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996732 | Prove that [imath] x^xy^y=z^z [/imath] has infinite integral solutions
Show that there exist an infinite number of solutions for [imath] x^xy^y=z^z [/imath] where [imath]x,y,z \gt 1[/imath] & [imath]x,y,z\in \mathbb Z[/imath] I don't know how to even start,infact I am not able to find a particular solution of this problem let alone infinite.How to approach? | 497702 | Do there exist an infinite number of integer-solutions [imath](x,y,z)[/imath] of [imath]x^x\cdot y^y=z^z[/imath] where [imath]1\lt x\le y[/imath]?
Question : Do there exist an infinite number of integer-solutions [imath](x,y,z)[/imath] of [imath]x^x\cdot y^y=z^z[/imath] where [imath]1\lt x\le y[/imath] ? Motivation : After struggling to find a solution, I've just got one solution, which is [imath](x,y,z)=(1679616, 2985984, 4478976).[/imath] In the following, I'm going to write how I got this solution. Letting [imath]d[/imath] be the greatest common divisor of [imath]x,y,z[/imath], we can represent [imath]x=ad, y=bd, z=cd[/imath] where [imath]a,b,c[/imath] are coprimes with each other. Then, we get [imath]d^{a+b-c}\cdot a^a\cdot b^b=c^c.[/imath] In the following, let's consider without the condition [imath]x\le y[/imath]. Here, I suppose [imath]a=2^m, b=3^n, a+b-c=1.[/imath] (As a result, this supposition works.) Then, we get [imath]d=\frac{c^c}{2^{ma}\cdot 3^{nb}}.[/imath] Hence, letting [imath]c=2^k\cdot 3^l[/imath], if [imath]kc\ge ma=m\cdot 2^m, lc\ge nb=n\cdot 3^n,[/imath] then [imath]d[/imath] is an integer. Since [imath](m,n)=(4,2)[/imath] satisfies the above conditions, then we get [imath]d=2^8\cdot 3^6=186624.[/imath] Hence we can get [imath]x=9d=2^8\cdot 3^8=1679616, y=16d=2^{12}\cdot 3^6=2985984, z=24d=2^{11}\cdot 3^7=4478976.[/imath] Note that here I interchanged [imath]x[/imath] and [imath]y[/imath]. P.S : I was surprised to get this solution because I got this almost by chance. So, I don't know the other solutions. If you have any helpful information, please teach me. |
997278 | Showing an algebraic element of [imath]\bar{\mathbb{Q}}[/imath] is in [imath]\bar{\mathbb{Q}}[/imath]
This question really has me stumped. We define [imath]\bar{\mathbb{Q}}[/imath] to be the set of elements in [imath]\mathbb{C}[/imath] that are algebraic over [imath]\mathbb{Q}[/imath]. I have shown that this is a field. Now, I'm assuming that we have some element [imath]b \in \mathbb{C}[/imath] that is algebraic over [imath]\bar{\mathbb{Q}}[/imath]. I want to show [imath]b \in \bar{\mathbb{Q}}[/imath]. My attempt: Since [imath]b[/imath] is algebraic over [imath]\bar{\mathbb{Q}}[/imath], we know the extension [imath][\bar{\mathbb{Q}}(b) : \bar{\mathbb{Q}}] = n < \infty[/imath]. Also, there's a polynomial [imath]f \in \bar{\mathbb{Q}}[x][/imath] s.t [imath]f(b) = 0[/imath]. So, I've let [imath]f = x^n + a_{n-1} x^{n-1} + ... + a_0[/imath], where the [imath]a_i \in \bar{\mathbb{Q}}[/imath]. In particular, we know that these [imath]a_i[/imath] are algebraic over [imath]\mathbb{Q}[/imath]. Next, we consider the extension [imath][\mathbb{Q}(a_0, a_1, ... , a_{n-1}) : \mathbb{Q}][/imath]. Since all the [imath]a_i[/imath] are algebraic over [imath]\mathbb{Q}[/imath], this extension is finite. Now, we consider [imath][\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}][/imath]. By Tower Law, this equals [imath][\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}(b)] \times [\mathbb{Q}(b) : \mathbb{Q}][/imath]. Both of these are finite extensions, so [imath][\mathbb{Q}(b, a_0, a_1, ... , a_{n-1}) : \mathbb{Q}][/imath] is finite. And this is where I get stuck. I'm not quite sure where to go from here and I just feel kinda confused. I'm not sure if this is even taking me towards the answer, but I'm not sure what to do. Any nudges would be great! EDIT: Corrected typos listed below. | 71267 | On the meaning of being algebraically closed
The definition of algebraic number is that [imath]\alpha[/imath] is an algebraic number if there is a nonzero polynomial [imath]p(x)[/imath] in [imath]\mathbb{Q}[x][/imath] such that [imath]p(\alpha)=0[/imath]. By algebraic closure, every nonconstant polynomial with algebraic coefficients has algebraic roots; then, there will be also a nonconstant polynomial with rational coefficients that has those roots. I feel uncomfortable with the idea that the root of a polynomial with algebraic coefficients is again algebraic; why are we sure that for every polynomial in [imath]\mathbb{\bar{Q}}[x][/imath] we could find a polynomial in [imath]\mathbb{Q}[x][/imath] that has the same roots? I apologize if I'm asking something really trivial or my question comes from a big misunderstanding of basic concepts. |
443642 | Prove that every group of order [imath]4[/imath] is abelian
How can I show that every group of order [imath]4[/imath] is abelian? Let's denote [imath]e,a,b,c[/imath] as the four elements of the group. Since [imath]e[/imath] is identity, we have [imath]e*x=x*e[/imath] for every [imath]x[/imath]: [imath]\begin{array}{|c|cccc|} \hline & e & a & b & c \\\hline e & e & a & b & c \\ a & a & & & \\ b & b & & & \\ c & c & & & \\\hline \end{array}[/imath] Now [imath]a*a=[/imath]? We have several possibilities. If I choose [imath]a*a=b[/imath] I can show [imath]a*b=a*(a*a)=(a*a)*a=b*a[/imath]. But so far I have shown only that [imath]a[/imath] and [imath]b[/imath] commute, there are also other pairs. And I have only discussed the choice [imath]a*a=b[/imath], there are also other possibilities. Is there a simpler way to do this? | 1622878 | If [imath]G[/imath] is a finite group s.t. [imath]|G|=4[/imath], is it abelian ?
If [imath]G[/imath] is a finite group s.t. [imath]|G|=4[/imath], is it abelian ? To me it's isomorphic to [imath]\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z[/imath] or [imath]\mathbb Z/4\mathbb Z[/imath], but a friend of me said that they are not the only group of order 4, and there exist some non-abelian. Do you have such example ? |
997396 | Proving an identity from a dilogarithm function.
If [imath]\def\Li{\operatorname{Li}}\Li'_{2}(z) = - \displaystyle\frac{\ln(1-z)}{z}[/imath], how does one get the identity, [imath] \Li_{2}\left(- \frac{1}{z}\right) + \Li_{2}(-z) + \displaystyle\frac{1}{2}(\ln(z))^2 = C, [/imath] where [imath]C[/imath] is the constant of integration and [imath]\Li_{2}(z)[/imath] is the dilogarithm function. I have been staring at this for awhile and do not understand how to use what is given. I know that the dilog function has some unusual identities, one of which is this one. I would greatly appreciate the help because then I could make sense of the other identities using a similar method. | 995741 | The dilogarithm function. Question on an identity of it
Upon reading a journal article about manipulating series using the dilogarithm function, I have a few questions. But before I ask them, let me give the information the article provides. Consider the series [imath]\displaystyle\sum \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...[/imath] The related power series is [imath]f(z) = \displaystyle\sum \frac{1}{k^2}z^k = z + \frac{z^2}{4} + \frac{z^3}{9} + ...[/imath] The derivative of this is [imath]f'(z) = \displaystyle\sum \frac{1}{k}z^{k-1} = 1 + \frac{z}{2} + \frac{z^2}{3} + ... = -\frac{\ln(1-z)}{z}[/imath] The series we have provided is just the derivative of the dilogarithm function, and so [imath]Li'_{2}(z) = -\displaystyle\frac{\ln(1-z)}{z}[/imath]. Now there is an identity that says [imath]Li_{2}( -\displaystyle\frac{1}{z}) + Li_{2}(-z) + \displaystyle\frac{1}{2}(\ln(z))^2 = C[/imath] where [imath]C[/imath] is a constant. From here, taking [imath]z = 1[/imath] gives us [imath]C = 2Li_{2}(-1) = 2(-1 + \displaystyle\frac{1}{4} - \displaystyle\frac{1}{9} + \displaystyle\frac{1}{16} - ...)[/imath]. This can be related to [imath]Li_{2}(1)[/imath] by using the fact that the even terms of [imath]\displaystyle\sum \frac{1}{k^2} = \sum \frac{1}{(2k)^2} = \frac{1}{4}Li_{2}(1)[/imath]. And so the odd terms must be [imath]\displaystyle\frac{3}{4}Li_{2}(1)[/imath] and so the alternating sum I gave earlier is the difference of the even terms and the odd terms which is [imath]-\displaystyle\frac{1}{2}Li_{2}(1)[/imath]. This shows that [imath]C = -Li_{2}(1)[/imath]. Here are my questions now. Where on earth does [imath]Li_{2}( -\displaystyle\frac{1}{z}) + Li_{2}(-z) + \displaystyle\frac{1}{2}(\ln(z))^2 = C[/imath] come from? I do not see that at all. I don't see how [imath]-(\displaystyle\frac{1}{2})Li_{2}(1)[/imath] is the alternating series. Theres no way to generate [imath]-1[/imath] from it which is the first term of the alternating series. |
999066 | troubles proving [imath] \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=1}^{n-1} A_k\bigr)[/imath] for sets [imath]A_i[/imath].
How can we show [imath] \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=1}^{n-1} A_k\bigr)[/imath] for any family of sets? So let [imath]x\in \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=1}^{n-1} A_k\bigr)=\bigcup_{n\in\mathbb N}\bigl(A_n\cap\bigcap_{k=1}^{n-1} A_k^c\bigr)[/imath]. So [imath]\exists n: x\in A_n\cap\bigcap_{k=1}^{n-1} A_k^c [/imath] and therefore [imath]x\in A_n \wedge x\in A_1^c\wedge\dots\wedge x\in A_{n-1}^c [/imath]. So we get [imath]x\in \bigcup_{n\in\mathbb N}A_n.[/imath] What happens if [imath]A_i[/imath] isn't pairwise disjoint? And how do you get the other way? | 635042 | Sequences of sets property
I'm having trouble to prove the following question: Supose [imath]\{A_n\}_{n\in\mathbb{N}}[/imath] is a family of sets such that [imath]A_1\subset A_2\subset A_3\subset\ldots[/imath] (it's possible to have [imath]A_n=A_{n+1}[/imath]). I need to prove that [imath]\lim_{n\to\infty}A_n = A_1\cup\bigcup_{n=2}^\infty(A_{n}\backslash A_{n-1}).[/imath] I'm not sure if this is useful, but I proved that [imath]A_{n}\backslash A_{n-1}[/imath] and [imath]A_{n'}\backslash A_{n'-1}[/imath] are disjoint if [imath]n\neq n'[/imath]. Thank you. |
999227 | Let [imath]n[/imath] be a 3-digit number. Prove [imath]9\mid n[/imath] iff the digits of [imath]n[/imath] sum to a multiple of 9.
I have convinced myself that this true, however I'm at a loss of where I should start with this proof. Looking at a similar proof with 3 instead of 9, I saw the use of the basis representation theorem, but I'm not 100% comfortable with that, so it was hard for me to follow. Is there a way to do this with mod? | 539457 | Divisibility by [imath]9[/imath]
Suppose we have a natural number [imath]N[/imath] with decimal representation [imath]A_kA_{k-1}\ldots A_0[/imath]. How do I prove that if the [imath]\sum\limits_{i=0}^kA_i[/imath] is divisible by [imath]9[/imath] then [imath]N[/imath] is divisible by [imath]9[/imath] too? |
999253 | Is the set of matrices with rank at most [imath]r[/imath] closed?
The question is as follows: [imath]\DeclareMathOperator{\rank}{rank}[/imath] Is the set [imath]S_r = \{A \in \Bbb R^{n \times n}: \rank(A) \leq r\}[/imath] closed in [imath]\Bbb R^{n \times n}[/imath] in the Euclidean topology? I have a feeling that this is true, but I got stuck looking for a convincing argument. Certainly this is true for [imath]r = n-1[/imath]. Perhaps it can be shown that [imath]S_{r-1}[/imath] is a closed subset of [imath]S_r[/imath]? No neat tricks are coming to mind, but I would think that there must be one. This seems like the kind of thing that has a canonical answer, so links are welcome. | 43192 | How to prove that the collection of rank-k matrices forms a closed subset of the space of matrices?
Let [imath]M[/imath] be the space of all [imath]m\times n[/imath] matrices. And [imath]C=\{X\in M|rank(X)\leq k\}[/imath] where [imath]k\leq \min\{m,n\}[/imath]. How to proof that C is a closed subset of M? |
999513 | the inequality [imath]\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2[/imath]
How to show that [imath]\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2[/imath] for [imath]a,b,c>0[/imath]? I tried to prove [imath]\frac{a^4}{a^3+b^3}\ge \frac {5a}4+\frac{-3b}4[/imath] but could not continue. Give me ideas, please. | 340577 | Prove inequality: [imath]\sum \frac{a^4}{a^3+b^3} \ge \frac{a+b+c}{2}[/imath]
Prove inequality with [imath]a,b,c >0[/imath] [imath]\frac{a^4}{a^3+b^3} + \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2}[/imath] I tried the inequality: [imath]\sum \frac {a^4+b^4}{a^3+b^3} \ge \sum \frac{a+b}2=a+b+c[/imath], but seem like it doesn't help |
999606 | proof of weak convergence of probablity measure
Let [imath](\mu_n)_{n\in\Bbb{N}}[/imath] be a sequence of probability measure on [imath]\Bbb{R}[/imath] with characteristic functions [imath](\phi_n)_{n\in\Bbb{N}}[/imath]. Assume that [imath]\lim_{n\rightarrow\infty}\phi_n(t)=1[/imath] for all [imath]t\in[-\delta,\delta][/imath] for some [imath]\delta>0.[/imath] Prove that [imath]\mu_n[/imath] converges weakly to [imath]\delta_0[/imath], and hence [imath]\phi_n[/imath] converges uniformly to 1 on any bounded interval. My thought: I try to prove the property of tightness of [imath](\mu_n)[/imath]. Then if subsequence of [imath](\mu_n)[/imath] can be proved to weakly converge to [imath]\delta_0[/imath]. By the uniqueness, we are done. | 704142 | Convergence of characteristic functions to [imath]1[/imath] on a neighborhood of [imath]0[/imath] and weak convergence
Prove the following statement: [imath] X_n \Rightarrow 0 [/imath] (convergence in distribution) if and only if [imath] (\exists\; \epsilon>0: |t|<\epsilon) \;\; \phi_n(t) \rightarrow 1 [/imath], where [imath]\phi_n(t)[/imath] is the characteristic function of the random variable [imath]X_n[/imath]. We could write [imath]X_n \rightarrow 0 [/imath] in probability since convergence in probability to a constant and convergence in distribution to a constant are the same concept. I guess Lévy's continuity theorem implies the 'only if' part. Thank you very much for your help in advance! |
999555 | Orbit Stabilizer Theorem? Question on HW?
I have received this question for a homework assignment, and I'm getting stuck at the last part :) So I've proved that [imath]H = \{f_m,0 \vert m \in \mathbb{R}^*\}[/imath] is a non-normal subgroup of [imath]A_1[/imath] ( the group of affine functions, [imath]f=mx+b[/imath]). I've also proved that the right cosets of [imath]H[/imath] are all in the form [imath]R_t = \{f \in A_1 \vert f(0) = t\}[/imath] But, how would I go about proving that the left cosets of [imath]H[/imath] are in the form [imath]L_t = \{ f \in A_1 \vert f(t) = 0 \}[/imath]? Also, would it be right to say that [imath]H=\rm{Stab}(0)[/imath]? Let A1 be the group of affine functions on R. We let the group operation on A1 be ◦, where (f ◦ g)(x) = g(f(x)) | 999893 | Homework Question about Orbit Stabilizer Theorem?
I have received this question for my homework assignment, and I'm getting stuck at the last part :) So I've proved that [imath]H=\{f_{m,0}\ | \ m\in\mathbb{R}_{*}\}[/imath] is a non-normal subgroup of [imath]A_{1}[/imath] ( the group of affine functions, [imath]f_{m, b}(x)=mx+b[/imath]). I've also proved that the right cosets of [imath]H[/imath] are all in the form [imath]Rt=\{f\in A_{1}\ | \ f(0)=t\}[/imath] But, how would I go about proving that the left cosets of H are in the form [imath]L_t=\{f\in A_{1}\ | \ f(t)=0\}[/imath]? Also, would it be right to say that H=Stab(0)? Let [imath]A_{1}[/imath] be the group of affine functions on [imath]\mathbb{R}[/imath]. We let the group operation on [imath]A_1[/imath] be [imath]\circ[/imath], where [imath](f \circ g)(x) = g(f(x))[/imath]. |
1000112 | Nonlinear continuous and unbounded operator
Let [imath]X[/imath] be an infinite-dimensional Banach space, and let [imath]B=\{x\in X: \|x\|\leq 1\}[/imath] be its closed unit ball. Does there exist a continuous mapping [imath]F: X\to X[/imath] such that the set [imath]F(B)=\{F(x): x\in B\}[/imath] is unbounded? Of course, such [imath]F[/imath] cannot be a linear operator, so [imath]F[/imath] must be nonlinear. Thank you very much in advance! | 29498 | Nonlinear function continuous but not bounded
I would like an example of a map [imath]f:H\rightarrow R[/imath], where [imath]H[/imath] is a (infinite dimensional) Hilbert space, and [imath]R[/imath] is the real numbers, such that [imath]f[/imath] is continuous, but [imath]f[/imath] is not bounded on the close unit ball [imath]\{ x\in H : \|x\| \leq 1\}[/imath]. Actually, [imath]H[/imath] could be replaced by any Banach space (but not just a normed space-- that's too easy). My motivation is that if [imath]f[/imath] is linear, this is impossible; but I have next to no intuition about non-linear functions. Edit: Here's an example for [imath]c_0[/imath] which is even differentiable (disclaimer: I found it here: http://www.ms.uky.edu/~larry/paper.dir/korea.ps). Define [imath]f:c_0\rightarrow F[/imath] (where F is your field, real or complex) by [imath] f(x) = \sum_{n=1}^\infty x_n^n \qquad (x=(x_n)). [/imath] You can estimate the sum by a geometric progression, so it does converge. A bit of checking shows that f is Frechet differentible (so certainly continuous). But [imath]f(1,1,\cdots,1,0,\cdots)=n[/imath] (if there are [imath]n[/imath] ones) so [imath]f[/imath] is not bounded on the closed unit ball. What I don't immediately see is how to adapt this to [imath]\ell^2[/imath], say. |
1000465 | Is a stably free module always free?
Let [imath]R[/imath] be a commutative ring with unity and [imath]M[/imath] an [imath]R[/imath]-module. If [imath]M\oplus R^m\cong R^n[/imath] for some integers [imath]m,n \geq 1[/imath] then must [imath]M[/imath] be finitely generated and free? Can somebody help me with this? Regards. | 58025 | Direct summand of a free module
Let [imath]M[/imath], [imath]L[/imath], [imath]N[/imath] be [imath]A[/imath]-modules and [imath]M=N\oplus L[/imath]. If [imath]M[/imath] and [imath]N[/imath] are free, is [imath]L[/imath] necessarily free? |
671465 | Reflexive relation on set of [imath]n[/imath] elements
How many reflexive relations are there on a set of [imath]n[/imath] elements? I did the problem and I got the answer [imath]2 ^ {n ^ 2}[/imath]. Is it correct? Thanks for the help..!! | 88032 | Number of reflexive relations defined on a set A with n elements
Problem: If a set [imath]A[/imath] has [imath]n[/imath] elements in it, how many reflexive relations can be defined on it? My solution Is the answer summation of (n^2 - n)C(i) for i=0 to n^2 -n [imath]\sum_{i=0}^{n^2-n} C(n^2-n,i) = \sum_{i=0}^{n^2-n} \binom{n^2-n}i[/imath] How? well if i make a matrix [imath]n\times n[/imath] now the diagonal elements have to be selected,out of remaining [imath]n^2-n[/imath] any number of elements can be selected. |
1001267 | What is the value of [imath]\cos\left(\frac{2\pi}{7}\right)[/imath]?
What is the value of [imath]\cos\left(\frac{2\pi}{7}\right)[/imath] ? I don't know how to calculate it. | 38414 | Exact values of [imath]\cos(2\pi/7)[/imath] and [imath]\sin(2\pi/7)[/imath]
What are the exact values of [imath]\cos(2\pi/7)[/imath] and [imath]\sin(2\pi/7)[/imath] and how do I work it out? I know that [imath]\cos(2\pi/7)[/imath] and [imath]\sin(2\pi/7)[/imath] are the real and imaginary parts of [imath]e^{2\pi i/7}[/imath] but I am not sure if that helps me... |
1001399 | Show that the metric [imath]d_1[/imath] on [imath]C([0,1])[/imath] does not give rise to a complete metric space.
Show that the metric [imath]d_1[/imath] on [imath]C([0,1])[/imath] does not give rise to a complete metric space. [imath] d_1(f,g) = \int_0^1 |f(s)−g(s)| \, ds [/imath] | 152233 | Showing [imath](C[0,1], d_1)[/imath] is not a complete metric space
I am completely stuck on this problem: [imath]C[0,1] = \{f: f\text{ is continuous function on } [0,1] \}[/imath] with metric [imath]d_1[/imath] defined as follows: [imath]d_1(f,g) = \int_{0}^{1} |f(x) - g(x)|dx [/imath]. Let the sequence [imath]\{f_n\}_{n =1}^{\infty}\subseteq C[0,1][/imath] be defined as follows: [imath] f_n(x) = \left\{ \begin{array}{l l} -1 & \quad \text{ [/imath]x\in [0, 1/2 - 1/n][imath]}\\ n(x - 1/2) & \quad \text{[/imath]x\in [1/2 - 1/n, 1/2 +1/n][imath]}\\ 1 & \quad \text{ [/imath]x\in [1/2 +1/n, 1][imath]}\\ \end{array} \right. [/imath] Then [imath]f_{n}[/imath] is cauchy in [imath](C[0,1], d_1)[/imath] but not convergent in [imath]d_1[/imath]. I have proved that [imath]f_{n}[/imath] is not convergent in [imath](C[0,1])[/imath] since it is converging to discontinuous function given as follows: [imath] f_n(x) = \left\{ \begin{array}{l l} -1 & \quad \text{ [/imath]x\in [0, 1/2 )[imath]}\\ 0 & \quad \text{[/imath]x = 1/2[imath]}\\ 1 & \quad \text{ [/imath]x\in (1/2 , 1][imath]}\\ \end{array} \right. [/imath] I am finding it difficult to prove that [imath]f_{n}[/imath] is Cauchy in [imath](C[0,1], d_1)[/imath]. I need help to solve this problem. Edit: I am sorry i have to show [imath]f_n[/imath] is cauchy Thanks for helping me. |
1002097 | A question about prime and factorization
Find a prime number [imath]p[/imath] and an integer [imath]b < p[/imath] such that [imath]p|(b^{p−1} − 1)[/imath]. Need help guys | 1002073 | Find a prime number [imath]p[/imath] and an integer [imath]b such that p divides b^{p−1}−1.[/imath]
Find a prime number [imath]p[/imath] and an integer [imath]b<p[/imath] such that [imath]p[/imath] divides [imath]b^{p−1}−1[/imath]. First I think of long divisions but it didn't work out. Now I'm stuck.. |
1001751 | How to prove that [imath]m+(4/m^2)>=0[/imath] for every m greater or equal to 0
I have a problem with proving this simple theorem. I've already figured out that the best strategy is to factorise is, so that to get eg. a square or another expression that from the definition has to be non-negative, but have no idea how exactly should I do it. Could you help me guys? | 1001834 | Prove that [imath]m+\frac{4}{m^2}\geq3[/imath] for every [imath]m > 0[/imath]
How to deal with [imath]m+\frac{4}{m^2}\geq3[/imath] for every [imath]m > 0[/imath] ? I multiplied both sides by [imath]m^2[/imath] and got [imath]m^3+4-3m^2\geq0[/imath] and have no idea how to continue with this |
1002337 | Prove that for any positive integer [imath]n[/imath], [imath]2\sqrt{n+1}-2\le 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}}\le 2\sqrt{n}-1[/imath]
Prove that for any positive integer n,[imath] 2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1 [/imath] Could anyone give me a hint on this question? Does it have something to do with Riemann Sums? | 995110 | How to prove the inequality [imath]2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1[/imath]?
Prove that for any positive integer [imath]n[/imath], [imath]2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1[/imath] Progress I think Riemann sum should be used for the middle term. I got the limit as [imath]n\to \infty[/imath] of the function inside using Riemann sum but what I cannot get are the lower and upper bound. |
1001804 | Existence of a sequence of polynomials that approximate a holomorphic function uniformly on the closed unit disk
Let [imath]f:D(0,1) \to \mathbb C[/imath] be continuous on the closed unit disc [imath]D(0,1)[/imath] and holomorphic on the open unit disc. Show that there exist a sequence of polynomial that converge uniformly in the closed unit disc to [imath]f[/imath]. Any help please? | 295509 | polynomial converges uniformily to analytic function
Let [imath]D=\{z:|z|<1\}[/imath]. suppose [imath]f(z)[/imath] is analytic on [imath]D[/imath] and continuous on [imath]\{z:|z|\leqslant1\}[/imath]. prove that there exists polynomial sequence [imath]\{P_n(z)\}[/imath] such that [imath]\{P_n(z)\}[/imath] converges uniformily to [imath]f(z)[/imath] on [imath]\{z:|z|\leqslant1\}[/imath] |
1002595 | How to prove that [imath]\forall n\in \mathbb{N}[/imath], [imath]\sum ^{n}_{i=1}i^{3}=\frac {n^{2}(n+1)^{2}}{4}[/imath]?
Use mathematical induction to prove that [imath]\forall n\in \mathbb{N}[/imath], [imath]\sum ^{n}_{i=1}i^{3}=\dfrac {n^{2}(n+1)^{2}}{4}[/imath] [imath]\begin{align*} \sum_{k=1}^{n+1} k^3 &= \sum_{k=1}^{n} k^3 + (n+1)^2 \stackrel{\rm(IH)}{=} \dfrac {n^{2}(n+1)^{2}}{4} + (n+1)^2 \\ &= \dfrac {n^{2}(n+1)^{2}+4(n+1)^2}{4} \end{align*}[/imath] Is it true? What to do next? | 341178 | Proving by induction that [imath]1^3 + 2^3 + 3^3 + \ldots + n^3 = \left[\frac{n(n+1)}{2}\right]^2[/imath]
Need guidance on this proof by mathematical induction. I am new to this type of math and don't know how exactly to get started. [imath] 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left[\frac{n(n+1)}{2}\right]^2 [/imath] |
1002811 | proof that p implies q entails not p or q
I could easily prove [imath]\neg P \lor Q[/imath] entails [imath]P \rightarrow Q[/imath]. It is well known that [imath]P \rightarrow Q[/imath] entails [imath]\neg P \lor Q[/imath] but I couldn't find a way to prove it. Although there is the same question; How to prove that [imath]P \rightarrow Q[/imath] is equivalent with [imath]\neg P \lor Q [/imath]?; it's a little bit confusing and I need to see step by step solution. Could you show me the way by using Natural Deduction? | 935640 | How to prove that [imath]P \rightarrow Q[/imath] is equivalent with [imath]\neg P \lor Q [/imath]?
In my book about Logic, which is called 'Language, Proof and Logic', by the way, there is explained that the conditional [imath] P \rightarrow Q [/imath] is equivalent with [imath]\neg P \lor Q[/imath]. There is another answer on math.stackexchange that gives as answer: 'just compare the truth tables and you can see that they are equivalent'. However, I would like to know how to prove this using formal propositional logic. This should be possible, right? |
216343 | Does Pi contain all possible number combinations?
I came across the following image: Which states: [imath]\pi[/imath] Pi Pi is an infinite, nonrepeating [imath]([/imath]sic[imath])[/imath] decimal - meaning that every possible number combination exists somewhere in pi. Converted into ASCII text, somewhere in that infinite string of digits is the name of every person you will ever love, the date, time and manner of your death, and the answers to all the great questions of the universe. Is this true? Does it make any sense ? | 1682029 | Pi's Recursiveness
I don't know if this will make sense, but: If [imath]\pi[/imath] is infinite and contains all strings of numbers including those of infinite length, then it must contain [imath]\sqrt2[/imath], and if [imath]\sqrt2[/imath] is infinite and contains all strings of numbers including those of infinite length then [imath]\sqrt2[/imath] contains [imath]\pi[/imath]. This means that there is [imath]\pi[/imath] inside of [imath]\pi[/imath], and therefore it's recursive. I probably went wrong somewhere (I am a secondary school student) and I don't know everything about these numbers. I am assuming that [imath]\pi[/imath] is infinite and contains all strings although I know this has been called into question. So can anyone help me out, because this just seems wrong yet the sum of all natural numbers is [imath]-1/12[/imath] so, anything is possible. |
1002534 | Kernel of an evaluation homomorphism [imath]\mathbb{C}[X_1,\dots, X_n] \rightarrow \mathbb{C}[/imath]
Let [imath]R:=\mathbb{C}[X_1,\dots, X_n][/imath], [imath]a=(a_1,\dots, a_n)\in \mathbb{C}^n[/imath] and [imath]\phi_a:R\rightarrow \mathbb{C}[/imath], [imath]\phi_a(f)=f(a)[/imath]. I want to show that [imath]\ker(\phi_a)=(X_1-a_1,\dots, X_n -a_n)[/imath]. I know that to be true for [imath]n=1[/imath] and I also know that [imath]\ker(\phi_a)[/imath] is a maximal ideal in [imath]R[/imath]. Moreover, [imath](X_1-a_1,\dots, X_n -a_n) \subseteq \ker(\phi_a)[/imath] is obvious. I am note sure if I can do an induction in this case. Thank you. | 500153 | Proving that kernels of evaluation maps are generated by the [imath]x_i - a_i[/imath]
I want to prove that the kernel of the evaluation map [imath]s_a : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto a_i[/imath] where [imath]a = (a_1,\dots, a_n) \in \mathbb{C}^n[/imath] is the ideal generated by [imath]\{x_1 - a_1, \dots, x_n - a_n\}[/imath]. The proof in the book first shows it easily for the case [imath]a = 0[/imath], and says that by substitutions of variables [imath]x_i' = x_i - a_i[/imath], you can show it for the rest of the cases. I know that [imath]f \in \ker s_a[/imath] iff [imath]f(a) = 0[/imath]. And [imath]\phi : \mathbb{C}[x_1,\dots, x_n] \rightarrow \mathbb{C}[x_1 - a_1, \dots, x_n-a_n], x_i \mapsto x_i - a_i[/imath] is a surjective ring homomorphism. Please give a hint and not the full answer. Thanks. |
1002725 | Why diagonal map is closed?
can you help me?... Let X [imath]\subseteq \mathbb{A}^n[/imath] be an algebraic set, and [imath]f, g:X\rightarrow k[/imath] two regular functions. If [imath]X[/imath] is irreducible and an open [imath]U \subseteq X[/imath] exists such that [imath]f_{|U} = g_{|U}[/imath], prove that [imath]f=g[/imath]. I have been able to prove that: Let E be the subset where [imath]f = g[/imath] so E is closed (Why? I know that [imath]E = (f,g)^{-1}(\Delta_{k})[/imath] but... why is [imath](\Delta_{k})[/imath] closed?). So [imath]U \subseteq E \Rightarrow \bar{U} \subseteq \bar{E}=E[/imath] (beacuse E is closed) [imath]\Rightarrow X \subseteq E[/imath], because [imath]\bar{U}=X \Rightarrow E = X[/imath] and [imath]f=g[/imath] in X. | 1002462 | 3 questions about Algebraic Geometry and Zariski topology
I've read this exercise on my book, but I don't know how to prove it... Can someone help me? Let X [imath]\subseteq \mathbb{A}^n[/imath] be an algebraic set, and [imath]f, g:X\rightarrow k[/imath] two regular functions. a) If we see [imath]f[/imath] as an application [imath]X \rightarrow \mathbb{A}^1(k)[/imath], prove that [imath]f[/imath] is continuous in the Zariski topology. b) If [imath]X[/imath] is irreducible and an open [imath]U \subseteq X[/imath] exists such that [imath]f_{|U} = g_{|U}[/imath], prove that [imath]f=g[/imath]. c) Give an example of [imath]X, f[/imath] and [imath]g[/imath] such that [imath]f \neq g[/imath] but [imath]f_{|U} = g_{|U}[/imath] in an open [imath]U \subseteq X[/imath]. Thank you! For b) I have been able to prove that: Let E be the subset where [imath]f = g[/imath] so E is closed (Why? I know that [imath]E = (f,g)^{-1}(\Delta_{k})[/imath] but... why is [imath](\Delta_{k})[/imath] closed?). So [imath]U \subseteq E \Rightarrow \bar{U} \subseteq \bar{E}=E[/imath] (beacuse E is closed) [imath]\Rightarrow X \subseteq E[/imath], because [imath]\bar{U}=X \Rightarrow E = X[/imath] and [imath]f=g[/imath] in X. |
1003221 | How to prove [imath]n!\leq(\frac{n+1}{2})^n[/imath]
Prove that for [imath]n\in\mathbb{N}[/imath] [imath]n!\leq(\frac{n+1}{2})^n.[/imath] I've solved base case for [imath]n=1[/imath] [imath]1\leq(\frac{1+1}{2})^1=1[/imath] The second step I've made was that I assumed that [imath]n!\leq(\frac{n+1}{2})^n[/imath] And then I have [imath](\frac{n+2}{2})^{n+1}[/imath]. What do I need to do now? | 992056 | How can I show that [imath]n! \leqslant (\frac{n+1}{2})^n[/imath]?
Show that [imath]n! \leqslant (\frac{n+1}{2})^n \quad \hbox{for all } n \in \mathbb{N}[/imath] I know that it can be done by induction but I always find line where I do not know what to do next. |
1003210 | Find an n such that the equation [imath]x^2+x=0[/imath] in mod n has 4 solutions
So this is my question and I also have to do it so it has 8 solutions, my problem is that I don't really understand how to approach the problem at all. Any type of help would be appreciated thanks | 1000339 | Number of solutions of polynomials in a field
Consider the polynomial [imath]x^2+x=0[/imath] over [imath]\mathbb Z/n\mathbb Z[/imath] a)Find an n such that the equation has at least 4 solutions b)Find an n such that the equation has at least 8 solutions My idea is to check individual n, and I found the answer for a is n=6 at x={0,2,3,5} f(x)={0, 6, 12, 30) Is there any shorter way? |
1004814 | Why is this [imath]0 = 1[/imath] proof wrong?
[imath]0 = 0 + 0 + 0 + ...[/imath] [imath]0 = (1 - 1) + (1 - 1) + (1 - 1) + ...[/imath] [imath]0 = 1 - 1 + 1 - 1 + 1 - 1 + ...[/imath] [imath]0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + ...[/imath] [imath]0 = 1 + 0 + 0 + 0 + ...[/imath] [imath]0 = 1[/imath] I can't really tell what is obviously wrong with this. It seems to use the same logic as we see in the derivation of things like [imath]\sum_{k=1}^{\infty} k = -\frac{1}{12}[/imath] which appears to be a quirky but accepted fact in mathematics. | 1394490 | "The God Proof" - what's wrong with it?
I stumbled upon "the God proof" which goes: [imath]0 = 0 + 0 + 0...[/imath] [imath] = (1-1) + (1-1) + (1-1) + ...[/imath] [imath]= 1 - 1 + 1 - 1 + 1 - 1 + ...[/imath] [imath]= 1 + (-1+1) + (-1+1) + (-1+1) + ...[/imath] [imath]= 1[/imath] Even though this result is obviously wrong, I can't quite pinpoint exactly what the 'illegal' operation here is? I know that it's possible to make a conditionally convergent infinite series converge to any value, so that can't be the issue (?). It makes me think there's something dangerous about representing numbers by infinite series that aren't absolutely convergent, but I haven't been able to find something that specifically addresses this, if someone could tell me where to look for more information, I'd be grateful. [background: undergrad student] |
1004565 | Can A be singular?
Let [imath]A\in \mathbb{C}^{n\times n}[/imath] satisfy [imath]A^{2}+A+I=0 [/imath] Can A be singular? So I have: [imath] (A-I)(A^{2}+A+I)=0\\ A^{3} = I \\ (\det A^{3}) = \det(I) \\ (\det A)^{3} = 1\\ \det A\neq 0 [/imath] So [imath]A[/imath] is not singular. Is this correct the way I've done this? | 997425 | Can [imath]A[/imath] be singular
[imath]A^2 + A + I= 0[/imath] Can [imath]A[/imath] be singular? Justify your answer. I do not know where to start. |
1005437 | Extended Euclid Algorithm
A Linear Diophantine Equation is of the following form: [imath]Ax+By+C=0[/imath], where [imath]x_1\le x\le x_2[/imath] and [imath]y_1\le y\le y_2[/imath]. If the value of [imath]A[/imath], [imath]B[/imath], [imath]C[/imath], [imath]x_1[/imath], [imath]x_2[/imath], [imath]y_1[/imath], [imath]y_2[/imath] are given and [imath]x_1\le x_2[/imath] and [imath]y_1\le y_2[/imath], then how many solutions can be found? How can I find out the total number of solutions according to the above condition? I asked the above question in this link:Solving a Linear Diophantine Equation before. But there I didn't give sample input and output. I liked the second answer in the above link but when I checked the sample input and output according to the second answer it gives me wrong result It is worthy of mention that the reasoning of the second answer based on [imath]A[/imath], [imath]B\gt 0[/imath], but my question is that what I have to do if [imath]A[/imath], [imath]B\lt 0[/imath] and how will I match sample input and output using the given answer. Here, I have included sample input and output: I need a method that satisfies the above input and output. I also need better explanation of the method with example. | 959029 | Solving a Linear Diophantine Equation
A Linear Diophantine Equation is of the following form: [imath]Ax+By+C=0[/imath], where [imath]x_1 \leq x \leq x_2[/imath] and [imath]y_1 \leq y \leq y_2[/imath]. If the value of [imath]A,B,C,x_1,x_2,y_1,y_2[/imath] are given and [imath]x_1 \leq x_2[/imath] and [imath]y_1 \leq y_2[/imath], then how many solutions can be found? How can I find out the total number of solutions according to the above condition. |
1006131 | How Would I Compute This Product?
[imath]f[/imath] is irreducible, How do I compute this [imath]\prod_{d\mid n}\prod_{f \in \mathbb{Z_p[x]},\, \deg(f) = d}f[/imath] I tried small examples for this: So [imath]n = 1, p = 2[/imath]. We have that this product is: [imath]x(x + 1) = x^2 + x = x^2 - x[/imath] [imath]n = 2, p = 2[/imath] We have that this product is : [imath]x(x + 1)(x^2 + x + 1) = (x^2 + x)(x^2 + x + 1) = x^4 + x^2 + x + x^2 + 2 = x^4 + x = x^4 - x[/imath] I'm looking at this and I'm guessing that this product is: [imath]x^{p^n} - x[/imath] but I have no idea how to show this in a general case. Would anyone have any suggestions how to do this? | 106721 | Product of all monic irreducibles with degree dividing [imath]n[/imath] in [imath]\mathbb{F}_{q^n}[/imath]?
In the finite field of [imath]q^n[/imath] elements, the product of all monic irreducible polynomials with degree dividing [imath]n[/imath] is known to simply be [imath]X^{q^n}-X[/imath]. Why is this? I understand that [imath]q^n=\sum_{d\mid n}dm_d(q)[/imath], where [imath]m_d(q)[/imath] is the number of irreducible monic polynomials with degree [imath]d[/imath], and that each element of the field satisfies [imath]X^{q^n}-X[/imath]. How does the conclusion follow from this? I tried substituting the exponent to [imath]X^{\sum_{d\mid n}dm_d(q)}-X[/imath] but this doesn't seem to do much good. |
70410 | Compact operator maps weakly convergent sequences into strongly convergent sequences
I found the following property of compact operators in a proof, and I can't prove it. Prove that if [imath]T \in \mathcal{L}(E,F)[/imath] is compact, and if [imath]u_n \rightharpoonup u[/imath] (the sequence converges weakly to [imath]u[/imath] in [imath]\sigma(E,E^*)[/imath]) then [imath]Tu_n \to Tu[/imath] strongly in [imath]F[/imath]. I was able to prove that [imath]Tu_n[/imath] has a convergent subsequence ([imath]u_n \rightharpoonup u[/imath] implies that [imath](u_n)[/imath] is bounded in [imath]E[/imath]. Then, because [imath]T[/imath] is compact it must follow that [imath](Tu_n)[/imath] must contain some strong convergent subsequence), but didn't manage to finalize the proof. Any reference or hints are welcome. | 1142451 | Compact Operators: Weak Convergence
Problem Given Banach spaces [imath]X[/imath] and [imath]Y[/imath]. Consider a compact operator [imath]C\in\mathcal{C}(X,Y)[/imath]. Then weak convergence is turned into strong convergence: [imath]x_n\rightharpoonup x\implies Cx_n\to Cx[/imath] I'd like to try proving it but would need some hints. Attempt Denote the sup-norm by: [imath]F:\Omega\to E:\quad\|F\|_\Omega:=\sup_{\omega\in\Omega}\|F(\omega)\|_E[/imath] Weak convergence is preserved under continuous operators: [imath]\|l(Cx_n-Cx)\|=\|(C'l)(x_n-x)\|\to0[/imath] By uniform boundedness principle weak convergence implies boundedness: [imath]x_n\rightharpoonup x:\quad\|l(x)\|_\mathbb{N}<\infty\implies\|x\|_\mathbb{N}<\infty[/imath] Hence one can exploit compactness of the operator: [imath](x_n)_{n\in\mathbb{N}}\text{ bounded}\implies C(x_n)_{n\in\mathbb{N}}\text{ precompact}[/imath] And one obtains strongly convergent subsequences. Should I combine these now and how? |
1006201 | An Inequality with complex numbers and [imath]1/\pi[/imath]
Let [imath]\displaystyle \{z_1,z_2, \ldots, z_n\}[/imath] be [imath]n[/imath] complex numbers such that: [imath]\displaystyle \sum\limits_{k=1}^n|z_k| = 1[/imath] Then we have to show that, there is a subset [imath]S[/imath] of [imath]\{1,2,\ldots,n\}[/imath], such that, [imath]\displaystyle \left|\sum\limits_{k \in S} z_k \right| \ge \frac{1}{\pi}[/imath]. Owing to the symmetry of the expression its easy to see: [imath]\displaystyle \left|\sum\limits_{k \in S} z_k \right| \ge \frac{1}{4}[/imath] Since, [imath]1 = \sum\limits_{k=1}^{n} |z_k| \le \sum\limits_{\mathfrak{Re}(z) \ge 0} |\mathfrak{Re} (z_k)|+\sum\limits_{\mathfrak{Re}(z) < 0} |\mathfrak{Re} (z_k)|+\sum\limits_{\mathfrak{Im}(z) \ge 0} |\mathfrak{Im} (z_k)|+\sum\limits_{\mathfrak{Im}(z) < 0} |\mathfrak{Im} (z_k)|[/imath], one of the four summations on the RHS must exceed the others, say wlog it is [imath]\mathfrak{Re}(z) \ge 0[/imath], then, [imath]\left|\sum\limits_{\mathfrak{Re}(z_k) \ge 0} z_k \right| \ge \left|\sum\limits_{\mathfrak{Re}(z_k) \ge 0} \mathfrak{Re}(z_k) \right| \ge \frac{1}{4}[/imath]. (1) But how do we reach the lower bound [imath]\displaystyle \frac{1}{\pi}[/imath] ? If we remove the homogeneity condition, we are required to prove, [imath]\displaystyle \left|\sum\limits_{k \in S} z_k \right| \ge \frac{1}{\pi}\sum\limits_{k=1}^n|z_k|[/imath] for a subset [imath]S[/imath] of [imath]\{1,2,\ldots,n\}[/imath]. (2) How do we determine if [imath]1/\pi[/imath] is the best constant here ? | 91939 | Inequality with complex numbers
Consider the following problem. Fix [imath]$n \in \mathbb N$[/imath]. Prove that for every set of complex numbers [imath]$\{z_i\}_{1\lei\le n}$[/imath], there is a subset [imath]$J\subset \{1,\dots , n\}$[/imath] such that [imath]\left|\sum_{j\in J} z_j\right|\ge \frac{1}{4\sqrt 2} \sum_{k=1}^n |z_k|.[/imath] I believe I proven have a stronger statement. Is this proof correct, and if so, what is the optimal constant? My proof. Consider all the [imath]z_i[/imath] with positive real part. Call the real part of the sum of these numbers [imath]X^+[/imath]. In a similar way, form [imath]X^-[/imath], [imath]Y^+[/imath], and [imath]Y^-[/imath]. Without loss of generality, let [imath]X^+[/imath] have the greatest magnitude of these. Note that because [imath]|\operatorname{Re}(z)|+|\operatorname{Im}(z)|\ge |z|[/imath], we have [imath] \left(\sum_{k=1}^n |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| \right) \ge \sum_{k=1}^n |z_k|.[/imath] But note that [imath]\sum \limits_{k=1}^n |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| = X^+ + |X^-|+ Y^+ +|Y^-|[/imath], so we have [imath] 4X^+ \ge \sum_{k=1}^n |z_k|.[/imath] By choosing [imath]J[/imath] to be the set of complex number with positive real part, this proves a stronger statement, because the factor of [imath]1/\sqrt 2[/imath] isn't needed. |
1006678 | Inverse image of a maximal ideal under a morphism of finitely generated [imath]\mathbb{C}[/imath]-algebras.
Let [imath] f: A\to B [/imath] be a morphism of finitely generated [imath]\mathbb{C}[/imath]-algebras, suppose [imath]\mathfrak{m}\unlhd B[/imath] is a maximal ideal, I want to show that [imath]f^{-1}(\mathfrak{m})[/imath] is a maximal ideal of [imath]A[/imath]. Consider the morphism [imath] A\to B/\mathfrak{m}, [/imath] [imath] a\mapsto f(a)+\mathfrak{m}, [/imath] this has kernel [imath]f^{-1}(\mathfrak{m})[/imath], by the first isomorphism theorem we get [imath] A/ f^{-1}(\mathfrak{m}) \cong \frac{f(A)\cap \mathfrak{m}}{\mathfrak{m}}, [/imath] so, RHS is a field, so LHS is a field and [imath]f^{-1}(\mathfrak{m})[/imath] is maximal. I think that my proof is wrong, probably I got RHS wrong and I am not using the fact that these are finitely generated or over [imath]\mathbb{C}[/imath] at all. What is the correct proof? | 348796 | If [imath]B[/imath] is finitely generated as a [imath]k[/imath]-algebra, and [imath]\phi:A\to B[/imath] is a [imath]k[/imath]-algebra map, is [imath]\phi^{-1}(M)[/imath] maximal for any maximal [imath]M\subset B[/imath]?
Suppose that [imath]A[/imath] and [imath]B[/imath] are commutative rings containing a field [imath]k[/imath], and [imath]B[/imath] is finitely generated [imath]k[/imath]-algebra. Let [imath]\phi: A\rightarrow B[/imath] be a ring homomorphism with [imath]\phi|_k =\mathrm{Id}[/imath]. I am trying to prove that if [imath]M\subset B[/imath] is a maximal ideal, then [imath]\phi^{-1}(M)[/imath] is a maximal ideal of [imath]A[/imath]. The case when [imath]A \subset B[/imath] is an integral extension of rings is well-known. I think I can also prove the result when [imath]\phi[/imath] is surjective. Inverse Image of Maximal Ideals discusses the case when [imath]B[/imath] is a finitely generated [imath]\mathbb{Z}[/imath]-algebra but I am not sure how to generalize this. |
1006821 | The ring [imath]\mathbb{Z}_2 \times \mathbb{Z}_2[/imath] is a domain
True\False :The ring [imath]\mathbb{Z}_2 \times \mathbb{Z}_2[/imath] is a domain solution True A commutative ring with identity is said to be an integral domain if it has no zero divisors. | 844337 | Prove [imath]R \times R[/imath] is NOT an integral domain
I have a question, and this is it in entirety: Let R be a non-zero ring(that is, R contains at least one element other than the zero element). Prove that [imath]R \times R[/imath] is NOT an integral domain. Now I haven't done a problem like this before, and it seems to be(from the Wikipedia definition) that this ring [imath]R[/imath] needs to be commutative to be an integral domain. Which it isn't stated to be here. What exactly am I meant to do? I haven't been given anything else than the area in grey. |
1006755 | Prove regarding number of distinct polynomial roots
Prove that for any g(x) in a [imath]\mathbb{Z}/p\mathbb{Z}[x][/imath] field, the degree of [imath]\gcd{(x^p-x,g(x))}[/imath] is equal to the number of distinct roots of g(x) First of all, I am assuming that the "number of distinct roots" refer to the number of distinct roots of g(x) in the mod p field Secondly, where should I start to go about proving this? | 1003508 | Number of roots of a polynomial over a finite field
For any [imath]g[/imath] in [imath]\mathbb{Z}/p\mathbb{Z}[x][/imath] prove that the degree of [imath]f = \gcd(x^p - x, g(x))[/imath] is exactly the number of distinct roots of [imath]g[/imath] in [imath]\mathbb{Z}/p\mathbb{Z}[/imath]. My main problem is that I do not really understand how to approach this problem at all. I am very new to solving proofs and any type of help would be appreciated thank you. |
1004628 | If [imath]\gcd(p,q+Ar)>1[/imath] for all [imath]A\in \mathbb{Z}[/imath], then [imath]\gcd(p,q,r)>1[/imath]
Prove that if [imath]\gcd(p,q+Ar)>1[/imath] for all integers [imath]A[/imath], then [imath]\gcd(p,q,r)>1[/imath]. I let [imath]A=0[/imath] to get that [imath]\gcd(p,q)>1[/imath]. Then I noticed that [imath]\gcd(p,q,r)=\gcd(\gcd(p,q),r)[/imath]. Another way to say it: Prove that if [imath]\gcd(ag,bg+Ar)>1[/imath] for all integers [imath]A[/imath] ([imath]\gcd(a,b)=1[/imath]), then [imath]\gcd(g,r)>1[/imath]. | 647600 | Prove: If [imath]\gcd(a,b,c)=1[/imath] then there exists [imath]z[/imath] such that [imath]\gcd(az+b,c) = 1[/imath]
I can't crack this one. Prove: If [imath]\gcd(a,b,c)=1[/imath] then there exists [imath]z[/imath] such that [imath]\gcd(az+b,c) = 1[/imath] (the only constraint is that [imath]a,b,c,z \in \mathbb{Z}[/imath] and [imath]$c\neq 0)$[/imath] |
1007552 | [imath]D_6 \cong D_3\times \mathbb Z_2[/imath]
im trying to show that [imath]D_6 \cong D_3\times \mathbb Z_2[/imath] as groups. [imath]D_n[/imath] is the is the dihedral of order 2n I know that it is possible to form all of the points in [imath]D_6[/imath] and in [imath]D_3[/imath] as compositions of [imath]R[/imath] the rotation operation (such that every vertex move one to the left) and T the reflection on one of the symmetry axis, so i tried to write it done and joining point that act the same, but for no use. | 322685 | Let [imath]n \geq 1[/imath] be an odd integer. Show that [imath]D_{2n}\cong \mathbb{Z}_2 \times D_n[/imath].
Let [imath]n \geq 1[/imath] be an odd integer. Show that [imath]D_{2n}\cong \mathbb{Z}_2 \times D_n[/imath]. I define a map [imath]\phi:D_{2n} \rightarrow \mathbb{Z}_2 \times D_n[/imath] by [imath]R \mapsto (0,r^{\frac{n+1}{2}})[/imath] and [imath]M \mapsto(1,m)[/imath]. Then I am stuck at showing the map is bijective. By the way, do we need to show that [imath](0,r^{\frac{n+1}{2}})(1,m)(0,r^{\frac{n+1}{2}})^{-1}=(1,m)[/imath] ? If we can prove this, then what can we conclude? |
1007595 | Radical of an ideal in a finitely generated ring over [imath]k[/imath] is the intersection of maximal ideals containing it.
From Matsumura p.34 Let [imath]k[/imath] be a field, [imath]A[/imath] a ring which is finitely generated over [imath]k[/imath], and [imath]I[/imath] a proper ideal of [imath]A[/imath]; then the radical of [imath]I[/imath] is the intersection of all maximal ideals containing [imath]I[/imath]. I understand that the radical of [imath]I[/imath] is contained in this intersection but I'm having troublr understanding the other direction. The proof says this follows from part 2 of Hilbert's Nullstellensatz Given a subset [imath]\Phi[/imath] of [imath]k[X_1, ... ,X_n][/imath] and an element [imath]f \in k[X_1, ... ,X_n][/imath] suppose that [imath]f[/imath] vanishes at every algebraic zero of [imath]\Phi[/imath]. Then some power of [imath]f[/imath] belongs to the ideal generated by [imath]\Phi[/imath]. So in order for the theorem to follow from this...don't we need [imath]k[/imath] to be algebraically closed? Because otherwise the maximal ideals are not necessarily of the form [imath](X_1 - a_1, ... , X_n - a_n)[/imath]. So we can't say that f vanishes at [imath](a_1, ... , a_n)[/imath]. | 812391 | Every prime ideal of a finitely generated [imath]\mathbb{R}[/imath]-algebra is an intersection of maximal ideals?
Why must every prime ideal of a finitely generated [imath]\mathbb{R}[/imath]-algebra (e.g. [imath]\mathbb{R}[X_1,X_2][/imath]) be the intersection of the maximal ideals containing it? This doesn't follow from the version of the Nullstellensatz I've seen: if [imath]k[/imath] is an algebraically closed field and [imath]S[/imath] is a finitely generated [imath]k[/imath]-algebra and [imath]P[/imath] is a prime ideal of [imath]S[/imath], then [imath]P[/imath] is the intersection of all maximal ideals containing [imath]P[/imath] ... since [imath]\mathbb{R}[/imath] isn't algebraically closed! But maybe the same method of proof could work? Many thanks in advance for any help with this! |
1007726 | Independence of multivariate random variables
Let [imath]X=(X_1,X_2,...,X_n)[/imath] be an independent multivariate random variable, i.e. [imath] F_X(x_1, x_2,...,x_n)=\prod_{i=1}^{n}F_{X_i}(x_i) [/imath] This implies that if we take 2 components [imath]X_i, X_j[/imath], these are independent as well: [imath] F_{(X_i, X_j)}(x,y)=F_{X_i}(x) \cdot F_{X_j}(y) [/imath] I need to prove that the reverse statement is not true, i.e this is false: [imath] F_{(X_i, X_j)}(x,y)=F_{X_i}(x) \cdot F_{X_j}(y) \implies F_X(x_1, x_2,...,x_n)=\prod_{i=1}^{n}F_{X_i}(x_i) [/imath] A theoretical proof or a counter-example are both acceptable. Thanks, András | 584670 | Pairwise independence of Random variables does not imply indendence
Show by a counterexample that for a family [imath](X_i)_{i\in I}[/imath] of random variables the independence of all pairs [imath](X_i,X_j)[/imath] with [imath]i,j\in I, i\neq j[/imath] does not imply the independence of the family (It is enough to consider [imath]\mbox{card}(I)=3[/imath].). Hello, I am searching for a simple example. The professor gave the hint that one might consider random variables which only can take two different values, but I do not know exactly what is meant. By the way: We call the family [imath](X_i)_{i\in I}[/imath] of random variables independent, if [imath](\sigma(X_i))_{i\in I}[/imath] is independent, whereat we call a family [imath](\mathcal{E}_i)_{i\in I}[/imath] with [imath]\mathcal{E}_i\subset\mathcal{A}[/imath] independent, if for any finite subset [imath]I_0\subset I[/imath] and any choice [imath]E_i\in\mathcal{E}_i, i\in I_0[/imath], it is [imath] \mathbb{P}(\bigcap_{i\in I_0}E_i)=\prod_{i\in I_0}\mathbb{P}(E_i). [/imath] |
1007736 | A question about Infimum.
Let [imath]A[/imath] be an infinite set that includes Real numbers and its not bounded. Let [imath]B[/imath] be a set of Real numbers [imath]x[/imath] s.t. the intersection [imath]A\cap[x,\infty)[/imath] is [imath]empty[/imath] or includes [imath]finite[/imath] number of elements. prove or disprove the existence of [imath]\inf B[/imath] if [imath]A[/imath] is not bounded. Actually this is the third part of the question that I didn't solve. I think there is more than one situation in this case. can someone give me a hint ? | 1001474 | Question about Infimum. (conclusion about intersection)
Let [imath]A[/imath] be an infinite set that includes Real numbers and is bounded. Let [imath]B[/imath] be a set of Real numbers [imath]x[/imath] s.t. the intersection [imath]A\cap[x,\infty)[/imath] is empty or includes [imath]finite[/imath] number of elements. prove that [imath]\inf B[/imath] exists. prove or disprove [imath]\inf B=\min B[/imath] prove or disprove that [imath]\inf B[/imath] exists if we don't ask for [imath]A[/imath] to be bounded. I started searching for examples to see the whole picture, like sequences of the form "[imath]\frac{1}{n}"[/imath] and others, but I didn't know to to do the intersection and why I must be sure that B has an infimum. its easier if someone can give an example of two sets [imath]A[/imath] and [imath]B[/imath], and make it clear why the intersection must be a finite set. (after choosing "x" from set B) |
1008466 | Prove that harmonic numbers satisfy the equality.
The [imath]k[/imath]th harmonic number is defined to be [imath]H_k = 1 + 1/2+ 1/3 + · · · + 1/k .[/imath] Prove that harmonic numbers satisfy the equality [imath]H_1 + H_2 + · · · + H_n = (n + 1)H_n − n[/imath] for all [imath]n \in\Bbb N.[/imath] | 994483 | Using induction to prove an equality in harmonic numbers
Question: Prove that harmonic numbers satisfy the equality using induction [imath] H_{1}+ H_{2} + · · · + H_{n} = (n + 1)H_{n} − n. [/imath] I have done the basis step: [imath](1 + 1)H_{1} − 1 = 1[/imath]. Correct. Done the inductive hypothesis: [imath](k + 1)H_{k} − k[/imath] and started solving: [imath] (k + 2)H_{k+1} − (k+1) = (k + 1)H_{k} − k + \frac{1}{k+1}. [/imath] What do I do once I get to this part? |
1000996 | How to prove [imath]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2[/imath]?
Prove the inequality for a triangle with sides [imath]a,b,c[/imath] we have [imath]\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2[/imath] Trial: Since [imath]a,b,c[/imath] are sides of a triangle I know [imath]a+b>c,b+c>a,a+c>b[/imath] | 54627 | Inequality involving sides of a triangle
How can one show that for triangles of sides [imath]a,b,c[/imath] that [imath]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2[/imath] My proof is long winded, which is why I am posting the problem here. Step 1: let [imath]a=x+y[/imath], [imath]b=y+z[/imath], [imath]c=x+z[/imath], and let [imath]x+y+z=1[/imath] to get [imath]\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}<2[/imath] Step 2: consider the function [imath]f(x)=\frac{1-x}{1+x}[/imath], and note that it is convex on the interval (0,1), so the minimum of [imath]\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}[/imath] is reached when the function takes the extreme points. i.e. [imath]x=y=0, z=1[/imath]. But going back to the fact that this is a triangle, we note that [imath]x=y=0 \implies a=0[/imath] which is not possible, so the inequality is strict. |
1008659 | Why does a ring homomorphism induce a continuous map between spectra?
Let [imath]\varphi: A \rightarrow B[/imath] be a ring homomorphism. Let [imath]f =\mathrm{Spec}(\varphi) : \mathrm{Spec}(B) \to \mathrm{Spec}(A)[/imath] be the map associated to [imath]\varphi[/imath]. Why is the map [imath]f[/imath] is continuous? I think we need to use [imath]IB[/imath] (the ideal of [imath]B[/imath] generated by [imath]\varphi(I)[/imath]). Then we got [imath]f^{-1}(V(I)) = V(IB)[/imath]. (why?) Could someone tell the details of this proof? Thank you :) | 764796 | A homomorphism induces a continuous map from [imath]{\rm Spec}(A') \to {\rm Spec}(A)[/imath].
Let [imath]A, A'[/imath] be commutative rings with [imath]1 \neq 0[/imath]. Let [imath]h : A \to A'[/imath] be such that [imath]h(1) = 1[/imath]. Then [imath]f: {\rm Spec}(A') \to {\rm Spec}(A)[/imath] defined by [imath]f(\mathfrak{p}') = h^{-1}(\mathfrak{p}')[/imath] is continous with respect to the spectral topologies. |
713199 | On [imath]\lfloor\sqrt n \rfloor+ \sum_{j=1}^n \lfloor n/j\rfloor[/imath]
How do we prove that [imath]\Big[\sqrt n \Big]+ \sum_{j=1}^n \bigg[ \dfrac nj\bigg][/imath] is an even integer for all [imath] n \in \mathbb N[/imath] ? (where [imath]\Big[ \space \Big][/imath] denotes the "greatest integer" function) | 661010 | How to show that [imath]\lfloor n/1\rfloor+\lfloor n/2 \rfloor+....+\lfloor n/n\rfloor+\lfloor{\sqrt{n}}\rfloor[/imath] is even?
Let [imath]n[/imath] is a natural number. Prove that [imath]\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+....+\left\lfloor\frac{n}{n}\right\rfloor+\left\lfloor{\sqrt{n}}\right\rfloor[/imath] is even. I used induction and the inequality [imath]x-1<\lfloor{x}\rfloor\le{x}[/imath] to prove it. Is there any other, nicer way to do it? |
1009634 | How to determine this integral
Let's take: [imath]\int \frac{1}{x\sqrt{x+1}} \ dx [/imath] I tried solve this by four hour, so I am asking for help | 690419 | Calculus 2 integral [imath]\int {\frac{2}{x\sqrt{x+1}}}\, dx[/imath]
Make a substitution to express the integrand as a rational function and then evaluate the integral [imath]\int {\frac{2}{x\sqrt{x+1}}}\, dx[/imath] What is the substitution that I have to make? |
1009454 | Dense convex proper subset of the Hilbert space [imath]l_2[/imath]: [imath]\{x|\sum x_i=0\}[/imath]
Let's consider the space [imath]l_2[/imath] (all sequences [imath]x[/imath] with [imath]\sum x_i^2 < +\infty[/imath]) and its subset [imath]Z = \{x|\sum x_i = 0\}[/imath]. I want to prove that the closure of [imath]Z[/imath] is [imath]l_2[/imath], but I can't. I tried to build some constructions to get [imath]Z[/imath] elements [imath]\epsilon[/imath]-close to an arbitrary [imath]l_2[/imath] element, but nothing comes out. I ask for some help. Is there any theorem or a method which might help me? | 997021 | Showing a certain subspace of Hilbert space is dense
Let H be the Hilbert space of square-summable sequences of reals. A few years ago I thought I had proved that the subspace Z of real sequences with only finitely many nonzero terms, such that they sum to zero, is dense in H. (Since then I've seen this confirmed in Rudin's text, Functional Analysis, 2nd ed., but only as a teensy subquestion in a terminally hairy exercise.) Now I can't quite reproduce my proof, so it may have been wrong. Can someone please tell me how to prove that. (What I tried was taking an arbitrary point c = (c1,...,ck,...) of H, and defining the point d(N) in H as follows for N [imath]\ge[/imath] 1: First set AN = (c1+...+cN) / N. Then set d(N)k = ck - AN for k [imath]\le[/imath] N, and set d(N)k = 0 for k > N. Clearly, the element d(N) of H lies in the subspace Z defined above. The squared Hilbert norm of its difference with c is of form N(AN)2 + T(N), where T(N) is just the squared norm of the tail of c, and so goes to 0 as N[imath]\to\infty[/imath]. I'm left with the expression N(AN)2, which so far I haven't been able to prove [imath]\to[/imath] 0 as N[imath]\to\infty[/imath]. I suspect this is true, and it works on all the examples I've tried so far.) NOTE: I'm only interested in a down-and-dirty proof that doesn't invoke anything but simple inequalities. Thanks for any help you can offer. |
1010045 | A curve is contained in a circle
I need to prove that if [imath]\alpha: I\rightarrow \mathbb{R}^{2}[/imath] is regular with curvature [imath]\kappa[/imath], then [imath]\alpha[/imath] is on a circle with radius [imath]r>0[/imath] if and only if [imath]|\kappa(t)|=\dfrac{1}{r}[/imath]. | 385006 | if the curvature is constant and positive, then it is on the circunference
I'm trying to prove that if [imath]\alpha(t)=(x(t),y(t))[/imath] is a [imath]C^2[/imath] regular curve [imath](\alpha'\neq0)[/imath] with constant and positive curvature, then [imath]\alpha[/imath] is on the circunference and if [imath]\alpha[/imath] is the circunference, then its curvature is constant. I solved the second part since [imath]\alpha[/imath] is the circunference, then [imath]\alpha(t)=(r\cos\theta,r\sin\theta)[/imath], so [imath]|\alpha'(t)|=r[/imath], I need help in the first part. Thanks a lot. |
1007404 | Problem on a subfield being dense in [imath]\mathbb C[/imath]
Let [imath]K[/imath] be a subfield of [imath]\mathbb C[/imath] not contained in [imath]\mathbb R[/imath]. Is [imath]K[/imath] always dense in [imath]\mathbb C[/imath]? I have studied to show a set [imath]A[/imath] is dense in [imath]B[/imath] we will have to show that for any element [imath]x[/imath] in [imath]B[/imath] there exists a sequence in [imath]A[/imath] converging to [imath]x[/imath].But dont know what should be my approach here? | 1010467 | For a subfield [imath]K[/imath] of [imath]\Bbb C[/imath] with [imath]K\nsubseteq \Bbb R[/imath], show [imath]K[/imath] is dense in [imath]\Bbb C[/imath].
Let [imath]K [/imath] be a subfield of [imath]\mathbb C[/imath] not contained in [imath]\mathbb R[/imath]. Is [imath]K[/imath] dense in [imath]\mathbb C[/imath]? My problem is I have never used the concept of dense set in algebra and neither have any idea about using sequences here. Please give some hints |
1010742 | Taylor Series Expansion for [imath]\frac1{(1+z^2)}[/imath]
Given [imath]f(z) = \frac{1}{1+z^2}[/imath], I want to find the Taylor series of [imath]f(z)[/imath] about [imath]z_0 = 0[/imath]. Intuitively, and based on the formation of a standard power series I have [imath]f(z) = \sum (-1)^n(z^2)^n[/imath]. What I need now is to find the power series for [imath](\frac{1}{1+z^2})^2[/imath] and [imath](\frac{1}{1+z^2})^3[/imath]. I assume I'll need to use partial fractions here but I'm a little stuck at the start. Tips would be appreciated. | 1010582 | Find a formula for Taylor series of [imath]\left(\frac{1}{1+z^2}\right)^n[/imath]
So the way I think I should approach this is by getting a result for [imath]n=1,2,3...[/imath] and then examine them. I could easily get the Taylor series expansion for [imath]n=1[/imath], but then I don't really know how to proceed, because I can't always break down a power of a polynomial. And a power of a summation doesn't seem as a legit answer either. Can anyone help? Thanks |
1011667 | Evaluate [imath]\int_{0}^{1} \log^2(x)\log^2(1-x)\,dx[/imath] using non-elementary methods.
The task is to evaluate, using the specific given hint, and only real-analysis methods, no complex analysis is allowed in the problem. [imath]\int_0^1 \log^2(x)\log^2(1-x)\, dx[/imath] The hint given is: [imath] \sum_{k=1}^{\infty} (H_k)(x^k) = (-)\frac{\log(1-x)}{1-x},[/imath] where [imath]H_k[/imath] is the harmonic number. If we integrate both sides we get. [imath] (2)\cdot\sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{(k+1)} = \log^2(1-x)[/imath] Recalling the hint: [imath]\sum_{k=1}^{\infty} (H_k)(x^k) = (-)\frac{\log(1-x)}{1-x}[/imath] Substitute [imath]x \to 1-x[/imath] [imath]\sum_{k=1}^{\infty} (H_k)((1-x)^k) = \frac{-\log(x)}{x}[/imath] Integrate both sides to get: [imath](2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{((1-x)^{k+1})}{k+1} = \log^2(x)[/imath] I think it is a good idea to evaluate the sums differently. Or simply, [imath]\log^2(x)\log^2(1-x) = \left[(2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{((1-x)^{k+1})}{k+1}\right]\left[(2)\cdot\sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{(k+1)}\right][/imath] But that would result in quadruple sums... Any ideas? | 959701 | Evaluate [imath]\int^1_0 \log^2(1-x) \log^2(x) \, dx[/imath]
I have no idea where to even start. WolframAlpha cant compute it either. [imath]\int^1_0 \log^2(1-x) \log^2(x) \, dx[/imath] I think it can be done with series, but I am not sure, can someone help a little so I can get a start?? Thanks! |
1012238 | Why is this algebraic curve irreducible?
I have found this in a book: but I don't know why the curve [imath]x^m-y^n=0[/imath] is irreducible when n, m are coprimes. Can someone help me? | 652392 | [imath]X^n-Y^m[/imath] is irreducible in [imath]\Bbb{C}[X,Y][/imath] iff [imath]\gcd(n,m)=1[/imath]
I am trying to show that [imath]X^n-Y^m[/imath] is irreducible in [imath]\Bbb{C}[X,Y][/imath] iff [imath]\gcd(n,m)=1[/imath] where [imath]n,m[/imath] are positive integers. I showed that if [imath]\gcd(n,m)[/imath] is not [imath]1[/imath], then [imath]X^n-Y^m[/imath] is reducible. How to show the other direction. Please help. |
1012501 | Cardinality of Cartesian product of two sets
If the set [imath]S[/imath] has cardinality [imath]\#S[/imath] and the set [imath]T[/imath] has cardinality [imath]\#T[/imath], what is the cardinality of [imath]S × T[/imath]? Ive tried [imath]S=\{S\}, T=\{T\} = S \times T= \{S,T\}[/imath] therefore the cardinality is [imath]2[/imath]. | 398850 | Cardinality of Cartesian Product of finite sets.
If [imath]a = \{1,2,3\}[/imath] and [imath]b = \{a,b,c\},\;[/imath] FIND [imath]\;n(a\times b)[/imath] Or is it impossible to multiply these sets? What will be the answer? |
1012515 | Proving [imath]\sup X=-\inf(-X)[/imath]
I would appreciate help with the following question Let [imath]X\subseteq\mathbb{R}[/imath] be a bounded set. Define [imath]-X=\{x\in\mathbb{R}\mid-x\in X\}.[/imath] Prove [imath]\sup X=-\inf (-X)[/imath] | 947035 | How to show [imath]-\sup(-A)=\inf(A)[/imath]?
Let [imath]\emptyset\neq A\subseteq\mathbb{R}[/imath] a bounded set. Consider [imath]-A=\{-a:a\in A\}[/imath]. I want to prove that [imath]-\sup(-A)=\inf(A)[/imath]. It is easy to see that [imath]-\inf(A)[/imath] is an upper bound of [imath]-A[/imath], so [imath]\sup(-A)\le -\inf(A)[/imath], then [imath]-\sup(-A)\ge \inf(A)[/imath]. How can we prove that [imath]-\sup(-A)\le \inf(A)[/imath]? Thanks. |
1012528 | A compact group with a finite dimensional faithful representation
Theorem: If [imath]G[/imath] a compact group has a finite dimensional faithful representation [imath]W[/imath], then any irreducible representation [imath]V[/imath] is contained in [imath]W(k,l) = W^{\otimes k} \otimes (W^*)^{\otimes l}[/imath] for some [imath]k, l[/imath]. How general is this result? How is it proved? The proof was sketched in a lecture, but I didn't understand it. I hope to use this with Stone-Weierstrass to prove the Peter-Weyl theorem for compact groups that come with a finite dimensional faithful representation. | 26909 | Does every irreducible representation of a compact group occur in tensor products of a faithful representation (and its dual)?
Let [imath]G[/imath] be a compact (Hausdorff) group and [imath]V[/imath] a faithful (complex, continuous, finite-dimensional) representation of it. (Hence [imath]G[/imath] is a Lie group.) Is it true that every irreducible representation of [imath]G[/imath] occurs as a summand of [imath]V^{\otimes n} \otimes (V^{\ast})^{\otimes m}[/imath] for some [imath]m, n[/imath]? (The original question only asked for summands of [imath]V^{\otimes n}[/imath] and, as anon mentions below, [imath]\text{U}(1)[/imath] is an easy counterexample.) I know that the corresponding result is true for finite groups, but the proof I know doesn't seem to easily generalize. It seems we ought to be able to apply Stone-Weierstrass: the characters you get from summands of [imath]V^{\otimes n} (V^{\ast})^{\otimes m}[/imath] form an algebra of class functions closed under addition, multiplication, and complex conjugate, so if we know that they separate points, they ought to be dense in the space of class functions. But 1) I'm not sure if we can show that these functions separate points, and 2) I'm not sure if the space of conjugacy classes (with the quotient topology from [imath]G[/imath]) is even Hausdorff. Motivation: I was looking for cheap ways to set up the representation theory of [imath]\text{SU}(2)[/imath]. In this rather special case the character of the defining representation [imath]V[/imath], which is self-dual, already separates conjugacy classes, and I think the above argument works. Then Clebsch-Gordan allows me to quickly classify the irreducible representations of [imath]\text{SU}(2)[/imath] without using Lie algebras. |
809071 | Find a formula for [imath]1 + 3 + 5 + ... +(2n - 1)[/imath], for [imath]n \ge 1[/imath], and prove that your formula is correct.
I think the formula is [imath]n^2[/imath]. Define [imath]p(n): 1 + 3 + 5 + \ldots +(2n − 1) = n^2[/imath] Then [imath]p(n + 1): 1 + 3 + 5 + \ldots +(2n − 1) + 2n = (n + 1)^2[/imath] So [imath]p(n + 1): n^2 + 2n = (n + 1)^2[/imath] The equality above is incorrect, so either my formula is wrong or my proof of the implication is wrong or both. Can you elaborate? Thanks. | 117780 | Proof by Induction: Solving [imath]1+3+5+\cdots+(2n-1)[/imath]
The question asks to verify that each equation is true for every positive integer n. The question is as follows: [imath]1+ 3 + 5 + \cdots + (2n - 1) = n^2[/imath] I have solved the base step which is where [imath]n = 1[/imath]. However now once I proceed to the inductive step, I get a little lost on where to go next: Assuming that k is true (k = n), solve for k+1: (2k - 1) + (2(k+1) - 1) (2k - 1) + (2k+2 - 1) (2k - 1) + (2k + 1) This is where I am stuck. Do I factor these further to obtain a polynomial of some sort? Or am I missing something? |
1011878 | A thinking problem of limit from my teacher.
Please find the limit:[imath]\mathop {\lim }\limits_{n \to \infty } n\left[ {{{\left( {\frac{1}{\pi }\left( {\sin \left( {\frac{\pi }{{\sqrt {{n^2} + 1} }}} \right) + \sin \left( {\frac{\pi }{{\sqrt {{n^2} + 2} }}} \right) + \cdots+ \sin \left( {\frac{\pi }{{\sqrt {{n^2} + n} }}} \right)} \right)} \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right].[/imath] | 1010468 | How find this limits with hardly form?
show that: [imath]\lim_{n\to\infty}n\left[\left(\dfrac{1}{\pi}\left(\sin{\left(\dfrac{\pi}{\sqrt{n^2+1}}\right)}+ \sin{\left(\dfrac{\pi}{\sqrt{n^2+2}}\right)}+\cdots+\sin{\left(\dfrac{\pi}{\sqrt{n^2+n}}\right)} \right)\right)^n-\dfrac{1}{\sqrt[4]{e}}\right]=-\dfrac{1}{\sqrt[4]{e}}\left(\dfrac{31}{96}+\dfrac{\pi^2}{6}\right)[/imath] I only solve this :[imath]\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\dfrac{1}{\sqrt[4]{e}}\tag{1}[/imath] use [imath]\sin{x}\approx x,\Longrightarrow \dfrac{1}{\pi}\sin{(\dfrac{\pi}{\sqrt{n^2+i}})}=\dfrac{1}{\sqrt{n^2+i}}+o(\dfrac{1}{\sqrt{n^2+i}})\to \dfrac{1}{n}\left(1+\dfrac{i}{n^2}\right)^{-\frac{1}{2}},n\to\infty[/imath] and note [imath](1+x)^{-1/2}=1-\dfrac{1}{2}x+o(x)[/imath] so [imath]\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1}{n}\sum_{i=1}^{n}\left(1-\dfrac{i}{2n^2}+o(i/n^2)\right)\right)^n=e^{-\frac{1}{4}}[/imath] But I can't solve my problem,Thank you |
1013296 | [imath]|G|=p^3[/imath], prove that [imath]p[/imath] divides |Z(G)|.
Suppose that a group [imath]G[/imath] has order [imath]p^{3}[/imath] where [imath]p[/imath] is prime. How would I prove that [imath]p[/imath] divides [imath]|Z(G)|[/imath]? | 982985 | Part of simple proof of nontrivial center in p-group
I'm trying to understand the proof of a Burnside theorem (as stated in Beachy's Abstract Algebra p. 328): Let [imath]p[/imath] be prime number. The center of any [imath]p[/imath]-group is nontrivial. Now, In the proof they say that if we let [imath]G[/imath] be a [imath]p[/imath]-group, then in the class equation [imath]|G| = |Z(G)|+\sum [G:C(x)][/imath] for all [imath]x[/imath] that is not in the center and represent a conjugacy class, we see that every term in [imath]\sum [G:C(x)][/imath] is divisible by [imath]p[/imath] since [imath]x\not\in Z(G) \implies [G:C(x)]>1[/imath]. This last statement is what I do not understand, how do we know that [imath]p \mid [G:C(x)][/imath] for any conjugacy class? I know that the elements in the conjugacy class of [imath]x[/imath] is in bijection with the cosets of [imath]C(x)[/imath], i.e. [imath][G:C(x)][/imath], but how can we be certain that the number of elements in a conjugacy class of [imath]x[/imath]/cosets of the centralizer of [imath]x[/imath] is divisible by [imath]p[/imath]? Best regards. |
486126 | Proving Cantor's theorem
[imath]\DeclareMathOperator{\card}{card}[/imath]From Problem-Solvers Topology Prove the following: CANTOR'S THEOREM If [imath]A[/imath] is a set, then [imath]\card A < \card \mathcal{P}(A)[/imath] where [imath]\card A[/imath] stands for the cardinality of set [imath]A[/imath]. My Answer If [imath]A = \emptyset[/imath], then [imath]\card A =0[/imath] and [imath]\card \mathcal{P}(A)=1[/imath]. If [imath]A = \{a\}[/imath], then [imath]\card A=1[/imath] and [imath]\card \mathcal{P}(A)=2[/imath]. So suppose that [imath]A[/imath] has at least two elements. Define a function [imath]f: A \rightarrow \mathcal{P}(A)[/imath] such that [imath]f(x) = \{x\}[/imath] for all [imath]x \in A[/imath]. Then [imath]f[/imath] is injective. But it cannot be surjective, because for any two distinct elements [imath]a,b \in A[/imath], there is no element in [imath]A[/imath] that is sent to the set [imath]\{a,b\}[/imath] in [imath]\mathcal{P}(A)[/imath]. Therefore, there is no bijection, and [imath]\card A < \card\mathcal{P}(A)[/imath]. Do you think my answer is correct? Thanks in advance | 1471291 | Show that no set [imath]S[/imath] is equinumerous with [imath]P(S)[/imath].
Looking for a proof check. I've written out the argument as it reads most easily to my own eyes. Let [imath]S[/imath] be a set and suppose by contradiction that [imath]\exists f:S \to P(S)[/imath] with f surjective. Define [imath]B = \{x \in A: x \notin f(n)\}[/imath]. [imath]f[/imath] surjective [imath]\implies B = f(x_0)[/imath] for some [imath]x_0[/imath]. [imath]x \in B \iff x \notin f(x)[/imath] but [imath]B = f(n_0)[/imath] so [imath]n_0 \in f(n_0) \iff n_0 \notin f(n_0)[/imath] which is a contradiction. Therefore [imath]\nexists f[/imath]. |
964381 | If you have two envelopes, and ...
Suppose you're given two envelopes. Both envelopes have money in them, and you're told that one envelope has twice as much money as the other. Suppose you pick one of the envelopes. Should you switch to the other one? Intuitively, you don't know anything about either envelope, so it'd be ridiculous to say that you should switch to the other envelope to maximize your expected money. However, consider this argument. Let [imath]x[/imath] be the amount of money in the envelope you picked. If [imath]y[/imath] is the amount of money in the other envelope, then the expected value equals [imath]E(y) = \frac{1}{2}\left(\frac{1}{2}x\right) + \frac{1}{2}\left(2x\right) = \frac{5}{4} x[/imath] But [imath]5x/4 > x[/imath], so you should switch! The Wikipedia article says that [imath]x[/imath] stands for two different things, so this reasoning doesn't work. I say this is not a valid resolution. Consider opening up the envelope that you pick, and finding [imath]\[/imath]10$ inside. Then you can run the expected value calculation to get [imath]E(y) = \frac{1}{2} \cdot \$5+\frac{1}{2} \cdot \$20 = \$12.50[/imath] This means that if you open one of the envelopes and find $\[imath]10[/imath], you should switch to the other envelope. The [imath]\[/imath]10[imath] doesn't stand for two different things, it literally just means [/imath]\[imath]10[/imath]. But you don't have to open up the envelope to run this calculation, you can just imagine what's inside, and run the calculation based on that. This is what "Let [imath]x[/imath] be the amount in the envelope" means. The problem with the argument is not that [imath]x[/imath] stands for two different things. So what is the problem? Previous questions on stack exchange have given the resolution that I just said I wasn't satisfied by, so please don't mark this as a duplicate. I want a different resolution, or a more satisfying explanation of why [imath]x[/imath] does stand for two different things. Apparently there is still research being published about this problem - maybe it isn't so obvious? I think there's something subtle wrong with the premise. Because there's no uniform probability distribution on [imath]\mathbb{R}[/imath], statements like "random real number" are not well-defined. Likewise, I think "one envelope has twice as much money as the other" assumes some probability distribution on [imath]\mathbb{R}[/imath], and perhaps our expected value calculation assumes that this distribution is uniform, which it cannot be ... | 1905572 | Why is this argument incorrect for the Envelope Paradox
Consider the envelope paradox problem: There are two envelopes both of which contain some money. One envelope contains twice the amount of the other (but other than that, you do not know how much is in them). You select one of the envelopes randomly and see the amount of money inside. You can opt to either keep the money in this envelope, or switch envelopes, which do you choose? A (false) argument for switching envelopes: Let the amount of money in the envelope you selected be [imath]X[/imath]. There is a [imath]50\%[/imath] chance that the envelope with more money is selected, and a [imath]50\%[/imath] chance that the envelope with less money is selected. Therefore, there is a [imath]50\%[/imath] chance that there is [imath]0.5X[/imath] in the other envelope, and a [imath]50\%[/imath] chance that there is [imath]2X[/imath] in the other envelope. So the expected payout for switching is [imath]{2X + 0.5X\over2} = 1.25X[/imath] which is better than the payout of [imath]X[/imath] for not switching. Therefore you should switch. This argument is wrong... where exactly does it fall? |
1014042 | Trigonometric limit [imath]\lim_{x\to 0} \frac{\cos(a+2x) - 2\cos(a+x) + \cos(a)}{x^2}[/imath] without l'Hospital
Compute the following limit: [imath]\lim_{x\to 0} \frac{\cos(a+2x) - 2\cos(a+x) + \cos(a)}{x^2}\ .[/imath] How would I go about solving this problem? I can't use l'Hospital. | 1014005 | Solving Trigonometric Limits
compute the following limit: [imath]\lim \frac{\cos(a+2x) - 2cos(a+x) + \cos(a)}{x^2}\ \mbox{as}\ x\to0.[/imath] How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse... but without used ´Hospital |
1006825 | Equivalent forms of Jensen's diamond principle
I try to prove that these four statements are equivalent: [imath]\Diamond[/imath] There are [imath]A_\alpha\subseteq \alpha\times\alpha[/imath] for [imath]\alpha<\omega_1[/imath] s.t. for all [imath]A\subset \omega_1\times\omega_1[/imath], [imath]\{\alpha<\omega_1:A\cap\alpha\times\alpha=A_\alpha\}[/imath] is stationary. There are [imath]f_\alpha:\alpha\to\alpha[/imath] [imath]\alpha<\omega_1[/imath] s.t. for each [imath]f:\omega_1\to\omega_1[/imath], [imath]\{\alpha:f\upharpoonright\alpha=f_\alpha\}[/imath] is stationary. There are [imath]f_\alpha:\alpha\to\alpha[/imath] [imath]\alpha<\omega_1[/imath] s.t. for each [imath]f:\omega_1\to\omega_1[/imath], there is [imath]\alpha>0[/imath] such that [imath]f\upharpoonright \alpha=f_\alpha[/imath]. I prove that 1, 2 and 3 are equivalent and 3 implies 4, but I don't know how to prove the equivalence of 4 and 1. I try to prove that 4 implies 3 or [imath]\lnot 2[/imath] implies [imath]\lnot 3[/imath] however I have a struck. Thanks for any hints. | 450842 | Equivalences of [imath]\diamondsuit[/imath]-principle.
I'm working on exercises from Kunen and I'm stuck. I must proof that the following are equivalent: There exists a sequence [imath]\langle A_\alpha:\alpha<\omega_1\rangle[/imath] such that [imath]\forall\alpha(A_\alpha\subset\alpha)[/imath] and for all [imath]A \subset \omega_1[/imath] the set [imath]\{\alpha \in \omega_1:A\cap\alpha=A_\alpha\}[/imath] is stationary. ([imath]\diamondsuit[/imath]-principle) There exists a sequence [imath]\langle A_\alpha:\alpha<\omega_1\rangle[/imath] such that [imath]\forall\alpha(A_\alpha\subset\alpha\times\alpha)[/imath] and for all [imath]A \subset \omega_1\times\omega_1[/imath] the set [imath]\{\alpha \in \omega_1:A\cap\alpha\times\alpha=A_\alpha\}[/imath] is stationary. There exists a sequence [imath]\langle f_\alpha:\alpha<\omega_1\rangle[/imath] such that [imath]\forall\alpha(f_\alpha:\alpha\rightarrow\alpha)[/imath] and for all [imath]f : \omega_1\rightarrow\omega_1[/imath] [imath]\exists\alpha(f|\alpha=f_\alpha\wedge\alpha>0)[/imath] There exists a sequence [imath]\langle f_\alpha:\alpha<\omega_1\rangle[/imath] such that [imath]\forall\alpha(f_\alpha:\alpha\rightarrow\alpha)[/imath] and for all [imath]f : \omega_1\rightarrow\omega_1[/imath] the set [imath]\{\alpha \in \omega_1:f|\alpha=f_\alpha\}[/imath] is stationary. I have proved that [imath]1\implies 2\implies 4\implies 3[/imath] and that [imath]4\implies 1,[/imath] but I'm having problems on proving that [imath]3\implies[/imath] "something", because it only talks about one [imath]\alpha[/imath]. Can someone help me? |
1011912 | does [imath]\int_\gamma ln(z) dz[/imath] depend on the branch cut I choose?
I would like to know why do I get two different results when I calculate [imath]\int_\gamma ln(z) dz[/imath] ,with [imath]\gamma[/imath] the unit cercle, when using two different branch cuts [imath][0,+\infty[[/imath] and [imath]]-\infty,0][/imath] ? Thanks a lot, | 1012986 | Why can I get two different results when calculating [imath]\int \ln(z) \,dz[/imath]?
When I take as a branch cut [imath]]\!\,-\!\infty,0][/imath] for the complex logarithm, I get: [imath]\int_{\lvert z\rvert<1}\ln(z)\,dz=\int_{-\pi}^{\pi}\ln(e^{i\theta})\,i\theta \,d\theta=-2i\pi[/imath] whereas, when I take [imath][0,+\infty[[/imath] as a branch cut, I get: [imath]\int_{\lvert z\rvert<1}\ln(z)dz=\int_{0}^{2\pi}\ln(e^{i\theta})\,i\theta \,d\theta=2i\pi[/imath] Does the integral of [imath]\ln(z)[/imath] depend on the branch cut we choose? |
1014409 | Test Convergence of the series [imath]\sum^\infty_{n=1}\frac{2n^2}{5n^2+2n+1}[/imath]
Is [imath]\sum^\infty_{n=1}\frac{2n^2}{5n^2+2n+1}[/imath] Convergent or Divergent? If convergent, find the sum. If Divergent, explain why. Since it is continuous, positive, and decreasing I used the integral test: [imath]\int^\infty_1\frac{2n^2}{5n^2+2n+1}\Rightarrow \int^\infty_1 \frac25dn +\int^\infty_1ndn+2\int^\infty_12n^2[/imath] [imath]\frac25\bigg[n\bigg]^\infty_1+\frac25\bigg[\frac{n^2}{2}\bigg]^\infty_1+2\bigg[\frac{n^3}{3}\bigg]^\infty_1[/imath] [imath]\infty+\infty+\infty=\infty; \text{Divergent by the Integral Test}[/imath] Did I do this correctly? | 1014384 | Is this convergent or divergent: [imath]\sum _{n=1}^{\infty }\:\frac{2n^2}{5n^2+2n+1}[/imath]?
Is this convergent or divergent? if convergent, find the sum. If divergent explain why. [imath]\sum _{n=1}^{\infty }\:\frac{2n^2}{5n^2+2n+1}[/imath] I want to use the divergent test which is [imath]\mathrm{If\:}\lim _{n\to \infty }a_n\ne 0\mathrm{\:then\:}\sum a_n\mathrm{\:diverges}[/imath] [imath]\frac{2n^2}{5n^2+2n+1}[/imath] I would assume to take out the largest [imath]n[/imath] in both the top and bottom [imath]\lim _{n\to \infty }\left(\frac{n^2}{n^2}\cdot \frac{2}{5+\frac{2}{n}+\frac{1}{n^2}}\right)\:[/imath] having the [imath]n^2[/imath] cancel [imath]\lim _{n\to \infty }\frac{2}{5+\frac{2}{n}+\frac{1}{n^2}}[/imath] after taking the limit [imath]\frac{2}{5+0+0}[/imath] so [imath]\frac{2}{5}[/imath][imath]\ne [/imath] [imath]\:0\:[/imath] so it's divergent? |
1013772 | Eigenvalues of a tridiagonal matrix
I am currently studying this type of Jacobi matrix for all n Example of the type of matrix [imath]n=4[/imath] [imath]\mathbf{X}=\left[\begin{array}{*{20}{c}} {0}&{1}&{0}&{0}\\ {1}&{0}&{1}&{0}\\ {0}&{1}& {0} &{1}\\ {0}&{0}&{1}&{0} \end{array}\right][/imath] I am struggling to prove that all eigenvalues for all n will be; [imath] \lambda \leq |2|[/imath] | 818362 | How to figure out the spectral radius of this matrix
[imath]A=\begin{array}{ccc} 0 & 1/2 & 0 & \cdots & 0 \\ 1/2 & 0 & 1/2 &\cdots& 0\\ 0 & 1/2 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots &\ddots & 1/2\\ 0 & 0 & 0 & 1/2 & 0 \end{array}[/imath] I need to know the spectral radius of [imath]A[/imath] in function of the dimension of the matrix [imath]n[/imath]. I know only one way to know the spectral radius, using the definition I find all eigenvalues and then find the maximum. but it very hard if [imath]n[/imath] is changing. Some help please! I did some iteration by [imath]det(\lambda I-A)[/imath] [imath]n=2 \quad r=\frac{\sqrt{1}}{2}[/imath] [imath]n=3 \quad r=\frac{\sqrt{2}}{2}[/imath] for [imath]n=2[/imath] and [imath]n=3[/imath] are correct. and if we take this rule, for [imath]n=5\quad \quad r=\frac{\sqrt{4}}{2}[/imath] but this isn´t true since I know radius for this matrix should be [imath]< 1[/imath] and if we take this rule for [imath]n=5[/imath] we have [imath]\frac{\sqrt4}{2}=1[/imath] Some hint? |
1014776 | Prove that [imath]\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}[/imath]
[imath]\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+v)}=\frac{1}{vv!}[/imath] I am struggling to find a solution for this but no luck yet. How can I analyze it to get to second part? | 1013030 | How to prove that [imath]\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}[/imath] for every [imath]k\geqslant1[/imath]
Does anyone have any idea how to prove that [imath]\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}[/imath] |
1014829 | Is it possible to compute these probabilities concerning a 6-digit password using theory of Markov chains?
Consider the situation of decoding a 6-digit password that consists of the symbols A to Z and 0 to 9, where all possible combinations are tried randomly and uniformly. Consider the following decoding method: At first a combination is chosen randomly and uniformly. At the next trial a digit from this combination is chosen uniformly at random and its entry is substituted by a uniformly randomly chosen element from [imath]\left\{A,...,Z,0,...,9\right\}[/imath]. This procedure is repeated until the password is found. (a) What is the probability that the correct password will never be entered? (b) What is the probability that eventually the same combination will be entered two consecutive times? I think it is very difficult to compute these probabilities by pure combinatorical thoughts, isn't it? As you can see here: 6-digit password - a special decoding method So I asked myself if it is possible and smart to see this decoding strategy as a kind of Markov chain maybe? Maybe it is easier then to solve? This task is asked in the context of a lecture concerning Markov chains, so it is at least not devious to have this assumption. Hope you can help me. | 1012620 | 6-digit password - a special decoding method
Consider the situation of decoding a 6-digit password that consists of the symbols A to Z and 0 to 9, where all possible combinations are tried randomly and uniformly. Consider the following decoding method: At first a combination is chosen randomly and uniformly. At the next trial a digit from this combination is chosen uniformly at random and its entry is substituted by a uniformly randomly chosen element from [imath]\left\{A,...,Z,0,...,9\right\}[/imath]. This procedure is repeated until the password is found. (a) What is the probability that the correct password will never be entered? (b) What is the probability that eventually the same combination will be entered two consecutive times? I already asked how to get the anwers to (a) and (b) without using the here mentioned special decoding method and I got great help, see Probability concerning a 6-digit password. Now I have to answer (a) and (b) using the decoding method and again I have enormous problems! Combinatorical thoughts are not my favourite business. Nevertheless I tried to find the probabilities in an analog way as it was shown to me in the linked thread. Additionally, I wonder if this task now maybe has something to do with Markov chains because the lecture this task is from is about Markov chains. I think there are (at least) the two following ways to understand the described decoding strategy, which sense is meant? Sense 1 We choose (randomly and uniformly) one of the [imath]36^6[/imath] possible combinations. If it is the right, we stop. Otherwise we then choose (randomly and uniformly) one of the 6 digits and substitute it (randomly and uniformly) by a symbol out of the alphabet. Then we choose (randomly and uniformly) one of the remaining 5 digits and subtitute it and so on until we substituted all the 6 digits. If this was the right password, we stop. If not, we again start substituting the 6 digits (the digits of the combination that we chosed at the beginning, that is, we stick to this combination). Sense 2 Same as above with the difference that after we substituted all 6 digits and saw that it is the wrong password, we choose (randomly and uniformly) another combination out of the [imath]36^6[/imath] possibilities (it can be the same as before) and then substitute the digits of this new combination. I think that sense 1 is meant and thus I considered the task in this sense. (a) Anyway, my result here is [imath]0[/imath], because as far as I see the probability not to have reached the right password after n passages is [imath] \left(1-\frac{1}{36^6}\right)\cdot\left(1-\prod_{k=1}^6\frac{1}{k\cdot 36}\right)^n [/imath] and this tends to [imath]0[/imath] as [imath]n\to\infty[/imath]. Remark: If we decode without this special method (see the linked thread) then the probability of (a) is 0, too. (b) The probability that we have eventually one pair of consecutive equal guesses is - to my results - [imath] \sum_{n=0}^{\infty}\left(1-\frac{1}{36^6}\right)\cdot\left(\prod_{k=1}^6\frac{35}{k\cdot 36}\right)\cdot\left(\prod_{k=1}^6\frac{34}{k\cdot 36}\right)^n\left(\prod_{k=1}^6\frac{1}{k\cdot 36}\right) [/imath] Edit I think my last result for (b) was not correct, I think instead it has to be [imath] \sum_{n=0}^{\infty}\left(1-\frac{1}{36^6}\right)\cdot\left(1-\prod_{k=1}^k\frac{1}{k\cdot 36}\right)\cdot\left(1-\prod_{k=1}^{6}\frac{2}{k\cdot 36}\right)^n\cdot\left(\prod_{k=1}^{6}\frac{1}{k\cdot 36}\right). [/imath] If I know compute the geometrical series and use that [imath]1-\prod_{k=1}^{6}\frac{1}{k\cdot 36}\approx 1[/imath], then I get that (with [imath]p:=\frac{1}{36^6}[/imath]) the probability of (b) is [imath] \approx \left(\frac{1}{2}\right)^6\cdot (1-p) [/imath] Remark: When decoding without this method (see the linked thread) then the probability of (b) is [imath]\frac{1}{2}\cdot (1-p)[/imath]. That is, using the method, if my result is correct, we have a much smaller probability for (b). So this decoding method is more efficient. Would be great to get a feedback from you to know if I am right. And, as mentioned, I am interested to know if the task has something to do with the context of Markov chains. Ciao & greetings Salamo |
1014081 | high school senior A division contest question
The product of my age 7 years ago and my age 7 years from now is a positive perfect square. Compute my age now. I set up the equation, [imath](x+7)(x-7)=a^2[/imath], so [imath]x^2 -49= a^2[/imath]. So I know that x has to be bigger than 7. But I don't know what to do next... Please help me! Thank you very much! | 1008106 | [imath](n-7)(n+7)=[/imath] some perfect square
This put in the context of a age problem will be: the product of my age seven years ago and seven years later is some perfect square Since this is a age problem that perfect square has to be between 1 and 100. I've try every factor of number between 1 and 100 inclusive, and found none that matches. Can there be another way besides trial and error? |
1008063 | Prove that for each [imath]\sigma \in Aut(S_n)[/imath] [imath]\sigma(1,2)=(a,b_2),\sigma(1,3)=(a,b_3), ...., \sigma(1,n)=(a,b_n)[/imath]
Prove that for each [imath]\sigma \in Aut(S_n)[/imath] [imath]n \neq 6[/imath]. [imath]\sigma(1,2)=(a,b_2),\sigma(1,3)=(a,b_3), ...., \sigma(1,n)=(a,b_n)[/imath] for some distinct integers [imath]a,b_1,....b_n \in \{1,...,n\}[/imath]. I was trying the action of [imath]\sigma[/imath] in [imath]S_n=<(i,i+1>[/imath] [imath]i=1(1)n[/imath]. But can't get any positive result. So need some help to prove it. | 1015224 | Prove that for each [imath]\sigma \in Aut(S_n)[/imath] [imath]n \neq 6[/imath]. [imath]\sigma(1,2)=(a,b_2),\sigma(1,3)=(a,b_3), ...., \sigma(1,n)=(a,b_n)[/imath]
I am reposting it getting insufficient help from the previous post (Although I got some hint) Prove that for each [imath]\sigma \in Aut(S_n)[/imath] [imath]n \neq 6[/imath]. [imath]\sigma(1,2)=(a,b_2),\sigma(1,3)=(a,b_3), ...., \sigma(1,n)=(a,b_n)[/imath] for some distinct integers [imath]a,b_1,....b_n \in \{1,...,n\}[/imath]. I was trying the action of [imath]\sigma[/imath] in [imath]S_n=<(i,i+1>[/imath] [imath]i=1(1)n[/imath]. But can't get any positive result. So need some help to prove it. I have proved earlier that when [imath]n≠6[/imath], the size of the conjugacy class of [imath](1,2)[/imath] is different from that of any other conjugacy class of elements of order [imath]2[/imath] in Sn[This is true because conjugacy classes contains elements of same cycle structure]. That proves that [imath]σ(1,2)=(a,b)[/imath] for some [imath]a,b[/imath]. But I didn't know that it will be needed (Thanks to Derek holt). Now what can I do? Because the elements [imath](1,2),(1,3),....,(1,n)[/imath] can go to any transposition. So what to do next? Can anyone please give me any elaborate proof? |
1015421 | [imath]\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}[/imath] solve for positive integers.
Solve for positive integers [imath]x,y,z[/imath] [imath]\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}[/imath] I tried to solve it by some generel work but it didn't help. | 1015428 | Solve for positive integers: [imath]\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}[/imath]
Solve for [imath]x,y,z\in\mathbb{N}[/imath] [imath]\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}[/imath] I tried by some general methods but they didn't help me. |
1015825 | let [imath]f,g:R\to \mathbb{Q}[/imath], show that [imath]f=g[/imath].
Let [imath]R[/imath] be a ring, [imath]\mathbb{Q}[/imath] be the field of rational numbers, and [imath]f,g: R \to \mathbb{Q}[/imath] be homomorphisms such that [imath]f(n)=g(n)[/imath] for all [imath]n\in \mathbb{Z}[/imath], where [imath]\mathbb{Z}[/imath] is the ring of integers, show that [imath]f=g[/imath]. (Hint: first show [imath]f(1/n)= g(1/n)[/imath] for all [imath]n \in \mathbb{Z}[/imath]) | 63018 | Showing two ring homomorphisms that agree on the integers must agree on the rationals
I have two ring homomorphisms [imath]f,g\colon \mathbb{Q}\to X[/imath]. I know that [imath]f=g[/imath] on the integers. How can I show that [imath]f[/imath] and [imath]g[/imath] agree on the rationals? My attempt: let [imath]x,y \in \mathbb{Z}[/imath], [imath]y\neq 0[/imath]. [imath]\begin{align*} f(x/y)&=f(x)f(1/y) &&\text{because }f\text{ is a ring homomorphism}\\ &=g(x)f(1/y) &&\text{because }f\text{ and }g\text{ agree on the integers} \end{align*}[/imath] I do not know how to rewrite [imath]f(1/y)[/imath] so that I can replace it with [imath]g(1/y)[/imath]. Any hints will be much appreciated. Thanks. |
1015366 | Alpha and omega limit sets
What are all the possible [imath]α[/imath]- and [imath]ω[/imath]-limit sets of points for the linear system of equations [imath]x'=Ax[/imath] given by the following matrices [imath]A[/imath]: I guess I don't really know how to get started, Could I have some guidance? So I was told to calculate the eigenvalues (fine) and eigenvectors (I kind of forgot how to do that...) [imath] A= \left[ {\begin{array}{cc} -4 & -2 \\ 3 & 11 \\ \end{array} } \right] [/imath] [imath]λ_1=-10[/imath] [imath]λ_2=-5[/imath] for [imath]λ_1=-10[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} 1 \\ 3 \\ \end{array} } \right] [/imath] for [imath]λ_2=-5[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} 2 \\ 1 \\ \end{array} } \right] [/imath] [imath] A= \left[ {\begin{array}{cc} 1 & 1 \\ 3 & 1 \\ \end{array} } \right] [/imath] [imath]λ_1=4[/imath] [imath]λ_2=-2[/imath] for [imath]λ_1=4[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} 1 \\ 1 \\ \end{array} } \right] [/imath] for [imath]λ_2=-2[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} -1 \\ 1 \\ \end{array} } \right] [/imath] [imath] A= \left[ {\begin{array}{cc} 4 & 5 \\ -5 & -2 \\ \end{array} } \right] [/imath] [imath]λ_1=1+4i[/imath] [imath]λ_2=1-4i[/imath] for [imath]λ_1=1+4i[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} -3/5-(4i)/5 \\ 1 \\ \end{array} } \right] [/imath] for [imath]λ_1=1-4i[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} -3/5+(4i)/5 \\ 1 \\ \end{array} } \right][/imath] | 1015541 | Alpha and Omega limit sets (dynamical systems)
What are all the possible [imath]α[/imath]- and [imath]ω[/imath]-limit sets of points for the linear system of equations [imath]x'=Ax[/imath] given by the following matrices [imath]A[/imath]: I guess I don't really know how to get started, Could I have some guidance? So I was told to calculate the eigenvalues (fine) and eigenvectors (I kind of forgot how to do that...) I know that the [imath]ω[/imath]-limit sets of points are the set of points that the system of equation approach as time goes to infinity, and the [imath]α[/imath]-limit sets of points are the points approached as t goes to negative infinity. [imath] A= \left[ {\begin{array}{cc} -4 & -2 \\ 3 & 11 \\ \end{array} } \right] [/imath] [imath]λ_1=-10[/imath] [imath]λ_2=-5[/imath] Since both eigenvalue are negative, this is a stable node, and the [imath]ω[/imath]-limit set=0. for [imath]λ_1=-10[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} 1 \\ 3 \\ \end{array} } \right] [/imath] for [imath]λ_2=-5[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} 2 \\ 1 \\ \end{array} } \right] [/imath] [imath] A= \left[ {\begin{array}{cc} 1 & 1 \\ 3 & 1 \\ \end{array} } \right] [/imath] [imath]λ_1=4[/imath] [imath]λ_2=-2[/imath] Since one eigenvalue is positive and one is negative, it is not stable, so I can't use the same reasoning that I used above to find the [imath]ω[/imath]-limit set... so I am not sure how to continue for [imath]λ_1=4[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} 1 \\ 1 \\ \end{array} } \right] [/imath] for [imath]λ_2=-2[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} -1 \\ 1 \\ \end{array} } \right] [/imath] [imath] A= \left[ {\begin{array}{cc} 4 & 5 \\ -5 & -2 \\ \end{array} } \right] [/imath] [imath]λ_1=1+4i[/imath] [imath]λ_2=1-4i[/imath] for [imath]λ_1=1+4i[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} -3/5-(4i)/5 \\ 1 \\ \end{array} } \right] [/imath] for [imath]λ_1=1-4i[/imath]: eigenvector [imath] = \left[ {\begin{array}{cc} -3/5+(4i)/5 \\ 1 \\ \end{array} } \right][/imath] |
1016080 | Show f(x):=sqrt(x) is uniformly continuous on [0,1]
Show [imath]f(x):=\sqrt{x}[/imath] is uniformly continuous on [imath][0,1][/imath]. What I did: Need to show that [imath]\forall \varepsilon>0: \exists\delta>0[/imath] such that [imath]\forall x,y\in(0,1): |x-y|<\delta\Rightarrow|f(x)-f(y)|<\varepsilon[/imath] Choose [imath]\delta=\varepsilon^2[/imath]. then for [imath]x,y\in(0,1)[/imath] with [imath]|x - y| < \delta[/imath], [imath]|f(x)-f(y)|=|\sqrt{x}-\sqrt{y}|[/imath] [imath]|\sqrt{x}-\sqrt{y}|\le|\sqrt{x}+\sqrt{y}|[/imath] [imath]|\sqrt{x}-\sqrt{y}|^2\le|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}| =|x-y|< \varepsilon^2[/imath] [imath]|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}<\sqrt{\varepsilon^2}=\varepsilon[/imath] | 569928 | [imath]\sqrt x[/imath] is uniformly continuous
Prove that the function [imath]\sqrt x[/imath] is uniformly continuous on [imath]\{x\in \mathbb{R} | x \ge 0\}[/imath]. To show uniformly continuity I must show for a given [imath]\epsilon > 0[/imath] there exists a [imath]\delta>0[/imath] such that for all [imath]x_1, x_2 \in \mathbb{R}[/imath] we have [imath]|x_1 - x_2| < \delta[/imath] implies that [imath]|f(x_1) - f(x_2)|< \epsilon.[/imath] What I did was [imath]\left|\sqrt x - \sqrt x_0\right| = \left|\frac{(\sqrt x - \sqrt x_0)(\sqrt x + \sqrt x_0)}{(\sqrt x + \sqrt x_0)}\right| = \left|\frac{x - x_0}{\sqrt x + \sqrt x_0}\right| < \frac{\delta}{\sqrt x + \sqrt x_0}[/imath] but I found some proof online that made [imath]\delta = \epsilon^2[/imath] where I don't understand how they got? So, in order for [imath]\delta =\epsilon^2[/imath] then [imath]\sqrt x + \sqrt x_0[/imath] must [imath]\le[/imath] [imath]\epsilon[/imath] then [imath]\frac{\delta}{\sqrt x + \sqrt x_0} \le \frac{\delta}{\epsilon} = \epsilon[/imath]. But then why would [imath]\epsilon \le \sqrt x + \sqrt x_0? [/imath] Ah, I think I understand it now just by typing this out and from an earlier hint by Michael Hardy here. |
1016401 | Number of matrices satisfying a given property
Let [imath]A[/imath] be a [imath]2\times 2[/imath] matrix with complex entries.How to find the number of [imath]2\times 2[/imath] matrices [imath]A[/imath] satisfying the equation [imath]A^3=A[/imath] ? | 143341 | [imath]2\times 2 [/imath] matrices over [imath]\mathbb{C}[/imath] that satisfy [imath]\mathrm A^3=\mathrm A[/imath]
Let [imath]\mathrm A \in \mathbb C^{2 \times 2}[/imath]. How many [imath]2 \times 2[/imath] matrices [imath]\mathrm A[/imath] satisfy [imath]\mathrm A^{3} = \mathrm A[/imath]. Infinitely many? If it is [imath]3 \times 3[/imath] matrix then by applying Cayley-Hamilton theorem I could have said that given matrix is diagonalizable. Also zero is eigenvalue of [imath]\mathrm A[/imath]. So it would be collection of all singular diagonalizable matrices. But how to count them? Here I have [imath]2 \times 2[/imath] matrices i feel I can't apply the Cayley-Hamilton Theorem here? I am stuck with these thoughts? |
1015468 | Prove (x3−2) is maximal ideal of Q[x] using isomorphism theorem for rings.
Prove [imath](x^3−2)[/imath] is a maximal ideal of [imath]\mathbb{Q}[x][/imath] using isomorphism theorems for rings. I tried using the second isomorphy theorem for rings, to use that [imath]( x ^ 3-2)[/imath] is maximal if and only if [imath]\mathbb{Q} [x ] / ( x ^ 3-2)[/imath] is field . I want to show [imath]\mathbb{Q} [x ] / ( x ^ 3-2)[/imath] is isomorphic to a field related to [imath]\mathbb{Q}[/imath]. I have [imath]A = \mathbb{Q} [x ][/imath] , [imath]I = ( x ^ 3-2)[/imath] but not [imath]B[/imath] to apply [imath]( B + I) / I \simeq B / (B ∩ I )[/imath] field. | 1015052 | Prove that [imath](x^3-2)[/imath] is a maximal ideal of [imath]\Bbb Q[x][/imath]
Prove [imath](x^3-2)[/imath] is maximal ideal of [imath]\Bbb Q[x][/imath] using isomorphism theorems for rings. I tried using the second isomorphism theorem for rings, to use that [imath]( x ^ 3-2)[/imath] is maximal if and only if [imath]\mathbb Q [x ] / ( x ^ 3-2)[/imath] is field . Seeing that [imath]\mathbb Q [x ] / ( x ^ 3-2)[/imath] is isomorphic to a field related to [imath]\mathbb Q[/imath]. I have [imath]A = \mathbb Q [x ][/imath] , [imath]I = ( x ^ 3-2)[/imath] but not to [imath]B[/imath] take to apply [imath]( B + I) / I [/imath] isomorphic [imath]B / (B ∩ I )[/imath] field. |
1017528 | For any [imath]a\in S_\Omega[/imath], either it has an immediate predecessor or there is an increasing sequence in [imath]S_\Omega[/imath] whose l.u.b. is [imath]a[/imath].
Specifically I would like to show: Let [imath]S_\Omega[/imath] be the minimal uncountable well-ordered set. For any [imath]a\in S_\Omega[/imath], either [imath]a[/imath] has an immediate predecessor in [imath]S_\Omega[/imath], or there exists an increasing sequence [imath]a_i[/imath] in [imath]S_\Omega[/imath] with [imath]a=\sup\{a_i\}[/imath]. The part that I don't understand is how to find such a sequence [imath]\{a_i\}[/imath]. I understand that if [imath]a[/imath] does not have an immediate predecessor, then for any [imath]x\in S_\Omega[/imath] and [imath]x<a[/imath], there exists [imath]y\in S_\Omega[/imath] s.t. [imath]x<y<a[/imath]. By this one can inductively find a sequence [imath]\{a_i\}[/imath] s.t. [imath]a_0<a_1<\cdots[/imath] and [imath]a_i<a[/imath] for all [imath]i[/imath]. But how can one guarantee that [imath]\sup\{a_i\}=a[/imath] in this case? I encountered this question while proving something else (Munkres Topology, 2/e p.159 12(c)). | 389420 | Why does every countable limit ordinal have cofinality [imath]\omega[/imath]?
According to Wikipedia, if [imath]\alpha[/imath] is a countable limit ordinal, then [imath]\mathrm{cf}(\alpha)=\omega[/imath]. It is intuitively clear to me that it should be so. Certainly the cofinality of such an ordinal must be [imath]\geq\omega[/imath]. We can now look at what the ordinal ends with. If it has a single [imath]\omega[/imath] at the end, then that is a cofinal subset and we are done. It may, however, not end in [imath]\omega[/imath]. It could, for example, end in [imath]\omega[/imath] [imath]\omega[/imath]s or [imath]\omega^2[/imath]. Then we can pick one element of each [imath]\omega[/imath] constituting the [imath]\omega^2[/imath], and the set of these elements is cofinal in [imath]\alpha[/imath], and its type is [imath]\omega.[/imath] If [imath]\alpha[/imath] ends in [imath]\omega^3[/imath], we can take one element of each [imath]\omega^2[/imath] constituting the [imath]\omega^2[/imath], and get a cofinal subset of type [imath]\omega.[/imath] It seems to me that this reasoning should work. It should probably be done inductively. How should I do it? |
1017441 | Why is empty product defined to be [imath]1[/imath]?
For example [imath]\prod_{2 \le j < 1} 2^j= 1.[/imath] How does that happen? | 110546 | What is the product of the empty set?
Give: [imath]fn(S)=\prod_{x\in S}x[/imath] what is: [imath]fn(\emptyset)[/imath] I can see reason that it would be defined as 1 or 0. Note: I thought about restricting the domain of [imath]S[/imath] but that would make the problem less general. If there is no general answer then answers for restricted domains would be valid. |
1017111 | Given a particular triangle that has been constructed, I want to prove that one of the angles must be [imath]> 45[/imath] degrees.
Suppose you are given an acute triangle [imath]XYZ[/imath] with the following properties: At [imath]\angle XZY[/imath], the [imath]\angle[/imath] bisector is drawn and extended all the way to [imath]XY[/imath]. Lets call the point where it intersects [imath]A[/imath]. From [imath]\angle ZXY[/imath] we draw a line to [imath]ZY[/imath] s.t. it cuts [imath]ZY[/imath] into two equal pieces. i.e. the point that intersects [imath]ZY[/imath] is the midpoint call it [imath]B[/imath]. And lastly, from [imath]\angle ZYX[/imath] we drop the altitude onto [imath]XZ[/imath]. Lets call the point it intersects [imath]C[/imath]. Now this triangle was constructed s.t. [imath]ZA \cap XB \cap YC = P[/imath], a point. Suppose we let [imath]\angle XZY = \phi[/imath]. How would we prove that [imath]\angle XZY > 45[/imath] degrees for this specific kind of triangle? Obviously my first thought was to suppose that it was [imath]\leq 45[/imath] degrees, and then break it up into two cases. Even then I am unsure. The help would be appreciated! | 914662 | The concurrence of angle bisector, median, and altitude in an acute triangle
[imath]ABC[/imath] is an acute triangle. The angle bisector [imath]AD[/imath], the median [imath]BE[/imath] and the altitude [imath]CF[/imath] are concurrent. Prove that angle [imath]A[/imath] is more than [imath]45[/imath] degrees. Here [imath]D,E,F[/imath] are points on [imath]BC,CA,AB[/imath] respectively. This question was asked in 4th All-Soviet Union Mathematics Competition in 1970 and the wording of the problem is same. |
1014156 | How to calculate [imath]\int_0^\infty \frac{dx}{1+x^6}[/imath]
Whenever I tried to do, it failed. Is there anyone to help? [imath]\int_0^\infty \frac{dx}{1+x^6}[/imath] | 658637 | Integrating [imath]\int_0^\infty\frac{1}{1+x^6}dx[/imath]
[imath]I=\int_0^\infty\frac{1}{1+x^6}dx[/imath] How do I evaluate this? |
1014790 | Connected space such that (almost) no subspace is connected
Is there a connected space [imath](X,\tau)[/imath] such that [imath]X[/imath] has more than [imath]2[/imath] points and the only proper connected subsets of [imath]X[/imath] are the singletons? | 634787 | Is there an infinite connected topological space such that every space obtained by removing one point from it is totally disconnected?
The particular point topology on any set is connected, but on removing the particular point, the complement is discrete, and hence totally disconnected. Although this is not even [imath]T^1[/imath], Cantor's leaky tent also has this property, and is a subspace of [imath]\mathbb{R}^2[/imath]. Is there an infinite connected topological space such that the complement of every point is totally disconnected? Or in other words, is there an infinite connected space such that every proper subset of at least two points is disconnected? |
1013455 | Determining which intervals decrease and increase and concavity
Given the function [imath]ƒ (x)=\frac{x^2}{(x−2)^2}[/imath] [imath]\bullet[/imath] Find the vertical and horizontal asymptotes of [imath]ƒ (x)[/imath]. My solution: The asymptotes are [imath]x=2[/imath] and [imath]y=1[/imath] [imath]\bullet[/imath] Find the intervals on which [imath]ƒ (x)[/imath] is increasing or decreasing. My solution: [imath](-inf,0)[/imath]--decreasing [imath]0[/imath]--min [imath](0,2)[/imath]--increasing [imath](2,inf)[/imath]---decreasing [imath]\bullet[/imath] Find the critical numbers and specify where local maxima and minima of [imath]ƒ (x)[/imath] occur. My solution: [imath]f'(x) = -\frac{4x}{(x-2)^3}[/imath] [imath]f''(x) = \frac{8(x+1)}{(x-2)^4}[/imath] critical point is [imath]x=0[/imath] domain is [imath]x[/imath] cant equal [imath]2[/imath] [imath]\bullet[/imath] Find the intervals of concavity and the inflection points of [imath]ƒ (x)[/imath]. My solution: [imath](-inf,-1)[/imath]--concave up [imath]-1[/imath]--inflection point [imath](-1,inf)[/imath]--concave up | 1012415 | Find min/max along with concavity, inflection points, and asymptotes
For the function [imath]ƒ (x)=\frac{x^2}{(x−2)^2}[/imath] I know the derivative is [imath]f'(x)=-\frac{4x}{\left(x-2\right)^3}[/imath] and [imath]f''(x) =\frac{8\left(x+1\right)}{\left(x-2\right)^4}[/imath] Critical point is [imath]x=-1[/imath] which is also the min. I think possible inflection point are [imath]x=1[/imath] and [imath]x=0[/imath] I need to find: 1)Find the vertical and horizontal asymptotes of [imath]ƒ (x)[/imath]. 2)Find the intervals on which [imath]ƒ (x)[/imath] is increasing or decreasing. 3)Find the critical numbers and specify where local maxima and minima of [imath]ƒ (x)[/imath] occur. 4)Find the intervals of concavity and the inflection points of [imath]ƒ (x)[/imath]. sorry for the long problem. Hope you can help |
1018256 | How do we solve quadratic congruences such as [imath]X^2+ 3X \equiv -5 \mod{7}[/imath]?
How do we solve quadratic congruences such as: [imath] X^2+ 3X \equiv -5 \mod{7} [/imath] | 44773 | How to solve polynomial equations in a field and/or in a ring?
I'm studying for my exam, and I stuck on solving polynomials in a field and/or in a ring. Let me give you some examples: (1) Solve equation [imath]x^2+4x+3=0[/imath] in field [imath]\mathbb{Z}_5[/imath], [imath]\mathbb{Z}_8[/imath] and in ring [imath]\mathbb{Z}_{12}[/imath] and [imath]\mathbb{Z}_5 \times \mathbb{Z}_{13}[/imath] (2) Solve equation [imath]x^2+2x=1[/imath] in field [imath]\mathbb{Z}_7[/imath], [imath]\mathbb{Z}_{11}[/imath] and in ring [imath]\mathbb{Z}_{12}[/imath] (3) Find ring [imath]\mathbb{Z}_n[/imath], in which equation [imath]x^2+2x=3[/imath] has at least 3 solutions. I'm not asking for solutions for these particular problems. I want to understand this topic, so I would appreciate example solutions of similar problems. Also, you're aware of some study materials - please list tutorials, books etc., about solving polynomial equations in different rings and fields (and similar topics). Thanks for help. |
1004459 | Integral of a geometric Brownian motion
I would like to compute [imath]G[/imath] defined as follows [imath]G(t):= \exp(-\int _0^t h_s~ ds )[/imath] with [imath]h[/imath] being a geometric Brownian Motion. For that I would need first to compute [imath]\int_0^t e^{(\mu-\sigma^2/2)s+ \sigma W_s }~ ds[/imath] and know its law. All I could figure out (that was not much) is that one can obtain all the moments by deriving the Laplace Transform and making the Laplace coefficient tends to zero witch is pretty straightforward Could someone give me a help with that ? Any advice is appreciated. Note that this question is not a duplicate since my question is to compute the exponential of an integral of a geometric Brownian motion and to compute the integral of the GBM itself.There is no expected value involveld Many thanks | 170008 | Laplace transform of integrated geometric Brownian motion
Is there any closed form of the Laplace transform of an integrated geometric Brownian motion ? A geometric Brownian motion [imath]X=(X_t)_{t \geq 0}[/imath] satisifies [imath]dX_t = \sigma X_t \, dW_t[/imath] where [imath]W=(W_t)_{t \geq 0}[/imath] denotes a Brownian motion and the associated integrated Brownian motion is [imath]\int_0^t X_s \, ds[/imath]. The Laplace transform of an integrated gometric Brownian motion is thus [imath] \mathcal{L}(\lambda) = \mathbb{E}\left[e^{-\lambda \int_0^t X_s ds } \right][/imath] |
950676 | Corollary of Kolmogorov zero-one law
Here is another corollary of the theorem: Kolmogorov zero-one law given in my textbook (Probability path). How can I apply the said theorem given that if [imath]X_n[/imath] are independent random variables, [imath][\sum_nX_n \mbox{ converges}][/imath] has probability 0 or 1. thanks :-) | 952496 | Application of Kolmogorov's Zero-One Law
Most of the books I read, they only state some examples of tail events. One of them is [imath][\sum_n X_n \ converges][/imath]. My main problem is to show that this is indeed a tail event. |
1018975 | "Proof" that [imath]0=1[/imath] using an integral.
I saw the following: [imath]\begin{align} \int \tan x \ \mathrm{d}x &= \int \sin x \sec x \ \mathrm{d}x \\ \int \tan x \ \mathrm{d}x &= -\cos x \sec x - \int - \cos x \sec x \tan x \ \mathrm{d}x \\ \int \tan x \ \mathrm{d}x &= -1 + \int \tan x \ \mathrm{d}x \\ 0 &= -1 \end{align}[/imath] I figure the mistake has to do with constants of integration, but I can't quite point it. Can someone explain to me what happens, please? I browsed a bit around here, looking for [imath]0=1[/imath] and the tag fake-proofs, but I didn't found it, so I apologize in advance in case this is a duplicate of something (just provide me the link and I'll delete this, no problems). Thanks! | 806254 | Using Integration By Parts results in 0 = 1
I've run into a strange situation while trying to apply Integration By Parts, and I can't seem to come up with an explanation. I start with the following equation: [imath]\int \frac{1}{f} \frac{df}{dx} dx[/imath] I let: [imath]u = \frac{1}{f} \text{ and } dv = \frac{df}{dx} dx[/imath] Then I find: [imath]du = -\frac{1}{f^2} \frac{df}{dx} dx \text{ and } v = f[/imath] I can then substitute into the usual IBP formula: [imath]\int udv = uv - \int v du[/imath] [imath]\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx[/imath] [imath]\int \frac{1}{f} \frac{df}{dx} dx = 1 + \int \frac{1}{f} \frac{df}{dx} dx[/imath] Then subtracting the integral from both sides, I've now shown that: [imath]0 = 1[/imath] Obviously there must be a problem in my derivation here... What wrong assumption have I made, or what error have I made? I'm baffled. |
1018810 | Why [imath]\mathbb{Z}[\sqrt{-3}][/imath] is not a Euclidean domain?
I need to prove that [imath]\mathbb{Z}[\sqrt{-3}][/imath] is not a Euclidean domain. I tried to show that [imath]\mathbb{Z}[\sqrt{-3}][/imath] is not a P.I.D. but all ideals that I generate by two elements, turn out to be principal. I already appreciate your help in advance. | 115934 | Is [imath]\mathbb Z[\sqrt{-3}][/imath] Euclidean under some other norm?
I know that [imath]\mathbb Z[\sqrt{-3}][/imath] is not a Euclidean domain under the usual norm [imath]N(x + y\sqrt{-3}) = x^2 + 3y^2[/imath], but that does not necessarily mean that it can't be a Euclidean domain. Is it possible to define some norm that could make it into a Euclidean domain? |
1019152 | Complex Analysis, showing a function is zero
Let [imath]\Omega[/imath] be the right half plane excluding the imgainary axis and [imath]f\in H(\Omega)[/imath] such that [imath]|f(z)|<1[/imath] for all [imath]z\in\Omega[/imath]. If there exists [imath]\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})[/imath] such that [imath]\lim_{r\rightarrow\infty}\frac{\log|f(re^{i\alpha})|}{r}=-\infty[/imath] prove that [imath]f=0[/imath]. The hint is define [imath]g_n(z)=f(z)e^{nz}[/imath], then by previous exericise [imath]|g_n|<1[/imath] for all [imath]z\in\Omega[/imath]. | 1008424 | Complex analysis, showing a function is constant
Let [imath]\Omega[/imath] be the right half plane excluding the imgainary axis and [imath]f\in H(\Omega)[/imath] such that [imath]|f(z)|<1[/imath] for all [imath]z\in\Omega[/imath]. If there exists [imath]\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})[/imath] such that [imath]\lim_{r\rightarrow\infty}\frac{\log|f(re^{i\alpha})|}{r}=-\infty[/imath] prove that [imath]f=0[/imath]. The hint is define [imath]g_n(z)=f(z)e^{nz}[/imath], then by previous exericise [imath]|g_n|<1[/imath] for all [imath]z\in\Omega[/imath]. |
1006156 | Why are some convergent Lebesgue integrals 'undefined'?
I sometimes read statements such as The integral [imath]\int_0^{\infty} dx \, \frac{\sin x}{x} [/imath] does not exist as a Lebesgue integral, because it is not absolutely convergent. But according to my understanding, the integral [imath]\int_0^{R} dx \, \frac{\sin x}{x} [/imath] exists as a Lebesgue integral for every [imath]R>0[/imath]. Why can't we simply define [imath] \int_0^{\infty} dx \, \frac{\sin x}{x} = \lim\limits_{R \rightarrow \infty} \int_0^{R} dx \, \frac{\sin x}{x}, [/imath] and hence give meaning to the former integral as a Lebesgue integral? Isn't this also how one defines improper Riemann integrals, as a limit of proper integrals? Please point out any misunderstandings. | 27775 | Why do we restrict the definition of Lebesgue Integrability?
The function [imath]f(x) = \sin(x)/x[/imath] is Riemann Integrable from [imath]0[/imath] to [imath]\infty[/imath], but it is not Lebesgue Integrable on that same interval. (Note, it is not absolutely Riemann Integrable.) Why is it we restrict our definition of Lebesgue Integrability to absolutely integrable? Wouldn't it be better to extend our definition to include ALL cases where Riemann Integrability holds, and use the current definition as a corollary for when the improper integral is absolutely integrable? |
1019608 | Showing [imath]\prod_{n < \omega} n = 2^{\aleph_0}[/imath]
I have to show that [imath]\prod_{n < \omega} n = 2^{\aleph_0}[/imath]. I'm having trouble getting started. I know that [imath]2^{\aleph_0}[/imath] is the set of binary sequences, or the space of functions from [imath]\mathbb{N}[/imath] to [imath]\{0, 1\}[/imath], but I'm having trouble parsing [imath]\prod_{n < \omega} n[/imath]. My best guess is that it's the infinite Cartesian product [imath]0 \times 1 \times 2 \times 3 \times \dots[/imath] but because [imath]0[/imath] is [imath]\emptyset[/imath], then so is this product. Alternatively, [imath]\prod_{n < \omega} n = \{f : \omega \rightarrow \bigcup n | (\forall n) f(n) \in n\}[/imath] Thus has the same problem, though, that this would be an empty space of functions because none of them can map [imath]0[/imath] anywhere. | 90191 | What is the product of all nonzero, finite cardinals?
To be specific, why does the following equality hold? [imath] \prod_{0\lt n\lt\omega}n=2^{\aleph_0} [/imath] |
1019892 | Show that [imath]a^m[/imath] is in [imath]H[/imath] for every [imath]a[/imath] in [imath]G[/imath]
Let [imath]H[/imath] be a normal subgroup of [imath]G[/imath], and let [imath]m=(G:H)[/imath]. Show that [imath]a^m \in H[/imath] for every [imath]a \in G[/imath]. I have been thinking about this question for a few days but I get something informal. What am I missing? There are [imath]m[/imath] [imath]H[/imath]-cosets. If we consider the cosets of [imath]e[/imath], [imath]a[/imath], [imath]a^2[/imath], [imath]\cdots[/imath], [imath]a^{m-1}[/imath], we get [imath]m[/imath] distinct cosets(actually I claim this without a valid proof, any hints?) Then the coset of [imath]a^m[/imath] should be one the above coset. (Then why must it be [imath]eH[/imath]?) Thanks in advance. | 1013646 | Let [imath]H[/imath] be a normal subgroup of index [imath]n[/imath] in a group [imath]G[/imath]. Show that for all [imath]g \in G, g^n \in H[/imath]
I am having a lot of trouble understanding the solution to this problem. [imath](gH)^n = g^nH \implies g^n \in H[/imath] Why does [imath]H^n[/imath] just turn into the identity? I am very confused, any help is appreciated. |
1019780 | Example of a non-linear, non-continuous mapping from [imath]\mathbb R[/imath] onto [imath]\mathbb R[/imath] whose graph is closed
Give an example of a non-linear mapping from [imath]\mathbb R[/imath] onto [imath]\mathbb R[/imath] whose graph is closed although it is not continuous. | 98526 | Discontinuous functions with closed graphs
I tried looking up a question regarding graphs of continuous functions on this site, but all the ones I found consider functions from [imath]\mathbb{R}[/imath] into [imath]\mathbb{R}[/imath]. I have been pondering the following question: given a general topological spaces [imath]X, Y[/imath], and a function [imath]f: X\to Y[/imath], when does [imath]Graph(f)[/imath] closed in [imath]X\times Y[/imath] imply that [imath]f[/imath] is continuous. By the closed graph theorem, this is true whenever [imath]X[/imath] and [imath]Y[/imath] are both Banach spaces. Also, it is fairly easy to prove that whenever [imath]Y[/imath] is a Hausdorff space and [imath]f[/imath] is continuous, then [imath]Graph(f)[/imath] is closed, but I do not think that the converse is true, so I am trying to find an example where [imath]X[/imath] is some topological space, [imath]Y[/imath] a Hausdorff space, [imath]f: X\to Y[/imath] a function with a closed graph in [imath]X\times Y[/imath], but who fails to be continuous. As of yet I have not been able to find such a counterexample, partially because I have no clue where to look for such a counterexample. I would really appreciate getting some directions to go in. |
1020275 | Lower Bound for subsets of [imath]\mathbb{N}[/imath]
Let [imath]P = \{A \in \mathbb{N} : A \neq \emptyset, |A|<\infty, |A| = 2k, k\in \mathbb{Z}^{+}\}[/imath]. Consider [imath]X = \{ A \in P: 1 \in P \}[/imath]. Does [imath]X[/imath] have a lower bound in [imath](P,\subseteq)[/imath]. We can see that [imath]X\subseteq P[/imath], which contains the element [imath]1[/imath] of [imath]P[/imath]. By definition, an element [imath]p\in P[/imath] is a lower bound for [imath]X[/imath] iff [imath]\forall x \in X[/imath] we have [imath]p\subseteq x[/imath]. Let [imath]p = \{ 1 , a_2, a_3,\cdots,a_{2k}\}[/imath] be the elements of [imath]P[/imath] which contain [imath]1[/imath] and are organized in ascending order, and let [imath]x = \{ 1 , b_2, b_3,\cdots,b_{2m}\}[/imath], where [imath]m\in \mathbb{Z}^+[/imath], and organized in ascending order. By comparing terms, it is not necessarily that [imath]a_{i} = b_{i}[/imath], then, there does not exist a [imath]p\in P: p\subseteq x, \forall x\in X[/imath], so [imath]X[/imath] does not have a lower bound. I am right or wrong, Thanks | 1018962 | Does X have a lower bound?
Let P={A [imath]\subseteq[/imath] [imath]\mathbb N[/imath]: A is non-empty, finite and has an even number of elements}. Consider X={A [imath]\in[/imath] P : 1 [imath]\in[/imath] A}. Does X have a lower bound in (P,[imath]\subseteq[/imath])? I think that there is a lower bound... |
1009511 | Primes dividing [imath]11, 111, 1111, ...[/imath]
How can I prove that every prime except 2 and 5 divide infinitely many of the following integers [imath]11, 111, 1111, ...[/imath] ? | 1034192 | Show that for any integer not divisible by 2 or 5, there is a multiple of it which is a string of 1s.
Given that a number [imath]n \equiv\{1,3,7,9\} \pmod{10} [/imath] show that there is a multiple of [imath]n[/imath], [imath]q[/imath] that is a string of consectutive [imath]1[/imath]s. |
451146 | If [imath]f[/imath] is integrable then [imath]|f|[/imath] is also integrable
Show that if [imath]f[/imath] is integrable on [imath][a, b][/imath] then [imath]\lvert f \rvert[/imath] is also integrable. Hint: Show that [imath]U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).[/imath] I have: [imath]U (P , \lvert f \rvert) \ge U (P , f )[/imath] and [imath] L(P , \lvert f \rvert) \ge L(P,f)[/imath] Thus, [imath]U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).[/imath] | 2760971 | Prove that if [imath]f[/imath] is integrable on [imath][a ,b][/imath] then [imath]|f(x)|[/imath] is also integrable on [imath][a ,b][/imath] and [imath]|∫_a^b f(x)\,\mathrm dx|[/imath] ≤ [imath]∫_a^b|f(x)|\,\mathrm dx[/imath]
Prove that if [imath]f[/imath] is integrable on [imath][a ,b][/imath] then [imath]|f(x)|[/imath] is also integrable on [imath][a ,b][/imath] and [imath]|∫_a^b f(x)\,\mathrm dx|[/imath] ≤ [imath]∫_a^b |f(x)|\,\mathrm dx[/imath] |
709777 | The Line with two origins
I have seen descriptions of the "line with two origins" using quotient spaces. My professor has defined it in an alternate way. However, I can not wrap my head around how the following descriptions forms a line with two origins. Consider [imath]X=\mathbb{R} \setminus \{0\} \cup \{p,q\}[/imath], that is [imath]X[/imath] is the union of the reals minus [imath]0[/imath], and two points. Consider sets of the type [imath]U_a = (-a,0) \cup \{p\} \cup (0,a)[/imath] [imath]V_a=(-a,0) \cup \{q\} \cup (0,a)[/imath] where [imath]a >0[/imath]. And let [imath]\mathcal{B}=\{U_a\}_{a>0} \cup \{V_a\}_{a >0} \cup \{ \text{all open intervals of} \hspace{2mm} \mathbb{R} \hspace{2mm} \text{not containing the origin} \}[/imath] Then [imath]\tau=\{\bigcup_{\alpha} B_{\alpha} \big | B_{\alpha} \in \mathcal{B} \}[/imath]. How is this a line with two origins? | 2152272 | What exactly is "the unit interval with two origins"?
This answer says the following: The converse need not be true: Take [imath]X=I\cup\{0'\}[/imath], a unit interval with two origins, where the neighborhood base of [imath]0'[/imath] is formed by intervals [imath][0',\epsilon):=\{0'\}\cup(0,\epsilon)[/imath]. Where exactly is [imath]0'[/imath] on the line? From the rest of the answer, it seems like we require [imath]0' \in \mathbb{R}\setminus I[/imath], but I'm honestly not sure why it deserves to |
1021007 | Mathematical Induction Proof 1
Prove that for every integer [imath] n \geq 1[/imath], we have [imath]\displaystyle \sum_{j=1}^n j^3 = \left(\dfrac{n(n+1)}{2}\right)^2[/imath] I know how to prove an induction proof, but I just can't get the algebra down on this problem. Can anyone help? | 1020970 | Proof involving Induction
Prove that for every integer n ≥ 1, we have [imath] \sum_{i=1}^ni^3=\left(\frac{n(n+1)}2\right)^2 [/imath] Solve using Mathematical Induction, include the Inductive Step Base Case is that both the left and right side [imath]=1[/imath] when [imath]n=1[/imath]. and the Inductive Hypothesis is [imath]1^3+2^3+\dots +k^3=\frac{\left( k(k+1)\right)^2}2[/imath] |
1020796 | Prove or disprove: [imath]\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3[/imath]
Is it true that [imath]\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3[/imath] for any positive integer [imath]n[/imath]? We cannot prove the statement using induction as it is, because the left-hand side is increasing while the right-hand side stays constant. So we need to modify the right-hand side to something like [imath]3-\dfrac{1}{n}[/imath]. But it is hard to quantify how much the left-hand side increases when we go from [imath]n[/imath] to [imath]n+1[/imath]. | 498774 | Improving bound on [imath]\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}[/imath]
An old challenge problem I saw asked to prove that [imath]\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3[/imath]. A simple calculation shows the actual value seems to be around [imath]2.8[/imath], which is pretty close to [imath]3[/imath] but leaves a gap. Can someone find [imath]C < 3[/imath] and prove [imath]\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C[/imath]? |
1020429 | [imath]f(b)-f(a) =((b-a)/2)\cdot(f'(a)+f'(b))-((b-a)³/12)\cdot f'''(c)[/imath]
Let [imath]f[/imath] be three times differentiable on [imath][a,b][/imath]. [imath]f'''[/imath] is continuous. Show that there is a [imath]c\in[a,b][/imath] such that [imath]f(b)-f(a) =((b-a)/2)\cdot(f'(a)+f'(b))-((b-a)³/12)\cdot f'''(c)[/imath] This looks like a combination of Mean Value Theorem and Taylor's series. but how to solve it, i don't know. Thanks for any answers! | 1008049 | Approximating a three-times differentiable function by a linear combination of derivatives
Let [imath]\,f[/imath] be three times differentiable on the interval [imath][a, b][/imath]. Show that there is a [imath]c[/imath] such that: [imath]f(b) - f(a) = \frac{b-a}{2}(f'(a)+f'(b)) - \frac{(b-a)^3}{12}f'''(c).[/imath] Answered in the comments. |
1021487 | Is [imath]f : A \to P(A), a \mapsto \{a\}[/imath] injective or surjective?
Given an arbitrary set [imath]A[/imath], let [imath]f:A \to P(A)[/imath] be the function defined for all [imath]a \in A[/imath] by "[imath]f(a) = \{a\}[/imath]". How would you prove that [imath]f[/imath] is injective or surjective? | 1016687 | Confused on Injection and Surjection Question - Not sure how to justify
Given an arbitrary set [imath]A[/imath], let [imath]f : A \rightarrow\wp(A)[/imath], be the function defined for all [imath]a\in A[/imath] by "[imath]f(a) = \{a\}[/imath]" Is [imath]f[/imath] injective? Is [imath]f[/imath] surjective? I am struggling with this question. How am i able to identify if the function is injective? Can somebody give an example? What confuses me is the function "f(a) = {a}" |
1021391 | Finding limit without using L'Hopital rule
How to prove [imath]\lim_{x \to \infty} \frac {\ln(x)}{x}=0[/imath] without using L'Hospital Rule. Just by using some basis limit properties. | 579491 | Limit of [imath]n/\ln(n)[/imath] without L'Hôpital's rule
I am trying to calculate the following limit without L'Hôpital's rule: [imath]\lim_{n \to \infty} \dfrac{n}{\ln(n)}[/imath] I tried every trick I know but nothing works. You don't have to prove it by definition. |
1021728 | Does each irreducible polynomial over the integers represent at least one prime?
There's not much more to the question: If [imath]f(x) \in \mathbb{Z}[x][/imath] is an irreducible polynomial, is there a simple proof that there must be some [imath]x[/imath] for which [imath]f(x)[/imath] is a prime number (positive or negative, e.g. [imath]f(x)=-13[/imath] is OK here)? By irreducible I mean to imply that [imath]f(x)[/imath] is not [imath]0[/imath] or a unit of [imath]\mathbb{Z}[x],[/imath] as some texts demand of an irreducible element. But to simplify things, just assume [imath]f[/imath] has degree at least [imath]1[/imath] [the degree 0 irreducibles being integer primes anyway, which do represent at least one prime]. UPDATE: Note that an example of Hurkyl's gives an irreducible [imath]f[/imath] not representing any primes. [imath]f(x)=x^2+x+4.[/imath] This was pointed out by Lucien in a comment, and that comment has a link to the question with Hurkyl's polynomial in an answer. | 337546 | How often must an irreducible polynomial take a prime value?
Suppose [imath]f(x)[/imath] is an irreducible polynomial over [imath]\mathbb Z[/imath] of degree [imath]n[/imath]. Is it always the case that there exist distinct [imath]x_1,\ldots,x_{2n+1}\in \mathbb Z[/imath] such that [imath]f(x_1),\ldots,f(x_{2n+1})[/imath] are all prime? So far I've tested a few irreducible polynomials using mathematica, and this has held. The motivation is to use this as a test for irreducibility, as if [imath]f(x)[/imath] is not irreducible then it takes on at most [imath]2n[/imath] prime values, as either factor takes on [imath]1[/imath] or [imath]-1[/imath] at most [imath]2n[/imath] times. (I'd also be interested in learning if this bound can be improved). This could potentially be expended to other rings of integers, i.e. [imath]\mathbb Z[i][/imath], but in that case the minimal number of primes needed to certify that [imath]f(x)[/imath] is irreducible would increase as more units appear. |
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