath] do all occurrence of 1 lie on the diagonal of the multiplication table of [imath]\mathbb{Z}_m[/imath]? I was think i have got a prove and find all the [imath]m[/imath] is [imath]1,2,3,4,6,8,12[/imath] and [imath]24[/imath]. but after a while, there is a bug in my prove. suppose [imath]m>10[/imath] and [imath]<10[/imath] part is too simple first [imath]m[/imath] cannot be odd, or [imath](2,m)=1[/imath] and there is x such that [imath]2x\equiv1\mod{m}[/imath], so [imath]x\ne2[/imath] for [imath]m>10[/imath] and this will against the 1 lie on diagonal. if [imath](3,m)=1[/imath], then [imath]3^2\equiv1\pmod{m}[/imath], then [imath]m\mid8[/imath] impossible too. so [imath]3\mid m[/imath]. if [imath](5,m)=1[/imath], using the same strategy, [imath]m|24[/imath], with [imath]m>10[/imath] verify [imath]\mathbb{Z}_{12}[/imath] and [imath]\mathbb{Z}_{24}[/imath] is a solution. so we just consider [imath]5\mid m[/imath]. I also verified [imath]7\mid m[/imath], so [imath]m=210\cdot Q[/imath], [imath]Q\in\mathbb{N}[/imath]. then, if [imath]x\in\mathbb{N}[/imath] and [imath]x<\sqrt{m}[/imath], that is [imath]x^2<m[/imath], so [imath]x^2\not\equiv1\pmod{m}[/imath].... so how to continue... | 2091280 | Find all positive integers [imath]n[/imath] such that [imath]a \equiv a^{-1} \pmod{n}[/imath] for all invertible [imath]a[/imath] modulo [imath]n[/imath]
Find all positive integers [imath]n[/imath] such that [imath]a \equiv a^{-1} \pmod{n}[/imath] for all invertible [imath]a[/imath] modulo [imath]n[/imath]. I found that [imath]n = 1,2,4,6,8,12,24[/imath] satisfy this. How can we prove that these are all of them? |
2810924 | Why is [imath]f(z-ia)= f(z+ia)[/imath]
Let [imath]f[/imath] be an entire function on [imath]\mathbb{C}[/imath]. Let [imath]g(z)=\overline{f(\bar z)}[/imath]. Then which of the following is/are true: [imath](1)[/imath] If [imath]f(z) \in \mathbb{R} [/imath] [imath]\forall z\in \mathbb {R}[/imath] then [imath]f(z)=g(z)[/imath] [imath](2)[/imath] If [imath]f(z)\in \mathbb {R} [/imath] [imath]\forall z\in \{z:im(z)=0\} \cup \{z:im(z)=a\}[/imath] for some [imath]a\gt[/imath]0 then [imath]f(z-ia)=f(z+ia)[/imath] [imath]\forall z\in\mathbb {C}[/imath] [imath](3)[/imath] If [imath]f(z)\in \mathbb {R} [/imath] [imath]\forall z\in \{z:im(z)=0\} \cup \{z:im(z)=a\}[/imath] for some [imath]a\gt[/imath]0 then [imath]f(z+2ia)=f(z)[/imath] [imath]\forall z\in\mathbb {C}[/imath] [imath](4)[/imath] If [imath]f(z)\in \mathbb {R} [/imath] [imath]\forall z\in \{z:im(z)=0\} \cup \{z:im(z)=a\}[/imath] for some [imath]a\gt[/imath]0 then [imath]f(z+ia)=f(z)[/imath] [imath]\forall z\in\mathbb {C}[/imath] What I have tried so far: For [imath](1)[/imath] it is easy to see that the set [imath]\{z \in \mathbb{C} : f(z)=g(z)\}[/imath] contains [imath]\mathbb{R}[/imath] and therefore by Identity Theorem, [imath]f(z)=g(z)[/imath]. And for part [imath](4)[/imath] Take a function [imath]f(z)=e^z[/imath] and then we see that [imath](4)[/imath] is NOT true. Am I correct? But I have no idea how to proceed for [imath](2)[/imath] & [imath](3)[/imath]? Any help would be appreciated. | 2766014 | Question on entire function
Question: let [imath]f[/imath] be an entire function on [imath]\mathbb{C}[/imath] and let [imath]g(z)=\overline{f(\bar{z})}[/imath] then which of the following is/are correct? (a) if [imath]f(z) ∈\mathbb{R}[/imath] for all [imath]z∈\mathbb{R}[/imath] then [imath]f=g[/imath] (b) if [imath]f(z)∈\mathbb{R}[/imath] for all [imath]z∈\{z : Im\text{ [/imath]z[imath] }=0\}∪\{z : Im\text{ [/imath]z[imath] }=a\}[/imath] for some [imath]a>0[/imath] then [imath]f(z+ ia)=f(z-ia)[/imath] for all [imath]z∈\mathbb{C}[/imath] (c) if [imath]f(z)∈\mathbb{R}[/imath] for all [imath]z∈\{z : Im\text{ [/imath]z[imath] }=0\}∪\{z : Im\text{ [/imath]z[imath] }=a\}[/imath] for some [imath]a>0[/imath] then [imath]f(z+ 2ia)=f(z)[/imath] for all [imath]z∈\mathbb{C}[/imath] (d) if [imath]f(z)∈\mathbb{R}[/imath] for all [imath]z∈\{z : Im\text{ [/imath]z[imath] }=0\}∪\{z : Im\text{ [/imath]z[imath] }=a\}[/imath] for some [imath]a>0[/imath] then [imath]f(z+ ia)=f(z)[/imath] for all [imath]z∈\mathbb{C}[/imath] My attempt: clearly, if [imath]z∈\mathbb{R}[/imath] then [imath]\bar{z}=z∈\mathbb{R}[/imath]. Hence if [imath]f(z)∈\mathbb{R}[/imath] then [imath]f(\bar{z})=f(z)∈\mathbb{R}[/imath] so that [imath]g(z)=\overline{f(\bar{z})}=f(\bar{z})=f(z)[/imath] for [imath]z[/imath] in [imath]\mathbb{R}[/imath] hence (a) is correct. But as beginner i am unable to prove or discard (b), (c), (d). I am stuck on this from hours, Please help me.. |
2810976 | sequence of random variables choosen from the interval [imath][0,1][/imath]
Suppose I have [imath]X_1\sim Uniform[0,1][/imath]. Let us define a sequence of random variables as follows: We will choose a random point from the interval [imath][0,X_1][/imath] and mark it by [imath]X_2[/imath]. then we will choose a random point from the interval [imath][0,X_2][/imath] and mark it by [imath]X_3[/imath] and so on. 0 I need to find a formula for the density function [imath]f_{{X}_n}[/imath]. I could see that [imath]X_n[/imath] has more probability to be near 0. I tried to express [imath]F_{X_n}[/imath] in terms of [imath]F_{X_{n-1}}[/imath] and [imath]F_{X_{n-2}}[/imath] but got only to this equation: [imath]F_{X_{n}}(x_{n+1})=\frac{F_{X_{n+1}}(x_{n+2})}{F_{X_{n-1}}(x_n)}[/imath] that leads me nowhere even if I trying to derivate it. Thank you, Michael | 2810381 | pdf of a member of a sequence of dependent random variables
I would very much appreciate a hint for the following problem Let [imath]\left(X_n\right)_{n=1}^\infty[/imath] be a sequence of random variables s.t.: [imath]X_1 \sim U_{[0,1]}[/imath] and for all [imath]n>1[/imath]: [imath]X_n \sim U_{[0,X_{n-1}]}.[/imath] Give a general expression for [imath]f_{X_n}[/imath] the pdf of [imath]X_n[/imath]. Thanks |
2811009 | Determine number of ways
How can we find with how many ways we can choose three subsets [imath]A_1, A_2, A_3[/imath] of [imath]\{1,2,3, \dots, 9\}[/imath] such that [imath]A_1 \cap A_2 \cap A_3= \varnothing[/imath] ? (their pairwise intersection can be non-empty) Can you give me a hint? | 1704153 | How many ways are there to choose 3 subsets [imath]A,B,C[/imath] of a set [imath]\{1,2,3...n\}[/imath] such that [imath] A \cap B \cap C = \emptyset [/imath]
I believe that this is a pretty simple question in terms of combinatorics. Assuming we have the following set [imath]\{ {1,2,3...n} \}[/imath] In how many ways can we choose 3 subsets (A,B,C) such that the intersection between all them is empty? [imath] A \cap B \cap C = \emptyset [/imath] Well, what I have so far is the following: For A we have [imath] 2^n [/imath] options (where n is the size of the set) Then, for B, the remaining options are exactly [imath] 2^{n-|A|} [/imath] since we can choose only from subsets who does not have elements in common with A. And finaly, for C, since we already have an empty intersection, we can choose whatever subset we want and get the expected result [imath] \emptyset [/imath]. But for some reason it feels like the numbers are too big. If we take for example the set {1,2}, then, according to my "formula", the number of options should be: [imath] \binom{4}{1} * \binom{2}{1} * \binom{2}{1} = \frac{4!}{3!} * \frac{2!}{1!} * \frac{2!}{1!} = 16 [/imath] Does it make sense? |
2810476 | How to prove [imath]1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5\theta)}{\sin\theta} [/imath]?
I need help to prove that the following is true: [imath]1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5 \theta)}{\sin\theta}[/imath] I realize that I must evaluate the real part of this, but whatever I get I am not quite sure how to get to the required expression. I have multiplied the numerator and denominator of the result of the geometric sum by the conjugate of [imath] e^{2i\theta} [/imath] and still have no luck. [imath]\sum_{i=0}^4 e^{2ni\theta}[/imath] (Apologies for poor formatting) | 2618462 | Proving complex series [imath]1 + \cos\theta + \cos2\theta +... + \cos n\theta [/imath]
So I have this result [imath]1 + z + z^2 + ... + z^n = \frac{z^{n+1}-1}{z-1}[/imath] which I proved already. Now I am supposed to use that result and De Moivre's formula to establish this identity [imath]1 + \cos\theta + \cos2\theta +... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n+\frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}[/imath] Can anyone help me? |
2810179 | Group isomorphic to all its proper quotients, and not simple
Let [imath]G[/imath] be a group such that for any proper normal subgroup [imath]H \subset G[/imath], we have [imath]G/H \cong G[/imath] (of course this isomorphism may not be given by the projection [imath]G \to G/H[/imath]). Does it follow that [imath]G[/imath] is a simple group? What can we say about such a group [imath]G[/imath] in general? Notice that [imath]G/H \cong G[/imath] doesn't imply [imath]H = \{e\}[/imath], for instance [imath]z \mapsto z^2[/imath] is a surjective morphism [imath]\Bbb C^{\times} \to \Bbb C^{\times}[/imath] of kernel [imath]\{\pm 1\}[/imath]. | 768561 | Sort-of-simple non-Hopfian groups
A finite simple group is one which has no homomorphic images apart from itself and the trivial group. However, the simple-groups tag does not include the condition "finite". My question is the following. Is the following true? Claim: A simple group is one where the only homomorphic images are itself and the trivial group. However, I am asking this question not because of the tag wiki, but because I think any counter-example would be interesting, especially if it was finitely generated (and especially especially if it was finitely presented). That is, Does there exists a (finitely presented) group [imath]G[/imath] which is not simple but where the only homomorphic images are itself and the trivial group? Such a counter-example would be non-Hopfian, and would be sort-of-simple, by the definition of a simple group that we want to exist. Thus the title of the question. |
2808045 | How to solve [imath]\dot x(t) = x(2t)[/imath]
During my recent work I encountered a weird type of differential equation containing scalar factors. In order to understand them better I thought it would make sense to look at a simpler example like: [imath] \dot x(t) = x(2t)[/imath] How does one solve this equation? Does it even have a solution? It seems kinda related to Functional Differential Equations which I am not familiar with. | 1365327 | What function satisfies [imath]F'(x) = F(2x)[/imath]?
The exponential generating function counting the number of graphs on [imath]n[/imath] labeled vertices satisfies (and is defined by) the equations [imath] F'(x) = F(2x) \; \; ; \; \; F(0) = 1 [/imath] Is there some closed form or other nice description of this function? Does it have a name? Of course, the series itself does not converge for any nonzero [imath]x[/imath], but like the Lambert W function (counting trees) it has combinatorial meaning. And the Lambert W function has a nice description as the inverse of [imath]x e^x[/imath]; maybe there is a similar description of [imath]F[/imath]? |
2811487 | Find the fundamental solution of [imath]x^2 - 95y^2 = 1[/imath] and of [imath]x^2 - 74y^2 = 1[/imath].
I'm trying to practice finding the fundamental solution of Pell Equations, but have not found efficient ways of finding fundamental solutions. However, it seems that using continued fractions would provide an effective way of finding these solutions, according to the link. To better understand how to actually use continued fractions through examples (with numbers), How would you find the fundamental solution of [imath]x^2 - 95y^2 = 1[/imath] and of [imath]x^2 - 74y^2 = 1[/imath] using continued fractions? | 1045127 | How to find a fundamental solution to Pell's equation?
It is quite straightforward to find the fundamental solutions for a given Pell's equation when [imath]d[/imath] is small. But I am unable to solve this equation, as I'm unable to find the fundamental solutions: Solve: [imath]x^2-29y^2=1[/imath] and [imath]x^2-29y^2=-1[/imath] with [imath]y\not=0[/imath]. Could you please guide me through the solution |
1927147 | Probability of getting 2 tails on 3 coin tosses?
I just learned in class that if you toss 3 coins (a quarter, a dime and a nickel), the probability of getting 2 tails only is [imath]\frac14[/imath] and I don't really understand why. My first intuition is that: 0 heads => 1 tail 1 head => 2 tails 2 heads => 1 tail 3 heads => 0 tails Therefore, it should be [imath]\frac14[/imath]. I understand that the number of permutations is 8, thus: HHH THT HHT THH HTT TTH TTT HTH And if you take those with 2 tails, it's [imath]\frac38[/imath]. But where is the flaw in my first reasoning? Why do I care about which coin gets Tails when I only want to know the probability of getting 2 tails? Thanks for your help. | 2158911 | Possible outcomes on a three coin flip.
If you flip a coin three times, the possible outcomes are [imath]\left \lbrace HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \right \rbrace[/imath] What is the probability of getting two tails? |
2811610 | Finitely generated field is a finite field
A ring [imath]R[/imath] is said to be finitely generated if there exists a finite subset [imath]A[/imath] of [imath]R[/imath] such that for all rings [imath]R'\subset R[/imath] with [imath]A\subset R'[/imath] we have [imath]R'=R[/imath]. ([imath]A[/imath] is called the generating set of [imath]R[/imath]) Following is the statement that I want to prove: If [imath]F[/imath] is a field which is finitely generated as a ring, then [imath]F[/imath] is a finite field. This is supposed to follow as a corollary to the general Nullstellensatz (Eisenbud, Theorem 4.19) Theorem: Let [imath]R[/imath] be a Jacobson ring. If [imath]S[/imath] is a finitely generated [imath]R[/imath]-algebra with the ring homomorphism [imath]\varphi: R\to S[/imath], then [imath]S[/imath] is a Jacobson ring. Further, if [imath]J\subset S[/imath] is a maximal ideal, then [imath]I = \varphi^{-1}(J)[/imath] is a maximal ideal of [imath]R[/imath], and [imath]S/J[/imath] is a finite field extension of [imath]R/I[/imath]. If [imath]F[/imath] is a field of characteristic [imath]p[/imath] then the result follows by taking [imath]R=\mathbb{F}_p[/imath] and [imath]S=F[/imath] in the above theorem. But, how to show that [imath]F[/imath] can't be of characteristic [imath]0[/imath]? | 103537 | A slick proof that a field which is finitely generated as a ring is finite
It is a known fact that if [imath]k[/imath] is a field that is finitely generated as a ring, which is the same as having a surjective ring homomorphism [imath]f:\mathbb{Z}[x_1,\dots,x_n]\to k[/imath] for some [imath]n\in \mathbb{N}[/imath], then [imath]k[/imath] must be finite. Since finite generation as a ring implies finite generation over the prime field ([imath]\mathbb Q[/imath] or [imath]\mathbb F_p[/imath]), by Noether normalization it follows that [imath]k[/imath] must be a finite extension of its prime field. In positive characteristic this finishes the job and in zero characteristic, should lead quickly to a contradiction, though I don't see immediately how. Is there an elementary/slick proof of this fact? |
782690 | Simple module over matrix rings
I'm trying to prove that if [imath]M[/imath] is a simple module over [imath]M_n(D)[/imath] where [imath]D[/imath] is a division algebra, then [imath]M\cong D^n[/imath]. I know that if [imath]M[/imath] is a simple module over [imath]R[/imath] then it is isomorphic to [imath]Rv=\{rv|r\in R\}[/imath] for all [imath]v\in M[/imath]. I'm trying to find a [imath]M_n(D)[/imath] module isomorphism [imath]f:M_n(D)v\rightarrow D^n[/imath]. I've been told to try the map that maps [imath]Xv\mapsto Xe_1[/imath], where [imath]e_1[/imath] is the column vector with 1st entry 1 and the rest 0. I can prove that this is a homomorphism but I'm stuck on the injectivity. Any help would be well appreciated! | 874233 | Simple [imath]M_n(D)[/imath]-module with [imath]D[/imath] a division ring
Define [imath]D[/imath] to be a division algebra over a field [imath]k[/imath] and [imath]R=M_n(D)[/imath] the [imath]n\times n[/imath] matrix ring over [imath]D[/imath]. A simple [imath]R[/imath]-module [imath]M[/imath] is the quotient of [imath]R[/imath]. I can write [imath]R=\bigoplus_j I_j[/imath] where [imath]I_j[/imath] is the subring of [imath]R[/imath] all of whose columns except the [imath]j[/imath]-th are zero. Is it true that the induced map [imath]R\to M[/imath] induces an isomorphism between [imath]M[/imath] and some [imath]I_j[/imath]? Many thanks! |
2810316 | [imath]M[/imath] is flat [imath]\Leftrightarrow S^{-1}M[/imath] is a flat [imath]S^{-1}R[/imath]-module
Hi I have the following problem: Let [imath]R[/imath] be a ring and [imath]S\subset R[/imath] multiplicatively closed and [imath]M[/imath] be an [imath]R[/imath]-module. Show: [imath]M[/imath] is flat [imath]\Leftrightarrow S^{-1}M[/imath] is a flat [imath]S^{-1}R[/imath]-module. I think I can show [imath]"\Rightarrow[/imath]" but for [imath]"\Leftarrow"[/imath] I don't know how to start. Can someone help me please? | 2803986 | Localization and flatness
Let [imath]M[/imath] be a flat [imath]S^{-1}A[/imath]-module. I want to show that [imath]M[/imath] is flat as an [imath]A[/imath]-module. [imath]\newcommand{\tp}{\otimes}[/imath] My "best" attempt is the following: Let [imath]0 \to N_1 \to N_2 \to N_3 \to 0[/imath] be exact (of [imath]A[/imath]-modules). Then, [imath]0 \to S^{-1}N_1 \to S^{-1}N_2 \to S^{-1}N_3 \to 0[/imath] is exact, and as [imath]M[/imath] is flat, [imath]0 \to S^{-1}N_1 \tp_{S^{-1}A} M \to S^{-1}N_2 \tp_{S^{-1}A} M \to S^{-1}N_3 \tp_{S^{-1}A} M \to 0[/imath] is flat, and due to [imath]S^{-1}N_i \tp_{S^{-1}A} M = S^{-1}N_i \tp_{S^{-1}A} S^{-1}M = S^{-1}(N_i \tp_A M)[/imath] [imath] 0 \to S^{-1}(N_1 \tp_A M) \to S^{-1}(N_2 \tp_A M) \to S^{-1}(N_3 \tp_A M) \to 0 [/imath] is exact. But I don't know why [imath] 0 \to N_1 \tp_A M \to N_2 \tp_A M \to N_3 \tp_A M \to 0 [/imath] is exact, which would imply the flatness of [imath]M[/imath] as an [imath]A[/imath]-module. |
2811965 | How to show infinite solutions of linear equation system?
For example we have α, β belongs to [imath]\Bbb R[/imath], then how to show that the following system of linear equations will have infinite many solutions whenever β belongs to [imath][−\sqrt 2,\sqrt2][/imath]? [imath]\left\{ \begin{align} -x + \sin \alpha\,y - \cos\alpha\,z=0,\\ \beta\, x + \sin \alpha\,y + \cos\alpha\,z = 0,\\ x + \cos\alpha\,y + \sin \alpha\,)z= 0. \end{align}\right.[/imath] | 2810570 | Prove that a system of linear equations, will have infinitely many solutions whenever there is a specific value.
I have a question where [imath]\alpha[/imath] and [imath]\beta[/imath] are elements of [imath]\Bbb R[/imath], and I have to show that a system of linear equations will have infinitely many solutions whenever [imath]\beta[/imath] is an element of [imath]\bigl[ -\sqrt 2, \sqrt 2\bigr][/imath]. There are three variables that have values like [imath]\sin\alpha,\cos\alpha, \beta x[/imath], etc. Now I know how to solve regular multi-variable linear equations to check if a linear system has infinite solutions or not, but this one is really tricky.... I don't even know where to start. PS. I am really bad at math, I don't even know how to read notations. Here's the actual question: Let [imath]\alpha[/imath], [imath]\beta[/imath] [imath]\in[/imath] R; then show that the following system of linear equations, will have infinite many solutions whenever [imath]\beta[/imath] [imath]\in[/imath] [- [imath]\sqrt 2[/imath], [imath]\sqrt 2[/imath] ]; [imath]\ - x + (\sin \alpha )y - (\cos \alpha )z = 0\\ \beta x + (\sin \alpha )y + (cos\alpha )z = 0\\ x + (cos\alpha )y + (sin\alpha )z = 0 [/imath] |
1858939 | Find all integer roots of: [imath]x^2(y-1)+y^2(x-1)=1[/imath]
Find all integer roots of: [imath]x^2(y-1)+y^2(x-1)=1[/imath] Obviously [imath](2,1)[/imath] and [imath](1,2)[/imath] are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers! I also tried to view it as a quadratic equation in terms of [imath]x[/imath], but forming the Delta of that equation didn't help much other than if [imath]y=1[/imath] then certainly Delta would be a square and equation would have answer for [imath]x[/imath]... | 3054222 | Integer solutions to [imath]x^2(y-1)+y^2(x-1)=1[/imath]
Find all values of [imath](x,y)[/imath] where [imath]x,y[/imath] belongs to integers if: [imath]x^2(y-1)+y^2(x-1)=1[/imath] I m a beginner so I do need some help. |
2811973 | On the space of continuous functions of a second-countable space
Let [imath]X[/imath] be a locally compact, Hausdorff, second-countable space, and let [imath]C(X,\mathbb{C})[/imath] the space of continuous, complex-valued functions over X endowed with the uniform norm. Is [imath]C(X,\mathbb{C})[/imath] separable? | 374474 | [imath]X[/imath] second countable locally compact Hausdorff implies [imath]C(X)[/imath] separable?
In the post When is [imath]C_0(X)[/imath] separable? , it is argued that if [imath]X[/imath] is second countable and locally compact Hausdorff, then [imath]C_{0}(X)[/imath] is separable. Is it also true that [imath]C(X)[/imath] is separable under the same hypotheses? |
2686944 | Is the space of continuous bounded real valued functions on a Polish space separable?
Let [imath]X[/imath] be a complete and separable metric space. Define \begin{equation} M := \{ f \, \mid \, f : X \rightarrow \mathbb{R}, \, f \text{ bounded and continuous } \} \end{equation} Is M separable with respect to supremum norm? | 301110 | [imath]X[/imath] metric separable then [imath]C(X)[/imath] separable
Is it true, that if [imath]X[/imath] is a separable metric space, then the space of all continuous functions on [imath]X[/imath] with the supremum metric is also separable? |
2812320 | differentiability of [imath]|xy|[/imath] at [imath](0,0)[/imath]
Let, [imath]f(x,y)=|xy|[/imath], Show that [imath]f[/imath] is differentiable at [imath](0,0)[/imath] I started with by trying to prove [imath]\lim_{h\to 0,k\to 0}\dfrac{f(0+h,0+k)-f(0,0)}{hk}[/imath] exists. Then, [imath]\lim_{h\to 0,k\to 0}\dfrac{|hk|}{hk}=\lim_{h\to 0,k\to 0}\dfrac{|h||k|}{hk}=\lim_{h\to 0}\dfrac{|h|}{h}\lim_{k\to 0}\dfrac{|k|}{k}[/imath] Can I do this? Because, if I do this, [imath]f[/imath] may not be differentiable, but I have to show the converse. Thanks, for any help! | 1832065 | Is [imath]|xy|[/imath] differentiable in [imath](0,0)[/imath]?
Is [imath]f(x,y) = |xy|[/imath] differentiable in [imath](0,0)[/imath]? I have no idea how to approach this problem. |
2812295 | Sum of a matrix with its transpose
I've a question that many of yours could consider stupid: if i sum a matrix with its transpose, I obtain a particular result? E.g. [imath]A + A^T = B[/imath], [imath]B[/imath] has some particular properties? | 506424 | How to prove [imath]A+ A^T[/imath] symmetric, [imath]A-A^T[/imath] skew-symmetric.
Prove that if [imath]A[/imath] is a square matrix, then: a) [imath]A+ A^T[/imath] is symmetric. b) [imath]A-A^T[/imath] is skew-symmetric. c) Use part (a) and (b) to show [imath]A[/imath] can be written as the sum of a symmetric matrix [imath]B[/imath] and a skew-symmetric matrix [imath]C[/imath], with [imath]A = B + C[/imath]. |
2813183 | Evaluate the integral [imath]\int_{\left | z \right |=3} (e^{z}-1)dz/(z(z-1)(z-i))[/imath]
I want to evaluate the integral [imath]\int_{\left | z \right |=3} \frac{e^{z}-1}{z(z-1)(z-i)}\,dz[/imath] by using the following theorem: If a function [imath]f[/imath] is analytic everywhere in the finite plane except for a finite number of singularities interior to a positively oriented simple closed curve [imath]C[/imath], then [imath]\int_{C}f(z) dz=2\pi i\text{Res}\left ( \frac{1}{z^{2}}f(\frac{1}{z}), 0 \right ).[/imath] Since [imath]\frac{1}{z^{2}}f(\frac{1}{z})=\frac{1+\frac{1}{2!z}+\frac{1}{3!z^{2}}\cdot \cdot \cdot }{1-(1+i)z+iz^{2}},[/imath] so I did long division, but the coefficient of 1/z(the residue of [imath]\frac{1}{z^{2}}f(\frac{1}{z})[/imath]) is [imath]0[/imath]. I know the method of calculating the residues at each singularities [imath]0[/imath], [imath]1[/imath], [imath]i[/imath], respectively. This method gives me the answer [imath]2\pi i(0+\frac{e-1}{1-i}+\frac{e^{i}-1}{i(i-1)})[/imath] which is not [imath]0[/imath]. Why answers are different?? Why does not hold long division method? Any help please. | 2812219 | Evaluate [imath]\int_{\left | z \right |=3} (e^{z}-1)dz/(z(z-1)(z-i))[/imath]
I want to evaluate the integral [imath]\int_{\left | z \right |=3} \frac{e^{z}-1}{z(z-1)(z-i)}\,dz[/imath] by using the following theorem: If a function [imath]f[/imath] is analytic everywhere in the finite plane except for a finite number of singularities interior to a positively oriented simple closed curve [imath]C[/imath], then [imath]\int_{C}f(z) dz=2\pi i\text{Res}\left ( \frac{1}{z^{2}}f(\frac{1}{z}), 0 \right ).[/imath] I want to calculate this integral by calculating the residue at 0 of [imath]\frac{1}{z^{2}}f(\frac{1}{z})=\frac{1+\frac{1}{2!z}+\frac{1}{3!z^{2}}\cdot \cdot \cdot }{1-(1+i)z+iz^{2}}[/imath] So I did long division but the coefficient of 1/z is 0. I know the method of calculating the residues at each singularities 0, 1, i, respectively. This method gives me the answer [imath]2\pi i(0+\frac{e-1}{1-i}+\frac{e^{i}-1}{i(i-1)})[/imath] which is not 0. Why answers are different?? Why does not hold long division method? Any help please. |
2812723 | What is the expectation of ‘the number. of coin flips until any string of 4 is repeated’, but each pattern can overlap?
I asked this question but did not get a reply, I am very keen to know how its done. I think I worded it poorly. You can flip a coin unlimited times. There are ([imath]2^{4} = 16[/imath]) blocks of 4, eg. HHTT, HHHH. What is the expectation of ‘the number. of coin flips until any pattern is repeated’, but each pattern can overlap? So HHHHH , would count as (HHHH)H and H(HHHH) | 2734550 | Expected Number of Flips for a Sequence of 4 to Repeat
I recently had this question in an interview: You are flipping a fair coin until a sequence of four flips repeats itself. The sequences are allowed to overlap. What is the expected number of flips? For example, if you flip HTHTHT, the sequence HTHT appears in flips 1-4 and 3-6. In this case, we are done after 6 flips. As another example, if you flip HHTTHHTT, the sequence HHTT repeats in flips 1-4 and 5-8, and we are done after 8 flips. What is the expected number of flips? I've been thinking about this question for a few days now, and I haven't come to an answer. I've tried the simpler problem where we consider a sequence of two flips instead of four flips, but it is still rather difficult. I suspect that there is a nice recursive way to solve this problem, but I can't figure it out. I am also interested in generalizations of this problem. For example, what is the expected number of flips for a sequence of [imath]n[/imath] to repeat itself? What happens if the coin isn't fair? I have another interview in a few days, so I would very much like to see how to solve this problem in advance. Any help is appreciated. Edit: Based on computational evidence (assuming I didn't make a mistake in the code), it appears that the expected number of flips is about 9.81. I would like to know the exact answer, as well as an analytic solution to this problem. Edit 2: Another piece of information that may be of use: I had 30 seconds to answer the question. This makes me believe that there is some "easy" way to get the answer, or they were looking for an approximate answer. Edit 3: @r.e.s. has kindly provided exact solutions for [imath]n=1,2,3,4,5[/imath]. For [imath]n=6[/imath], numerical computation seems to indicate that the answer is around [imath]18.977[/imath] or [imath]18.978[/imath]. I will be updating periodically with approximations for other values of [imath]n[/imath]. Edit 4: [imath] \begin{array}{|c|c|} \hline n & E(L_{n}) \\ \hline 1 & \frac{5}{2}\\ \hline 2 & \frac{35}{8}\\ \hline 3 & \frac{435}{64}\\ \hline 4 & \frac{2513}{256}\\ \hline 5 & \frac{57922047}{4194304}\\ \hline 6 & \approx 18.9775\\ \hline 7 & \approx 25.928\\ \hline 8 & \approx 35.288\\ \hline \end{array} [/imath] |
2813483 | Tagged trees when 1,2,3 are leaves and the minum path between them >2
Let [imath]V=(1,2,...,n), n>5[/imath] be a set of vertices. How many tagged trees are there on V such that the vertices 1,2,3 are leaves and the minimum path length between every 2 of them is greater than 2? What I've tried: I said that without 1,2 and 3 there are [imath](n-3)^{n-5}[/imath] tagged trees on V without 1,2,3. So now we need to connect 1,2,3 to the tree with one edge each (so it will be a leaf) and it can't be the same vertex in the tree every time because this will result a length of 2. So this will give: [imath](n-3)^{n-5}\cdot(n-3)\cdot(n-4)\cdot(n-5)[/imath] Is this nonsense or somewhat correct? | 2811683 | How many trees over [imath]{1,2,3,...n}[/imath] with conditions
I’m stuck on this question in graph theory. The question is: How many labeled trees are there over [imath]V={0,1,2,...n}[/imath] with which vertices 1,2,3 are leaves, and distance between any two of these leaves is 3 or more. I tried using Cayley theorem but I don’t know how to apply it in this specific question. |
2809727 | How to find the self-intersection point of [imath]x^x=y^y(x,y>0)[/imath]?
As the figure below shows, the graph of the implicit function [imath]x^x=y^y,(x,y >0)[/imath] composes of a straight line and an arc, which of the two have an intersection point [imath]P[/imath]. How to find the coordinates [imath](x_p,y_p)[/imath] of [imath]P[/imath]? Does there exist a closed-form solution? | 2251318 | Where does the curve [imath]x^y = y^x[/imath] intersect itself?
This problem is quite easily solved by using logarithms and derivative and forming the function [imath]f(x,y) = x^y - y^x[/imath], however there are assertions that this problem can be solved without using either of the two. I can not see how one would proceed to solve this problem in the absence of logarithmic manipulation and differentiation. Can anyone shed some light on this? |
2814833 | Sum of total variation over countable partition of interval equals variation over interval?
Let [imath]I=[a,b][/imath], and consider some [imath]f\in BV(I)[/imath] (the space of all functions with bounded total variation on [imath]I[/imath]). Let [imath]\mathcal Q=\{[a_k,b_k]:k\in\mathbb N,~\operatorname{cl}\left(\bigcup_{k=1}^\infty[a_k,b_k]\right)=I,~(a_k,b_k)\cap(a_j,b_j)=\emptyset~\forall i\neq j \}.[/imath] In their book, Laws of Chaos: Invariant Measures and Dynamical Systems in One Dimension, Gora and Boyarsky claim in a problem that it is possible for [imath]\sum_{k=1}^\infty V_{[a_k,b_k]}(f)<V_I(f).[/imath] I struggle to see how this is true, and believe I have a proof that [imath]\sum_{k=1}^\infty V_{[a_k,b_k]}(f)=V_I(f).[/imath] Now this question is pretty much the same as this one, but I do not think the accepted answer there is actually correct, as the example given shows that we can have [imath]V_{[a,c)}(f)+V_{[c,d]}(f)<V_I(f),[/imath] but the variation of a function over half open intervals is very different from that over closed intervals, so this does not actually address the question. In fact we know that we always have [imath]V_{[a,c]}(f)+V_{[c,d]}(f)=V_I(f).[/imath] My supposed proof for the countable case is a standard approximation argument showing inequalities both ways, but it is quite long to type out, and Boyarski and Gora are much wiser than me so my proof is most probably wrong. Thus before I go through the effort of typing it out I'd appreciate some input from the learned members here, because I might be being silly and missing an easy counterexample. EDIT: On inspecting the fantastic answer by mathcounterexamples.net I realised that I misinterpreted the given counterexample in the linked post, and it is actually shows perfectly what fails in the countable case. Thus I will be flagging this post as a duplicate, as it adds nothing new. | 874509 | variation of a function over countable intervals
Let [imath]f[/imath] be a function of bounded variation on [imath][0,1][/imath]. Let [imath]\{[a_n,b_n]\}_{n=1}^\infty[/imath] such that [imath](a_n,b_n)[/imath] are pairwise disjoint and [imath]\cup_{n=1}^\infty [a_n,b_n]=[0,1][/imath]. (for example, [imath][1/2, 1], [0,1/3], [1/3,1/3+1/3^2], \cdots[/imath] ) Can we write [imath] \operatorname{Var}_{[0,1]} f=\sum_{k=1}^\infty \operatorname{Var}_{[a_k,b_k]} f? [/imath] |
2815103 | If [imath]f: \mathbb{R}^2 \to \mathbb{R}[/imath] is polynomial in each argument, [imath]f[/imath] is polynomial
Given [imath]f: \mathbb{R}^2 \to \mathbb{R}[/imath] such that for all real [imath]y[/imath], [imath]t \mapsto f(t,y)[/imath] is a polynomial function and for all real [imath]x[/imath], [imath]t \mapsto f(x,t)[/imath] is a polynomial function, is it necessarily true that [imath](x,y) \mapsto f(x,y)[/imath] is a polynomial function? | 211297 | Is a bivariate function that is a polynomial function with respect to each variable necessarily a bivariate polynomial?
Let [imath] \mathbb{F} [/imath] be an uncountable field. Suppose that [imath] f: \mathbb{F}^{2} \rightarrow \mathbb{F} [/imath] satisfies the following two properties: For each [imath] x \in \mathbb{F} [/imath], the function [imath] f(x,\cdot): y \mapsto f(x,y) [/imath] is a polynomial function on [imath] \mathbb{F} [/imath]. For each [imath] y \in \mathbb{F} [/imath], the function [imath] f(\cdot,y): x \mapsto f(x,y) [/imath] is a polynomial function on [imath] \mathbb{F} [/imath]. Is it necessarily true that [imath] f [/imath] is a bivariate polynomial function on [imath] \mathbb{F}^{2} [/imath]? What if [imath] \mathbb{F} [/imath] is merely countably infinite? |
2815141 | Playing with automorphism over the rational functions
Let [imath]\mathbb F(t)[/imath] be the rational functions over the field [imath]\mathbb F[/imath] with [imath]t[/imath] as an argument, and Let [imath]\alpha,\beta \in Aut(\mathbb F(t))[/imath] such that [imath]\alpha(t)=\frac 1 t[/imath], [imath]\beta(t)=1-t[/imath]. (a) Define [imath]H=\langle \alpha,\beta \rangle \le Aut(\mathbb F(t))[/imath]. Show that [imath]H\cong S_3[/imath]. (b) Find [imath]y\in \mathbb F(t)[/imath] such that [imath]\mathbb F (x)^H=\mathbb F(y)[/imath]. For (a), It's not that hard to show directly that [imath]|\alpha \beta \alpha|=|\alpha|=|\beta|=2[/imath], and [imath]|\alpha \beta|=|\beta \alpha |=3[/imath], and so on, but I couldn't find something more elegant. For (b) I have no Idea. | 1413044 | Fixed Field of [imath]\sigma, \tau[/imath]
Let [imath]k[/imath] be a field and let [imath]K=k(x)[/imath] be the rational function field in one variable over [imath]k[/imath]. Let [imath]\DeclareMathOperator{\aut}{Aut}\sigma, \tau \in \aut(K)[/imath] s.t. [imath]\sigma\left(\frac {f(x)}{ g(x)}\right)=\frac {f(1/x)}{g(1/x)}\qquad \&\qquad \tau\left(\frac {f(x)}{ g(x)}\right)=\frac {f(1-x)}{g(1-x)}.[/imath] Find the fixed field [imath]F[/imath] of [imath]<\sigma, \tau>[/imath]. Determine [imath]\DeclareMathOperator{\gal}{Gal}\gal(k/F)[/imath]. Find an [imath]h \in F[/imath] so that [imath]F=k(h)[/imath]. Now for the first automorphism [imath]\sigma[/imath], I think if I concentrate on [imath]k[x][/imath] it will give me some idea. Now, if [imath]f(x)=a_0+a_1x+a_2x^2+ \cdots + a_nx^n \in k[x][/imath] then if [imath]x^n \sigma(f(x))=f(x)[/imath] the case will be [imath]a_i=a_{n-i}[/imath] [imath]\forall i=0(1)[n/2][/imath]. So all the funtions in [imath]K[/imath] which will be fixed by [imath]\sigma[/imath] are of the form [imath]\frac {f(x)}{ g(x)}[/imath] where [imath]f(x)=a_0+a_1x+a_2x^2+ \cdots + a_nx^n[/imath] and [imath]g(x)=b_0+b_1x+b_2x^2+ \cdots + b_nx^n[/imath] such that [imath]a_i=a_{n-i}[/imath] [imath]\forall i=0(1)[n/2][/imath] and [imath]b_i=b_{n-i}[/imath] [imath]\forall i=0(1)[n/2][/imath]. Please check whether am I right or not? And any ideas for the rest of the part? |
2815330 | Prove [imath]\sum_{k=0}^{n}k*p_n(k)=n![/imath]
I simplified the above problem to Prove[imath]\sum_{k=1}^n\left(\sum_{i=0}^{n-k}\frac{(-1)^i}{(k-1)!\,i!}\right)=1[/imath] Here [imath]p_n(k)[/imath] denotes number of permutations of [imath]\{1,2,\dots ,n\}[/imath] such that we have k fixed points. How to proceed? | 1985845 | If [imath]S=\{1,2,\dots,n\}[/imath] and [imath]P_n(k)[/imath] be the number of permutations of [imath]S[/imath] having [imath]k[/imath] fixed points, then [imath]\sum_{k=0}^nk.P_{n}(k)=n![/imath]
Let [imath]S = \{1,2,\dots,n\}[/imath]. Then [imath]i \in S[/imath] is said to be a fixed point of a permutation [imath]p[/imath] or [imath]S[/imath] if [imath]p(i) = i[/imath]. Let [imath]P_n(k) [/imath] be the number of permutations of [imath]S[/imath] which have [imath]k[/imath] fixed point. Prove that [imath]\sum_{k=0}^nk\cdot P_{n}(k) = n![/imath]. |
1424279 | To disprove a limit by [imath]\epsilon-\delta[/imath] criterion, how should we find an [imath]\epsilon[/imath] so that we arrive at contradiction?
Whenever we are supposed to prove that a limit for a given function at a given point doesn't exist, using the [imath]\epsilon-\delta[/imath] criterion, we assume that the limit exists, show that for ONE value of [imath]\epsilon[/imath], we arrive at a contradiction, and thus prove that the limit does not exist. I am having trouble choosing the value of [imath]\epsilon[/imath]. Supposing we know that the graph is discontinuous at the given point, what should we look for to assume the value of [imath]\epsilon[/imath]? Or is the value completely arbitrary? Or do we arrive at a contradiction for more than one values of [imath]\epsilon[/imath]? Just to make myself clear, here's an example: [imath]f(x)=x\;\;\;\;\;\;\;\;\;\;\;if\;x<1[/imath] [imath]\;\;\;\;\;\;\;\;=3-x\;\;\;\;\;\; if\;x\ge1[/imath] To prove that the limit doesn't exist at [imath]x=1[/imath], the value of [imath]\epsilon[/imath] taken was [imath]\frac12[/imath]. I won't go into the details of the solution, but we arrive at a contradiction as [imath]1<1[/imath]. How did we know that [imath]\epsilon=\frac12[/imath] would work? | 2347692 | How to prove that limit doesn't exist using epsilon-delta definition?
It is easy to prove the limit exists, all we have to show is there exists a relationship between [imath]\delta[/imath] and [imath]\epsilon[/imath]. But how are we supposed to prove limit doesn't exists? The problem is when we are proving for a limit we already know what the limit is and with that, algebra is all that's needed. Please show through an example (you may show that [imath]\lim_{x\rightarrow0} \frac{1}{x}[/imath] doesn't exist) If possible please use the explanation scheme that is used by this answer https://math.stackexchange.com/a/66552/335742 |
2815561 | Find the volume bounded by the [imath]xy[/imath] plane, cylinder [imath]x^2 + y^2 = 1[/imath] and sphere [imath]x^2 + y^2 +z^2 = 4[/imath]
Find the volume bounded by the [imath]xy[/imath] plane, cylinder [imath]x^2 + y^2 = 1[/imath] and sphere [imath]x^2 + y^2 +z^2 = 4[/imath]. I am struggling with setting up the bounds of integration. First, I will calculate the 'first-quadrant' piece of the volume. [imath]z[/imath] will traverse from [imath]0[/imath] to [imath]2[/imath]. [imath]x[/imath] should start from the cylinder and go to the edge of the current circle of the sphere: [imath]\sqrt{1-y^2} \le x \le \sqrt{4-x^2-z^2}[/imath] However, the same applies to [imath]y[/imath]: (I am only calculating half of the volume right now, where the smaller circle is the lower bound): [imath]\sqrt{1-x^2} \le y \le \sqrt{4-y^2-z^2}[/imath] However, this cannot work as both [imath]x[/imath] and [imath]y[/imath] are dependent. What is the error? | 2680630 | Find the volume of the region inside both the sphere [imath]x^2+y^2+z^2=4[/imath] and the cylinder [imath]x^2+y^2=1[/imath]
Find the volume of the region inside both the sphere [imath]x^2+y^2+z^2=4[/imath] and the cylinder [imath]x^2+y^2=1[/imath] [imath]2\int_0^{2\pi}\int_0^{\frac{\pi} 6}\int_0^{2}1.r^2\sin \varphi dr d\varphi d\theta + \int_0^{2\pi}\int_{\frac{\pi} 6}^{\frac{5\pi} 6}\int_0^{\frac 1 {\sin \varphi}}1.r^2\sin \varphi dr d\varphi d\theta [/imath] This question is in the lecture notes but I didn't understand how the answer is obtained. Could someone help? |
2815804 | Why do we define topology like this?
In the book "Topology" by James Munkres we define topology on a set [imath]X[/imath] like the following(page 76): [imath]\mathscr T[/imath] is a topology on [imath]X[/imath] if: [imath]\emptyset,X\in\mathscr T[/imath] [imath]T\subset\mathscr T\implies\bigcup T\in\mathscr T[/imath] [imath]T\subset\mathscr T\land T\mbox{ is finite}\implies \bigcap T\in\mathscr T[/imath] Why does the third property have the fact that [imath]T[/imath] is finite? What does it gives us? | 4114 | Why do we require a topological space to be closed under finite intersection?
In the definition of topological space, we require the intersection of a finite number of open sets to be open while we require the arbitrary union of open sets to be open. why is this? I'm assuming this has something to do with the following observation: [imath]\cap_{n=1}^{\infty} (-\frac{1}{n},\frac{1}{n}) = \{0\}[/imath] and there is some reason we don't want singletons to be considered open, I am wondering what this reason is. Am I thinking in the right direction here? Thanks :) |
2816024 | Order of arbitrary element in [imath]GL(2, q)[/imath]
If [imath]A[/imath] is an arbitrary element of the general linear group [imath]GL(2, q)[/imath], is there a way to determine the order of [imath]A[/imath] without having to compute matrix powers? The "brute force" method works fairly well. There are [imath]q (q + 1) (q - 1)^2[/imath] elements in [imath]GL(2, q)[/imath]. At worst, you could check all divisors of this. Using this question it's possible to limit the search to the divisors of [imath]q(q - 1)[/imath] and [imath](q - 1)(q + 1)[/imath]. However, I am slightly unsatisfied at having to compute the order this way. | 79047 | On the order of elements of [imath]GL(2,q)[/imath]?
There's a particular property of the elements of [imath]GL(2,q)[/imath], the general linear group of [imath]n\times n[/imath] matrices over a finite field of order [imath]q[/imath], that I don't understand. First, I know that the order of [imath]GL(2,q)[/imath] is [imath](q^2-1)(q^2-q)=q(q+1)(q-1)^2[/imath], since there are [imath]q^2-1[/imath] possible vectors for the first column, excluding the [imath]0[/imath] vector, and [imath]q^2-q[/imath] possible vector for the second column, excluding all multiples of the first. So the order of any element must divide [imath]q(q+1)(q-1)^2[/imath] by Langrange. Howeover, there is a further detail that any element must have order dividing [imath]q(q-1)[/imath] or [imath](q-1)(q+1)[/imath]. Is there a reason why one can narrow down the order to divide one of those smaller factors of [imath]q(q+1)(q-1)^2[/imath]? Thanks! |
2815730 | Finding points in common
So I had these two functions and the following exercise: [imath]f(x)=x^3-2x^2[/imath] [imath]g_p(x)=px[/imath] Determine all values of [imath]p[/imath] for which [imath]f[/imath] and [imath]g_p[/imath] have three points in common. Whats a good way to tackle this? I tried equaling them to each other but that doesn't seem to work. Any good methods? (No hints please). | 2815711 | Prove that [imath]g[/imath] is tangent to the graph of [imath]f[/imath]
So I had these two functions and the following exercise: [imath]f(x)=x^3-2x^2[/imath] [imath]g_p(x)=px[/imath] Prove that [imath]g_{-1}[/imath] is a tangent line to the graph of [imath]f[/imath]. How many points do [imath]g_{-1}[/imath] and [imath]f(x)[/imath] have in common? So yeah, the points in common are fairly simple to figure out: Points in common: [imath]x=1[/imath] and [imath]x=0[/imath] So to see if it's actually tangent to the graph, I took the derivative of both functions at the points where they meet. The result: [imath]f'(x)=3x^2-4x[/imath] [imath]g'_{-1}(x)=-1[/imath] So [imath]f'(1)=-1[/imath] and [imath]f'(0)=0[/imath] So I'm not sure why I get this answer? By the question, we are supposed to prove that its tangent to the graph of [imath]f[/imath] but by my calculation, it seems like only [imath]f(1)[/imath] is tangent with [imath]g_{-1}(1)[/imath]. Help is appreciated! (Please avoid hint answers). |
2815687 | Can this be computed analytically?
For even values of [imath]n[/imath], can the following be proven analytically? [imath](n+1)\binom{n}{\frac{n}2}\int_{\frac12}^1[x(1-x)]^{\frac{n}2}dx=\frac12[/imath] I cannot seem to compute this analytically. Is it possible? I can compute this numerically for various values of [imath]n[/imath]: In [202]: from sympy import binomial as bnm; from numpy import power as pw In [203]: wp=lambda n=10: (n+1)*bnm(n,int(n/2))*I(lambda x:pw(x*(1-x),int(n/2)),1/2,1) In [204]: wp(n=18),wp(20),wp(26),wp(32),wp(42),wp(48),wp(52) Out[204]: (0.500000000000000, 0.500000000000000, 0.500000000000000, 0.500000000000000, 0.500000000000000, 0.500000000000001, 0.500000000000000) Please note that this question is not a duplicate of Prove: [imath]\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx[/imath] for [imath]0 \leq k \leq n[/imath] The reason that this question is not a duplicate of that question is the different integration limits. The different integration limits completely alter the question as well as the solution. | 86542 | Prove: [imath]\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx[/imath] for [imath]0 \leq k \leq n[/imath]
I would like your help with proving that for every [imath]0 \leq k \leq n[/imath], [imath]\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx . [/imath] I tried to integration by parts and to get a pattern or to use the binomial formula somehow, but it didn't go well. Thanks a lot! |
2815845 | which is Greater [imath]31^{11}[/imath] or [imath]17^{14}[/imath]
Which is Greater [imath]31^{11}[/imath] or [imath]17^{14}[/imath] My try: Consider [imath]x=11 \ln(31)=11 \ln\left(32\left(1-\frac{1}{32}\right)\right)=55 \ln 2+11\ln\left(1-\frac{1}{32}\right)[/imath] hence [imath]x=55\ln2 -11\left(\frac{1}{32}+\frac{1}{2048}+\cdots \infty\right) \lt 55 \ln 2[/imath] Now consider [imath]y=14 \ln (17)=14 \ln\left(16 \left(1+\frac{1}{16}\right)\right)[/imath] hence [imath]y=56 \ln 2+14\ln \left(1+\frac{1}{16}\right) \gt 56 \ln 2[/imath] Now we have [imath]55 \ln 2 \lt 56 \ln 2[/imath] which implies [imath]11 \ln(31) \lt 14 \ln (17)[/imath] [imath]\implies[/imath] [imath]31^{11} \lt 17^{14}[/imath] is there any Elementary way to solve this? | 1591544 | How prove this inequality??
How prove : [imath] {31}^{11}<{17}^{14} [/imath] help please! |
2816497 | What is the coefficient of [imath]x^{11}[/imath] in the power series expansion of [imath]\frac{1}{1-x-x^4}[/imath]?
I am really stuck on this problem. I don't really understand power series expansions. However, I think this has to do with generating functions. | 1312695 | How to do power series expansion
What is the coefficient of [imath]x^{11}[/imath] in the power series expansion of [imath]\frac 1{1-x-x^4}[/imath]? How do I do power series expansions? |
2816662 | Integration question with square root and cubic function
I am faced with this question. I tried u-substitution and integration by parts but was not successful. A hint may also suffice. Thanks [imath]\int_{0}^{1} \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx[/imath] I took out x-1 common from both the numerator and denominator only to find myself stuck further than I anticipated. Substituting the radical as u , that is [imath]u=\sqrt{x^2-3x+2}[/imath] doesn't help either as then i cannot express the numerator in terms of u | 2815241 | Expanding with fractional powers
I've been given the following integral to evaluate [imath]\int_0^1(3x^3 - x^2 + 2x - 4)\frac{1}{\sqrt{x^2 - 3x + 2}} dx.[/imath] Can I expand fractional powers? |
2816909 | Evaluate [imath]\int_{0}^{\infty} \frac{x}{1+e^x}dx[/imath]
[imath]\int_{0}^{\infty} \frac{x}{1+e^x}dx=\int_{0}^{\infty} \frac{x(e^x+1-e^x)}{1+e^x}dx=\int_{0}^{\infty} x\cdot dx - \int_{0}^{\infty} \frac{x\cdot e^x}{1+e^x}[/imath] I can't think of a way to proceed further | 700659 | Computing [imath]\int_0^\infty\mathrm{d} x\frac{x}{e^x+1}[/imath] with contour integration
Let's set: [imath] \int_0^\infty\mathrm{d}x\frac{x}{e^x+1}=I. [/imath] I would like to compute it using, presumably, the methods of complex analysis and contour integration. |
2816989 | Can this fuctional equation be solved?
[imath]f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}[/imath] [imath]f(x^2)=xf(x)[/imath] [imath]f(x+1)=f(x)+1[/imath] Can this functional equation be solved ? | 109458 | [imath]f(x^2) = xf(x)[/imath] implies that [imath] f(x) = mx[/imath]?
Suppose a function [imath]f : \mathbb{R} \to \mathbb{R} [/imath] satisfies the relation [imath]f(x^2) = xf(x) \ \ \forall x[/imath] Does this imply [imath]f[/imath] must be a straight line, [imath]f(x) = mx[/imath]? If so, why? If not, are there other such functions? |
2817011 | How to find basis for a subspace?
So i have the following subspace r − s − 2t − u −r + 2s + 5t + 2u s + 3t + u 3r + s + 5u [imath]\begin{bmatrix}1 & -1 & -2 & -1 \\ -1 & 2 & 5 & 2 \\ 0 & 1 & 3 & 1 \\ 3 & 1 & 0 & 5 \\ \end{bmatrix}[/imath] These four equations are all put into square brackets. There are no zero's in the questions equal to the equations. This is a past exam question . And i need to find the basis. I know that r,s,t,u are all scalars. I was thinking of decomposing these equations into variables and numbers, but i am not sure. How would i find the basis ? Could somebody help me ? I was thinkig it might have something to do with the span. YES i already asked the question. Please to not mark it as a duplicate.. I asked it again so more people can see it and maybe help me. | 2816988 | How to find a basis for a subspace?
So I have the following subspace [imath]r − s − 2t − u=0[/imath] [imath]−r + 2s + 5t + 2u=0[/imath] [imath]s + 3t + u=0[/imath] [imath]3r + s + 5u=0[/imath] And I need to find the basis. I know that [imath]r,s,t,u[/imath] are all scalars. I was thinking of decomposing these equations into variables and numbers, but i am not sure. How would i find the basis ? Could somebody help me? I was thinkig it might have something to do with the span. |
2817783 | Two question on inequality
While doing inequality problems from a exercise I stuck on the following two problems. [imath]1.[/imath] If [imath]a,b,c[/imath] be positive real numbers, such that the sum of any two is greater than the third, prove that [imath]a^2 (p-q)(p-r)+b^2 (q-p)(q-r)+c^2 (r-p)(r-q)>0[/imath] for all real [imath]p,q,r.[/imath] [imath]2.[/imath] If [imath]a,b,c[/imath] be positive real numbers, not all equal, and [imath]n[/imath] is a negative rational number, prove that [imath]a^n (a-b)(a-c)+b^n (b-a)(b-c)+c^n (c-a)(c-b) >0[/imath] No need of a full solution. Just give me some hint. I want to do this myself | 1918278 | prove that [imath]a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0[/imath]
If [imath]a,b,c[/imath] are the sides of a triangle and [imath]p,q,r[/imath] are positive real numbers then prove that [imath]a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0[/imath] After modification. I have to prove [imath](a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)[/imath] |
2817845 | Inequality by induction
Q: Prove that for arbitrary [imath]x_1,x_2,...,x_n[/imath]: [imath]\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+...+\frac{x_n}{1+x_1^2+x_2^2+...+x_n^2}\lt\sqrt{n}[/imath] I tried using mathematical induction. The case for [imath]n=1[/imath] is obvious. Assume it is true for [imath]n=k[/imath]. So for [imath]n=k+1[/imath]: [imath]\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+...+\frac{x_n}{1+x_1^2+x_2^2+...+x_n^2}+\frac{x_{n+1}}{1+x_1^2+x_2^2+...+x_{n+1}^2}\lt\sqrt{n+1}[/imath] [imath]\sqrt{n}+\frac{x_{n+1}}{1+x_1^2+x_2^2+...+x_{n+1}^2}\lt\sqrt{n+1}[/imath] by the induction hypothesis [imath]\frac{x_{n+1}}{1+x_1^2+x_2^2+...+x_{n+1}^2}\lt\sqrt{n+1}-\sqrt{n}[/imath] [imath](\sqrt{n+1}+\sqrt{n})x_{n+1}\gt{1+x_1^2+x_2^2+...+x_{n+1}^2}[/imath] How do I proceed from here? Am I even on the right track? | 998612 | Question regarding an inequality
How to prove that [imath] \frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+\frac{x_n}{1+x_1^2+\cdots+x_n^2}<\sqrt{n} [/imath] knowing that [imath](x_n)[/imath] is a positive sequence ? I looked up all kinds of inequalities such AM-GM, Chebyshev, Cauchy-Schwarz, but I couldn't manage to obtain anything useful.. Can anyone help ? |
2817879 | Bijective function from [imath]\Bbb R[/imath] to [imath]\Bbb R^2[/imath]
I know that [imath]\Bbb R[/imath] has the same cardinality as [imath]\Bbb R^2[/imath], but can someone give me a bijective or even just a surjective function [imath]f:\Bbb R \to \Bbb R^2[/imath]? | 838626 | Bijective mapping between [imath]\mathbb{R}^2[/imath] and [imath]\mathbb{R}[/imath]?
Please don't sign this as a duplicate of this, they aren't. I am interested about the actual algorithm, not a proof of the existence. Does a such mapping exists? I think, it must, because both of them have the same cardinality ([imath]\aleph_1[/imath]). My actual suggestion for a such mapping were simply merge the digits of the real numbers. For example, from [imath]\pi[/imath] and [imath]e[/imath] we could got [imath]3.1415...[/imath] and [imath]2.7182...[/imath] would lead to [imath]32.17411852...[/imath] . Although I am not sure it were ok. Maybe a better, clearer solution also exists? |
2818437 | Prove G is a group given properties of G
Let [imath]G[/imath] be a set with the operation [imath]*[/imath] such that: [imath]G[/imath] is closed with respect to [imath]*[/imath] [imath]*[/imath] is asociative There exists an element [imath]e \in G[/imath] such that [imath]e*x=x[/imath] for all [imath]x\in G[/imath] Given [imath]x \in G[/imath], there exists [imath]y\in G[/imath] such that [imath]y*x=e[/imath] The axioms of closure and asociativity are satisfied by the hypotesis 1 and 2. Now it is left to prove that [imath]x*e=x[/imath] and [imath]x*y=e[/imath] to conclude [imath]G[/imath] is a group. I've made some sustitutions to try to get the commutative property on those produts but I don't get a way to achieve it. These set of suppositions are then stronger than the group definition as they can be deduced from these ones right? What shoud I compute to prove [imath]G[/imath] is a group? | 756309 | How to prove that it is a group?
Let [imath]G[/imath] be a set with a binary operation *, associating to each pair of elements [imath]x[/imath] and [imath]y[/imath] of [imath]G[/imath] a third element [imath]x*y[/imath] of [imath]G[/imath]. Suppose that the following properties are satisfied: [imath](x*y)*z = x*(y*z)[/imath] for all elements [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath] of [imath]G[/imath] (the Associative Law); there exists an element [imath]e[/imath] of [imath]G[/imath] such that [imath]e*x = x[/imath] for all elements [imath]x[/imath] of [imath]G[/imath]; for each element [imath]x[/imath] of [imath]G[/imath] there exists an element [imath]x'[/imath] of [imath]G[/imath] satisfying [imath]x'*x = e[/imath]. How to prove i. [imath]x*e=x[/imath] and ii. [imath]x*x'=e[/imath], so that [imath]G[/imath] is a group? |
2719156 | Proof by Contrapositive help?
I am having trouble providing a proof by contrapositive here, can anyone provide a proper proof by contrapositive here? Let [imath]x \in \mathbb{Z}[/imath]. Prove that [imath]3x+1[/imath] is even if and only if [imath]5x-2[/imath] is odd. | 1539456 | [imath]3x + 1[/imath] is even iff [imath]5x-2[/imath] is odd
I'm asked to prove 'Let [imath]x \in Z[/imath]. [imath]3x + 1[/imath] is even iff [imath]5x-2[/imath] is odd'. I have the following proof techniques in my toolbox: trivial/vacuous proofs (not so relevant in this case), direct proof and proof via the contrapositive. 1) If [imath]3x + 1[/imath] is even, then [imath]5x- 2[/imath] is odd. 2) If [imath]5x - 2[/imath] is odd, then [imath]3x + 1[/imath] is even. Theorem 1a: An odd number multiplied by an odd number yields an odd number. Proof via direct proof. Let [imath]n, m \in Z[/imath]. Assume n and m are odd numbers, so they can be written as [imath]n = 2a + 1[/imath] and [imath]m = 2b + 1[/imath] for any [imath]a,b \in Z[/imath]. Therefore [imath]mn[/imath] = [imath](2a + 1)(2b + 1)[/imath] = [imath]4ab + 2a + 2b + 1[/imath] = [imath]2(2ab + a + b) + 1[/imath]. Since [imath]2ab + a + b + 1[/imath] is an integer, [imath]2(2ab + a + b) + 1[/imath] is odd. Theorem 1b: Adding and odd number to an odd number yields an even number. Proof via direct proof. Let [imath]n, m \in Z[/imath]. Assume n and m are odd numbers so they can be written as [imath]n = 2a + 1[/imath] and [imath]m = 2b + 1[/imath] for any [imath]a,b \in Z[/imath]. Therefore [imath]m + n[/imath] = [imath](2a + 1) + (2b + 1)[/imath] = [imath]2a + 2b + 2[/imath] = [imath]2(a + b + 1)[/imath]. Since [imath]a + b + 1[/imath] is an integer, [imath]2(a + b + 1)[/imath] is an even integer. Theorem 1: If [imath]3x + 1[/imath] is even, then [imath]5x- 2[/imath] is odd. Proof via direct proof. We know via 1a and 1b that for [imath]3x + 1[/imath] to be even [imath]x[/imath] has to be odd. Since [imath]x[/imath] is odd, [imath]x = 2k + 1[/imath] for any [imath]k \in Z[/imath]. Therefore [imath]5(2k + 1) - 2[/imath] = [imath]10k + 5 - 2[/imath] = [imath]10k + 3[/imath] = [imath]10k + 2 + 1[/imath] = [imath]2(5k + 1) + 1[/imath]. Since [imath]5k + 1[/imath] is an integer, [imath]2(5k + 1) + 1[/imath] is an odd integer. Theorem 2a: Subtracting an even number from an odd number yields an odd number. Proof via direct proof. Assume [imath]m,n \in Z[/imath], [imath]m[/imath] is odd, [imath]n[/imath] is even, then: [imath]m = 2a + 1[/imath] and [imath]n = 2b[/imath]. [imath]m - n[/imath] = [imath](2a + 1) - 2b[/imath] = [imath]2a - 2b + 1[/imath] = [imath]2(a - b) + 1[/imath]. Since [imath]a-b[/imath] is an integer, [imath]2(a - b) + 1[/imath] is an odd integer. Theorem 2: If [imath]5x - 2[/imath] is odd, then [imath]3x + 1[/imath] is even. We know via 1a and 2a that for [imath]5x - 2[/imath] to be odd, [imath]x[/imath] must be odd. So, [imath]x = 2k + 1[/imath] for [imath]k \in Z[/imath]. Therefore: [imath]3(2k + 1) + 1[/imath] = [imath]6k + 4[/imath] = [imath]2(3k + 2)[/imath]. Since [imath]3k + 2[/imath] is an integer, [imath]2(3k + 2)[/imath] must be an even integer. [imath]\blacksquare[/imath] Since this is the first time I've needed to establish intermediate results I was wondering if this approach is correct. |
2811672 | Prove that for any two sets [imath]A[/imath] and [imath]B[/imath], [imath]A\notin B[/imath] or [imath]B\notin A[/imath]
A quick preface to the below question: this is my first post on math.se! I am excited to begin participating in such a wonderful community. Any feedback as to how I can improve this or subsequent posts is certainly feedback from which I could benefit. (Exercise 3.2.2 of Tao's Analysis I) Use the axiom of regularity (and the singleton set axiom, which guarantees the existence of the singleton set) to show that if [imath]A[/imath] is a set, then [imath]A\notin A{. \kern 0.05em}^\text{1}[/imath] Furthermore, show that if [imath]A[/imath] and [imath]B[/imath] are sets, then [imath]A\notin B[/imath] or [imath]B\notin A[/imath]. Thoughts: Consider some set [imath]A[/imath] and the singleton [imath]\{A\}[/imath]. By the axiom of regularity, the only element of [imath]\{A\}[/imath], namely [imath]A[/imath], is not a set or disjoint from [imath]\{A\}[/imath]. Clearly, [imath]\{A\}\cap A=\emptyset[/imath] and thus, [imath]A\notin A[/imath]. Moreover, suppose [imath]B[/imath] is some other set such that [imath]\neg(A\notin B \lor B\notin A)[/imath], i.e. [imath]A\in B\land B\in A[/imath]. This implies, in very vauge notation, that [imath]A=\{\{A,\dots\},\dots\}[/imath] and [imath]B=\{\{B,\dots\},\dots\}[/imath]. Can I get a contradiction out of this? In the case of [imath]A[/imath], we get that [imath]x\in\{A,\dots\}\lor x\in \{a\}[/imath] for some [imath]x\in A[/imath] and all other objects [imath]a\in A[/imath]. How can I prove that there exists no element of [imath]A[/imath] (and [imath]B[/imath]) that is not a set unless it is disjoint from [imath]A[/imath]. I would greatly appreciate any hints as to how I can complete my argument. 1 This first problem has been discussed on this site times before (see here and here). I include a solution only because the context demands it, but to avoid a duplicate question, I ask that it not be discussed here. | 2946902 | In ZFC, is it possible for a set to be an element of one of its elements?
I'd like to know if [imath]x\in y \land y\in x[/imath] is possible under the ZFC set theory. I appreciate any help! |
2819640 | RSA decryption coefficient
Sorry, I know there are several threads about RSA encryption and how to calculate [imath]d[/imath]. But there is a thing I still don't understand. So you calculate [imath]d[/imath] by using the following expression (see here): [imath] e \cdot d\equiv 1(mod\ \varphi(N)) [/imath] The the usual way to proove this (based on [imath]e^{\varphi(N)}\equiv1 (mod\ N)[/imath]) is: [imath] m \equiv m^{k\cdot\varphi(N)+1} \equiv (m^{\varphi(N)})^k\cdot m \equiv 1^k\cdot m (mod\ N) [/imath] There are serveral things I don't understand: The Euler theorem only applies, if [imath]m[/imath] and [imath]\varphi(N)[/imath] are coprime (which is not the case here) [imath]m^{\varphi(N)} (mod\ N)[/imath] is defnitly not [imath]1[/imath] Nonetheless: For a defined [imath]k[/imath] the equation [imath]m^{k\cdot \varphi(N)} \equiv 1 (mod\ N)[/imath] seems to be correct. But why is that? And why can you just write [imath](m^{\varphi(N)})^k \equiv 1 (mod\ N)[/imath], if [imath]k[/imath] can't be an arbitrary number? | 87718 | RSA: How Euler's Theorem is used?
I'm trying to understand the working of RSA algorithm. I am getting confused in the decryption part. I'm assuming [imath]n = pq[/imath] [imath]m = \phi(n) = (p - 1)(q - 1)[/imath] E is the encryption key [imath]\gcd(\phi(n), E) = 1[/imath] D is the decryption key, and [imath]DE = 1 \mod \phi(n)[/imath] [imath]x[/imath] is the plain text Encryption works as ([imath]y = x^E \mod n[/imath]) and decryption works as ([imath]x = y^D \mod n[/imath]) The explanation for why the decryption works is that since [imath]DE = 1 + k\phi(n)[/imath], [imath]y^D = x^{ED} = x^{1 + k \phi(n)} = x(x^{\phi(n)})^k = x \mod n[/imath] The reason why last expression works is [imath]x^{\phi(n)} = 1 \mod n[/imath] ? According to Eulers theorem this is true only if [imath]x \text{ and }\phi(n)[/imath] are coprimes. But [imath]x[/imath] is only restricted to be [imath]0 < x < n[/imath] and [imath]\phi(n) < n[/imath]. So [imath]x[/imath] should be chosen to be coprime with [imath]\phi(n)[/imath]? Help me clear out the confusion! |
2819797 | The difference between continuity and uniform continuity.
I need someone's help to make me clear with both definition of continuity and uniform continuity. Let's say a function is continuous on a closed interval [imath][a,b].[/imath] So what I can relate from this case, if we use the definition of continuity then we can see that delta is depending on the epsilon and an element of the interval [imath][a,b].[/imath] However, if the function is uniformly continuous, the delta is only depending on the epsilon. Can anyone explain the difference between both definitions? Thanks. | 653100 | Difference between continuity and uniform continuity
I understand the geometric differences between continuity and uniform continuity, but I don't quite see how the differences between those two are apparent from their definitions. For example, my book defines continuity as: Definition 4.3.1. A function [imath]f:A \to \mathbb R[/imath] is continuous at a point [imath]c \in A[/imath] if, for all [imath]\epsilon > 0[/imath], there exists a [imath]\delta > 0[/imath] such that whenever [imath]|x-c| < \delta[/imath] (and [imath]x \in A[/imath]) it follows that [imath]|f(x)-f(c)| < \epsilon[/imath]. Uniform continuity is defined as: Definition 4.4.5. A function [imath]f:A \to \mathbb R[/imath] is uniformly continuous on [imath]A[/imath] if for every [imath]\epsilon > 0[/imath] there exists a [imath]\delta > 0[/imath] such that [imath]|x-y| < \delta[/imath] implies [imath]|f(x)-f(y)| < \epsilon[/imath]. I know that in Definition 4.3.1, [imath]\delta[/imath] can depend on [imath]c[/imath], while in definition 4.4.5, [imath]\delta[/imath] cannot depend on [imath]x[/imath] or [imath]y[/imath], but how is this apparent from the definition? From what appears to me, it just seems like the only difference between Definition 4.3.1 and Definition 4.4.5 is that the letter [imath]c[/imath] was changed to a [imath]y[/imath]. My guess is that the first definition treats [imath]c[/imath] as a fixed point and it is only [imath]x[/imath] that varies, so in this case, [imath]\delta[/imath] can depend on [imath]c[/imath] since [imath]c[/imath] doesn't change. Whereas for the second definition, neither [imath]x[/imath] or [imath]y[/imath] are fixed, rather they can take on values across the whole domain, [imath]A[/imath]. In this case, if we set a [imath]\delta[/imath] such that it depended on [imath]y[/imath], then when we pick a different [imath]y[/imath], the same [imath]\delta[/imath] may not work anymore. Is this somewhat a correct interpretation? Anymore clarifications, examples, would be appreciated. |
2820386 | The solvability in field of Prime Characteristic p
Let [imath]\mathbb{F}_p[/imath] be a field of prime characteristic [imath]p>0[/imath].The question is what we can say about the solvability of the equation [imath]x^2+y^2=-1[/imath] in [imath]\mathbb{F}_p[/imath] for every prime [imath]p[/imath]. I have found out that there are [imath]\frac{p-1}{2}[/imath] different squares in the field [imath]\mathbb{F}_p[/imath]. I was thinking about the circle of radius [imath]kp-1[/imath] has integer point for some [imath]k[/imath]. | 1614677 | The equation [imath]-1 = x^2 + y^2[/imath] in finite fields
In an ordered field we have [imath]x^2 \ge 0[/imath], hence the equation [imath]-1 = x^2 + y^2[/imath] has no solution. But what about finite fields in general? What is the solutions set [imath] -1 = x^2 + y^2 [/imath] of this equation? |
2820297 | What's equal this:[imath]{\frac 12 ^ \frac 13} ^{\cdots \frac1n}[/imath]?
I want to know wht's equal this number :[imath]{\frac 12 ^ \frac 13} ^{\cdots \frac1n}[/imath], however it's seems trivial that is converge to 1 , but i don't know how i can evaluate it , Really i have two question: the first is : what is the partial sum of it ? and what's equal for n large enough or go to infty ? | 497733 | Convergence of [imath]a_n=(1/2)^{(1/3)^{...^{(1/n)}}}[/imath]
The sequence [imath]a_n=(1/2)^{(1/3)^{...^{(1/n)}}}[/imath] doesn't converge, but instead has two limits, for [imath]a_{2n}[/imath] and one for [imath]a_{2n+1}[/imath] (calculated by computer - they fluctuate by about 0.3 at around 0.67). Why is this? |
2820640 | Average value of smallest dice rolled?
We roll a dice until we get a [imath]5[/imath]. What is the expected value of the smallest dice rolled? I am looking for a thought process/an idea. Seeing a full solution would be great as well. Calculating the probability of obtaining [imath]1,\dots, 5[/imath] is what I seem to struggle with. I thought it is simply a Bernoulli distribution, namely that the probability of getting [imath]k[/imath] in the first [imath]i[/imath] throws is simply [imath]1-\left (\frac{5}{6}\right)^{i}[/imath]. Intuitively, the more I throw, the more likely I am to get a [imath]1[/imath] or [imath]2[/imath] so these numbers should have a greater weight than [imath]3,4,5[/imath]. I can't seem to express this mathematically though. | 2796826 | You roll a die until you get a $5$, what is the expected value of the minimum value rolled?
I am struggling to work out a simple way to answer this, and a rationale behind this approach using tail sum (as I do not understand): [imath]$$E\left( x\right) =\sum ^{5}_{k=1}P\left( x\geq k\right)=\dfrac {1}{6}\sum ^{5}_{k=1}\left( \sum ^{\infty }_{i=0}\left( \dfrac {k}{6}\right) ^{i}\right) = \frac{137}{60}.$$[/imath] Does this always hold? [imath]E\left( x\right)=\sum ^{n}_{k=1}kP\left( x= k\right) =\sum ^{n}_{k=1}P\left( x\geq k\right)[/imath] I have never seen this formula, but working through it I understand. EDIT: I can get the right answer with a long winded method, calculating each probability separately, which I believe the Tails sum speeds up: This is my long winded approach. [imath]$$E\left( X_{\min }\right) = 5P\left( x= 5\right) +\ldots +1P(x=1)$$[/imath] [imath]$$=5\left( \dfrac {1}{6}\sum ^{\infty }_{i=0}\left( \dfrac {1}{6}\right) ^{i}\right) +4\left( \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {2}{6}\right) ^{i}- \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {1}{6}\right) ^{i}\right)+\ldots$$[/imath] [imath]$$+1\left( \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {5}{6}\right) ^{i}- \dfrac {1}{6}\sum ^{\infty }_{i=1}\left( \dfrac {4}{6}\right) ^{i}\right),$$[/imath] where [imath]\dfrac {1}{6}[/imath] represents getting a [imath]$5$[/imath], so the first sum is all the possibilities of getting repeated [imath]$6$[/imath]s and then a [imath]$5$[/imath], or just rolling a [imath]$5$[/imath]. The next sum is all the possibilities of getting a [imath]$4$[/imath] or a [imath]$6$[/imath], then a [imath]$5$[/imath], minus all the possibilities of just getting a [imath]$6$[/imath] then a [imath]$5$[/imath]. So it represents all the strings of just [imath]$4$[/imath]s or [imath]$6$[/imath]s before getting a [imath]$5$[/imath], so all the possibilities of minimum value being a [imath]$4$[/imath]. |
2820351 | Are the coefficients of [imath]n^{th}[/imath] order monic polynomial with distinct roots in [imath]\mathbb C[/imath] dense in [imath]\mathbb R^n[/imath]
Suppose for a [imath]n^{th}[/imath] order monic polynomial [imath]p(t) = t^n + \alpha_{n-1}t^{n-1} + \dots + \alpha_0[/imath], we identify the coefficients in [imath]\mathbb R^n[/imath], i.e., [imath]\alpha = (\alpha_{n-1} ,\dots, \alpha_0) \in \mathbb R^n[/imath]. Let [imath]\mathcal F = \{(b_{n-1}, \dots, b_0): p(t)=t^{n} + b_{n-1} t^{n-1} + \dots + b_0 \text{ has $n$ distinct roots in $\mathbb C$}\}.[/imath] I am wondering whether [imath]\mathcal F[/imath] is dense in [imath]\mathbb R^n[/imath]. | 2820324 | Is the set of real matrices diagonalizable in [imath]\mathcal M_n(C)[/imath] dense in the set of all companion matrices?
Let [imath]\mathcal E=\{A \in \mathcal M_n(\mathbb R): A \text{ is of companion form} \}[/imath]. That is, [imath]A[/imath] has the following form \begin{align*} A=\begin{pmatrix} 0 & \cdots & 0& -a_{0} \\ 1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}. \end{align*} Let [imath]\mathcal D = \{A \in \mathcal M_n(\mathbb R): A \text{ is diagonalizable in [/imath]\mathcal M_n(\mathbb C)[imath]}\}[/imath]. We know [imath]\mathcal D[/imath] is dense in [imath]\mathcal M_n(\mathbb R)[/imath]. I am wondering whether [imath]\mathcal E \cap \mathcal D[/imath] is dense in [imath]\mathcal E[/imath]. The set [imath]\mathcal E[/imath] is closed. It is not clear to me whether this statement is true. |
2820779 | Integration by u subsitution
So i have this integral [imath]\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}x^2 \ln(x^3-3)\,dx.[/imath] I was thinking of using u subsitution to make everything easier. I made [imath]u = x^3-3[/imath] and [imath]du = 3x^2dx[/imath]. So I would then re-write my integral as [imath]1/3\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}} \ln(x^3-3)\.[/imath] How would I proceed from here. Should I plug in the integral values? Wouldn't I need to integrate the [imath]\ln()[/imath]? Should I use u substitution again? Please help! I already asked this question, so don't mark it as a duplicate. Unfortunately I couldn't understand what other people were writing. | 2820682 | How to solve integration by parts
![][1] So i have this definite integral: [imath]\int_{\sqrt[3]{4}}^{\sqrt[3]{3+e}}x^2 \ln(x^3-3)\,dx.[/imath] I thought of using integration by parts. So \begin{align} u & =\ln(x^3-3) \\[8pt] du & =\frac{3x^2}{x^3-3} \\[8pt] dv & =x^2 \, dx \\[8pt] v & =\frac{x^3} 3 \end{align} |
2820927 | Solving: [imath]\int_{-\infty}^{+\infty}e^{-2i\pi xy-\pi x^2/r^2}dx[/imath]
I would like to solve:[imath]\int_{-\infty}^{+\infty}e^{-2i\pi xy-\pi x^2/r^2}dx[/imath] Where [imath]i[/imath] is the imaginary unit, and [imath]y,r[/imath] are constants. I've used Wolfram to get [imath]re^{-\pi r^2 y ^2}[/imath]. Any way of solving it step-by-step? | 1229096 | Applying Fourier transform to a gaussian
Let [imath]G_\beta(w) = e^{\beta w^2}[/imath] Now I get the process of applying a fourier transform (or inverse) to get a new gaussian: [imath]G_\beta(x) = G_\beta(0) e^{\frac{-x^2}{4\beta}}[/imath] but in doing the derivation the constant [imath]G_\beta(0)[/imath] never arose. I know how to calculate the constant, but where did it come from in the derivation? edit: Fourier form used: [imath]\int_{-\infty}^{\infty} f(w) e^{-iwx} dw[/imath] |
2789882 | Find the highest power of [imath]2[/imath] which divides the numerator of [imath]H_{k}=1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2k-1}[/imath]
Question: Let [imath]k[/imath] be give the postive integers,Find the highest power of [imath]2[/imath] which divides the numerator of [imath]H_{k}=1+\dfrac{1}{3}+\dfrac{1}{5}+\cdots+\dfrac{1}{2k-1}[/imath] I found some case when [imath]k=2[/imath],then [imath]H_{2}=1+\dfrac{1}{3}=\dfrac{4}{3}[/imath],because the numerator [imath]4[/imath] havethe highest power of is [imath]2[/imath] when [imath]k=3[/imath],then [imath]H_{3}=1+\dfrac{1}{3}+\dfrac{1}{5}=\dfrac{23}{15}[/imath]because the numerator [imath]23[/imath]have the highest power of is [imath]0[/imath] when [imath]k=4[/imath],then [imath]H_{4}=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}=\dfrac{176}{105}[/imath]because the numerator [imath]144[/imath] have the highest power of is [imath]4[/imath] So I conjecture this problem answer is [imath]2V_{2}(k)[/imath].then How to prove？ | 1860105 | 2-adic valuation of odd harmonic sums
I'm playing with p-adic valuations, and find that the odd harmonic sums, [imath]\tilde{H}_k=\sum_{i=1}^{k}\frac{1}{2i-1}[/imath], has 2-adic valuation [imath]||k^2||_2=2||k||_2[/imath]. E.g.) [imath]\tilde{H}_4=\frac{176}{85}[/imath] has [imath]||\tilde{H}_4||_2=||176||_2=4=2||4||_2[/imath]. [imath]\tilde{H}_6=\frac{6508}{3465}[/imath] has [imath]||\tilde{H}_6||_2=2[/imath]. [imath]||\tilde{H}_8||_2=||\frac{91072}{45045}||_2=6[/imath]. [imath]\cdots[/imath] How can I prove this? (Some simple observation I have tried: As [imath]||H_k||_2 =-r[/imath] for [imath]2^r\leq k <2^{r+1}[/imath], [imath]||H_k||_2=||H_{2k}||_2+1=||\frac{1}{2}H_k||_2[/imath]. Thus ultrametric ineq for [imath]\tilde{H}_k=H_{2k}-\frac{1}{2}H_k[/imath] doesn't help at all, yielding trivial result [imath]\tilde{H}_k\in \mathbb{Z}_2[/imath].) (That [imath]||\tilde{H}_k||_2=2||k||_2[/imath] for odd [imath]k[/imath] is almost trivial, so the problem can be reduced to show [imath]||\tilde{H}_{2k}||_2=||\tilde{H}_k||_2+2[/imath], still have no idea) (I've checked this holds up to [imath]\sim 1000[/imath] using Mathematica, and Mathematica spits the result almost immediately for [imath]n\sim 1000[/imath] and takes about a minute for [imath]n\sim 10000[/imath]. ) For anyone interested : I've used the following Mathematica code, try it yourself a = Table[ IntegerExponent[ Numerator[HarmonicNumber[2 n] - HarmonicNumber[n]/2], 2]/2, {n, 10000}]; b = Table [IntegerExponent[n, 2], {n, 10000}]; NonNegative[Min[a - b]] |
2820401 | Find the limit of the sequence {[imath]a_{n}[/imath], given by
Find the limit of the sequence {[imath]a_{n}[/imath]}, given by[imath] a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1[/imath] My try: [imath] a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54[/imath] that is the sequence is incresing and each term is positive. Let the limit of the sequence be [imath]x[/imath]. Then [imath] \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x[/imath] [imath] \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}[/imath] [imath]\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)[/imath] [imath]\Rightarrow x^3-2x+1=0[/imath] and this equation has three roots [imath]x=\dfrac {-1\pm \sqrt {5}}{2},1[/imath] So the limit of the sequence is [imath]\dfrac {-1 + \sqrt {5}}{2}[/imath]. What is the other way to find the limit of the sequence? | 2820464 | Find the limit of given sequence
Find the limit of the sequence {[imath]a_{n}[/imath]}, given by[imath] a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1[/imath] My try: [imath] a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54[/imath] that is the sequence is incresing and each term is positive. Let the limit of the sequence be [imath]x[/imath]. Then [imath] \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x[/imath] [imath] \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}[/imath] [imath]\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)[/imath] [imath]\Rightarrow x^3-2x+1=0[/imath] and this equation has three roots [imath]x=\dfrac {-1\pm \sqrt {5}}{2},1[/imath] So the limit of the sequence is [imath]\dfrac {-1 + \sqrt {5}}{2}[/imath]. how can i say that the limit is [imath]\dfrac {-1 + \sqrt {5}}{2}[/imath]? |
2809168 | measurable function that has non measurable derivative
A measurable function's derivative in the sense that [imath]f^{-1}(]-\infty,\alpha[) \in\mathcal M \ \forall \alpha\in\Bbb R[/imath] is measurable. But what about in the sense [imath]f^{-1}(\text{measurable set})=\text{measurable set}[/imath] ? The first statement can be seen by looking at [imath]\frac{f(x+1/n)-f(x)}{1/n}[/imath] which is measurable for all [imath]n[/imath]. (why does it imply it converges towards a measurable function?) | 639674 | Differentiable function has measurable derivative?
Let [imath]f:[0,T] \to \mathbb{R}[/imath] be a differentiable function. Is it true that [imath]f'[/imath] is measurable? If so, is this also true if [imath]f[/imath] is differentiable almost everywhere? Sorry for lack of effort but I don't have any clue about the answer. |
2821252 | prove that for [imath]0 , f'(a)=\frac{1}{r\pi}\int_{0}^{2\pi}P(\theta)e^{-i\theta}d\theta[/imath]
If function [imath]f(z)[/imath] is analytic and one valued in [imath]|z-a|<R[/imath] , prove that for [imath]0<r<R[/imath] [imath]f'(a)=\frac{1}{r\pi}\int_{0}^{2\pi}P(\theta)e^{-i\theta}d\theta[/imath] where [imath]P(\theta)[/imath] is real part of [imath]f(a+re^{i\theta})[/imath] EDIT : As pointed out by @Kavi Ram Murthy the question has been answered correctly here : Complex Analysis: Complex Integral Since the answer has not been accepted I will copy the answer and mark this as closed (with due credits to @Kavi Ram Murthy) My approach : We know that [imath]f'(a)=\oint\frac{f(z)}{(z-a)^2}dz[/imath] subsituting [imath]z=a+re^{i\theta}[/imath] and simplifying i get [imath]f'(a)=\frac{1}{2r\pi}\int_{0}^{2\pi}f(a+re^{i\theta})e^{-i\theta}d\theta[/imath] I am struck here I dont know how to procced further and prove first equation is equal to last equation . Any hints or sugestions ? (Also I may be be totally wrong in approaching the solution please let me know if I am | 2819061 | Complex Analysis: Complex Integral
If the function [imath]f(z)[/imath] is analytic and one valued in [imath]|z-a|<R[/imath], How do we prove that [imath]0<r<R[/imath], [imath]f^{'}(a)[/imath] = [imath]\frac{1}{{\pi}r}\int P(\theta)e^{-i\theta}d\theta[/imath], where [imath]P(\theta)[/imath] is the real part of [imath]f(a+re^{i\theta})[/imath]. I am guessing Cauchy's integral formula can be used but that has lead me to [imath]f^{'}(a)[/imath] = [imath]\frac{1}{2{\pi}r}\int f(a+ re^{i\theta})e^{-i\theta}d\theta[/imath]. How to proceed from here. |
2820894 | Meaning of [imath]\gtrless[/imath] in the context of multiple hypothesis testing
While reading a paper, I encountered the symbol [imath]\gtrless[/imath], which I had not seen before: Please point me to a reference or tell me what the symbol and associated symbols above and below means. | 608454 | What is [imath]\gtrless[/imath]?
I'm reading Papadimitriou & Steiglitz's Combinatorial Optimization and came across notation I'd never seen before and don't know what it means. The [imath]\LaTeX[/imath] markup for it is \gtrless ([imath]\gtrless[/imath]), which took me quite a while to find. It arises in the formulation of general linear programs in terms of the constraints on the variables: [imath] x_j \geq 0 \;\; j \in N\\ x_j \gtrless 0 \;\; j \in \bar{N} [/imath] It's not "not equals" because there's places in the text where the authors say [imath]x[/imath] can be zero. |
2821435 | How to find solutions of a matrix
I need to find the solutions of a matrix. So I have this system of equations [imath]2x+ky+2z=0[/imath] [imath]x-y+z=1[/imath] [imath]y-z=k[/imath] I augment it as to get: [imath]\begin{pmatrix}2 & k & 2 & 0\\ 1 & -1 & 1 & 1 \\ 0 & 1 & -1 & k \end{pmatrix}[/imath] If i reduced the latter in reduced echelon form I get: [imath]\begin{pmatrix} 1 & 0 &0 &k+1\\ 0 &1 &-1& k\\ 0 &0 &-k-2 & k²+2(k+1) \end{pmatrix}[/imath] I was thinking that the solutions for x,y,z are x=k+1 y= -2/k+2 z= -(k²+2k+2) / k+2 Are my solutions correct? | 2821391 | How to find values of a matrix to make it consistent
So I have this system of equations [imath]2x+ky+2z=0[/imath] [imath]x-y+z=1[/imath] [imath]y-z=k[/imath] I augment it as to get: [imath]\begin{pmatrix}2 & k & 2 & 0\\ 1 & -1 & 1 & 1 \\ 0 & 1 & -1 & k \end{pmatrix}[/imath] If i reduced the latter in reduced echelon form I get: [imath]\begin{pmatrix} 1 & 0 &0 &k+1\\ 0 &1 &-1& k\\ 0 &0 &-k-2 & k²+2(k+1) \end{pmatrix}[/imath] I need to find the k values for which the system is consistent. It seems to me that the system is consistent when [imath]k\neq -2[/imath]. However, when I plug this matrix in the ti nspire cx cas calculator, I get that the reduced echelon form is [imath]\begin{pmatrix} 1 &k/2 &1& 0\\ 0 &1 & -1 &k\\ 0 &0& 1 &-(k²+2k+2)/(k+2) \end{pmatrix}[/imath] The system for me is still consistent when [imath]k[/imath] is not equal to [imath]-2[/imath], because if it were equal to [imath]-2[/imath] then [imath]z= a[/imath] divided by 0, which makes it undefined. However, when I try to compute the solutions, meaning [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath], I get different solutions for the matrices. How come? Which one do you think is correct? E.g with MY REF I get that [imath]x= k+1[/imath]. With REF on the calculator I get that [imath]x=-y(k/2)-z[/imath]. Can somebody please explain the differences with the ccalculator's REF and my REF , or if maybe the soltuions are the same and i just need to expan better? Thanks! |
2618997 | Inverse of a particular matrix: [imath]A = D + aJ[/imath], all entries of [imath]J[/imath] are equal to [imath]1[/imath]
Let [imath]A = D + aJ[/imath] where [imath]D = \mbox{diag}(d_1,\ldots,d_{n})[/imath] and [imath]J[/imath] is [imath]n\times n[/imath] matrix of all ones. Is it possible to find [imath]A^{-1}[/imath] analytically? | 941291 | Inverse of a diagonal matrix plus a constant
I am looking for an efficient solution for inverting a matrix of the following form: [imath]D+aP[/imath] where [imath]D[/imath] is a (full-rank) diagonal matrix, [imath]a[/imath] is a constant, and [imath]P[/imath] is an all-ones matrix. Inverse of constant matrix plus diagonal matrix gives a solution to the special case where all diagonal entries of [imath]D[/imath] are the same. The Sherman-Morrison formula is also capable of providing a solution to this problem by setting appropriate [imath]u[/imath] and [imath]v[/imath], but it loses efficiency ([imath]O(n^3)[/imath] time complexity to compute) at high dimension. I am hoping to get a result in the same form so the space and time complexity are both [imath]O(n)[/imath]. |
2821224 | Find a counterexample that [imath]f(x)[/imath] is Gateaux differentiable and [imath]\lambda(x)[/imath] is not continous
I've learned that function [imath]f:\mathbb{R}^n\rightarrow \mathbb{R}^p[/imath], at [imath]x \in \mathbb {R}^n[/imath], [imath]f(x)[/imath] is Gateaux differentiable, then exits a linear operator [imath]\lambda (x):\mathbb{R}^n\rightarrow \mathbb{R}^p[/imath] such that [imath]D_uf(x)=\lim_{t \rightarrow 0}{\frac{f(x+tu)-f(x)}{t}}=\lambda(x)(u),\forall u \in \mathbb{R}^n[/imath]. If [imath]\lambda(x)[/imath] is continous, [imath]f(x)[/imath] is Frechet differentiable. I want to find a counterexample that [imath]f(x)[/imath] is Gateaux differentiable and [imath]\lambda(x)[/imath] is linear, not continous, but it's hard for me, help me, thank you! | 1401688 | What is an example of Gâteaux differentiable but not Fréchet differentiable at a point in a finite-dimensional space?
Let [imath]V,W[/imath] be nonzero normed spaces over [imath]\mathbb{K}[/imath] such that [imath]V[/imath] is finite-dimensional. Let [imath]E[/imath] open in [imath]\mathbb{K}[/imath] and [imath]p\in E[/imath]. Let [imath]f:E\rightarrow W[/imath] be Gâteaux-differentiable at [imath]p[/imath]. Is [imath]f[/imath] necessarily Fréchet-differentiable at [imath]p[/imath] in this case? I think this is not true in general, but cannot find a counterexample. What would be a counterexample? |
2821866 | Equivalence relation on topological space such that each equivalence class and the quotient space is connected
Let [imath]X[/imath] be a topological space and [imath]\sim[/imath] be an equivalence relation on [imath]X[/imath] such that the quotient space [imath]X/\sim[/imath] is connected and each equivalence class of [imath]\sim[/imath] is connected (as a subspace of [imath]X[/imath] ). Then how to prove that [imath]X[/imath] is connected ? I was trying to go by contradiction assuming [imath]X[/imath] has a non-trivial cl-open subset say [imath]A[/imath] and then trying to get a non-trivial cl-open subset of [imath]X/\sim[/imath] from that, but with no luck. Please help. | 2337195 | Connectedness in the Quotient Space
The problem says this: Let [imath]\sim[/imath] be an equivalence relation in [imath]X[/imath] and consider the quotient space [imath]X/\sim[/imath]. Prove that: [imath]X[/imath] connected [imath]\Rightarrow[/imath] [imath]X/\sim[/imath] connected. If [imath]X/\sim[/imath] is connected and every equivalence class [imath][x] = \left \{ y\in X : y \sim x \right \}[/imath] is connected then [imath]X[/imath] is connected. Proof 1) is easy and I know how to do it. But I'm stuck in the proof 2). That's what I have tried: Let's suppose [imath]X[/imath] isn't connected. Then, [imath]X=U\sqcup V[/imath] with [imath]U[/imath] and [imath]V[/imath] open. Let [imath]\pi : X\rightarrow X/\sim[/imath] be the application that sends every [imath]x\in X[/imath] into his class. I want to see that [imath]\pi (U)[/imath] and [imath]\pi(V)[/imath] are open and that [imath]\pi (U)\cap\pi (V)= \varnothing[/imath] to conclude that [imath]X/\sim[/imath] isn't connected (because [imath]\pi[/imath] is surjective). But that is absurd, so [imath]X[/imath] must be connected. I know how to prove that [imath]\pi (U)\cap\pi (V)= \varnothing[/imath], but I don't know how to show that [imath]\pi (U)[/imath] and [imath]\pi(V)[/imath] are open. Any advice? Thanks! |
2822255 | What is the inverse of [imath]y=x![/imath] for [imath]x>0[/imath]?
Trying to find the inverse of the function [imath]y=x![/imath] for domain restricted to [imath]x>0[/imath]. Does this function have an inverse for the given domain? If so, please include methods used for finding the inverse to help further my understanding of the solution. Thanks! | 2078997 | Inverse of a factorial
I'm trying to solve hard combinatorics that involve complicated factorials with large values. In a simple case such as [imath]8Pr = 336[/imath], find the value of [imath]r[/imath], it is easy to say it equals to this: [imath]\frac{8!}{(8-r)!} = 336.[/imath] Then [imath](8-r)! = 336[/imath] and by inspection, clearly [imath]8-r = 5[/imath] and [imath]r = 3[/imath]. Now this is all and good and I know an inverse function to a factorial doesn't exist as there is for functions like sin, cos and tan etc. but how would you possibly solve an equation that involves very large values compared to the above problem without the tedious guess and checking for right values. Edit: For e.g. if you wanted to calculate a problem like this (it's simple I know but a good starting out problem) Let's say 10 colored marbles are placed in a row, what is the minimum number of colors needed to guarantee at least [imath]10000[/imath] different patterns? WITHOUT GUESS AND CHECKING Any method or explanation is appreciated! |
2819391 | Proof: [imath]f(z)[/imath] is constant
Let [imath]A\subseteq C[/imath] be an open connected domain and [imath]f[/imath] a holomorphic in [imath]A[/imath]. I want to show that: if [imath]f(A)[/imath] is a subset of a line on the complex plane, then [imath]f[/imath] is constant. I began with expressing every [imath]z_0[/imath] on some line [imath]\gamma (t) =at+b[/imath] on the complex plane as: [imath]z_0=x_0+i[ax_0+b][/imath] So: [imath]f(z)=x+i[ax+b][/imath] , but I'm stack (pun intended) with one variable. I doubt that this syllogism is correct. Any help on that one? | 181311 | Holomorphic function mapping a set onto a straight line
I wonder if this is correct: there is a holomorphic function on an open connected subset [imath]G[/imath] of [imath]\mathbb{C}[/imath] which maps [imath]G[/imath] onto a subset of a straight line, and I have to show that the function is constant. I thought I can suppose that the straight line is the real axis (otherwise I can find a rotation and a traslation that will do so) and so using the Cauchy-Riemann equations I find that the function is constant since its imaginary part is zero. Is that correct? Thank you |
2822773 | Is the following inequality True or False?
Ι got a feeling that [imath]\sum_{x=1}^{\infty}\Big\lvert\sum_{k=0}^{\infty} \frac{x^{2k}}{(k+1)!}(-1)^{k} \Big\rvert \geq \sum_{n=1}^{\infty} \frac{1}{n} [/imath] because [imath]\sum_{x=1}^{\infty} \Big\lvert x-\frac{x^2}{3!}+\frac{x^4}{5!}... \Big\rvert \geq 1+\frac{1}{2}+\frac{1}{3}...[/imath] i feel that somehow terms will get canceled but i cant prove it!! **this came up as a part of problem i was solving . I got no idea if the above inequallity is true ** | 2805521 | Does [imath]\sum _{n=1}^{\infty }\frac{\left|\sin\left(n\right)\right|}{n}[/imath] converge?
I'm not sure whether the following series converges or diverges: [imath]\sum _{n=1}^{\infty }\frac{\left|\sin (n)\right|}{n}[/imath] I proved that the series [imath]\sum _{n=1}^{\infty }\frac{\sin^2 (n )}{n}[/imath] converge. Is there a way I can use that? I've tried using Dirichlet series test with the latter but didn't got nowhere since [imath]\frac{1}{\left|\sin x\right|}[/imath] is not monotone decreasing. |
2822913 | Condition of an integer divisible by [imath]3[/imath]
Prove that an integer [imath]n[/imath] is divisible by [imath]3[/imath] if and only the sum of it's digits is divisible by [imath]3[/imath]. This how I proceeded [imath]n=\overline{a_1a_2...a_k}[/imath] Now we get [imath]n=a_1\cdot10^k+a_2\cdot10^{(k-1)}+....+a_k[/imath] Then I find no clue. | 1457478 | Prove that a number is divisible by 3 iff the sum of its digits is divisible by 3
I'm trying to teach myself some number theory. In my textbook, this proof is given: Example (2.3.1) Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3. Let [imath]n=a_0a_1\ldots a_k[/imath] be the decimal representation of an integer [imath]n[/imath]. Thus [imath]n=a_k+a_{k-1}10+a_{k-2}10^2+\cdots+a_010^k[/imath] where [imath]0\le a_i<10[/imath] The key observation is that [imath]10\equiv1\pmod3[/imath], i.e., [imath][10]=[1][/imath]. Hence [imath][10^i]=[10]^i=[1][/imath], i.e., [imath]10^i\equiv1\pmod 3[/imath]. The assertion is an immediate consequence of this congruence. I don't understand the last statement. Why does it follow from that congruence? |
1382055 | Condition on p for convergence of [imath]\sum{\frac{1}{n(\log(n))^p}}[/imath]
For what values of [imath]p[/imath] is the series [imath]\sum{\frac{1}{n(\log(n))^p}}[/imath] divergent and for what values it is convergent? | 2823441 | Convergence of [imath]\sum\frac{1}{n(\log(n))^c}[/imath]
Analyze the convergence of [imath]\sum\frac{1}{n(\log(n))^c}.[/imath] I know that it diverges for [imath]c\leq1[/imath] by using comparision test and integral test for convergence through the [imath]\int_{2}^{\infty}\frac{1}{x\log x}dx= \log(\log(\infty))-\log(\log2)[/imath] I also have a gut feeling that it converges for [imath]c>1[/imath] but not able to prove so. Please try explaining through less exotic methods by not resorting to weird convergence test theorem. Thanks! |
2823569 | Sum function for power series
I have the series [imath]\sum_{n=0}^{\infty}\frac{n}{n+1}x^n[/imath]. It is convergent for [imath]-1<x<1[/imath]. I want to find the sum function. I've tried the following: \begin{align*} \sum_{n=0}^{\infty} x^n &\sim \frac{1}{1-x}\\ \sum_{n=0}^{\infty} \frac{x^{n+1}}{1+n}&\sim -\log(1-x)&\text{Integrating both sides}\\ \sum_{n=0}^{\infty} \frac{x^{n}}{1+n}&\sim-\frac{\log(1-x)}{x}&\text{Divide with [imath]x[/imath]}\\ \sum_{n=0}^{\infty}\frac{nx^{n-1}}{1+n}&\sim\frac{1}{(1-x)x}+\frac{\log(1-x)}{x^2}&\text{Derivative with respect to [imath]x[/imath]}\\ \sum_{n=0}^{\infty}\frac{nx^{n}}{1+n}&\sim\frac{1}{(1-x)}+\frac{\log(1-x)}{x}&\text{Multiply by [imath]x[/imath]} \end{align*} I can't decide on what to do from this point. I seem to miss the [imath]n[/imath] in the nominator, but changing index does not really help med out. Can you point me in the right direction? Edit: I used the tips and finished it. Thank you. | 1730216 | Find a function for the infinite sum [imath]\sum_{n=0}^\infty \frac{n}{n+1}x^n[/imath]
I need to find a function [imath]f(x)[/imath] which is equal to the sum [imath] \sum_{n=0}^\infty \frac{n}{n+1}x^n, [/imath] for every [imath]x\in \mathbb{R}[/imath] for which the series converge. Now, using WolframAlpha, I've found the answer [imath]f(x)=\frac{-x+x\ \text{log}(1-x)-\text{log}(1-x)}{(x-1)x},\ \ \ \text{when}\ |x|<1[/imath] However I'm trying to figure out how to find this function [imath]f(x)[/imath] on my own. I've tried to combine the Maclaurin series of [imath]\ln(1+x)[/imath], [imath]\sin(x),[/imath] and several other functions but I don't seem to succeed. Any suggestions how to work out this problem from scratch? |
2822715 | Does the series [imath]\sum \frac{\sin p}{p}[/imath], where [imath]p[/imath] prime, converge?
Question. Assume that [imath]\{p_n\}[/imath] is the sequence of prime numbers, in increasing order. Does the series [imath] \sum_{n=1}^\infty \frac{\sin p_n}{p_n} [/imath] converge? The only criterion to establish convergence that I can think of, which might be of any use, is the Abel's Test, which requires that the partial sums of the sequence [imath]b_n=\sin p_n[/imath] are bounded, i.e., that there exists an [imath]M>0[/imath], such that [imath] |\sin p_1+\cdots+\sin p_n|\le M, \quad\text{for all $n\in\mathbb N$}. [/imath] Any ideas? | 2272248 | Regarding the sum [imath]\sum_{p \ \text{prime}} \sin p[/imath]
I'm very confident that [imath]\sum_{p \ \text{prime}} \sin p [/imath] diverges. Of course, it suffices to show that there are arbitrarily large primes which are not in the set [imath]\bigcup_{n \geq 1} (\pi n - \epsilon, \pi n + \epsilon)[/imath] for sufficiently small [imath]\epsilon[/imath]. More strongly, it seems that [imath]\sin p[/imath] for prime [imath]p[/imath] is dense in [imath][-1,1][/imath]. This problem doesn't seem that hard though. Here's something that (to me) seems harder. If [imath]p_n[/imath] is the nth prime, what is [imath]\limsup_{n \to +\infty} \sum_{p \ \text{prime} \leq p_n} \sin p?[/imath] What is [imath]\sup_{n \in \mathbb{N}} \sum_{p \ \text{prime} \leq p_n} \sin p? [/imath] Of course, we can ask analogous questions for [imath]\inf[/imath]. I'm happy with partial answers or ideas. For example, merely an upper bound. |
2823831 | Transpose matrix and inner product
In one of the proofs in class there was given the equality for the dot product: [imath]\langle Ax, Ax\rangle = \langle x, A^tAx\rangle[/imath] I don't understand why this is correct. Is there a way to show this without explicitly looking at the multiplications and sums? thanks. | 1684589 | Property of the conjugate transpose matrix with inner product
I'm trying to prove that for a certain matrix [imath]A[/imath], and its conjugate transpose [imath]A^*[/imath], we have [imath]⟨Ax,y⟩=⟨x,A^*y⟩[/imath], where [imath]⟨⟩[/imath] represent the inner product. So here it's simply the dot product in [imath]R^n[/imath]. Does anyone have an idea of how to prove this? Thank you! |
2823817 | Is a left and right simple ring with unity a division ring?
By a left simple ring I mean a ring with no proper, nontrivial left ideal. [imath]R[/imath] be such a ring. Let [imath]u(≠0)\in R[/imath]. Then [imath]Ru=R[/imath] (since [imath]1u=u≠0[/imath]). Now [imath]Ru=[/imath]{[imath]ru:r\in R[/imath]}. So there is an [imath]r\in R[/imath] such that [imath]ru=1[/imath]. Right simplicity of [imath]R[/imath] implies that there is an [imath]s\in R[/imath] such that [imath]us=1[/imath]. [imath](ru)s=r(us) \implies 1s=r1 \implies s=r=u^{-1}.[/imath] So each nonzero element has a multiplicative inverse. Thus [imath]R[/imath] is a division ring. Is the claim and/or proof correct? | 1821467 | Division ring if and only if it has no proper left ideals.
Let [imath]R[/imath] be an associative ring with [imath]1[/imath]. Prove that [imath]R[/imath] is a division ring if and only if [imath]R[/imath] has no proper left ideals. Clearly, if [imath]R[/imath] is a division ring and [imath]I\neq\{0\}[/imath] is a left ideal, then every nonzero [imath]z\in I[/imath] satisfies [imath]1=z^{-1}z\in I[/imath], hence [imath]I=R[/imath]. Now if [imath]z\in R[/imath] is nonzero and [imath]R[/imath] has no proper left ideals, then [imath]I:=\{rz:r\in R\}[/imath] is equal to [imath]R[/imath] and there exists [imath]r\in R[/imath] such that [imath]rz=1[/imath]. However, we still need to prove [imath]zr=1[/imath]. How can we do this? Thank you. |
2823853 | Proving [imath]n! for positive integer n>1 by induction[/imath]
I am trying to solve this excercise for induction: Prove that for every positive integer [imath]n > 1[/imath], [imath]n! < n^n[/imath]. The first thing that I did is to prove [imath]P(2)[/imath]: [imath] P(2):\ 2 × 1 <2^2, [/imath] And this is true because [imath]2 < 4[/imath]. Assuming that [imath]P(k)[/imath] is true for some [imath]k[/imath] in general, I do not know how to prove it for [imath]k + 1[/imath]. Can you help me please? | 2775451 | Induction principle for [imath]n![/imath]
How can I prove that [imath]n!<n^n[/imath] for every [imath]n>1[/imath] using the induction principle? If I put [imath]n=2[/imath] I get [imath]2<4[/imath] so I know that [imath]n!<n^n[/imath] is true. Now I don't know how to prove that [imath](n+1)!<(n+1)^{n+1}[/imath]. I can write [imath](n+1)!=n!(n+1)[/imath] but then I don't know how to go on. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.