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1315033 | what is the way to reach apropriate answer about my question about group in algebra
Assume there exist [imath]2[/imath] natural numbers that are coprime ([imath]m[/imath] and [imath]n[/imath] such that [imath](m,n)=1[/imath]) such that for each [imath]g[/imath], [imath]h \in G[/imath] we have [imath]g^m h^m = h^m g^m[/imath] and [imath]g^n h^n = h^n g^n[/imath]. Then [imath]G[/imath] is abelian group. (That means [imath]ab=ba[/imath].) Hint: [imath](m,n)=1[/imath] then there exist [imath]c,d\in\mathbb{Z}[/imath] s.t. [imath]cm+dn=1[/imath]. thus [imath]gh = g^{cm+dn} h^{cm+dn} = g^{cm} g^{dn} h^{cm} h^{dn} = (g^m)^c (g^n)^d (h^m)^c (h^n)^d = (g^n)^d (g^m)^c (h^m)^c (h^n)^d[/imath] Use property in hypothesis to reach [imath]gh=hg[/imath]. Please help me to continue. | 326702 | A group such that [imath]a^m b^m = b^m a^m[/imath] and [imath]a^n b^n = b^n a^n[/imath] ([imath]m[/imath], [imath]n[/imath] coprime) is abelian?
Let [imath](G,.)[/imath] be a group and [imath]m,n \in\mathbb Z[/imath] such that [imath]\gcd(m,n)=1[/imath]. Assume that [imath] \forall a,b \in G, \,a^mb^m=b^ma^m,[/imath] [imath]\forall a,b \in G, \, a^nb^n=b^na^n.[/imath] Then how prove [imath]G[/imath] is an abelian group? Some context: Some of these commutation relations often imply that [imath]G[/imath] is abelian, for example if [imath](ab)^i = a^i b^i[/imath] for three consecutive integers [imath]i[/imath] then [imath]G[/imath] is abelian, or if [imath]g^2 = e[/imath] for all [imath]g[/imath] then [imath]G[/imath] is abelian. This looks like another example of this phenomenon, but the same techniques do not apply. |
1318660 | An integration question
Suppose that [imath]-1<a<1[/imath] and consider [imath]g(a)=\int_0^{π/2}\ln\left(\frac{1+a\cos x}{1-a\cos x}\right)\frac{\mathrm dx}{\cos x}.[/imath] How can I find the value of [imath]g(a)[/imath]? I believe the denominator, [imath]\cos x[/imath], may be problematic when it equals [imath]0[/imath]. How would I deal with it? | 571570 | Evaluating the integral [imath]\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx[/imath]
How can I evaluate the following integral? [imath] \int_0^{\pi/2} \log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\, \frac{1}{\cos\left(x\right)}\,{\rm d}x\,, \qquad\left\vert\,a\,\right\vert \le 1[/imath] I tried differentiating under the integral with respect to the parameter [imath]a[/imath], and I also tried expanding the log term in a Taylor series and then switching the order of integration and summation. I ran into difficulties with both approaches. |
1314005 | Solving nonlinear first-order difference equation [imath] d_m = p_0 + p_1d_{m-1} + p_2(d_{m-1})^2 [/imath] (extinction problem)
The steady-state equilibrium is [imath] d^* = \frac{1-p_1-\sqrt{(p_1-1)^2-4p_0p_2}}{2p_2} [/imath]. Based on a plot, I guessed the solution [imath] d_m = d^*(1-e^{-\alpha m}) [/imath], which is pretty close but not correct. What other method can I use to solve this? Context: This is to calculate the extinction probability of a branching process at each step, as in http://en.wikipedia.org/wiki/Branching_process#Example_of_extinction_problem. | 1157670 | Survival probability up to time [imath]n[/imath] in a branching process.
Let [imath]\{Z_n : n=0,1,2,\ldots\}[/imath] be a Galton-Watson branching process with time-homogeneous offspring distribution [imath]\mathbb P(Z_{n,j} = 0) = 1-p = 1 - \mathbb P(Z_{n,j}=2), [/imath] where [imath]0<p<1[/imath]. That is, [imath]Z_0 = 1[/imath] and [imath]Z_n = \sum_{j=1}^{Z_{n-1}}Z_{n,j}[/imath] for [imath]n\geqslant 1[/imath]. Let [imath]T=\inf\{n : Z_n=0\} [/imath] be the extinction time of the process. I want to find [imath]t_n := \mathbb P(T>n), [/imath] i.e. the probability that the process survives up to time [imath]n[/imath]. I found the following recurrence: [imath]t_{n+1} = p(2t_n - t_n^2)[/imath] (with [imath]t_0=1[/imath]) from this mathoverflow question: https://mathoverflow.net/questions/87199/branching-process-survival-probability I checked the recurrence for small values of [imath]n[/imath], and confirmed that [imath]t:=\lim_{n\to\infty} t_n[/imath] satisfies [imath] t = p(2t - t^2),[/imath] both in the case where [imath]t=0[/imath] [imath]\left(p\leqslant\frac12\right)[/imath] and where [imath]t=2-\frac1p[/imath] [imath]\left(p>\frac12\right)[/imath]. So I'm fairly confident this recurrence is valid. However, I have no idea how to solve it. For context, this problem comes from Adventures in Stochastic Processes by Sidney Resnick: From @Did's comment, it appears to be intractable to find a closed form for [imath]\mathbb P(T>n)[/imath]. I find it curious that the question would be asked were it not, though. |
1318512 | prove that for every integer [imath]a>0[/imath] there is a unique representation [imath]a=r*s^2[/imath]
I need to prove that for every integer [imath]a>0[/imath] there is a unique representation [imath]a=r*s^2[/imath] where [imath]r[/imath] is not dividable by any square: there is no [imath]d>1[/imath] such that [imath]d^2|r[/imath] What I tried is to show a as a unique multiplication of primes and then show the case that a is odd or not, but didn't get anywhere.. any help will be appriciated | 139737 | Prove that every positive integer [imath]n[/imath] is a unique product of a square and a squarefree number
I am trying to prove that for every integer [imath]n \ge 1[/imath], there exists uniquely determined [imath]a > 0[/imath] and [imath]b > 0[/imath] such that [imath]n = a^2 b[/imath], where [imath]b[/imath] is squarefree. I am trying to prove this using the properties of divisibility and GCD only. Is it possible? Let me assume that [imath]n = a^2 b = a'^2b'[/imath] where [imath]a \ne a'[/imath] and [imath]b \ne b[/imath]'. Can we show a contradiction now? |
811387 | Characterization of positive definite matrix with principal minors
A symmetric matrix [imath]A[/imath] is positive definite if [imath]x^TAx>0[/imath] for all [imath]x\not=0[/imath]. However, such matrices can also be characterized by the positivity of the principal minors. A statement and proof can, for example, be found on wikipedia: http://en.wikipedia.org/wiki/Sylvester%27s_criterion However, the proof, as in most books I have seen, is very long and involved. This makes sense in a book where you wanted to prove the other theorems anyway. But there has to be a much better way to prove it. What is the "proof from the book" that positive definite matrices are characterized by their [imath]n[/imath] positive principal minors? | 1343912 | show the following equivalence
Let [imath]A = \left(a_{ij}\right) \in C^{n \times n}[/imath] be a self-adjoint matrix (that is, a matrix such that [imath]A^\ast = A[/imath]. Show that [imath]A[/imath] is positive definite if and only if the determinant of the matrix [imath]\begin{bmatrix}a_{11} & . ..& a_{1k}\\ .. & .. & .. \\ a_{k1} & .. & a_{kk} \end{bmatrix}[/imath] is positive for all positive integers [imath]1\le k \le n[/imath]. I was thinking of constructing a basis [imath]b_1,b_2,...,b_n[/imath] and using induction but i am not quite sure |
1318126 | lipschitz continuity from a constant c?
Let [imath]f : [1,\infty) \to \mathbb R[/imath] be uniformly continuous. Show that there exists a number [imath]C > 0[/imath] such that [imath]|f(x)|\le Cx[/imath] for all [imath]x \ge 1[/imath]. | 539998 | [imath]f[/imath] defined on [imath][1,\infty )[/imath] is uniformly continuous. Then [imath]\exists M>0[/imath] s.t. [imath]\frac{|f(x)|}{x}\le M[/imath] for [imath]x\ge 1[/imath].
[imath]f[/imath] defined on [imath][1,\infty )[/imath] is uniformly continuous. Then [imath]\exists M>0[/imath] s.t. [imath]\frac{|f(x)|}{x}\le M[/imath] for [imath]x\ge 1[/imath]. I know f uniformly continuous [imath]\implies \forall\varepsilon > 0s.t.\forall x,y\in [1,\infty)[/imath], [imath]|f(x)-f(y)|<\varepsilon \space\space\forall |x-y|<\delta[/imath] By Mean value theorem, [imath]\exists c\in (x,y)\forall x,y\in [1,\infty)[/imath] such that [imath]\frac{f(x)-f(y)}{x-y}=f'(c)[/imath] [imath]\frac{|f(x)-f(y)|}{|x-y|}=|f'(c)|<M[/imath] for some [imath]M>0[/imath] [imath]|f(x)-f(y)|<M|x-y|<M\delta [/imath] How can I get an expression for [imath]|f(x)|/x[/imath]? |
1319337 | Prove [imath]\lim_{x \to c}{x^2}=c^2[/imath] where [imath]c[/imath] is a real number
Prove [imath]\lim_{x \to c}{x^2}=c^2[/imath] where [imath]c[/imath] is a real number with the [imath](\epsilon, \delta)[/imath] definition. I know that you need to assume a value for [imath]\delta[/imath]. However, I don't understand how that one assumption, which only represents one case, implies that the limit is always true. Please explain as you prove the limit. | 330297 | Prove that [imath]\lim_{x\to 2}x^2=4[/imath] using [imath]\epsilon-\delta[/imath] definition.
Prove that [imath]\lim_{x\to 2}x^2=4[/imath] using [imath]\epsilon-\delta[/imath] definition. By the mean of [imath]\epsilon-\delta[/imath] definition, [imath]|x-2|\le \delta,|x+2|\le \delta+4[/imath] then [imath]|x-2||x+2|\le \delta(\delta+4),|x^2-4|\le \delta^2+4\delta[/imath]. Assign [imath]\epsilon=\delta^2+4\delta,|x^2-4|\le\epsilon[/imath]. Q.E.D. Is this method correct? If yes, why I always see people do this by putting [imath]\delta = \min(1,\epsilon/5)[/imath]...? Thanks. |
126927 | If [imath]f'[/imath] is bounded, then [imath]f[/imath] is uniformly continuous
Let [imath]f\colon(a,b) \to \mathbb R[/imath] be differentiable and suppose that there exists an [imath]M>0[/imath] such that [imath]|f'(x)| \leq M[/imath] for all [imath]x[/imath] in [imath](a,b)[/imath]. Prove that [imath]f[/imath] is uniformly continuous. | 1581293 | Using Bounded Derivative to Show Uniform Continuity
Suppose [imath]f: (a,b) \to \mathbb{R}[/imath] is differentiable on [imath](a,b)[/imath] and [imath]|f '(x)|\leq M[/imath] for all [imath]x \in (a,b)[/imath]. Prove that [imath]f[/imath] is uniformly continuous on [imath](a,b)[/imath]. My attempt: I started by observing the definition of uniform continuity, which is for all [imath]\epsilon \gt 0[/imath] there exists [imath]\delta \gt 0[/imath] such that for all [imath]x,y \in (a,b)[/imath], [imath]|x-y|\lt \delta[/imath] implies [imath]|f(x) - f(y)|\lt \epsilon[/imath]. Using this I began by trying to find [imath]\delta[/imath]. My first thought was to use the information about the derivative. I wrote out that $\lim_{x\to y}[imath]|[/imath](f(x) - f(y) \over x - y)$|[imath]\leq M[/imath]. Using this, [imath]-M[/imath]($\lim_{x\to y}(x - y))[imath]\leq[/imath]\lim_{x \to y}(f(x) - f(y))[imath]\leq[/imath]M(\lim_{x \to y}(x - y))$. Now this implies that $|\lim_{x\to y}f(x) - f(y)|[imath]\leq[/imath]M(\lim_{x \to y}(x - y))$. I'm thinking that this is what will get me to the [imath]\delta[/imath] that I'm looking for. But the limit on the right hand side approaches 0, so this should imply that [imath]|\lim_{x \to y}f(x) - f(y)|[/imath][imath]\leq[/imath] [imath]0[/imath]. Is this the result that will show uniform continuity? |
1315974 | Solving a little Diophantine equation:[imath](n-1)!+1=n^m[/imath]
How can I solve this Diophantine equation: [imath](n-1)!+1=n^m[/imath] with [imath]n,m[/imath] positive integers? From Wilson's theorem we can note that [imath]n[/imath] is a prime number. I proved to rewriting the equation as:[imath](n-2)!=n^{m-1}+n^{m-2}+....+1[/imath] but in vain. I proved to solve also through the theorem LTE bur I analysed only some cases without obtain a general solution. | 805068 | To find all [imath]n[/imath] such that [imath](n-1)!+1[/imath] is a perfect power of [imath]n[/imath]
How to find all positive integers [imath]n[/imath] such that [imath](n-1)!+1[/imath] can be written as [imath]n^k , k\in \mathbb Z^+[/imath] ? |
1318603 | How do I find the value of n in the following equation
This is from the controversial GCSE question in the UKs recent exams. The orginal question is thus: There are [imath]n[/imath] sweets in a bag. [imath]6[/imath] of the sweets are orange, the rest are yellow. Hannah takes at random a sweet from the bag and eats it. She then takes another sweet at random from the bag and eats it. The probability that hannah eats two orange sweets is 1/3 a) show that [imath]n^2 - n - 90 = 0[/imath] This in itself isn't too difficult: [imath] \frac{6}{n}\frac{5}{n-1} = \frac{30}{n^2 - n} = \frac{1}{3} \Rightarrow \frac{90}{n^2 -n} = 1 [/imath] so that [imath] n^2 - n -90 = 0 [/imath] But, what I'm trying to figure out is the value of n, the only way to do this I have thought of is trial and error, quickly finding the answer to be 10. Is there a different option here? | 1313076 | EdExcel GCSE question about Hannah and the sweets: show that [imath]n^2-n=90[/imath]
This is my reconstruction of the EdExcel GCSE question that has caused such a Twitter storm in the UK in the last 24 hours, along with its solution. Hannah has a bag containing [imath]n[/imath] sweets, 6 of which are orange. She eats two sweets at random from the bag. The probability that the two sweets Hannah eats are both orange is [imath]\frac{1}{3}[/imath]. Show that [imath]n^2 - n = 90[/imath]. Possible follow-up (I don't know if this was part of the original): how many sweets were there in total in the bag? |
1319372 | Basis for the linear spacer [imath]\ell^p[/imath]
Is there any well known basis (Hamel basis) for the vector space [imath]\ell^p[/imath]? And what about the cardinality of such basis? Is it countable? | 194189 | A Hamel basis for [imath]\ell^p[/imath]?
I am looking for an explicit example for a Hamel basis for [imath]\ell^{p}[/imath]?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one can express any element of the vector space as a finite linear combination of these. After some trying I could not write one explicitly. A quick google search did not reveal anything useful except for the proof of uncountability of a an infinite Hamel basis. Maybe I am being a bit silly but I don't think the answer is as obvious as for a Schauder basis for the same case. So, what is an explicit example for a Hamel basis for [imath]\ell^{p}[/imath]?? |
1319515 | Low-rank matrix space
Let [imath]M(m,n,r)[/imath] be the matrix space of real matrices [imath]m\times n[/imath] with [imath]rank \leq r[/imath]. Is [imath]M(m,n,r)[/imath] an open set? or closed set? or Does it have some property? Regards | 1311549 | How to prove that given set is a connected 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|\operatorname{rank}(X)\leq k\}[/imath] where [imath]k\leq \min\{m,n\}[/imath]. Check whether the set [imath]C[/imath] is: Closed Connected Compact Open What are some other good properties of the set [imath]C[/imath],for example is it a manifold? Clearly the set [imath]C[/imath] is closed if someone is interested a good proof can be found here, hence [imath]C[/imath] is not open. Also as [imath]C[/imath] is unbounded therefore not compact. How to check whether the set [imath]C[/imath] is connected or not? |
1319539 | A question regarding power series expansion of an entire function
Let [imath]f[/imath] be an entire function and let for each [imath]a\in \mathbb R[/imath], there exists at least one coefficient [imath]c_n[/imath] in [imath]f(z)=\sum\limits_{n=0}^{\infty}c_n(z-a)^n[/imath], which is zero. Then [imath]f^{(n)}(0)=0[/imath] for infinitely many [imath]n\geq 0[/imath] [imath]f^{(n)}(0)=0[/imath] for every [imath]n\geq 0[/imath] [imath]f^{(2n+1)}(0)=0[/imath] for every [imath]n\geq 0[/imath] There exists [imath]k\geq 0[/imath] such that [imath]f^{(n)}(0)=0[/imath] for all [imath]n\geq k[/imath] We know that [imath]c_n=\frac{f^{(n)}(a)}{n!}[/imath] for all [imath]n\in \{0,1,2\ldots\}[/imath]. Thus for [imath]a=0[/imath], [imath]c_n=\frac{f^{(n)}(0)}{n!}[/imath]. By hypothersis, atleast one [imath]c_n=0[/imath]. After that I could not do anything. Please help! | 976276 | Vanishing of Taylor series coefficient
I am solving previous year question paper some competitive exam. Give me some hint to solve the following problem. Let [imath]f[/imath] be an entire function. Suppose for each [imath]a \in \mathbb{R} [/imath] there exists at least one coefficient [imath]c_n[/imath] in [imath]f(z) = \sum_{n=0}^{\infty} c_n (z-a)^n[/imath] which is zero. Then, a) [imath]f^{(n)}(0) = 0[/imath] for infinitely many [imath]n \ge 0[/imath] b) [imath]f^{(2n)}(0)=0[/imath] [imath] \forall n \ge 0[/imath] c) [imath]f^{(2n+1)}(0)=0[/imath] [imath] \forall n \ge 0[/imath] d) [imath]f^{(n)}(0) = 0[/imath] for all sufficiently large n. Thanks in advance. |
1320203 | Number theory: show 5777 cannot be written as [imath]p+2k^2[/imath]
Exercise: show that we cannot write [imath]5777=p+2k^2[/imath] where [imath]p[/imath] is a prime or [imath]1[/imath] and [imath]k\ge0[/imath]. I'd like your answer/hint won't be "Test all the first [imath]53[/imath] squares", namely the brute force. What do you suggest? | 1320012 | Elementary number theory: sums of primes and squares
Show that we cannot write:[imath]5777=p+2a^2[/imath] where [imath]p[/imath] is a either 1 or a prime and [imath]a\ge0[/imath]. Can you give me a hint? ...i'd better not test all primes :) |
1320399 | Split [imath]\mathbb{N}[/imath] into a countable union of countable sets.
A friend thought of this problem and I found it interesting to think about so I want to share it with you. I am intrigued how you will solve the problem. Find countably infinite many sets [imath]U_i[/imath] such that [imath]\mathbb{N}=\cup_{i=1}^{\infty}U_i,[/imath] with [imath]U_i[/imath] disjunct and countably infinite. | 847465 | Expressing [imath]\Bbb N[/imath] as an infinite union of disjoint infinite subsets.
The title says it. I thought of the following: we want [imath]\Bbb N = \dot {\bigcup_{n \geq 1} }A_n[/imath] We pick multiples of primes. I'll add [imath]1[/imath] in the first subset. For each set, we take multiples of some prime, that hasn't appeared in any other set before. Then [imath]\begin{align} A_1 &= \{1, 2, 4, 6, 8, \cdots \} \\ A_2 &= \{3, 9, 15, 21, 27, \cdots \} \\ A_3 &= \{5, 25, 35, 55, \cdots \} \\ A_4 &= \{7, 49, 77, \cdots \} \\ &\vdots \end{align} [/imath] I'm heavily using the fact that there are infinite primes. I think these sets will do the job. Can someone check if this is really ok? Also, it would be nice to know how I could express my idea better, instead of that hand-waving. Alternate solutions are also welcome. Thank you! Edit: the subsets must be also infinite. |
1320082 | [imath]G[/imath] is a group and [imath]N,M[/imath] are normal subgroups of [imath]G[/imath]. Prove that [imath]nm=mn[/imath] for all [imath]n\in N,m\in M[/imath].
My problem is the following [imath]G[/imath] is a group and [imath]N,M[/imath] are normal subgroups of [imath]G[/imath]. [imath]N\cap M = \{e\}[/imath]. Prove that [imath]nm = mn[/imath] for every [imath]n\in N,m\in M[/imath]. What i did - I know that [imath]gng^{-1}\in[/imath] N for all g [imath]\in G[/imath] So also [imath]mnm^{-1}\in N [/imath] Now if [imath]nm =mn[/imath] then i can make the above be [imath]n \in N[/imath]. but if not .. then I can't find a contradiction. Any help will be appreciated. | 773405 | If [imath]H,K⊲G[/imath] and [imath]H∩K = \{1_G\}[/imath], then all elements in [imath]H[/imath] commute with all elements in [imath]K[/imath]
Let [imath]H,K⊲G[/imath] be two normal subgroups such that [imath]H∩K = \{1_G\}[/imath]. Prove all elements in [imath]H[/imath] commute with all elements in [imath]K[/imath]. I have no idea how to do this. |
1318373 | Prove that every integer is a sum of two squares and a cube of integer
Prove that every integer is a sum of two squares and a cube of integer. Probably the only connected result I know, is the Fermat's theorem on sums of two squares. But I don't know how to prove that in other cases (for numbers not being of the form [imath]x^2+y^2=x^2+y^2+0^3[/imath]) one can choose such two squares that the remainder will be a cube. | 1039106 | How prove this diophantine equation [imath]x^2+y^2+z^3=n[/imath] always have integer solution
show that: For any postive ineteger [imath]n[/imath],then the equation [imath]n=x^2+y^2+z^3[/imath] always have integer solution My idea: such as [imath]n=1[/imath],then we have [imath]1=0^2+0^2+1^3[/imath] [imath]2=0^2+1^2+1^3[/imath] [imath]3=1^2+1^2+1^3[/imath] [imath]4=2^2+0^2+0^3[/imath] [imath]5=1^2+2^2+0^3[/imath] [imath]6=1^2+2^2+1^3[/imath] [imath]7=2^2+2^2+(-1)^3[/imath] [imath]8=0^2+0^2+2^3[/imath] [imath]9=1^2+0^2+2^3[/imath] [imath]10=1^2+1^2+2^3[/imath] [imath]\cdots\cdots\cdots[/imath] But for general [imath]n[/imath], How prove it? |
1320703 | The product of xy of two real numbers x and y is irrational then at least one of the x or y must be irrational.
Prove if true or find a counterexample.... The product of [imath]x y [/imath] of two real numbers [imath]x[/imath] and [imath]y[/imath] is irrational then at least one of the [imath]x[/imath] or [imath]y[/imath] must be irrational. | 507865 | Check if this proof about real numbers with an irrational product is correct.
Can anyone confirm if my proof is correct, please? Claim:- “If [imath]x[/imath] and [imath]y[/imath] are real numbers and their product is irrational, then either [imath]x[/imath] or [imath]y[/imath] must be irrational.” Proof:- Assume that both [imath]x[/imath] and [imath]y[/imath] are rational. Now, let [imath]x = \dfrac pq[/imath] and [imath]y = \dfrac mn[/imath] since both of them are rational. [imath]xy =\dfrac pq * \dfrac mn = \dfrac{pm}{qn}[/imath] Thus, if the product [imath]xy[/imath] can be written as a fraction, it's not a irrational number. Therefore if one of [imath]x[/imath] and [imath]y[/imath] is not irrational, then the product is not irrational. By the principle of proof by contraposition, If [imath]x[/imath] and [imath]y[/imath] are real numbers and their product is irrational, then either [imath]x[/imath] or [imath]y[/imath] must be irrational. |
1308820 | Euler-Lagrange equation
This is PDE Evans, 2nd edition: Chapter 8, Exercise 2: Find [imath]L=L(p,z,x)[/imath] so that the PDE [imath]-\Delta u + D\phi \cdot Du = f \quad \text{in }U[/imath] is the Euler-Lagrange equation corresponding to the functional [imath]I[w] := \int_U L(Dw,w,x) \, dx[/imath]. (Hint: Look for a Lagrangian with an exponential term.) First, one must note that [imath]U \subset \mathbb{R}^n[/imath] and [imath]L(p,z,x)=L(Du,u,x)[/imath], where [imath]Du=\nabla u =\sum_{i=1}^n u_{x_i}[/imath]. One should also note that the Euler-Lagrange equation associated with the energy functional [imath]I[w]:=\int_U L(Dw(x),w(x),x) \, dx[/imath] is [imath]-\sum_{i=1}^n (L_{p_i}(Du,u,x))_{x_i}+L_z(Du,u,x) = 0 \quad \text{in }U.[/imath] Here is my work so far: I set [imath](L_{p_i})_{x_i}:= (p_i)_{x_i}-\phi_{x_i} p_i[/imath] and [imath]L_z:=-f(x)[/imath] so that the LHS of the Euler Lagrange PDE is \begin{align} 0 &=-\sum_{i=1}^n (L_{p_i}(Du,u,x))_{x_i}+L_z(Du,u,x) \\ &=-\sum_{i=1}^n\left[(p_i)_{x_i} - \phi_{x_i} p_i \right] - f(x) \\ &= -\sum_{i=1}^n \left[u_{x_ ix_i} - \phi_{x_i} u_{x_i} \right] - f(x) \\ &= -\sum_{i=1}^n u_{x_ ix_i} + \sum_{i=1}^n \phi_{x_i} u_{x_i} - f(x) \\ &= -\Delta u + D\phi \cdot Du - f(x). \end{align} This gives us the PDE [imath]-\Delta u + D\phi \cdot Du = f[/imath] as written in Exercise 2. Now, how can I write [imath]L(p,z,x)[/imath]? I was thinking [imath]L(p,z,x)=\frac 12|p|^2 - \qquad \qquad - zf(x),[/imath] so that, upon taking the derivative with respect to [imath]p_i[/imath] or with respect to [imath]z[/imath], we would obtain both the [imath]L_{p_i}[/imath] and [imath]L_z[/imath] that I have, respectively. However, I have trouble writing the middle term, which is why I left a blank space. The textbook author did say "look for a Lagrangian with an exponential term". | 270110 | A problem from Evans' PDEs book: find a Lagrangian for a given Euler-Lagrange equation
Find [imath]L=L(p,z,x)[/imath] so that the PDE: [imath]-\Delta u +D\varphi \cdot Du =f [/imath] is the Euler-Lagrange equation corresponding to the functional [imath]I[w]:=\int_UL(Dw,w,x)dx[/imath]. (Hint,:Look for a Lagrangian with an exponential term), We can easily to calculate an equation's Euler-Lagrange equation, while the inverse calculation I feel hard for me ,and I still don't know how to use the Hint! Thanks a lot! |
1319832 | For finite [imath]\Sigma[/imath], if [imath]\Sigma \vdash A[/imath] then [imath]\Sigma \models A[/imath]
In my book I have a theorem called "Soundness theorem" and it says: For finite set [imath]\Sigma[/imath], if [imath]\Sigma \vdash A[/imath] then [imath]\Sigma \models A[/imath] Can someone tell me what the symbol [imath]\vdash[/imath] and the symbol [imath]\models[/imath] means? | 1115066 | Meaning of symbols [imath]\vdash[/imath] and [imath] \models[/imath]
I'm confused about the use of symbols [imath]\vdash[/imath] and [imath] \models[/imath]. Reading the answers to Notation Question: What does [imath]\vdash[/imath] mean in logic? and What is the meaning of the double turnstile symbol ([imath]\models[/imath])? I see that: the turnstile symbol [imath] \vdash [/imath] denotes syntactic implication. Then [imath] S \vdash \psi[/imath] means that [imath]\psi[/imath] can be derived from the formulae in [imath]S[/imath] the double turnstile [imath] \models [/imath], denotes semantic implication. Then the notation [imath]A \models B[/imath] means simply that B is true in every model in which A is true. So I understand that, e.g., [imath](ZFC+ A) \vdash B[/imath] means that [imath]A \Rightarrow B[/imath] is a theorem in ZFC and, in this sense [imath]\vdash[/imath] is a mathematical symbol for a theorem. While [imath](ZFC+ A) \models B[/imath] means that [imath] B[/imath] is true in every model of ZFC in which [imath]A[/imath] is true, and, in this sense [imath]\models[/imath] is not a mathematical but a metamathematical symbol. I understand correctly? And those are the only uses of the two symbols? |
1320841 | Finding the integral of [imath]\sin(x)\,dx[/imath] using Riemann sums?
I'd like to know how to show that [imath]\int_{0}^{\pi/2} {\sin(x)\;dx}=1[/imath] using Riemann sums. If someone could please show me step by step and explain I would be very grateful! Edit: I did see the other question but I don't think I need to use trig identities on this. I'm still confused about how to go further than the setup. | 525495 | [imath]\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi} {\sin(x)\;dx}[/imath] using Riemann sums?
How to find the integral [imath]\int_{\frac{1}{3}\pi}^{\frac{2}{3}\pi} {\sin(x)\;dx}=1[/imath] using Riemann sums? |
1321589 | Number of elements in the quotient ring [imath]\mathbb{Z}_6 [x]/\langle 2x +4\rangle[/imath]
I am confused about this quotient ring. I know that [imath]\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle = \{ f(x) + \langle x^2 + 1\rangle \mid f(x) \in \mathbb{Z}_7[x] \}[/imath]. Here [imath]x^2 + 1[/imath] is a zero element in [imath]\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle[/imath], so if we replace [imath]x[/imath] by [imath]i[/imath] we get [imath]\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle =\mathbb{Z}_7 (i) [/imath]. I know the distinct element of [imath] \mathbb{Z}_6 [x]/\langle 2x +4\rangle[/imath] are [imath]\langle 2x +4\rangle,\;x + \langle 2x +4\rangle, \;x^2 + \langle 2x +4\rangle, \ldots[/imath] but I do not know how to prove this. Please tell what could we do similar here in case of [imath] \mathbb{Z}_6 [x]/\langle 2x +4\rangle[/imath]? Since [imath]1[/imath] is the zero element of the polynomial [imath]2x +4 [/imath]. Why we can not replace here [imath]x[/imath] by [imath]1[/imath]? Any help would be appreciated. Thank you. | 1296379 | Cardinality of quotient ring [imath]\mathbb{Z_6}[X]/(2X+4)[/imath]
Let [imath]R[/imath] be the ring obtained by taking the quotient of [imath]\mathbb{Z_6}[X][/imath] by principal ideal [imath](2X+4)[/imath]. Then 1) [imath]R[/imath] has infinite elements 2) [imath]R[/imath] is field 3) [imath]5[/imath] is unit in [imath]R[/imath] 4) [imath]4[/imath] is unit in [imath]R[/imath]. My Attempt: \begin{equation*} \mathbb{Z_6}[X]/(2X+4) = \mathbb{Z_6}[X]/(2(X+2)) = \mathbb{Z_3}[X], \end{equation*} as [imath]X+2[/imath] has root in [imath]\mathbb{Z_6}[/imath]. According to me 1) and 3) are correct but I'm not sure about my answer. Help me. I am not more familiar with polynomial and quotient ring. I searched this problem here but couldn't get this here. If this problem is already asked here then how can I get this? Thank you in advance. |
1305375 | What is the next number having the same number of bit 1s?
You are given a number, [imath]A[/imath], and you have to determine a number, [imath]B[/imath], such that [imath]B>A[/imath] and the number of [imath]1's[/imath] in the binary representation of [imath]A =[/imath] number of [imath]1's[/imath] in the binary representation of [imath]B[/imath]. What is the smallest [imath]B[/imath]? | 1304731 | Computing the [imath]n^{\textrm{th}}[/imath] permutation of bits.
I've seen this post about the [imath]n^{\textrm{th}}[/imath] permutation of a set but that is not what I need. If you have a bit string (ones and zeros only) there are algorithms to quickly permute the NEXT lexicographically ordered bit permutation. For example take [imath]000111 \rightarrow 001011 \rightarrow 001101\rightarrow\cdots[/imath] etc. If the string is long then there are going to be approximately one Bajillion of these things. And I want to know how to compute the [imath]n^{\textrm{th}}[/imath] guy without doing the exhaust. Backgound: This is for a parallel computing job where I need to farm out the search space of elements of a set. Foreground: I own a copy of Knuth's 4th volume of "The Art of Computer Programming" and I think the answer is in there but I can't seem to find it. (It's like 900 pages). I'm posting here in the hopes that someone has knowledge of this (obviously). Even if you can point me to a source, say the part in Knuth's book where he describes this problem I would be most grateful. |
1321510 | If [imath]0\le x_{n+m}\le x_n\cdot x_m[/imath], show that [imath]\lim (x_n)^{1/n}[/imath] exists
Suppose there is a sequence [imath]\{x_n\}_{n\in\Bbb N}[/imath] s.t. [imath]0\le x_{n+m} \le x_n\cdot x_m\quad\forall m,n\in\Bbb N[/imath] show that [imath]\lim_{n\to\infty}\left(x_n\right)^{1/n}=\xi\in\Bbb R[/imath] Seeing the "[imath]\forall n,m\in\Bbb N[/imath]", the first thing to pop into my mind is Cauchy's criterion for sequential convergence, but this approach has got me nowhere, because there seems to be nothing I can do to control [imath]|(x_n)^{1/n}-(x_m)^{1/m}|[/imath]. Another thing to be noticed is that in my text book this exercise belongs to the "infinite sum" category instead of "sequence", thus it is my notion that maybe there is a very tricky method to connect this sequence to a certain infinite sum and finish the proof? Any help or hint will be appreciated. Best regards! | 350699 | If [imath]x_{m+n} \le x_n+x_m[/imath], then [imath]\lim x_n/n[/imath] exists and is equal to [imath]\inf x_n/n[/imath]
Let [imath](x_n)_{n \ge 1}[/imath] be a sequence of real numbers satisfying [imath]x_{m+n} \le x_n+x_m[/imath] [imath]m,n \ge 1[/imath]. Show that [imath]\lim \limits_{n \to \infty} \dfrac{x_n}{n}[/imath] exists and is equal to [imath]\inf \left \{\dfrac{x_n}{n}; n\ge 1 \right \}[/imath]. |
1322566 | why is [imath]\omega^n \neq 0[/imath] for a nondegenerate 2 form [imath]\omega[/imath].
Let [imath]\omega[/imath] be a nondegenerate alternating [imath]2[/imath]-form on an [imath]2n[/imath]-dimesional Vectorspace [imath]V[/imath], meaning that for all nonzero [imath]v \in V[/imath] the map [imath]w \mapsto \omega(v,w)[/imath] is not identically zero. Why is the nth power [imath]\omega^n = \omega \wedge \dots \wedge \omega \neq 0[/imath]? Every textbook i have looked at says this is very easy, but i do not see it. Thanks for your help. | 417365 | How to show [imath]\omega^n[/imath] is a volume form in a symplectic manifold [imath](M, \omega)[/imath]?
I have the following problem: Let [imath](M, \omega)[/imath] be a symplectic manifold. How can I show [imath]\omega^n=\underbrace{\omega\wedge \ldots\wedge \omega}_{n-times},[/imath] satisfies [imath]\omega^n(p)\neq 0[/imath] for all [imath]p\in M[/imath]. I believe that is not too dificult but I'm not used with exterior product.. Any help will be valuable... This is important because the nondegeneracy condition on [imath]ω[/imath] is equivalent to the condition that M has an even dimension [imath]2n[/imath] and the top wedge product [imath]\omega^n[/imath] is nowhere vanishing on M, i.e., [imath]ω^n[/imath] is a volume form. In particular, [imath]M[/imath] must be orientable and is canonically oriented by [imath]\omega^n[/imath]. |
257442 | [imath]d(x_{n+2},x_{n+1})≤rd(x_{n+1},x_n)[/imath] gives convergent sequence
Let [imath](X,d)[/imath] be a complete metric space, [imath]r\in (0,1)[/imath] and [imath](x_n)[/imath] be a sequence in [imath]X[/imath] such that [imath]d(x_{n+2},x_{n+1})≤rd(x_{n+1},x_n)[/imath] for every [imath]n\in N[/imath]. How can we show that [imath](x_n)[/imath] is a convergent sequence? | 255183 | Show that {[imath]x_n[/imath]} is a convergent sequence if [imath]d(x_{n+2}, x_{n+1})≤ rd(x_{n+1}, x_n)[/imath]
Let [imath](X, d)[/imath] be a complete metric space, [imath]r∈ (0,1)[/imath] and [imath]\{x_n\}[/imath] be a sequence in [imath]X[/imath] such that [imath]d(x_{n+2}, x_{n+1})≤ rd(x_{n+1}, x_n),[/imath] for every [imath]n∈ℕ[/imath]. Show that [imath]\{x_n\}[/imath] is a convergent sequence. |
1141862 | The set of all bijections from N to N is infinite, but not countable
Let [imath]N=\{0,1,2,3,...\}[/imath] be the set of all non-negative integers and [imath]A[/imath] the set of all bijections from [imath]N[/imath] to itself. Prove that [imath]i)[/imath] [imath]A[/imath] is an infinite set. [imath]ii)[/imath] There exist no bijection from [imath]N[/imath] to [imath]A[/imath] | 1321653 | Is symmetric group on natural numbers countable?
I guess it is too difficult a question to ask about the cardinality of [imath]S_{\mathbb{N}}[/imath] so I would like to ask whether it is countable or not. I tried to prove it is uncountable somewhat mimicking the Cantor's diagonal argument but failed. |
1304229 | Compute the maximum of [imath]|f(z)|[/imath] when [imath]|z| \leq 1[/imath] and [imath]f(z)=\sin (z)[/imath]
Compute the maximum of [imath]|f(z)|[/imath] when [imath]|z| \leq 1[/imath] and [imath]f(z)=\sin (z)[/imath] So since [imath]f[/imath] is holomorphic on [imath]|z| \leq 1[/imath], we know we'll find the max of [imath]|f(z)|[/imath] on [imath]|z|=1[/imath]. So: [imath]|f(z)|=|\sin(z)|=|\frac{e^{iz}-e^{-iz}}{2i}|=|\frac{e^{iz}}{2i}-\frac{e^{-iz}}{2i}|\leq |\frac{e^{iz}}{2i}|+|\frac{e^{-iz}}{2i}|=\frac{|e^{iz}|}{2}+\frac{|e^{-iz}|}{2}=\frac{e^{|iz|}}{2}+\frac{e^{|-iz|}}{2}=e[/imath] I'm not so certain about this solution. Is there another way? Is using the triangle inequality safe when looking for a maximum? | 1322938 | Maximum of [imath]|\sin(z)|[/imath] as [imath]\{z: |z| \leq 1 \} [/imath]
Maximum of [imath]|\sin(z)|[/imath] as [imath]\{z: |z| \leq 1 \} [/imath] So according to the Maximum Principle, the maximum is when [imath]|z|=1[/imath]. I tried using the fact that [imath]\sin z= \dfrac {e^{iz}-e^{-iz}}{2i}[/imath], but didn't know how to continue from there. Setting [imath]z=e^{it}[/imath] doesn't help much. I saw this question was answered before, however the answer uses hyperbolic functions ([imath]\sinh, \cosh[/imath]) which I haven't actually studied. Thanks in advance for any assistance! |
219109 | Convergence of an infinite product [imath]\prod_{k=1}^{\infty }(1-\frac1{2^k})[/imath]?
Problem: I want to prove that the infinite product [imath]\prod_{k=1}^{\infty }(1-\frac{1}{2^{k}})[/imath] does not converge to zero. It doesn't matter to find the value to which this product converges, but I am still curious to know if anybody is able (if possible of course) to find the value to which this infinite product converges. I appreciate any help. I tried the following trick: [imath]\prod_{k=1}^{n}(1+a_{k})\geq 1+\sum_{k=1}^{n}a_{k}[/imath] which can be easily proven by inudction, where [imath]a_{k}>-1[/imath] and they are all positive or negative. In this case, [imath]a_{k}=-\frac{1}{2^{k}}[/imath], but I get : the infinite product is greater than or equal to zero. | 1322245 | Is the product [imath]\prod_{k=1}^\infty \frac{2^k-1}{2^k}[/imath] necessarily [imath]0[/imath]?
I have the product [imath]\prod_{k=1}^\infty \frac{2^k-1}{2^k}[/imath]. I know that every successive partial product will necessarily be smaller than the last, as we are multiplying always by a number smaller than 1. But, as the expression keeps getting bigger as k does, and it does so very fast, does this product actually converge on a constant different than [imath]0[/imath]? I tried calculating the first partial products but just saw that the terms kept approaching [imath]0.28[/imath]. It looks like a very random number, is this just a case of slow convergence to [imath]0[/imath]? I have no idea of how to calulate this, any help would be appreciated. |
1323346 | Isomorphism of U(10) and U(5)
I was reading through my text and an example was given which states that [imath]U(10) \simeq \mathbb{Z}_{4} \simeq U(5)[/imath]. How should I see that this is true? U(10) = {1,3,7,9), U(5) = {1,2,3,4} and [imath]\mathbb{Z}_{4} = {0,1,2,3}[/imath] Since U(10) implies the set {1,3,7,9} under the multiplication modulo 10, we see that (1.9)mod(10) = 9 but 9 isn't in [imath]\mathbb{Z}_{4}[/imath] | 639220 | Isomorphisms between the groups [imath]U(10), U(5)[/imath] and [imath]\mathbb{Z}/4\mathbb{Z}[/imath]
I think its silly question but I have nt this in my mind at this time. Any one can help How we can see [imath]U(10) \overset{def}= \{1,3,7,9\}[/imath], [imath]\mathbb Z/4\mathbb Z \overset{def}= \{0,1,2,3\}[/imath], [imath]U(5) \overset{def}= \{1,2,3,4\}[/imath] are isomorphic groups where [imath]U(10)[/imath] and [imath]U(5)[/imath] are groups under multiplication? |
1323738 | Verifying that [imath]d(x,y)= \sum_{i=1}^n (x-y)^2%[/imath] satisfies the triangle inequality
How can I prove that the function [imath]d(x,y) = \displaystyle\sum_{i=1}^n (x_i-y_i)^2[/imath] satisfies the triangle inequality [imath] d(x,z) \leq d(x,y)+d(y,z)?[/imath] The book from which this problem came suggests that the Cauchy-Schwarz inequality of the form [imath]\displaystyle\sum_{i=1}^n (x_iy_i) \leq (\displaystyle\sum_{i=1}^n x_i^2 \displaystyle\sum_{i=1}^n y_i^2)^{1/2}[/imath] may be helpful. Also, [imath]d: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}[/imath]. | 75207 | Euclidean distance proof
How can I show that the Euclidean distance satisfies the triangle inequality? Where the Euclidean distance is given by: [imath]d(p,q) = \sqrt{(p_1-q_1)^2 + \cdots + (p_n-q_n)^2}[/imath] Triangle Inequality: [imath]\forall x,y,z\Bigl( d(x,z) \leq d(x,y) + d(y,z)\Bigr)[/imath]. |
1323996 | For which prime numbers [imath]p[/imath] there exist [imath]x,y\in \Bbb{Z}[/imath] such that [imath]p=x^2+2y^2[/imath]?
For which prime numbers [imath]p[/imath] there exists [imath]x,y\in \Bbb{Z}[/imath] such that [imath]p=x^2+2y^2[/imath]? I guess I am to use continued fraction, but I am not sure how. I know how to find solutions for defined numbers but I can't find sets of [imath]p[/imath] that satisfy the above. It will be much appreciated if you have any lead or hint as for how to do it. | 236087 | Numbers representable as [imath]x^2 + 2y^2[/imath]
I need to describe all numbers of the form [imath]x^2 + 2y^2[/imath]. So far I've reduced the problem to primes, and showed p=2 satisfies it. I've also shown that any primes mod 5 or 7 can't be a written in this form. How do I proceed to show that it holds for all primes mod 1 or 3? I think I'll need to use quadratic reciprocity somehow (only since it is the topic of the homework). I have a hunch the supplementary quadradic recprocity laws will be important. Any help would be appreciated. Thanks! |
1324159 | Integral of a non negative function over an infinite interval.
If [imath]f(x) \geq 0[/imath] and for every [imath]x \in [0, \infty)[/imath], [imath]f(x)[/imath] is Riemann integrable on [imath][0,n][/imath] for all n. If [imath]\int_{0}^{\infty}f(x)dx < \infty[/imath] then is it true that [imath]\lim_{x \rightarrow \infty}f(x) = 0[/imath]. If yes then how? | 370317 | Does an absolutely integrable function tend to [imath]0[/imath] as its argument tends to infinity?
Suppose that [imath]f:[0,\infty)\rightarrow\mathbb{R}[/imath] is continuous. Is it true that [imath]\int_{0}^\infty|f(t)|dt<\infty\Rightarrow \lim_{t\rightarrow\infty}f(t)=0?[/imath] If so can you provide a proof, otherwise a counter example. Thank you. |
1309306 | Accumulation points of [imath] \{x_n \in \mathbb{R}, n \in \mathbb{N} \ \ | \ x_n = n\sin(n) \}[/imath]?
A younger student asked me: What are accumulation points of the following set? [imath] \{x_n \in \mathbb{R}, n \in \mathbb{N} \ \ | \ x_n = n\sin(n) \}[/imath] I really can't answer this question, could anyone help me? | 221018 | Is [imath]n \sin n[/imath] dense on the real line?
Is [imath]\{n \sin n | n \in \mathbb{N}\}[/imath] dense on the real line? If so, is [imath]\{n^p \sin n | n \in \mathbb{N}\}[/imath] dense for all [imath]p>0[/imath]? This seems much harder than showing that [imath]\sin n[/imath] is dense on [-1,1], which is easy to show. EDIT: This seems a bit harder than the following related problem, which might give some insight: When is [imath]\{n^p [ \sqrt{2} n ] | n \in \mathbb{N}\}[/imath] dense on the real line, where [imath][\cdot][/imath] is the fractional part of the expression? I am thinking that there should be some probabilistic argument for these things. EDIT 2: Ok, so plotting a histogram over [imath]n \sin n[/imath] is similar to plotting [imath]n \sin(2\pi X)[/imath] where [imath]X[/imath] is a uniform distribution on [imath][-1,1].[/imath] This is not surprising, since [imath]n[/imath] mod [imath]2\pi[/imath] is distributed uniformly on [imath][0,2\pi].[/imath] Now, the pdf of [imath]\sin(2\pi X)[/imath] is given by [imath]f(x)=\frac{2}{\pi \sqrt{1-x^2}}[/imath] in [imath](-1,1)[/imath] and 0 outside this set. The pdf for [imath]n \sin(2\pi X)[/imath] is [imath]g_n(x)=\sum_{k=1}^n \frac{1}{nk} f(x/k)[/imath] so the limit density is what we get when [imath]n \rightarrow \infty.[/imath] (This integrates to 1 over the real line). Now, it should be straightforward to show that for any interval [imath][a,b],[/imath] [imath]\int_a^b g_n(x) dx \rightarrow 0[/imath] as [imath]n \rightarrow \infty.[/imath] Thus, the series [imath]g_n[/imath] is "too flat" to be able to accumulate positive probability anywhere. (The gaussian distribution on the other hand, has positive integral on every interval). |
265714 | What is the dual space of [imath]C([0,T];X)[/imath] ([imath]X[/imath] Hilbert space)?
What is the dual space of [imath]C([0,T];X)[/imath], where [imath]X[/imath] is a Hilbert space? Is it [imath]\operatorname{BV}([0,T]; X^*)[/imath]? As we know, for [imath]C([0,T])[/imath], the dual space is [imath]\operatorname{BV}([0,T])[/imath], but when it is continuous with vector values in [imath]X[/imath], is it still true? | 13473 | Riesz representation and vector-valued functions
A version of the Riesz Representation Theorem says that a continuous linear functional on the space of continuous real-valued mappings on a compact metric space, [imath]C(X)[/imath], can be identified with a signed Borel measure on the set [imath]X[/imath]. Are there any similar results when we replace [imath]C(X)[/imath] by the space of continuous functions of [imath]X[/imath] (compact metric) into [imath]Y[/imath] when (1) [imath]Y=R^N[/imath] or in general (2) [imath]Y[/imath] is a Banach space? I suspect the answer is yes, but I would like to find the right reference to start looking at. Thanks. |
1325099 | Justifying the integral identity “[imath]\int \mathrm{e}^{cx} \sin bx\, \mathrm{d}x = \frac{\mathrm{e}^{cx}}{b^2 + c^2}(c \sin bx - b \cos bx)[/imath]”
I suck at integrals. Here on this Wiki list, it says that [imath]\int \mathrm{e}^{cx} \sin bx \,\mathrm{d}x = \frac{\mathrm{e}^{cx}}{b^2 + c^2}(c \sin bx - b \cos bx),[/imath] which is easy to check – but how could I have derived that? | 540295 | Integrate [imath]e^{ax}\sin(bx)?[/imath]
Is there a general formula for finding the primitive of [imath]e^{ax}\sin(bx)?[/imath] I've done this manually with [imath]a=9[/imath] and [imath]b=4[/imath] using Euler's formulas. But it takes a bit of time. Is there a pattern here? |
1325324 | Why are these representations of e the same?
I heard that [imath]e[/imath] can be defined as the limit as n approaches infinity of [imath](1 + 1/n)^n[/imath], but I also heard that [imath]e[/imath] is also defined as the sum of the reciprocals of the factorials from [imath]0[/imath] to [imath]\infty[/imath]. How are these the same? | 99016 | Proof that [imath]e=\sum\limits_{k=0}^{+\infty}\frac{1}{k!}[/imath]
How can it be proved that the Euler constant equals the limit of the sum of all [imath]\frac{1}{k!}[/imath] when [imath]k[/imath] goes from [imath]0[/imath] to [imath]+\infty[/imath] ? |
1326909 | A calculator is broken so that the only keys that still work are the basic trigonometric and inverse trigonometric functions
A calculator is broken so that the only keys that still work are the [imath]\sin[/imath], [imath]\cos[/imath], [imath]\tan[/imath], [imath]\cot[/imath], [imath]\sin^{-1}[/imath], [imath]\cos^{-1}[/imath], and [imath]\tan^{-1}[/imath] buttons. The display initially shows 0. (Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.) (a) Find, with proof, a sequence of buttons that will transform [imath]x[/imath] into [imath]\frac{1}{x}[/imath]. (b) Find, with proof, a sequence of buttons that will transform [imath]\sqrt x[/imath] into [imath]\sqrt{x+1}[/imath]. (c) Prove that there is a sequence of buttons that will produce [imath]\frac{3}{\sqrt{5}}[/imath]. This is a continuation of a closed problem a while ago. I have solved parts, a and b but c is a challenge. (a) We know that [imath]\tan(\arctan(x))= x[/imath] so inversing the equation, you get [imath]\frac{1}{\tan(\arctan(x))}=\boxed{\cot(\arctan(x))}[/imath] (b) We can solve this using right-triangle trigonometry. With legs, 1 and [imath]\sqrt{x}[/imath], the hypotenuse would be [imath]\sqrt{x+1}[/imath]. To have [imath]\frac{1}{\sqrt{x+1}}[/imath], you would write it as [imath]\cos(\arctan(\sqrt{x}))[/imath]. To transform [imath]\sqrt{x}[/imath] ot [imath]\sqrt{x+1}[/imath], you would have [imath]\frac{1}{\cos(\arctan(\sqrt{x}))}[/imath]. From part (a), we can find the reciprocal of anything. All we need is the reciprocal which is [imath]\boxed{\cot(\arctan(\cos(\arctan(\sqrt{x})))}[/imath] How can we solve c when the initial display is 0? | 1102231 | Calculating certain functions if only certain buttons on a calculator are permitted
A calculator is broken. The only keys that work are [imath]\sin, \cos, \tan, \cot, \arcsin, \arccos[/imath], and [imath]\arctan[/imath] buttons. The original display is [imath]0[/imath]. In this problem, we will prove that given any positive rational number [imath]q[/imath], show that pressing some finite sequence of buttons will yield [imath]q[/imath]. Functions are always in radian form. (a) Find and prove that there exists a sequence of buttons that will turn [imath]\sqrt x[/imath] into [imath]\sqrt{x+1}[/imath]. For this, I got lucky and tried out [imath]\sec (\arctan(x))[/imath] on my calculator and got [imath]\sqrt{x^2+1}[/imath], so I just used the reciprocal and got [imath]\cos (\arctan (x)) = \frac{1}{\sqrt{x^2+1}}[/imath]. However, how I can actually prove this? (b) Prove that there exists a sequence of buttons that will yield [imath]\frac{3}{\sqrt{5}}[/imath]. I know that to go from [imath]x[/imath] to [imath]\frac{1}{x}[/imath] it is [imath]\cot (\arctan (x))[/imath], and I will have to use part a) to get to part b). How do I utilize this? I also know that [imath]\sqrt{\frac{9}{5}}[/imath] = [imath]\frac{3}{\sqrt{5}}[/imath], but after that, I'm stuck. Any hints? |
1327575 | How to find [imath]\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n } [/imath]?
How to find [imath]\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n } [/imath] ? I tried taking using logarithm to bring the expression to sum form and then tried L Hospital's Rule.But its not working.Please help! This is what wolfram alpha is showing,but its not providing the steps! BTW if someone can tell me a method without using integration, I'd love to know! | 2042903 | Evaluating [imath]\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}}[/imath]
I'm trying to find and prove the value of[imath]\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}}[/imath] I was thinking that since [imath]\frac{1}{n}(n!)^{\frac{1}{n}} = \frac{1}{n} \left[ (1)^{\frac{1}{n}}(2)^{\frac{1}{n}}...(n)^{\frac{1}{n}} \right][/imath] and we know that [imath]\lim_{n\to \infty}n^{\frac{1}{n}} = 1[/imath] and [imath]k^{\frac{1}{n}} \leq \ n^{\frac{1}{n}} \ \ \ \ \forall k \leq n[/imath] So [imath](n!)^{\frac{1}{n}} \leq 1 \ \forall n \ [/imath]* Then it is bounded. We also know that [imath]\lim \frac{1}{n} =0[/imath], therefore [imath]\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}} = 0 **[/imath] I'm pretty sure this line of reasoning is ok. Now proving it is another thing. Any suggestions? *I realize now that this statement is not true, but that did not get me any closer to solving the problem. I do believe that it is bounded though. **LOL, I don't believe this to be true anymore either, a wild guess tells me that the solution may be [imath]\frac{1}{e}[/imath] |
1328665 | Is it a composite number?
How do I prove [imath]19\cdot8^n+17[/imath] is a composite number? Or is that number just a prime? So I tried to find a divisor in the cases [imath] n = 2k [/imath] and [imath] n = 2k + 1 [/imath]. But I had no success. Do you have any ideas? | 1153333 | Prove that the number [imath]19\cdot8^n+17[/imath] is not prime, [imath]n\in\mathbb{Z}^+[/imath]
On a Google search, I've notice that often the "trick" involves factoring the expression (and thus showing that the expression isn't prime), but I can't see it. Is this how you would go about it in this case, or would that be a dead end? Any hints? |
567707 | Power series method to solve Airy’s differential equation
Using power series method, solve Airy’s equation [imath]y′′+ xy = 0.[/imath] How do I start solving this? Thanks in advance! | 1197528 | Solution of [imath]y''+xy=0[/imath]
The differential equation [imath]y''+xy=0[/imath] is given. Find the solution of the differential equation, using the power series method. That's what I have tried: We are looking for a solution of the form [imath]y(x)= \sum_{n=0}^{\infty} a_n x^n[/imath] with radius of convergence of the power series [imath]R>0[/imath]. Then: [imath]y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n[/imath] [imath]y''(x)= \sum_{n=1}^{\infty} (n+1) n a_{n+1} x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n[/imath] Thus: [imath]\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n+ x \sum_{n=0}^{\infty} a_n x^n=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=0}^{\infty} a_n x^{n+1}=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=1}^{\infty} a_{n-1} x^n=0 \\ \Rightarrow 2a_2+\sum_{n=1}^{\infty} \left[ (n+2) (n+1) a_{n+2}+ a_{n-1}\right] x^n=0[/imath] So it has to hold: [imath]a_2=0 \\ (n+2) (n+1) a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots[/imath] For [imath]n=1[/imath]: [imath]3 \cdot 2 \cdot a_3+ a_0=0 \Rightarrow a_3=-\frac{a_0}{6}[/imath] For [imath]n=2[/imath]: [imath]4 \cdot 3 \cdot a_4+a_1=0 \Rightarrow a_4=-\frac{a_1}{12}[/imath] For [imath]n=3[/imath]: [imath]5 \cdot 4 \cdot a_5+a_2=0 \Rightarrow a_5=0[/imath] For [imath]n=4[/imath]: [imath]6 \cdot 5 \cdot a_6+a_3=0 \Rightarrow 30 a_6-\frac{a_0}{6}=0 \Rightarrow a_6=\frac{a_0}{6 \cdot 30}=\frac{a_0}{180}[/imath] For [imath]n=5[/imath]: [imath]7 \cdot 6 \cdot a_7+ a_4=0 \Rightarrow 7 \cdot 6 \cdot a_7-\frac{a_1}{12}=0 \Rightarrow a_7=\frac{a_1}{12 \cdot 42}[/imath] Is it right so far? If so, how could we find a general formula for the coefficients [imath]a_n[/imath]? EDIT: Will it be as follows: [imath]a_{3k+2}=0[/imath] [imath]a_{3k}=(-1)^k \frac{a_0}{(3k)!} \prod_{i=0}^{k-1} (3i+1)[/imath] [imath]a_{3k+1}=(-1)^k \frac{a_1}{(3k+1)!} \prod_{i=0}^{k-1} (3i+2)[/imath] If so, then do we have to write seperately the formula for the coefficients of [imath]x^0, x^1[/imath], because otherwhise the sum would be from [imath]0[/imath] to [imath]-1[/imath] ? |
1329344 | Prove that [imath]\sin(n\pi x)[/imath] weakly converges to [imath]0[/imath] in [imath]L^2(0,1)[/imath]
Let [imath]f_n(x):=\sin(n\pi x)\;\;\;\text{for }x\in (0,1)[/imath] and [imath]\langle f,g\rangle:=\int_{(0,1)}fg\;d\lambda^1\;\;\;\text{for }f,g\in L^2(0,1)[/imath] I want to show, that [imath](f_n)_{n\in\mathbb{N}}[/imath] weakly converges to [imath]0[/imath], i.e. [imath]\langle f_n,g\rangle\stackrel{n\to\infty}{\to}\langle 0,g\rangle=0\;\;\;\text{for all }g\in L^2(0,1)[/imath] Unfortunately, [imath]\left\|f_n\right\|_{L^2(0,1)}\stackrel{n\to\infty}{\to}\frac 12\ne 0\;,[/imath] i.e. [imath]f_n[/imath] doesn't converge in [imath]L^2(0,1)[/imath] to [imath]0[/imath] (which would immediately imply weak convergence too). So, it seems like I need to prove the statement from the definition above. However, I've absolutely no idea how to do so. | 990084 | Can we prove the Riemann-Lebesgue lemma by using the Weierstrass approximation theorem?
I'd like to prove the following version of the Riemann-Lebesgue lemma: Let [imath]f: [0,1] \to \mathbb R[/imath] be continuous. Then [imath]\int_0^1 f(x)\sin(nx) \, dx \xrightarrow{n \to \infty} 0[/imath] It's quite easy to show the lemma for the case [imath]f\in C^1[/imath]. So I was hoping that one can approximate the [imath]f\in C^0[/imath] with polynomials (by using the Weierstrass approximation theorem) and then apply the already shown case for [imath]f\in C^1[/imath]. Yet it didn't work out for me. Is this a feasible way? |
1328753 | The Fifteen Puzzle and [imath]S_n[/imath]
I was studying permutation groups from the book "Abstract Algebra and Applications" by Karlheinz Spindler in which page 553 I came across the following interesting problem. It is on the famous "The Fifteen Puzzle" for which I am unable to upload the picture but rather I would try to use mathematical notation. Imagine the fifteen puzzle grid is being presented by the following [imath]4\times 4[/imath] matrix given by [imath]\left[\begin{matrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12\\ 13 & 15 & 14 & \circ \end{matrix}\right][/imath] where [imath]\circ[/imath] is the empty place in the grid. We can slid the "numbered tiles" through blank space and are supposed to bring down the final structure as [imath]\left[\begin{matrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12\\ 13 & 14 & 15 & \circ \end{matrix}\right][/imath] Now the questions are Can some one tell me exactly what am I supposed to do now to solve the problem ? | 635188 | What is the parity of permutation in the 15 puzzle?
You might know the 15 puzzle: [imath]\hskip1.4in[/imath] Concerning the solvability, Wiki says: The invariant is the parity of the permutation of all 16 squares plus the parity of the taxicab distance (number of rows plus number of columns) of the empty square from the lower right corner. This is an invariant because each move changes both the parity of the permutation and the parity of the taxicab distance. In particular if the empty square is in the lower right corner then the puzzle is solvable if and only if the permutation of the remaining pieces is even. I don't get what exactly is meant with the parity of the permutation in this special case? |
1329387 | Calculation of an improper integral in the context of complex functions
I am facing the following improper integral: [imath]\int_0^\infty \frac{x^5\sin x}{(1+x^2)^3}dx.[/imath] Clearly the expression under the integral is a meromorphic function analytic on the nonnegative part of the real line and on the upper half plane (excluding the real line). I do not know how to proceed. I expect that we need to integrate over some larger path and use some residue formula. However we cannot integrate over a the border of a semicircle in the upper half plane increasing in radius because [imath]\sin x[/imath] becomes large as [imath]\operatorname{Im} x[/imath] becomes large. Moreover I do not see how to get from there to an explicit formula of the integral since through this method we would get the value of [imath]\int_{-\infty}^\infty \frac{x^5\sin x}{(1+x^2)^3}dx.[/imath] I do not really understand how I should proceed. I do not need a full answer but a hint (for example over which path to integrate or which formula or theory to use) should be enough. | 839426 | How do I evaluate the integral [imath]\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx[/imath]?
I have no idea how to start, it looks like integration by parts won't work. [imath]\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx[/imath] If someone could shed some light on this I'd be very thankful. |
1329340 | Does [imath]\lim_{x \to 4} \sqrt{x-4}[/imath] exist?
Does [imath]\lim_{x \to 4} \sqrt{x-4}[/imath] exist? The domain of the function is [imath][4,\infty)[/imath] so does the left hand limit exist? And does the general limit exist? | 193355 | Does [imath]\sqrt{x}[/imath] have a limit for [imath]x \to 0[/imath]?
I am taking a calculus course, and in one of the exercises in the book, I am asked to find the limits for both sides of [imath]\sqrt{x}[/imath] where [imath]x \to 0[/imath]. Graph for sqrt(x) from WolframAlpha: This is how I solved the exercise: For simplicity, I choose to disregard the negative result of [imath]\pm\sqrt{x}[/imath]. Since we are looking at limits for [imath]x \to 0[/imath], both results will converge at the same point, and will thus have the same limits. [imath]\sqrt{x}[/imath] = [imath]0[/imath] for [imath]x = 0[/imath]. [imath]\sqrt{x}[/imath] is a positive real number for all [imath]x > 0[/imath]. [imath]\displaystyle \lim_{x \to 0^+} \sqrt{x} = \sqrt{+0} = 0[/imath] [imath]\sqrt{x}[/imath] is a complex number for all [imath]x < 0[/imath]. [imath]\displaystyle \lim_{x \to 0^-} \sqrt{x} = \sqrt{-0} = 0 \times \sqrt{-1} = 0i = 0[/imath] The solution in the book, however, does not agree that there exists a limit for [imath]x \to 0-[/imath]. I guess there are three questions in this post, although some of them probably overlaps: Does [imath]\sqrt{x}[/imath] have a limit for [imath]x \to 0[/imath]? Are square root functions defined to have a range of only real numbers, unless specified otherwise? Is [imath]\sqrt{x}[/imath] continuous for [imath]-\infty < x < \infty[/imath]? WolframAlpha says the limit for x=0 is 0: limit (x to 0) sqrt(x) And also that both the positive and negative limits are 0: limit (x to 0-) sqrt(x) If my logic is flawed, please correct me. |
1330293 | Why is [imath]\sum_{n=0}^{\infty }\left ( \frac{1}{2} \right )^{n}= 2[/imath]?
I'm sorry if this is duplicated, but I can not find any answer to it. | 1306353 | Why [imath] \sum_{k=0}^{\infty} q^k [/imath] sum is [imath] \frac{1}{1-q}[/imath] when [imath]|q| < 1[/imath]
Why is the infinite sum of [imath] \sum_{k=0}^{\infty} q^k = \frac{1}{1-q}[/imath] when [imath]|q| < 1[/imath] I don't understand how the [imath]\frac{1}{1-q}[/imath] got calculated. I am not a math expert so I am looking for an easy to understand explanation. |
532672 | Proving subset is not connected iff there exist open sets in X
Prove that E [imath]\subseteq[/imath] X is not connected if and only if there exist open sets [imath]A, B \subseteq X[/imath] such that [imath]E \subseteq A ∪ B, A ∩ B[/imath] = [imath]\emptyset[/imath] and [imath]E ∩ A[/imath] and [imath]E ∩ B[/imath] are both nonempty. [imath]X[/imath] is a nonempty set equipped with a metric d. Having a hard time proving the ⇒ part of the proof. Help! | 688183 | Prove that [imath]E[/imath] is disconnected iff there exists two open disjoint sets [imath]A[/imath],[imath]B[/imath] in [imath]X[/imath]
Let [imath](X,d)[/imath] be a metric space. Prove that [imath]E[/imath] is disconnected iff there exists two open disjoint sets [imath]A[/imath],[imath]B[/imath] in [imath]X[/imath] such that [imath]E\cap A\neq\emptyset, E\cap B\neq \emptyset[/imath] and [imath]E\subset A\cup B[/imath]. I'm not sure how to begin, so let me just start by pointing some stuff out from the question that I noticed, and my knowledge as of now. (Hopefully it'll be useful.) There is [imath]a\in A, b\in B[/imath] which are both limit points of [imath]E[/imath] (I think) Since [imath]A[/imath] and [imath]B[/imath] are disjoint, [imath]A\cap B=\emptyset[/imath] The closure of a set is the "smallest" set which contains its limit points, (so it is a closed set as well). Both [imath]A[/imath] and [imath]B[/imath] are open so they are made up entirely of interior points, so they don't have any points on the boundary. I know that the definition of separated means that I have [imath]A,B\subset X[/imath], for which [imath]A\cap \bar{B}=\emptyset=\bar{A}\cap B[/imath], where the bar above the set denotes its closure. And I know that the definition of connected set means for [imath]E\subset X[/imath], [imath]E[/imath] is connected if [imath]E[/imath] is not the union of two non-empty separated sets, so a disconnected set would be for [imath]E\subset X[/imath], [imath]E[/imath] is disconnected if it is the union of two separated sets (which are both non-empty). Any hints as to how to begin would be appreciated. Thank you. |
1330493 | How do you prove [imath]\sum \frac {n}{2^n} = 2[/imath]?
How do you prove [imath]\sum_{n=1}^{\infty} \frac {n}{2^n} = 2\ ?[/imath] My attempt: I have been trying to find geometric series that converge to 2 which can bind the given series on either side. But I am unable to find these. Is there a general technique to find the sum? This is a high school interview question and must be easy enough to solve in a few minutes. Please give any hints for the first step towards a solution. | 1526014 | How do I compute [imath]\sum_{k=1}^{\infty} k \cdot p^k[/imath]
I have no idea how to compute this infinite sum. It seems to pass the convergence test. It even seems to be equal to [imath]\frac{p}{(1-p)^2}[/imath], but I cannot prove it. Any insightful piece of advice will be appreciated. |
1330589 | Deriving chromatic polynomials
How to derive the chromatic polynomial from a Cycle? I derived the chromatic polynomial for a triangle [imath] K_3[/imath] it's: [imath]t(t-1)(t-2)[/imath] But I don't understand how to get it for Cycles [imath]C_n[/imath]. | 91009 | Prove that the chromatic polynomial of a cycle graph [imath]C_{n}[/imath] equals [imath](k-1)^{n} + (k-1)(-1)^{n}[/imath]
This is a homework question. But I am completely stuck. My only intuition was to go about it inductively from a "greedy algorithm" maybe know as the deletion-contraction algorithm. And to somehow use the information about the jth cycle to solve the j+1th. But I'm not sure how I'd do it. Thank you very much for looking this over. |
1307699 | Short proof that [imath]\rho^\prime(x,y) = \min\{1,\rho(x,y)\}[/imath] is a metric
Let [imath](X,\rho)[/imath] be a metric space. Define [imath]\rho^\prime: X \times X \to \mathbf{R}[/imath] by [imath]\rho^\prime (x,y) = \min\{1,\rho(x,y)\}[/imath] for all [imath]x, y \in X[/imath]. Does anyone know of a short proof that [imath]\rho^\prime[/imath] satisfies the triangle inequality? It is easy (but tedious) to verify by checking cases. | 533795 | Show that [imath]d(x,y)=\min \{1,|x-y|\}[/imath] is a metric on [imath]\mathbb R[/imath]
Is the following just a matter of showing the 3 properties that make up a metric?? Define d on [imath]\Bbb R\times\Bbb R[/imath] by [imath]d(x,y)=\min \{1,|x-y|\}[/imath]. Show that [imath]d[/imath] is a metric on [imath]\Bbb R[/imath] [imath]d(x,y)=0[/imath] if [imath]x=y[/imath] [imath]d(x,y)=d(y,x)[/imath] for every [imath]x,y \in X[/imath] [imath]d(x,y)\le d(x,z)+d(z,y)[/imath] for every [imath]x,y,z \in X[/imath] |
1330687 | Can we prove that [imath]K_n[/imath] is Hamiltonian for all [imath]n\geq3[/imath]?
Can we prove that [imath]K_n[/imath] is Hamiltonian for all [imath]n\geq 3[/imath] ? I was unable to prove it. Please help me with this. | 458425 | Show that if [imath]n\geq 3[/imath], the complete graph on [imath]n[/imath] vertices [imath]K_n[/imath] contains a Hamiltonian cycle.
I'm asked the following quesiton: Show that if [imath]n\geq 3[/imath], the complete graph on [imath]n[/imath] vertices [imath]K_n[/imath] contains a Hamiltonian cycle. This seems obvious since [imath]K_n[/imath] contains a subgraph which is a cycle graph hitting all [imath]n[/imath] vertices. What am I missing here? |
1331147 | Prove that the 4 degree polynomial has at least two roots
In my assignment I have to prove that: Let [imath]P(x)=x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}[/imath]. Prove that if P has a root in [imath]x_{0}[/imath] and [imath]P'(x_{0})\ne0[/imath] then P has a least two roots. My solution has a problem in it, and I wondered if you can help me with it. Since the functoin is a polynomial then it is differentiable always. If we calculate the limit: [imath]\lim _{x\to \infty}P(x)=\lim_{x\to-\infty}P(x)=\infty[/imath] I tried to solve is by showing through Rolle's theorem by contradiction that the derivative is always [imath]\ne0[/imath] but no one told me that [imath]a_{i}\ne0[/imath] so I'm stuck. I know I should you IVT but I'm pretty much stuck. Your help is appriciated. | 1327925 | Proof that a degree 4 polynomial has at least two roots
Let [imath]P(x) = x^4+a_3x^3+a_2x^2+a_1x+a_0[/imath] [imath]P(x_0) = 0[/imath] [imath]P'(x_0) \not= 0[/imath] with [imath]x_0[/imath] and each [imath]a_i[/imath] real. Prove that [imath]P(x)[/imath] has a at least two real roots. I can't figure why this is true. |
1330668 | Central Limit theorem: Taylor series diverges for harmonics with higher number and those harmonics can't be neglected
I've read several proofs of Central Limit Theorem and they all seemed inaccurate to me, because they drop last members of Taylor series, whereas those members are not infinitesimal. The classical proof of Central Limit Theorem via characteristic functions is based on the fact that [imath]e^{-\frac{x^2}{2}}[/imath] is eigenfunction of Fourier transform. If [imath]\xi[/imath] is a random variable with [imath]E\xi=0[/imath] and [imath]Var(\xi)=1[/imath], we substitute it with another random variable [imath]\nu=\xi/\sqrt{n}[/imath]. We look into Fourier spectrum of its probability density function: [imath]\varphi_{\nu}(t)=\int\limits _{x=-\infty}^{\infty}e^{itx}f_{\nu}(x)dx=\int\limits _{x=-\infty}^{\infty}(e^{i\cdot0\cdot x}+itxe^{i\cdot0\cdot x}+\frac{i^{2}t^{2}x^{2}}{2!}e^{i\cdot0\cdot x}+\frac{i^{3}t^{3}x^{3}}{3!}e^{i\cdot0\cdot x}+...)f_{\nu}(x)dx=[/imath] [imath]=\int\limits _{x=-\infty}^{\infty}(e^{i\cdot0\cdot x}+itxe^{i\cdot0\cdot x}+\frac{i^{2}t^{2}x^{2}}{2!}e^{i\cdot0\cdot x}+\frac{i^{3}t^{3}x^{3}}{3!}e^{i\cdot0\cdot x}+...)f_{\xi}(\sqrt{n}x)\sqrt{n}dx=\int\limits _{y=-\infty}^{\infty}(e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+it\frac{y}{\sqrt{n}}e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+\frac{i^{2}t^{2}y^2}{2!\cdot n}e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+\frac{i^{3}t^{3}y^{3}}{3! \cdot n\sqrt{n}}e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+...)f_{\xi}(y)dy[/imath] And then into Fourier spectrum [imath]\hat{f}_{{S}_{\nu,n}}(t)[/imath] of p.d.s. of sum [imath]f_{S_{\nu,n}}(x)[/imath] of [imath]n[/imath] instances of [imath]\nu[/imath], which is n-th power of [imath]\varphi_{\nu}(t)[/imath]. Each [imath]t[/imath]-th harmonic of [imath]\hat{f}_{{S}_{\nu,n}}(t)[/imath] in the interval [imath]t=[-C,C][/imath] has amplitude, close to [imath]e^{-t^2/2}[/imath] (with allowed error [imath]\epsilon[/imath]) and that interval [imath]C(\epsilon, n)[/imath] grows infinitely with growing [imath]n[/imath]: To complete the proof, we need to show that starting from some [imath]n[/imath] amplitudes of harmonics outside that interval [imath]t=[-C,C][/imath] can be considered negligible and dropped without significantly affecting the calculation of Fourier synthesis [imath]f_{S_{\nu,n}}(x)[/imath]. In other words, this picture should not take place: I don't know hot to do that. I tried Parseval's identity for p.d.f. of [imath]\nu[/imath] and its characteristic function: [imath]\int\limits_{x=-\infty}^{\infty}f^2_\nu(x)dx = \int\limits_{t=-\infty}^{\infty}\hat{f}^2_\nu(t)dt[/imath]. As [imath]\sqrt{n}[/imath] grows larger, [imath]\nu[/imath] becomes more of a Dirac delta-function, thus its square integral gets infinitely large: At the same time, central part of the spectrum of [imath]\nu[/imath] is [imath]e^{-t^2/2n}[/imath], and its square integral on [imath][-\infty, \infty][/imath] converges to [imath]\int\limits_{-\infty}^{\infty}e^{-\frac{t^2}{n}}dt=\sqrt{\frac{n}{2}}[/imath]. Ok, this goes to infinity as n goes to infinity and allows remaining harmonics to have infinitesimal amplitudes, but doesn't guarantee that. Dead end. EDIT: As Yemon Choi suggested, Levy theorem proves that higher harmonics have infinitesimal amplitudes for large enough [imath]t[/imath] as [imath]n[/imath] grows. That's it. | 1329933 | Central Limit Theorem proof: Taylor series diverges for harmonics with higher number and those harmonics can't be neglected
I've read several proofs of Central Limit Theorem and they all seemed inaccurate to me, because they drop last members of Taylor series, whereas those members are not infinitesimal. The classical proof of Central Limit Theorem via characteristic functions is based on the fact that [imath]e^{-\frac{x^2}{2}}[/imath] is eigenfunction of Fourier transform. If [imath]\xi[/imath] is a random variable with [imath]E\xi=0[/imath] and [imath]Var(\xi)=1[/imath], we substitute it with another random variable [imath]\nu=\xi/\sqrt{n}[/imath]. We look into Fourier spectrum of its probability density function: [imath]\varphi_{\nu}(t)=\int\limits _{x=-\infty}^{\infty}e^{itx}f_{\nu}(x)dx=\int\limits _{x=-\infty}^{\infty}(e^{i\cdot0\cdot x}+itxe^{i\cdot0\cdot x}+\frac{i^{2}t^{2}x^{2}}{2!}e^{i\cdot0\cdot x}+\frac{i^{3}t^{3}x^{3}}{3!}e^{i\cdot0\cdot x}+...)f_{\nu}(x)dx=[/imath] [imath]=\int\limits _{x=-\infty}^{\infty}(e^{i\cdot0\cdot x}+itxe^{i\cdot0\cdot x}+\frac{i^{2}t^{2}x^{2}}{2!}e^{i\cdot0\cdot x}+\frac{i^{3}t^{3}x^{3}}{3!}e^{i\cdot0\cdot x}+...)f_{\xi}(\sqrt{n}x)\sqrt{n}dx=\int\limits _{y=-\infty}^{\infty}(e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+it\frac{y}{\sqrt{n}}e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+\frac{i^{2}t^{2}y^2}{2!\cdot n}e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+\frac{i^{3}t^{3}y^{3}}{3! \cdot n\sqrt{n}}e^{i\cdot0\cdot \frac{y}{\sqrt{n}}}+...)f_{\xi}(y)dy[/imath] And then into Fourier spectrum [imath]\hat{f}_{{S}_{\nu,n}}(t)[/imath] of p.d.s. of sum [imath]f_{S_{\nu,n}}(x)[/imath] of [imath]n[/imath] instances of [imath]\nu[/imath], which is n-th power of [imath]\varphi_{\nu}(t)[/imath]. Each [imath]t[/imath]-th harmonic of [imath]\hat{f}_{{S}_{\nu,n}}(t)[/imath] in the interval [imath]t=[-C,C][/imath] has amplitude, close to [imath]e^{-t^2/2}[/imath] (with allowed error [imath]\epsilon[/imath]) and that interval [imath]C(\epsilon, n)[/imath] grows infinitely with growing [imath]n[/imath]: To complete the proof, we need to show that starting from some [imath]n[/imath] amplitudes of harmonics outside that interval [imath]t=[-C,C][/imath] can be considered negligible and dropped without significantly affecting the calculation of Fourier synthesis [imath]f_{S_{\nu,n}}(x)[/imath]. In other words, this picture should not take place: I don't know hot to do that. I tried Parseval's identity for p.d.f. of [imath]\nu[/imath] and its characteristic function: [imath]\int\limits_{x=-\infty}^{\infty}f^2_\nu(x)dx = \int\limits_{t=-\infty}^{\infty}\hat{f}^2_\nu(t)dt[/imath]. As [imath]\sqrt{n}[/imath] grows larger, [imath]\nu[/imath] becomes more of a Dirac delta-function, thus its square integral gets infinitely large: At the same time, central part of the spectrum of [imath]\nu[/imath] is [imath]e^{-t^2/2n}[/imath], and its square integral on [imath][-\infty, \infty][/imath] converges to [imath]\int\limits_{-\infty}^{\infty}e^{-\frac{t^2}{n}}dt=\sqrt{\frac{n}{2}}[/imath]. Ok, this goes to infinity as n goes to infinity and allows remaining harmonics to have infinitesimal amplitudes, but doesn't guarantee that. Dead end. EDIT: As Yemon Choi suggested, Levy theorem proves that higher harmonics have infinitesimal amplitudes for large enough [imath]t[/imath] as [imath]n[/imath] grows. That's it. |
1331389 | Finding [imath]\sum\limits_{i=0}^{\infty}\frac{i}{4^i}[/imath]
I'm struggling with finding this sum: [imath]\sum\limits_{i=0}^{\infty}\frac{i}{4^i}[/imath] Any pointers would be greatly appreciated. I have not found any information of use elsewhere, but maybe I am using the wrong search terms. | 50919 | Calculate the sum of the infinite series [imath]\sum_{n=0}^{\infty} \frac{n}{4^n}[/imath]
A previous problem had us solving [imath]\sum_{n=0}^{\infty} \frac{1}{4^n}[/imath] which I calculated to be [imath]\frac{4}{3}[/imath] using a bit of mathematical manipulation. Wonderful. Thank you for all the prompt responses. Could anyone suggest an alternate technique that does not involve differentiation? |
594 | How do you prove that a prime is the sum of two squares iff it is congruent to 1 mod 4?
It is a theorem in elementary number theory that if [imath]p[/imath] is a prime and congruent to 1 mod 4, then it is the sum of two squares. Apparently there is a trick involving arithmetic in the gaussian integers that lets you prove this quickly. Can anyone explain it? | 2734861 | Prove that every prime [imath]p[/imath] of which [imath]-1[/imath] is a quadratic residue can be represented in the form [imath]x^2+y^2[/imath].
Prove that every prime [imath]p[/imath] of which [imath]-1[/imath] is a quadratic residue can be represented in the form [imath]x^2+y^2[/imath]. In a similar question the result is achieved by noting that every prime of the form [imath]p=4k+1[/imath] can be written as the sum of two squares and "it follows that -1 is a quadratic residue [imath]\pmod p[/imath] ". So, in my question, does it just suffice to show that [imath]p[/imath] is the sum of two squares if [imath]p \equiv 1 \pmod 4[/imath]? How should we proceed? |
1331444 | The generators of the group [imath]\langle\mathbb{Z}_n,\oplus\rangle[/imath] are all [imath]g \in \mathbb{Z}_n [/imath] for which [imath]\gcd(g,n)=1[/imath]
I'm trying to find a proof of this: The group [imath]\langle\mathbb{Z}_n,\oplus\rangle[/imath] is cyclic for every [imath]n[/imath], where [imath]1[/imath] is a generator. The generators of the group [imath]\langle\mathbb{Z}_n,\oplus\rangle[/imath] are all [imath]g \in \mathbb{Z}_n [/imath] for which [imath]\gcd(g,n)=1[/imath], as the reader can prove as an exercise. It is perfectly clear that [imath]1[/imath] generates all [imath]\mathbb{Z}_n[/imath], but I can't get myself to understand the second part or find a way to prove it. Thanks. | 786452 | How to find a generator of a cyclic group?
A cyclic group is a group that is generated by a single element. That means that there exists an element [imath]g[/imath], say, such that every other element of the group can be written as a power of [imath]g[/imath]. This element [imath]g[/imath] is the generator of the group. Is that a correct explanation for what a cyclic group and a generator are? How can we find the generator of a cyclic group and how can we say how many generators should there be? |
1331933 | Find x in degrees: (there souls be 2 answers from 0 degrees to 360 degrees) 5cos(x)=2sec(x)-3
[imath]2(1/\cos(x))-5\cos(x)-3=0[/imath] [imath](2/\cos(x))-5\cos(x)-3=0[/imath] [imath]2-5\cos(x)-3=\cos(x)[/imath] [imath]-5\cos(x)-1=\cos(x)[/imath] [imath]-6\cos(x)=1[/imath] [imath]\cos(x)=-1/6[/imath] [imath]x=99.6[/imath] Reference angle[imath]=80.4[/imath] [imath]180+80.4=260.4[/imath] [imath]x=99.6^o[/imath], [imath]260.4^o[/imath] I did this, but when I checked it, it didn't work. Is this correct? | 1331912 | Find [imath]x[/imath]: [imath]5\cos(x) = 2\sec(x)-3[/imath]
This was the last question on my exam and it's driving me nuts the answer has to be in degrees [imath]5\cos(x) = 2\sec(x)-3[/imath] [imath]5\cos^2(x)/2 = -3[/imath] [imath]\cos^2(x) = -6/5[/imath] Since [imath]-6/5[/imath] is negative The answer [imath]\sqrt{\cos^2(x)}[/imath] could not be correct |
1330632 | Integrate [imath]\frac{1}{\sqrt{4-x^2}}[/imath]
Evaluate [imath]\int\frac{1}{\sqrt{4-x^2}}dx[/imath] I had this question on my calc exam today, and I have no clue how it's done. I was trying to factor 4-x² to see if I could see any patterns but no luck. One thing I did notice was that [imath]\frac{d}{dx}(\arcsin(x)) =\frac{1}{\sqrt{1-x^2}} [/imath] | 596895 | Help with [imath]\int \frac 1{\sqrt{a^2 - x^2}} \mathrm dx[/imath]
[imath]\int \frac 1{\sqrt{a^2 - x^2}} \,dx[/imath] In the second passage it become [imath]\int \frac 1{a\sqrt{1 - \left(\frac xa\right)^2}}\,dx[/imath] So can someone explain me what kind operation is done at denominator? |
1331944 | Equivalence Relations OF sets
We define the relationship between the two sets [imath]S_1≡S_2[/imath] if and only if [imath]|S_1|=|S_2|[/imath]. How to show that [imath]≡[/imath] is an equivalence relation ? sorry I'm from Iran and Basic my English is poor. | 430493 | Is "have the same cardinality" a equivalence relation?
A relation is a subset of the Cartesian product of two sets, if "have the same cardinality", denoted as [imath]R[/imath], is a relation, then there must exist set [imath]A, B[/imath], such that [imath]R \subset A \times B[/imath]. What are [imath]A, B[/imath] then? They cannot be "set of all sets", because there is no such set according to axiom of regularity (ZFC set theory). Did I miss something? Thanks. |
1332082 | Proof related to matrix
Let [imath]A[/imath] and [imath]B[/imath] be [imath]n \times n[/imath] real matrices such that [imath]A^2 = I, B^2 = I[/imath] and [imath](AB)^2 = I[/imath]. Prove that [imath]AB = BA[/imath]. Someone help me with this problem | 388401 | If [imath]A^2 = I[/imath], [imath]B^2=I[/imath] and [imath](AB)^2=I[/imath], then [imath]AB = BA[/imath]
Matrix Question If [imath]A^2 = I[/imath], [imath]B^2=I[/imath] and [imath](AB)^2=I[/imath], then [imath]AB = BA[/imath] Basically, got up to [imath]A(BA-AB)B = 0[/imath] (by cancelling and equating terms from [imath]I^2 = I[/imath] and to [imath]A^2B^2 = A^2B^2[/imath] and using distributive laws), but that doesn't work out too well! Thanks for help in advance! |
1332282 | Combinatorial proof for [imath]\sum_{k = 0}^n \binom {r+k} k = \binom {r + n + 1} n[/imath]
I'm trying to figure out a combinatorial proof for: [imath]\displaystyle \sum_{k \mathop = 0}^n \binom {r+k} k = \binom {r + n + 1} n[/imath] I've tried the committee counting thing, but that didn't work. | 2209194 | Proving that [imath]\sum\limits_{k=0}^{n} {{m+k} \choose{k}} = { m+n+1 \choose n }[/imath]
I've been trying to prove this problem for awhile now, as preparation for a test, could anyone provide a solution I could follow? [imath]\sum\limits_{k=0}^{n} {{m+k} \choose{k}} = { m+n+1 \choose n }[/imath] for all nonnegative integers m and n. Any help will be appreciated. |
1331725 | Factor the polynomial
Factor the polynomial [imath]X^3-X+1[/imath] in [imath]F_{23}[/imath] and [imath]X^3+X+1[/imath] in [imath]F_{31}[/imath]. How can I know in which way to factor a polynomial mod [imath]p[/imath]? Is there some specified method to do that? Thanks. | 1073190 | How to factor polynomials in [imath]\mathbb{Z}_n[/imath]?
How to factor a certain polynomial over [imath]Zn[/imath]. for example factor the following polynomial into irreducible polynomials in [imath]Z5[/imath]: [imath]X^3+X^2+X-1[/imath] or factor the following polynomial into irreducible polynomials in [imath]Z2[/imath]: [imath]X^4+X+1[/imath] is there a certain method (algorithm) I can follow? Please help I'm stuck and i really need the help! thank you in advance!! |
1332465 | Convergence of a stochastic integral
Let [imath](B_t)[/imath] the standard Brownian Motion and [imath](H_t)[/imath] be an adapted continuous process. Show that [imath]\frac{1}{B_t}\int _0^tH_sdB_s [/imath] converge in probability. I guess that the limit is [imath]H_0[/imath] but I don't know how to prove this statement. How can we show that [imath]\mathbb{P}(|\frac{1}{B_t}\int _0^tH_s-H_0dB_s|>\epsilon )\to 0[/imath] when [imath] t\to 0 [/imath] ? I have tried to use Markov inequality then Cauchy Schartz inequality in order to make appear [imath]\mathbb{E}((\int H_s-H_0dB_s)^2)[/imath] but this didn't work because of the term [imath]\frac{1}{B_t}[/imath]. | 573000 | "Continuity" of stochastic integral wrt Brownian motion
I'd like to prove a nice property of a stochastic integral with respect to Brownian motion. Let [imath](H_t)_{t\geq0}[/imath] be a progressive and bounded process that is continuous at [imath]0[/imath] and [imath]B[/imath] a standard Brownian motion. Then [imath]\frac{1}{B_{\varepsilon}}\int_{0}^{\varepsilon}H_s\mathbb{d}B_s\rightarrow H_0[/imath] as [imath]\varepsilon\rightarrow0[/imath] in probability. Anyone got some hints ? I'm really puzzled and I don't know where to start. EDIT_2.0. Applying Ito-Isometry might be a bit tricky. This Brownian Motion in the denominator kinda troubles me :/ |
1332327 | A question on Hölder inequality
Let [imath]p, q > 1[/imath] such that [imath]\frac{1}{p} + \frac{1}{q} = 1[/imath]. Then [imath]|\sum\limits_{i = 1}^n x_i y_i| \leq ||x||_p ||x||_q, \;\; \forall x, y \in \mathbb{R}^n.[/imath] I have to prove it considering [imath]u = \frac{x}{||x||_p} \;\; \text{and} \;\; v = \frac{y}{||y||_q}[/imath] and using [imath]\bf{Young's}[/imath] inequality. Can someone, please, give me a hint? | 589791 | Proving Holder's inequality using Jensen's inequality
Let [imath]p[/imath] and [imath]q[/imath] be positive reals such that [imath]\frac{1}{p}+\frac{1}{q} = 1[/imath], so that [imath]p,q[/imath] in [imath](1,\infty)[/imath]. For [imath]\vec a[/imath] and [imath]\vec b \in \mathbb{R}^2[/imath] prove that [imath]|\vec a \cdot \vec b | \leq ||\vec a||_p|| \vec b||_q[/imath]. A hint was posted for using Jensen's inequality to use [imath]\phi(x) = ln(1 + e^x)[/imath]. But I don't know how I'd work that in. |
192938 | Probability Question: Would A always have a greater chance of [imath]A\cap B[/imath]?
My professor assigned a completely random question on our problem set. Basically, it goes: Sophie is 30 years old and majored in philosophy. As a student, she participated in anti-nuclear demonstrations and was deeply concerned with social justice. Which is more probable? 1. Sophie is a bank teller. 2. Sophie is a bank teller and is active in the feminist movement. I'm guessing the point the prof was trying to get across is that assuming you have two different events [imath]A[/imath] and [imath]B[/imath], [imath]A\cap B[/imath] would always have a smaller probability than either. Is this correct? | 954907 | The water heater problem ( mathematician or plumber)??
Isn't it absurd? [imath]\textbf{Problem-}[/imath] Suppose my water heater broke and heat in my apartment raised high. I went to a "person" to ask him to take a look at it, he came to my apartment, used a bunch of spare parts and then fixed it. I paid him for the repairs." Now what is more likely, [imath]1)[/imath] He is a mathematician, [imath]2)[/imath] He is a mathematician and plumber. Now look at this Image where, A (white) means he is both mathematician and plumber and B (yellow) means he is only a mathematician and not a plumber. Now probability A means that a mathematician plumber fixed my water heater, and probability B means a mathematician who is not a plumber fixed it, Now since A [imath]\leq[/imath] A+B, it is more likely that a mathematician fixed my water heater. Isn't it absurd. But if I denote B by plumbers who are not mathematicians, I get it is more likely a plumber fixed my water heater, so which one is more likely? |
1332581 | Roots of a 6-degree polynomial
Find the roots of the equation [imath]2000x^6+100x^5+10x^3+x-2=0.[/imath] I am struggling finding a root using rational root theorem. Even if I get a root, I have to find all the roots. Please don't use WolframAlpha, I am looking for an approach, etc. Thanks. | 651024 | Solve [imath]2000x^6+100x^5+10x^3+x-2=0[/imath]
One of the roots of the equation [imath]2000x^6+100x^5+10x^3+x-2=0[/imath] is of the form [imath]\frac{m+\sqrt{n}}r[/imath], where [imath]m[/imath] is a non-zero integer and [imath]n[/imath] and [imath]r[/imath] are relatively prime integers.Then the value of [imath]m+n+r[/imath] is? Tried to use the fact that another root will be [imath]\frac{m-\sqrt{n}}r[/imath] as coefficients are rational but there are six roots and using sum and product formulas would allow many variables in the equations. |
1330581 | Show that linear finite rank operators are open mappings
Suppose [imath]X[/imath] and [imath]Y[/imath] are topological vector spaces, [imath]dim(Y) < \infty[/imath], [imath]\Lambda : X \rightarrow Y[/imath] is linear, and [imath]\Lambda (X) = Y[/imath]. Prove that [imath]\Lambda[/imath] is an open mapping. Thanks in advance. Edit: Definition of TVS: Here, a Topological Vector Space is a vector space such that vector addition and scalar multiplication are continuous and also singletons are closed. | 425889 | Is this map to a finite dimensional topological vector space an open map?
Suppose [imath]X[/imath] and [imath]Y[/imath] are topological vector spaces, [imath]\dim (Y)<\infty[/imath], and [imath]f:X\to Y[/imath] is linear and surjective. Prove that [imath]f[/imath] is open and if the null space [imath]N[/imath] of [imath]f[/imath] is closed, then [imath]f[/imath] is also continuous. The second assertion is easier since the quotient space [imath]X/N[/imath] has the universal mapping property. |
1333272 | Continuous version of the Cantor-Schroder-Bernstein Theorem
Does the existence of continuous injections [imath]f: A\rightarrow B[/imath] and [imath]g: B\rightarrow A[/imath] imply the existence of a bicontinuous bijection between A and B (ie topological equivalence)? If not, what is a good counter example? | 985911 | Homeomorphism(topological spaces) version of Cantor–Bernstein–Schroeder theorem
Let [imath]A[/imath] , [imath]B[/imath] be topological spaces such that there for some subset [imath]D[/imath] of [imath]B[/imath] there is a homeomorphism form [imath]A[/imath] to [imath]D[/imath] and for some subset [imath]E[/imath] of [imath]A[/imath] there is a homeomorphism form [imath]B[/imath] to [imath]E[/imath] ; then must there exist a homeomorphism from [imath]A[/imath] to [imath]B[/imath] ? |
1333235 | What is [imath]\lim_{n \to \infty} n^3 a_n[/imath]?
[imath]a_n[/imath] is the Fourier coefficient of [imath]f(x) = \left(1 - \frac{|x|}{\pi}\right)^4[/imath] The answer is infinity, but can someone give an answer that doesn't require explicit computation of the [imath]a_n[/imath]? I'm looking for an explanation that applies convergence and divergence results, and maybe even growth and decay conditions of the coefficients, of the Fourier series. Thanks, Edit: Two things to note - 1) [imath]a_n[/imath] for [imath]f(x) = \left(1 - \frac{|x|}{\pi}\right)[/imath] decays like [imath]\frac{1}{n^2}[/imath]. Can I use this to say something about the decay of [imath]a_n[/imath] for [imath]f(x) = \left(1 - \frac{|x|}{\pi}\right)^4[/imath]? 2) [imath]f(x) = \left(1 - \frac{|x|}{\pi}\right)^4[/imath] is not differentiable at zero. | 1329705 | Finding the limit that involves Fourier coefficients,
Given the function [imath]f(x) = 1 - \dfrac{|x|}{\pi}[/imath], I had computed its Fourier coefficients, using integration by parts and got: [imath] a_n = \begin{cases} 0, & \text{for $n$ even}, \\[6pt] \dfrac{2}{n^2\pi^2}, & \text{for $n$ odd}, \\[6pt] \dfrac{1}{2}, & \text{for }n = 0. \end{cases} [/imath] So, I know the coefficients decay like [imath]\dfrac{1}{n^2}[/imath]. What if I now took [imath]f(x)[/imath] to be [imath]\left(1 - \dfrac{|x|}{\pi}\right)^4[/imath]. What is [imath]\lim\limits_{n\to\infty}[/imath] [imath]n^3a_n[/imath]? This is part 2 of a question that I am working on, and the question does not ask for a detailed proof but rather just the answer and the reasoning behind it. Thanks, Edit: Of course, if we iterate the integration by parts enough times, we can certainly compute the Fourier coefficients for the new function. But this is quite a bit of computation and I'm sure is not the point of the question - part 1 of the question already asked for explicit computation of the [imath]a_n[/imath]'s. I'm not sure how to proceed, without direct computation. Perhaps there are decay / growth conditions of the [imath]a_n[/imath]'s and convergence properties of the Fourier series that I am not aware of... |
1333603 | Quick way to solve simple probability questions like these?
Two dice are rolled. What is the chance of rolling at least ONE six? (The answer is [imath]\frac{11}{36}[/imath].) My solutions manual has a very elaborate table drawing method and I don't want to do that. I need something more quick. (I could have drawn the table in the time it took me to answer this question and so forth, but I'm keen on learning new methods!) | 598838 | If I roll two fair dice, the probability that I would get at least one 6 would be....
[imath]11[/imath] out of [imath]36[/imath]? I got this by writing down the number of possible outcomes ([imath]36[/imath]) and then counting how many of the pairs had a [imath]6[/imath] in them: [imath](1,6)[/imath], [imath](2,6)[/imath], [imath](3,6)[/imath], [imath](4,6)[/imath], [imath](5,6)[/imath], [imath](6,6)[/imath], [imath](6,5)[/imath], [imath](6,4)[/imath], [imath](6,3)[/imath], [imath](6,2)[/imath], [imath](6,1)[/imath]. Is this correct? |
1334099 | Dense subsets in tensor products of Banach spaces
Assume [imath]B_1[/imath] and [imath]B_2[/imath] are Banach spaces of univariate functions. Moreover, assume that the sets [imath]D_1 \subset B_1[/imath] and [imath]D_2 \subset B_2[/imath] are dense with respect to the respective norms [imath]\|\cdot\|_{B_1}[/imath] and [imath]\|\cdot\|_{B_2}[/imath]. Now we consider the algebraic tensor product [imath]B_1 \otimes_a B_2 := \{\sum_{i=1}^n f_i g_i \ :\ n \in \mathbb{N}, f_i \in B_1, g_i \in B_2\}[/imath] and define [imath]B := \overline{B_1 \otimes_a B_2}[/imath] as the completion with respect to some norm [imath]\|\cdot\|_B[/imath]. Now my question is: Is the tensor product [imath]D:=D_1 \otimes D_2[/imath] of the dense subsets also dense in [imath]B[/imath] with respect any choice of the norm [imath]\|\cdot\|_B[/imath]? Or does [imath]\|\cdot\|_B[/imath] have to fulfill certain conditions? | 1333190 | Dense subsets in tensor products of Banach spaces
Assume [imath]B_1[/imath] and [imath]B_2[/imath] are Banach spaces of univariate functions. Moreover, assume that the sets [imath]D_1 \subset B_1[/imath] and [imath]D_2 \subset B_2[/imath] are dense with respect to the respective norms [imath]\|\cdot\|_{B_1}[/imath] and [imath]\|\cdot\|_{B_2}[/imath]. Now we consider the algebraic tensor product [imath]B_1 \otimes_a B_2 := \{\sum_{i=1}^n f_i g_i \ :\ n \in \mathbb{N}, f_i \in B_1, g_i \in B_2\}[/imath] and define [imath]B := \overline{B_1 \otimes_a B_2}[/imath] as the completion with respect to some norm [imath]\|\cdot\|_B[/imath]. Now my question is: Is the tensor product [imath]D:=D_1 \otimes D_2[/imath] of the dense subsets also dense in [imath]B[/imath] with respect any choice of the norm [imath]\|\cdot\|_B[/imath]? Or does [imath]\|\cdot\|_B[/imath] have to fulfill certain conditions? I would argue: if [imath]\|\cdot\|_B[/imath] is a crossnorm, then [imath]D \subset B_1 \otimes_a B_2[/imath] is dense with respect to this crossnorm and since [imath]B_1 \otimes_a B_2[/imath] is dense in [imath]\overline{B_1 \otimes_a B_2}[/imath] the question holds true for all crossnorm? Is this correct? |
1334287 | Find all [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] for which [imath]f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))[/imath]
Problem Find all [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] which satisfy [imath]f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))[/imath] for all [imath]x,y\in\mathbb{R}[/imath]. This is a contest math problem, and I have very little experience with functional equations. Thus I sadly have no progress to show. Admittedly, I'm posting this just to get some input on good strategies and solutions. | 154301 | Strategies to find the set of functions [imath]f:\mathbb{R}\to\mathbb{R}[/imath] which satisfy a given functional equation
My question is as follows: What methods can be used to find the set of functions [imath]f:\mathbb{R}\to\mathbb{R}[/imath] satisfying a certain functional equation. An example of a case where this applies is the following: Find all functions [imath]f:\mathbb{R}\to\mathbb{R}[/imath] which satisfy the following equation: [imath]f(x^{3})+f(y^{3})=(x+y)(f(x^{2})+f(y^{2})-f(xy)):\forall x, y\in\mathbb{R}[/imath] I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis. Thanks in advance. |
1333233 | In OLS is the vector of residuals always 0?
I am trying to show that [imath]\sum_{i=1}^ne_i = 0[/imath] I have two hints, so to speak: [imath] HX = X[/imath] where [imath]H[/imath] is the hat matrix, and that [imath]\sum_{i=1}^ne_i = e'1[/imath] My solution is as follows: [imath]e'1 = Y'(I-H)1=[(X\beta)' - (X\beta)'H]1=(\beta'X' - \beta'X'X(X'X)^{-1}X')1 [/imath] [imath]=(\beta'X' - \beta'X')1 = 01 = 0 [/imath] Seems simple enough but this implies that [imath]e'[/imath] and by extension [imath]e[/imath] is always a vector of zeros, which seems counter-intuitive. | 1334282 | How do I show that the sum of residuals of OLS are always zero using matrices
I am trying to show that [imath]\sum_{i=1}^ne_i = 0[/imath] using matrices (or vectors). I have two hints, so to speak: [imath] HX = X[/imath] where [imath]H[/imath] is the hat matrix, and that [imath]\sum_{i=1}^ne_i = e'1[/imath] My previous solution, In OLS is the vector of residuals always 0?, is wrong since I expanded [imath]Y =X\beta[/imath], leaving out the error term. If I include it I end up with a tautology. |
1334443 | Showing that the product [imath]x*y := \frac{x+y}{xy+1}[/imath] is a group operation on [imath](-1, 1)[/imath]
I need to show that the following is an abelian group: [imath]x*y = \frac{x+y}{xy+1}[/imath] on the set [imath]\{x \in \Bbb R \,|\, -1 < x < 1\}[/imath]. I have been working on this problem, trying to show closure. I know that we need to show that [imath]|x+y|<|xy+1|[/imath] for all [imath]x, y \in (-1,1)[/imath]. Can I assume that the max value that the expression yields is [imath]1[/imath] if we take [imath]x=1[/imath] and [imath]y=1[/imath]? And the lowest value that is possible is when [imath]x=-1[/imath] and [imath]y=1[/imath]? Or am I going about this the wrong way? | 1328260 | Prove that [imath]x * y = \frac{x+y}{1+xy}[/imath] is a stable part of [imath]G=(-1, 1)[/imath]
I have to prove that the result of [imath]x * y \in G[/imath] so [imath]\frac{x+y}{1+xy} \in (-1, 1)[/imath]. So [imath]x > -1[/imath] and [imath]y > -1[/imath] at the same time [imath]x < 1[/imath] and [imath]y < 1[/imath]. If I multiply the first 2 expressions I obtain [imath]xy < 1[/imath] which is true, and if I sum up the last 2 I get [imath]x + y < 2[/imath]. At this point I am not sure if I am doing well. How should I continue this problem? |
1334687 | Span of Polynomials in [imath]\mathcal{C}(\mathbb{R})[/imath]
Let [imath]\mathcal{P}=\{1, x, x^2, x^3 \ldots\}[/imath]. Then pick out the correct statements. A) Span[imath]\mathcal{P}=\mathcal{C}(\mathbb{R})[/imath] B) Span[imath]\mathcal{P}[/imath] is a subspace of [imath]\mathcal{C}(\mathbb{R})[/imath] C) [imath]\{1, x, x^2, x^3 \ldots\}[/imath] is a linearly independent set. D) Trigonometric functions belong to Span[imath]\mathcal{P}[/imath]. C is obviously correct. The thing is need clarification is this. Is there a difference between [imath]Span\mathcal{P}[/imath] and closure of [imath]Span\mathcal{P}[/imath]. So my answer is only B and C are correct Is that alright? | 1334796 | [imath]\operatorname{span}(x^0, x^1, x^2,\cdots)[/imath] and the vector space of all real valued continuous functions on [imath]\Bbb R[/imath]
Let [imath]p_n(x)=x^n[/imath] for [imath]x\in\Bbb R[/imath] and let [imath]\mathcal P=span\{p_0,p_1,p_2,\cdots\}[/imath] . Then [imath]\mathcal P[/imath] is the vector space of all real valued continuous functions on [imath]\Bbb R[/imath]. [imath]\mathcal P[/imath] is the subspace of all real valued continuous functions on [imath]\Bbb R[/imath]. [imath]\{p_0,p_1,p_2,\cdots\}[/imath] is a linearly independent set in the vector space of all continuous functions on [imath]\Bbb R[/imath]. Trigonometric functions belong to [imath]\mathcal P[/imath] Not 1 since sinc function is real valued and continuous, and not 4, too. Can I conclude 2,3 are the correct ones? |
1335347 | Dirac Delta function inverse Fourier transform
We know that the Fourier transform of the Dirac Delta function is defined as [imath]\int_{-\infty}^{\infty} \delta(t) e^{-j\omega t} dt = 1,[/imath] and if I were to reconstruct the function back in time domain, the inverse Fourier transform is defined as [imath]\delta(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{j\omega t} d\omega.[/imath] How do compute I this integral analytically? | 1335309 | Dirac Delta function inverse Fourier transform
We know that the Fourier transform of the Dirac Delta function is defined as [imath]\int_{-\infty}^{\infty} \delta(t) e^{-i\omega t} dt = 1,[/imath] and if I were to reconstruct the function back in time domain, the inverse Fourier transform is defined as [imath]\delta(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega t} d\omega.[/imath] How do I compute this integral analytically? |
1334187 | Show that [imath]X-C[/imath] is connected
Let [imath]X[/imath] be a connected metric space, let [imath]A \subseteq X[/imath] be a connected set. Let [imath]\mathcal{C}[/imath] be a connected component of [imath]X-A[/imath]. Show that [imath]X-\mathcal{C}[/imath] is connected. Ok, so Im been dealing with this exercise for a while, and I am completely stucked. I've managed the following (which I don't know if it is even true.) Since [imath]X-A \subseteq \mathcal{C}[/imath] we have that [imath]X-\mathcal{C} \subseteq A[/imath]. Now since [imath]A[/imath] is a connected set, and [imath]\mathcal{C}[/imath] is closed (since it's a connected component), one way to prove the statement is to see that [imath]X-\mathcal{C}[/imath] is closed in [imath]A[/imath] because in this case, [imath]X-\mathcal{C}[/imath] would be open and closed in [imath]A[/imath], connected which implies [imath]A=X-\mathcal{C}[/imath] and therefore [imath]X-\mathcal{C}[/imath] is connected. However Im not being able to show that [imath]X-\mathcal{C}[/imath] is closed in [imath]A[/imath], (im not sure if in fact it is). If not, If its possible, I would like a hint, not a complete solution! | 157175 | If [imath]X[/imath] is a connected subset of a connected space [imath]M[/imath] then the complement of a component of [imath]M \setminus X[/imath] is connected
I have an exercise found on a list but I didn't know how to proceed. Please, any tips? Let [imath]X[/imath] be a connected subset of a connected metric space [imath]M[/imath]. Show that for each connected component [imath]C[/imath] of [imath]M\setminus X[/imath] that [imath]M\setminus C[/imath] is connected. |
1302635 | A finite group which has a unique subgroup of order [imath]d[/imath] for each [imath]d\mid n[/imath].
Problem Suppose G is a finite group of order [imath]n[/imath] which has a unique subgroup of order [imath]d[/imath] for each [imath]d\mid n[/imath]. Prove that [imath]G[/imath] must be a cyclic group. My idea: I try to prove it by induction. Let [imath]p|n[/imath] be a prime. Then by condition, there exists a unique subgroup [imath]H[/imath] of order [imath]n/p[/imath]. Since [imath]|gHg^{-1}|=|H|[/imath], we must have [imath]gHg^{-1}=H[/imath] by the uniqueness part of the condition. So, H is a normal subgroup. Now, [imath]|G/H|=p[/imath] and thus [imath]G/H=\langle x\rangle[/imath] where [imath]x^p \in H[/imath]. However, I cannot continue. My another idea is that first consider the case when [imath]|G|[/imath] is a power of some prime [imath]p[/imath]. But, it still doesn't work. | 1387863 | For finite group [imath]G[/imath] of order [imath]n[/imath] such that there exists a unique subgroup of order equal to each positive divisor of [imath]n[/imath]. Is it necessarily cyclic?
Let [imath]G[/imath] be a finite group of order [imath]n[/imath]. Suppose [imath]k[/imath] is a positive integer such that [imath]k|n[/imath], then there exists a unique subgroup of order [imath]k[/imath]. Is it necessarily cyclic? If [imath]G[/imath] is Abelian, then I can prove it, but if [imath]G[/imath] is nonAbelian, then it is difficult to me to prove or disprove. |
1335662 | If log8n=1/2p, log22n=q, and q-p=4, find n
I'm having a hard time finding the value of [imath]a[/imath] in this problem. My teacher was trying to explain to me the process in which to get it but I did not understand him. | 1335479 | Can anybody help me with this logarithm problem?
If [imath](\log_4 x)^2= (\log_2 x)(\log_a x)[/imath] (the [imath]4[/imath] is the little number next to log by the way) , find the value of [imath]a[/imath]. |
1335941 | Why not take the tensor product of two left modules in this way?
Let [imath]A,B[/imath] be two left [imath]R[/imath]-modules. I was wondering if we then can form the tensor product of [imath]A[/imath] and [imath]B[/imath] by the free abelian group on [imath]A \times B[/imath] divided out by the span of the following elements [imath](a+a',b) -(a,b) - (a',b), (a,b+b')-(a,b)-(a,b')[/imath] and [imath](ra,b)-(a,rb)[/imath]. In literature the construction is done for [imath]A[/imath] a right [imath]R[/imath] module and [imath]B[/imath] a left [imath]R[/imath] module. I don't see why we should distinguish between those two types. | 117773 | For [imath]M\otimes_R N[/imath], why is it so imperative that [imath]M[/imath] be a right [imath]R[/imath]-module and [imath]N[/imath] a left [imath]R[/imath]-module?
I was reading about the construction of a tensor product of any right [imath]R[/imath]-module [imath]M[/imath] and left [imath]R[/imath]-module [imath]N[/imath] over a ring [imath]R[/imath]. Why is it required that [imath]M[/imath] and [imath]N[/imath] be right and left modules, respectively? How does the construction not work otherwise? |
1334200 | All simple modules are projective [imath]\Rightarrow[/imath] semisimple
Let [imath]A[/imath] be a finite dimensional algebra over a field [imath]K[/imath]. It is clear that if [imath]A[/imath] is semisimple, then every simple module is projective. Does the converse hold ? It seems false, but I can't find a counterexample. A non-semisimple algebra with this property must have a non-projective indecomposable module, but that's as far as I could go. | 1172018 | Semisimplicity is equivalent to each simple left module is projective?
As it is well-known, a ring with unity [imath]R[/imath] is semisimple if and only if each left [imath]R[/imath]-module is projective. My question: Is semisimplicity of [imath]R[/imath] equivalent to each simple left [imath]R[/imath]-module being projective? I think it is true for a finite dimensional algebra [imath]R[/imath]. But, in general, I could not reach any conclusion. Any help would be thanked! |
1336192 | Fine the value of [imath]P(n+1)[/imath] given values of [imath]P[/imath] from 1 to [imath]n[/imath]
[imath]P(x)[/imath] is a polynomial of degree [imath]n[/imath] that satisfies [imath]P(k)=\frac{k}{k+1}[/imath] for [imath]k=0,1,2,3,...,n[/imath]. Find [imath]P(n+1)[/imath]. What have I tried: I have literally no idea how to do questions of this kind. Also, in general what is the method solving such questions where they say a polynomial behaves a certain way with a range of numbers and expect you to extrapolate the next value? Thank you! | 636919 | Suppose that [imath]P(x)[/imath] is a polynomial of degree [imath]n[/imath] such that [imath]P(k)=\dfrac{k}{k+1}[/imath] for [imath]k=0,1,\ldots,n[/imath]. Find the value of [imath]P(n+1)[/imath]
Suppose that [imath]P(x)[/imath] is a polynomial of degree [imath]n[/imath] such that [imath]P(k)=\dfrac{k}{k+1}[/imath] for [imath]k=0,1,\ldots,n[/imath]. Find the value of [imath]P(n+1)[/imath]. I could not relate this question with this one How to find [imath]P(n+1)[/imath], given [imath]P(x)[/imath] for [imath]x = 0,1,\ldots,n[/imath]?, so maybe it is not a duplicate I have absolutely no idea, how the function became a polynomial and how to approach it. Please help! |
1336487 | question on self adjoint operator
Suppose [imath]A[/imath] is a [imath]n\times n[/imath] matrix with complex entries and [imath]A^*A=A^2[/imath]. Does it imply [imath]A=A^*.[/imath] | 1302436 | [imath]A^2=A^*A[/imath]. Why is matrix [imath]A[/imath] Hermitian?
Let [imath]A[/imath] be [imath]n \times n[/imath] matrix and [imath]A^2=A^*A[/imath]. Why is [imath]A[/imath] a Hermitian matrix? |
1336601 | Does the minimum of two decreasing divergent series diverge?
If [imath]\displaystyle \sum_{n=1}^{\infty}a_n[/imath] and [imath]\displaystyle \sum_{n=1}^{\infty}b_n[/imath] are both divergent series with [imath]a_n\downarrow0[/imath] and [imath]b_n\downarrow0[/imath], [so [imath](a_n)[/imath] and [imath](b_n)[/imath] are decreasing sequences which converge to 0], and if [imath]c_n=\min\{a_n,b_n\},\;\;[/imath] does the series [imath]\displaystyle \sum_{n=1}^{\infty}c_n[/imath] necessarily diverge? (I was led to ask this question after reading this question and math110's solution to it: [imath]a_n\downarrow 0, \sum\limits_{n=1}^{\infty}a_n=+\infty, b_n=min\{a_n,1/n\}[/imath], prove [imath]\sum b_n [/imath] diverges..) | 12986 | find examples of two series [imath]\sum a_n[/imath] and [imath]\sum b_n[/imath] both of which diverge but for which [imath]\sum \min(a_n, b_n)[/imath] converges
Find examples of two series [imath]\sum a_n[/imath] and [imath]\sum b_n[/imath] both of which diverge but for which [imath]\sum \min(a_n, b_n)[/imath] converges. To make it more challenging, produce examples where [imath]a_n[/imath] and [imath]b_n[/imath] are positive and decreasing. Edit: This problem is taken verbatim from Exercise 2.7.11 on page 68 of Abbott's Understanding analysis. |
1336730 | Evaluate: [imath]\int_0^1 \frac{1-x}{(1+x)\log(x)}\, dx[/imath]
Evaluate: [imath]\int_0^1 \frac{1-x}{(1+x)\log(x)}\, dx[/imath] Wolfram alpha gives an answer of [imath]\log(2/\pi)[/imath], but I have been unable to prove this. Help appreciated. | 1332600 | Finding [imath]\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}[/imath]
The original question was to evaluate: [imath]\int_{0}^{1} \frac{1-x}{(1+x) \ln x}\,dx[/imath] Using the substitution [imath]x=\tan^2 \theta [/imath], I simplified it down to the integral [imath]\int_{0}^{\frac{\pi}{4}} \frac{\tan \theta - \tan^3 \theta}{\ln \tan \theta}[/imath]. From here, I am stuck and am not sure where to continue. |
1336818 | Calculate the sum [imath]\sum_{n=0}^\infty\frac{1}{(4n)!}[/imath]
How to determine the sum [imath]\sum_{n=0}^\infty\frac{1}{(4n)!}[/imath] ? Do I need to somehow convert (4n)! to (2n)! or in tasks like this, should I get the (4n)! after some multiplying? Thank you all for your time! | 221519 | Identify infinite sum: [imath]\sum\limits_{n=0}^{+\infty}\frac{x^{4n}}{(4n)!}[/imath]
Find [imath]f(x)[/imath], the unknown function satisfying [imath]f(x) = \sum\limits_{n=0}^{+\infty}\frac{x^{4n}}{(4n)!}[/imath] I'm looking for a direct solution which is different from mine, if possible. |
1337065 | Integral closure of [imath]\mathbb{Z}[/imath] in [imath]\mathbb{C}[/imath] is not finitely generated as a [imath]\mathbb{Z}[/imath]-module
Let [imath] \mathbb{Z}^{'}_{\mathbb{C}}[/imath] be the integral closure of [imath] \mathbb{Z} [/imath] in [imath] \mathbb{C} [/imath], i.e. [imath] \mathbb{Z}^{'}_{\mathbb{C}}[/imath] contains all complex roots of polynomials with integer coefficients and [imath]1[/imath] as the leading coefficient. Prove that [imath] \mathbb{Z}^{'}_{\mathbb{C}} [/imath] is not finitely generated as a [imath] \mathbb{Z} [/imath]-module. My attempt: Suppose [imath] \mathbb{Z}^{'}_{\mathbb{C}}=\mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} [/imath] for some [imath] \{ \alpha_{i}\}_{i=\overline{1,n}} \subset \mathbb{Z}^{'}_{\mathbb{C}} [/imath]. Let's take [imath] \alpha , \beta \in \mathbb{Z}^{'}_{\mathbb{C}} [/imath] with minimal polynomials of degrees [imath] m [/imath] and [imath] n [/imath] respectively. We can construct a monic polynomial with integral coefficients [imath] P(z)=\prod(z-(\alpha_{i}+\beta_{j})) [/imath] where [imath] \alpha_{i},\beta_{j} [/imath] are the conjugates of [imath]\alpha [/imath] , [imath]\beta [/imath] and therefore the minimal polynomial of [imath] \alpha + \beta [/imath] has degree [imath] \leq mn [/imath]. I was thinking that the degrees of the minimal polynomials of the elements of [imath] \mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} [/imath] could be somehow bounded by a certain constant, whereas [imath] \mathbb{Z}^{'}_{\mathbb{C}}[/imath] contains elements whose degrees of the minimal polynomials are arbitrarily large. I think this works when we are asked to prove that the set of all algebraic numbers [imath] \overline{\mathbb{Q}} [/imath] is not a finitely dimensional [imath]\mathbb{Q}[/imath]-vector space, but I don't know how to proceed in my case. I would really appreciate any ideas, hints or solutions. Thank you very much! | 1336994 | Integral closure of [imath]\mathbb{Z}[/imath] in [imath]\mathbb{C}[/imath] is not finitely generated as a [imath]\mathbb{Z}[/imath]-module?
Let [imath] \mathbb{Z}^{'}_{\mathbb{C}}=\{ z \in \mathbb{C} | \exists f \in \mathbb{Z}[X] \text{ monic such that } f(z)=0\} [/imath] be the integral closure of [imath] \mathbb{Z} [/imath] in [imath] \mathbb{C} [/imath]. Prove that [imath] \mathbb{Z}^{'}_{\mathbb{C}} [/imath] is not finitely generated as a [imath] \mathbb{Z} [/imath]-module. My attempt: Suppose [imath] \mathbb{Z}^{'}_{\mathbb{C}}=\mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} [/imath] for some [imath] \{ \alpha_{i}\}_{i=\overline{1,n}} \subset \mathbb{Z}^{'}_{\mathbb{C}} [/imath]. I know that given [imath] \alpha , \beta \in \mathbb{Z}^{'}_{\mathbb{C}} [/imath] with minimal polynomials of degrees [imath] m [/imath] and [imath] n [/imath] respectively, then we can construct a monic polynomial with integral coefficients [imath] P(z)=\prod(z-(\alpha_{i}+\beta_{j})) [/imath] where [imath] \alpha_{i},\beta_{j} [/imath] are the conjugates of [imath]\alpha [/imath] , [imath]\beta [/imath] and therefore the minimal polynomial of [imath] \alpha + \beta [/imath] has degree [imath] \leq mn [/imath]. I was thinking that then the degrees of the minimal polynomials of the elements of [imath] \mathbb{Z}\alpha_{1}+...\mathbb{Z}\alpha_{n} [/imath] could be somehow bounded by a certain constant, whereas [imath] \mathbb{Z}^{'}_{\mathbb{C}}[/imath] contains elements whose degrees of the minimal polynomials are arbitrarily large, take [imath] 2^{\frac{1}{n}} [/imath] for example,[imath] n\geq 2[/imath]. I think this works when we are asked to prove that the set of all algebraic numbers [imath] \overline{\mathbb{Q}} [/imath] is not a finitely dimensional [imath]\mathbb{Q}[/imath]-vector space, but I don't know how to proceed in my case. I would really appreciate any ideas, hints or solutions. Thank you very much! |
990953 | Multiplying and adding fractions
Why multiplying fractions is equal to multiply the tops, multiply the bottoms? [imath]\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b \times d},[/imath] And why [imath]\frac{a}{b}\times \frac{c}{c}=\frac{a}{b},[/imath] Also why [imath]\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}.[/imath] I understand it, but I want a mathematical approach as a math student proves it. Also I want to know the mathematics topic of this question (number theory, logic, etc). A full answer is not necessary. Just a reference. | 2374833 | Why are fractions "multiplied across"?
Suppose we have two rationals, [imath]\frac{a}b[/imath] and [imath]\frac{c}d[/imath]. I daresay anyone with some form of mathematical education would disagree that our result would be [imath]\frac{a}b\times \frac{c}d=\frac{ac}{bd}[/imath]. That is, we multiply our numerators to get the result numerator and we multiply the denominator to get the result denominator. However, mathematically speaking, why do we do this? If we consider the canonical definition of multiplication (i.e. repeated addition) in the case where we have two integers, say [imath]3[/imath] and [imath]4[/imath], we would come up with [imath](3\times 4) = (4 + 4 + 4)[/imath]. If this is so (and hopefully it is) what does it mean mathematically to add [imath]\frac{c}d[/imath] to itself [imath]\frac{a}b[/imath] times? After doing a bit of research, all I've managed to come up with is this image. Unfortunately, I don't see how this has anything to do with multiplication and seems to me like little more than a tool for teaching grade schoolers. This also led me to think about how we multiply decimal numbers. For example, [imath](0.2) \times (0.4) = 0.08[/imath] (i.e. [imath]2\times 4[/imath] with the decimal moved over a number of places equal to the number significant digits past the decimal). My intuition tells me that the answer to one of these will provide the answer to the other. Lastly, excuse me if this question seems silly, but I've pondered it quite a bit and can't come up with anything mathematically rigerous. EDIT: I changed all literal values to be arbitrary values (as they should have been originally). EDIT 2: Please note that I am PERFECTLY CAPABLE of multiplying fractions together in any way shape or form. This is NOT a post about how to multiply fractions. This post is about fundamentally understanding what it means to multiply a pair of fractions together. |
1206639 | Understanding simple transcendental field extensions
I would like to understand the definition of a simple transcendental extension and the theorem that states all such extensions are isomorphic. So for example, if [imath]K \subseteq \mathbb{C} [/imath] is any subfield and [imath]L:K[/imath] a field extension with [imath]\alpha \in L [/imath] then [imath]K(\alpha) \cong K(X)[/imath], where [imath]K(X)[/imath] is the field of rational functions in the indeterminate [imath]X[/imath]. First, what makes a field extension transcendental? And second, what is the explicit isomorphism between [imath]K(\alpha)[/imath] and [imath]K(X)[/imath]? I've heard this isomorphism called the "evaluation homomorphism" so could you please explain what this map is doing specifically? My understanding of this theorem is that in this context [imath]\alpha[/imath] (like [imath]X[/imath]) is purely a mathematical symbol, and is no longer considered as the root of some polynomial over [imath]K[/imath]. Thus, it is as if we are writing rational expressions in [imath]X[/imath] as those in [imath]\alpha[/imath]. However, this interpretation seems banal to me (why even write it in [imath]\alpha[/imath] if we have it in [imath]X[/imath]?) and what exactly makes [imath]K(X)[/imath] transcendental in the first place? | 1337221 | Transcendental Extensions. [imath]F(\alpha)[/imath] isomorphic to [imath]F(x)[/imath]
Let [imath]E[/imath] be an extension field of [imath]F[/imath] and [imath]\alpha \in E[/imath]. Then [imath]\alpha[/imath] is transcendental over [imath]F[/imath] if and only if [imath]F(\alpha)[/imath] is isomorphic to [imath]F(x)[/imath], the field of fractions of [imath]F[x][/imath]. This was a theorem in an abstract algebra textbook with a very brief proof. Can someone please explain why this theorem holds? I'm having difficulty grasping the concepts at hand. Thanks! |
1336964 | Why is is [imath]K(\alpha,\beta)/K(\alpha)[/imath] algebraic if [imath]K(\alpha,\beta)/K(\beta)[/imath] is algebraic?
Let [imath]K[/imath] be a field, and let [imath]\alpha[/imath] be transcendental over [imath]K[/imath] and algebraic over [imath]K(\beta)[/imath]. We have a Hasse diagram of field extensions diag http://i.picresize.com/images/2015/06/23/nHsFx.jpg Now, by reduction to absurdity [imath]\beta[/imath] must be transcendental over [imath]K[/imath] as well (this is asked but I'm not sure if it's relevant to the question). The problem is to prove that [imath]\beta[/imath] must be algebraic over [imath]K(\alpha)[/imath]. The first thing I tried is to take whatever polynomial annihilates [imath]\alpha[/imath]: [imath]a_0 + a_1\alpha\cdots + a_n\alpha^n = 0[/imath] with [imath]a_0,\dots,a_n\in K(\beta)[/imath]. If [imath]\beta[/imath] were algebraic over [imath]K[/imath], then I could rewrite this as [imath]\sum\nolimits_1^m\lambda_{i,0}\beta^i + \left(\sum\nolimits_1^m\lambda_{i,1}\beta^i\right)\alpha + \cdots + \left(\sum\nolimits_1^m\lambda_{i,n}\beta^i\right)\alpha^n[/imath] And this would of course give a polynomial that annihilates [imath]\beta[/imath] with coefficients in [imath]K(\alpha)[/imath] (aside from the fact that [imath]K(\alpha,\beta)/K(\alpha)[/imath] would be algebraic by mulitplicity of degree). The thing is, when it comes to transcendental extensions, I'm just not sure that [imath]\left\{\beta^k\right\}_{k\in\Bbb N}[/imath] is still a [imath]K[/imath]-basis for [imath]K(\beta)[/imath]. If this were true then I suppose my question is how to prove this. If not, what other method might I try? I also wonder whether or not this is a special case of [imath]E,L\supset K \,\wedge\, F/E \text{ algebraic }\stackrel ?\implies F/L\text{ algebraic }[/imath] My guess is no, but I also can't think of an counterexample because I only know of simple transcendental extensions (where it would be true by what I originally wanted to prove). | 730558 | Field extensions and algebraic/transcendental elements
Let [imath]E[/imath] be an extension of field [imath]F[/imath], and let [imath]\alpha, \beta \in E[/imath]. Suppose [imath]\alpha[/imath] is transcendental over [imath]F[/imath] but algebraic over [imath]F(\beta)[/imath]. Show that [imath]\beta[/imath] is algebraic over [imath]F(\alpha)[/imath]. Okay, first questions: What does the notation [imath]F(\alpha)[/imath] and [imath]F(\beta)[/imath] mean? And being transcendental means it solves no equations with rational coefficients, but what does it mean for a field? |
1337399 | Show that the order of [imath]a\times b[/imath] is equal to [imath]nm[/imath] if gcd(n,m)=1
[imath](G,*)[/imath] is an abelian group and [imath]a,b[/imath] are elements of G. Let [imath]n=ord(a)[/imath] and [imath]m=ord(b)[/imath] in [imath]G[/imath]. Show that the order of [imath]a*b[/imath] is equal to [imath]nm[/imath] if [imath]gcd(n,m)=1[/imath]. I have already proved that [imath]ord(a*b)|nm[/imath]. But the other part is a problem. Thanks | 424097 | Order of products of elements in a finite Abelian group
We want to show that if [imath]a,b\in G[/imath] where [imath]G[/imath] is a finite Abelian group, we have [imath]\operatorname{LCM}(|a|,|b|) = |ab|[/imath] given that [imath]ab \neq e[/imath]. How I approached this question was by saying let [imath]\operatorname{LCM}(|a|,|b|) = L[/imath]. Then if we show that [imath]L[/imath] divides [imath]|ab|[/imath] and [imath]|ab|[/imath] divides [imath]L[/imath] then [imath]|ab| = L[/imath]. Showing that [imath]|ab|[/imath] divides [imath]L[/imath] was fine. But then I am having trouble with the second part that is showing [imath]L[/imath] divides [imath]|ab|[/imath]. For this, so far I have: Consider [imath](ab)^{|ab|} = e = a^{|ab|}b^{|ab|} [/imath] since we know [imath]a \neq b^{-1}[/imath] by assumption then we can conclude that [imath]|a|[/imath] divides [imath]|ab|[/imath] and [imath]|b|[/imath] divides [imath]|ab|[/imath]. We know that [imath]L = \frac{|a||b|}{\operatorname{gcd}(|a|,|b|)}[/imath] from here can we conclude that [imath]L[/imath] divides [imath]|ab|[/imath]? |
1112536 | Spectral Measures: Unitary Map
This thread is a record. Given a Hilbert space [imath]\mathcal{H}[/imath]. Consider a normal operator: [imath]N:\mathcal{D}\to\mathcal{H}:\quad N^*N=NN^*[/imath] and its spectral measure: [imath]E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})[/imath] Regard a unitary transformation: [imath]U:\mathcal{H}\to\mathcal{K}:\quad U^*=U^{-1}[/imath] How to check that the transformed becomes: [imath]M:=UNU^{-1}=\int_\mathbb{C}\lambda\mathrm{d}UEU^{-1}(\lambda)=:\int_\mathbb{C}\lambda\mathrm{d}F(\lambda)[/imath] | 1112531 | Spectral Measures: Embedding
This thread is just a note! Given a Hilbert space [imath]\mathcal{H}[/imath]. Consider a normal operator: [imath]N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*[/imath] And its spectral measure: [imath]E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int_\Omega\lambda\mathrm{d}E(\lambda)[/imath] Regard an embedding: [imath]J\in\mathcal{B}(\mathcal{H}_0,\mathcal{H}):\quad J^*J=1_0[/imath] Suppose it reduces: [imath]P:=JJ^*:\quad PN\subseteq NP[/imath] Then one has: [imath]N_0:=J^*NJ:\quad N_0^*N_0=N_0N_0^*[/imath] And it holds: [imath]E_0:=J^*EJ:\quad N_0=\int_\Omega\lambda\mathrm{d}E_0(\lambda)[/imath] How to prove this from scratch? |
1337444 | A question on vector subspace
Let [imath]V[/imath] be the vector space of all functions [imath]f \colon \mathbb{R} \to \mathbb{R}[/imath] over [imath]\mathbb{R}[/imath], is the set of functions which are continuous a subspace? I think if you add functions which are continuous, the resultant function should also be continuous, but I'm not sure about it | 154386 | Subspaces of all real-valued continuous functions on [imath]\mathbb{R}^1[/imath]
I'll go ahead and give you the problem first, and then explain my trouble with it. Which of the following subsets are subspaces of the vector space C(-[imath]\infty[/imath],[imath]\infty[/imath]) defined as follows: Let V be the set of all real-valued continuous functions defnined on [imath]\mathbb{R}^1[/imath]. If [imath]f[/imath] and [imath]g[/imath] are in [imath]V[/imath], we define [imath](f + g)(t) = f(t) + g(t)[/imath]. If [imath]f[/imath] is in [imath]V[/imath] and [imath]c[/imath] is a scalar, we define [imath]c[/imath] · [imath]f[/imath] by [imath](c · f)(t) = cf(t)[/imath]. Then [imath]V[/imath] is a vector space, which is donatedby C(-[imath]\infty[/imath],[imath]\infty[/imath]). (a) All nonnegative functions (b) All constant functions There are more, but I have a feeling that after the (a) I'll be able to get the rest myself. Anyay, my problem is that the problem says it required calculus, and I don't understand how calculus even comes into it. As far as I can tell, (b) would be a subspace (because it's are closed under operations of V), but (a) would not because V could make a negative value positive. I may be completely wrong though; any suggestions? |
1337833 | Definition of lebesgue integral with respect to measure [imath]\mu[/imath]
In Rudin's Real and Complex Analysis, the Lebesgue integral is defined as: L et [imath](X,m,\mu)[/imath] be a measure space, where [imath]X[/imath] is a set, [imath]m[/imath] is a [imath]\sigma[/imath] algebra on [imath]X[/imath] and [imath]\mu[/imath] is a measure. Then, if [imath]f:X \to [0,\infty][/imath] and [imath]E \in m[/imath], we define [imath]\int_E f d\mu = \sup \int_E s d\mu \tag{1}[/imath] where the supremum is taken over all simple functions [imath]s, 0 \leq s \leq f[/imath] I do not have much background in measure theory, and I am wondering why we assume [imath]f[/imath] to be measurable to define its integral. EVEN IF [imath]f[/imath] is not a measuarable function, the above definition (1) would still be well-defined. Why do we define integral only for measurable [imath]f[/imath]? | 1081021 | Lebesgue Integral of Non-Measurable Function
In what follows I'm only considering positive real valued functions. Everywhere I look about the definition of the Lebesgue integral it is required to consider a measurable function. Why do we not define the integral for non-measurable functions? From what I see we require measurablility of the simple functions that approximate f, not f itself. The definition I'm considering is given a measure space [imath]X[/imath] with measure [imath]\mu[/imath] and a measurable function [imath]f[/imath] we define [imath] \int_E f \, \mathrm{d}\mu = \sup_{s \in S} \int_X s \,\mathrm{d}\mu [/imath] where [imath]S = \{ s : X \to [0, \infty) \mid 0 \le s \le f, s \text{ is simple, measurable} \}[/imath]. For example consider [imath]\mathbb{R}[/imath] with the sigma algebra [imath]\varnothing, \mathbb{R}[/imath] with measure [imath]\mu[/imath] given by [imath]\mu(\varnothing) = 0, \mu(\mathbb{R}) = 1[/imath] and consider [imath]f = \chi_{[0,1]}[/imath] then why can't we say that [imath] \int_{\mathbb{R}} f \,\mathrm{d} \mu = 0 [/imath] (since the only measurable simple function such that [imath]0\le s \le f[/imath] is [imath]s = 0[/imath]) which would follow the definition above? Is this not well defined? In general I'm struggling to see why measurable functions (other than measurable simple functions) are used. |
1337688 | Taylor series expansion of function [imath]f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}[/imath]
Determine the Taylor series expansion of function [imath]f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}[/imath]. [imath]f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}[/imath] It is known: (1.) [imath]\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}[/imath] (2.) [imath]\frac{d}{dx} (\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}) = \sum_{n=0}^\infty\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}[/imath] (3.) In third step I multiplied these two series, but I am not sure whether it is correct: [imath]\arcsin(x)\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}[/imath] EDIT: How did you get this sum [imath]\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}[/imath] ? And, in case I have to determine the product of the (some other) series, which one of these should I write? (a) [imath]\sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}[/imath] or (b) [imath] = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m-1)!!x^{2m+1}}{2^mm!(2m+1)})\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}[/imath] | 549028 | Deriving Maclaurin series for [imath]\frac{\arcsin x}{\sqrt{1-x^2}}[/imath].
Intrigued by this brilliant answer from Ron Gordon, I was attempting to find the Maclaurin series for [imath]f(x)=\frac{\arcsin x}{\sqrt{1-x^2}}=g(x)G(x)[/imath] with [imath]g(x)=\frac{1}{\sqrt{1-x^2}}[/imath] and [imath]G(x)[/imath] its primitive. So I attempted to multipy series, which yielded this: [imath]f(x)=\sum_{n=0}^{\infty}x^{2n+1} (-1)^n\sum_{k=1}^{n}\frac{1}{k+1} { -\frac{1}{2}\choose n-k}{ -\frac{1}{2}\choose k},[/imath] which I'm unable to simplify further. How to proceed? Or is this approach doomed? |
1338187 | “Odd” method of determening area of a triangle using (no trigonometry)
While learning some basic programming, I came across a (to me) very strange method of calculating the area of a triangle. In Javascript, this was expressed as: var perimeter = (side1 + side2 + side3)/2; var area = Math.sqrt(perimeter*((perimeter-side1)*(perimeter-side2)*(perimeter-side3))); This can be mathematically represented as following: [imath]A_\triangle=\sqrt{d\Biggl(( d-a )(d-b)(d-c) \Biggr) }[/imath] where [imath]d[/imath] equals to half of the sum of the length of the three sides of the triangle, and [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] represent the three different lengths respectively. I cannot, either by expanding or just 'looking' at the equation, figure out why this works (I assume it does so after testing with a few example triangles). If any kind (and clever) soul could explain why this works, It'd much appreciated. | 576831 | Proof of Heron's Formula for the area of a triangle
Let [imath]a,b,c[/imath] be the lengths of the sides of a triangle. The area is given by Heron's formula: [imath]A = \sqrt{p(p-a)(p-b)(p-c)},[/imath] where [imath]p[/imath] is half the perimeter, or [imath]p=\frac{a+b+c}{2}[/imath]. Could you please provide the proof of this formula? Thank you in advance. |
1289211 | Solving the congruence [imath]3x^2 + 6x + 1 \equiv 0 \pmod {19}[/imath]
I tried to solve this equation but without a success: [imath]3x^{2}+6x+1 \equiv 0 \pmod {19}[/imath] I concluded hat [imath]x(x+2)\equiv 6 \pmod{19}[/imath], the only way i think to solve this is by just trying all the options. But there must be a more efficiant way. I would like to get help with that, thanks | 1336592 | solve [imath]3x^2 + 6x +1 \equiv 0 \pmod {19}[/imath]
I need to solve [imath]3x^2 + 6x +1 \equiv 0 \pmod {19}[/imath] I saw the same problem here - Solving the congruence [imath]3x^2 + 6x + 1 \equiv 0 \pmod {19}[/imath] but didn't understand how he got to the conclusion [imath]x(x+2) \equiv 6 \pmod {19}[/imath] and anyway im trying to solve it how we learned in class - multiply both sides and the modulo by [imath]4a[/imath] then solve the two equations [imath]y^2 \equiv b^2 -4ac \pmod {4an}[/imath] [imath]2ax + b \equiv y \pmod {4an}[/imath] so I tried multiplying the hole thing by [imath]4a[/imath] , that is [imath]4 \times 3[/imath] and got to [imath](2 \times 3x + 6)^2 \equiv 36 -4 \times 3 \pmod {19 \times 4 \times 3}[/imath] now I am stuck . any help will be appreciated |
1338340 | Prove that [imath]e^x-x\geq 1[/imath].
Please prove that [imath]e^x - x[/imath] is always bigger than or equal to 1 | 1330815 | How to prove [imath]1+x \leq e^x~\forall x \in \mathbb{R}?[/imath]
How to prove [imath]1+x \leq e^x~\forall x \in \mathbb{R}[/imath] I'm stuck, I tried taking logs but didn't know how to proceed. |
1312210 | Show that representative functions on a profinite group factors.
Let [imath]G[/imath] be a compact group. A representative function [imath]f\in\mathcal{C}(G,\mathbb{K})[/imath] is a function such that [imath]\dim\left(\operatorname{span}\left(Gf\right)\right)< \infty[/imath]. Remark that the representative functions form a subalgebra of [imath]\mathcal{C}(G,\mathbb{K})[/imath]. I'm following the book "The Structure of Compact Groups" by Hofmann&Morris on this subject and stumbled upon a problem : I would like to be able to show that that a representative function [imath]f[/imath] on a profinite group [imath]G[/imath] factors as [imath]f=f\circ \pi[/imath] where [imath]\pi:G\rightarrow H[/imath] is a surjection onto a finite group. My goal would be to use this in order to find the set of the representative functions from [imath]\mathbb{Z}_p[/imath] to [imath]\mathbb{K}[/imath], denoted [imath]R(\mathbb{Z}_p,\mathbb{K})[/imath] for [imath]\mathbb{K}=\mathbb{C}[/imath] and [imath]\mathbb{K}=\mathbb{R}[/imath]. Yet I have no clue about how I could do this, I'm pretty new to profinite stuff and I can't "see" anything. Ps: I've asked this on math.stackexchange.com, but realized afterwards it may have been better to ask this here. | 1309570 | Show that a representative function on a profinite group factor through a surjection
Let [imath]G[/imath] be a compact group. A representative function [imath]f\in\mathcal{C}(G,\mathbb{K})[/imath] is a function such that [imath]\dim\left(\operatorname{span}\left(Gf\right)\right)< \infty[/imath]. Remark that the representative functions form a subalgebra of [imath]\mathcal{C}(G,\mathbb{K})[/imath]. I'm following the book "The Structure of Compact Groups" by Hofmann&Morris on this subject. I would like to be able to show that that a representative function [imath]f[/imath] on a profinite group [imath]G[/imath] factors as [imath]f=f\circ \pi[/imath] where [imath]\pi:G\rightarrow H[/imath] is a surjection onto a finite group. My goal would be to use this in order to find the set of the representative functions from [imath]\mathbb{Z}_p[/imath] to [imath]\mathbb{K}[/imath], denoted [imath]R(\mathbb{Z}_p,\mathbb{K})[/imath] for [imath]\mathbb{K}=\mathbb{C}[/imath] and [imath]\mathbb{K}=\mathbb{R}[/imath]. Yet I have no clue about how I could do this, I'm pretty new to profinite stuff and I can't "see" anything. |
1338616 | Why the complex number system is not an ordered field
In high school, we are taught that we do not have [imath]2i < 3i[/imath], i.e., the complex number system is not an ordered field. (Real number, for example, is an ordered field. For example, [imath]2 < 3[/imath]). Why? My comment to this is because in the complex corrdinate, in [imath]Re-Im[/imath] coordinate, the concept of complex number is somewhat a rotation around the origin. | 487997 | Total ordering on complex numbers
Show that there doesn't exist a relation [imath]\succ[/imath] between complex numbers such that (i) For any two complex numbers [imath]z,w[/imath], one and only one of the following is true: [imath]z\succ w,w\succ z,[/imath] or [imath]z=w[/imath] (ii) For all [imath]z_1,z_2,z_3\in\mathbb{C}[/imath] the relation [imath]z_1\succ z_2[/imath] implies [imath]z_1+z_3\succ z_2+z_3[/imath]. (iii) For all [imath]z_1,z_2,z_3\in\mathbb{C}[/imath] with [imath]z_3\succ 0[/imath], then [imath]z_1\succ z_2[/imath] implies [imath]z_1z_3\succ z_2z_3[/imath]. Suppose [imath]i\succ 0[/imath]. From (iii) we have [imath]i^2\succ 0[/imath], so [imath]-1\succ 0[/imath], so applying (ii) we get [imath]0\succ 1[/imath]. But repeating (iii) on [imath]-1\succ 0[/imath] we get [imath]1\succ 0[/imath], a contradiction. So either [imath]i=0[/imath] or [imath]0\succ i[/imath]. How can I proceed from here? |
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