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2707652 | Error while calculating the derivative of a function to the power of another function
I asked myself how one may calculate the derivative of a function that looks like this: [imath]h(x) = f(x)^{g(x)}[/imath] And used the following process of derivation [imath]h'(x) = \frac{d}{dx}(f(x) \cdot 1^{g(x)})[/imath] [imath]h'(x) = f(x) \cdot \frac{d}{dx}(1^{g(x)}) + 1^{g(x)} \cdot f'(x)[/imath] [imath]h'(x) = f(x) \cdot 1^{g(x)} \cdot ln(1) \cdot g'(x) + 1^{g(x)} \cdot f'(x)[/imath] [imath]h'(x) = f'(x)^{g(x)}[/imath] But when checking with a calculator it's clear than the result is wrong, I suspect my mistake is very obvious but I'm not able to find it. So the question is, could you tell me what I did wrong? | 533900 | Differentiating [imath]f(x)^{g(x)}[/imath]
Is there any general rule for what the derivative of [imath]f(x)^{g(x)}[/imath] (where [imath]f(x),g(x)[/imath] are differentiable functions) is in terms of [imath]f(x),g(x),f'(x),g'(x)[/imath]. In other words is there something analogous to product,chain and quotient rules for such expressions? |
2707796 | Suppose that [imath] a,b \in \mathbb{Z}[/imath] If 4 is divisible by [imath] a^2 + b^2[/imath], then [imath]a[/imath] and [imath]b[/imath] are not odd.
I am trying to prove old test exercises and this one is resisting me. I know you can believe that I meant [imath]a^ 2 + b ^ 2[/imath] is divisible by 4, but no, in the test it says that 4 is divisible by [imath]a ^ 2+ b^2[/imath]. Suppose that [imath] a,b \in \mathbb{Z}[/imath] If 4 is divisible by [imath] a^2 + b^2[/imath], then [imath]a[/imath] and [imath]b[/imath] are not odd. I started like this: Suppose that a and b are even, then [imath]a = 2m[/imath] and [imath]b = 2n[/imath] for some [imath]m,n \in \mathbb{Z} [/imath]. Thus [imath] a^2 = (2m)^2 \ \mbox{and}\ \ b^2= (2n)^2[/imath] [imath] a ^2 + b^2= (2m)^2 + (2n)^2[/imath] [imath] a^2 + b^2 = (4 m^2 + 4n^2) = 2(2m^2+2n^2) [/imath] Now [imath]2m ^ 2 + 2n^2\in \mathbb{Z} = k[/imath] Then [imath] a^2+b^2= 2k[/imath] Is this a correct proof?/ Any better ways to prove this. Sorry for the lousy English. Thanks. | 1290330 | Proving by Contradiction
Prove by Contradiction Suppose [imath]a, b \in Z[/imath]. If [imath]4|(a^2 + b^2)[/imath], then [imath]a[/imath] and [imath]b[/imath] are not both odd. So the contradiction: Assume [imath]4|(a^2 + b^2)[/imath], where [imath]a[/imath] and [imath]b[/imath] are both odd. Then [imath]a=2k+1[/imath], [imath]b=2l +1[/imath] And then I substitute it into the equation and the result is something like this? [imath]4|4(k^2 + k +l^2 + l) + 2[/imath] What do I do next to prove this? |
2708351 | We know [imath]\int x^2\,\mathrm dx = \frac13 x^3+C[/imath], so why isn't [imath]\int \sin^2x\,\mathrm dx =\frac13\sin^3x + C[/imath]?
At first I would like to apologize to the members who answered this. This question had considerable information so I asked different question. Actual question I wanted to ask, We know that [imath]\int (x + b)^n\,\mathrm dx =\frac{(x+b)^{n+1}}{n+1} + C[/imath] so while solving [imath]\int x^2\,\mathrm dx = \frac13x^3 + C[/imath] also, [imath]\int (x + 3)^4\,\mathrm dx = \frac15(x + 3)^5 + C[/imath] but for trig functions like, [imath]\int \sin^2x\,\mathrm dx \neq \frac13\sin^3x + C[/imath] Instead, we have to simplify it first algebraically. So why does the intuition fail here? I apologize once again. But still I learnt some new things. | 2708293 | Why does the basic intuition fail here? What is the reason behind it?
We know that [imath]\int f(ax + b)\,\mathrm dx = \frac1aF(ax + b) + C[/imath] so while solving [imath]\int x^2\,\mathrm dx = \frac13x^3 + C[/imath] also, [imath]\int (x + 3)^4\,\mathrm dx = \frac15(x + 3)^5 + C[/imath] but for trig functions like, [imath]\int \sin^2x\,\mathrm dx \neq \frac13\sin^3x + C[/imath] Instead, we have to simplify it first algebraically. So why does the intuition fail here? Update: I seem to be confused with f. Till now [imath]\int \sin^2(1*x+0)\,\mathrm dx \ [/imath], I have assumed f = (sin(x))^2 with a=1 and b=0. There seems something wrong with my understanding. |
2708326 | Find the limit of the sequence [imath]x_{n+1}=x_n-x_n^{n+1}[/imath] where [imath]0[/imath]
Find the limit of the sequence [imath]x_{n+1}=x_n-x_n^{n+1}[/imath] where [imath]0<x_1<1[/imath]. I think zero is the limit but how do I prove that? I prove that sequence is monotone decreasing and bounded so convergent. | 186293 | Limit of [imath](x_n)[/imath] with [imath]0 and x_{n + 1} = x_n - x_n^{n + 1}[/imath]
Let [imath]0 < x_1 < 1[/imath] and [imath]x_{n + 1} = x_n - x_n^{n + 1}[/imath] for [imath]n \geqslant 1[/imath]. Prove that the limit exists and find the limit in terms of [imath]x_1[/imath]. I have proved the existence but cannot manage the other part. Thanks for any help. |
2708623 | How to determine [imath]\prod_{g\in G}g[/imath]?
It’s an exercise in my textbook. Let [imath]G[/imath] be a finite Abelian group, then determine [imath]\prod\limits_{g\in G}g.[/imath] Actually, I do not quite get what it is asking. What does it mean by “determine”? What needs to be determined? Moreover, there’s a following task asking me to show with the help of my first question that [imath](p−1)! \equiv −1~{\rm mod}~p~~ (p~{\rm prime}).[/imath] I’m then getting more confused... Any help, hint or detail, would be greatly appreciated. Thanks! PS: They are exercises from my textbook, on page 49 of The Theory of Finite Groups, An Introduction (page 62 of the pdf). | 53185 | The product of all the elements of a finite abelian group
I'm trying to prove the following statements. Let [imath]G[/imath] be a finite abelian group [imath]G = \{a_{1}, a_{2}, ..., a_{n}\}[/imath]. If there is no element [imath]x \neq e[/imath] in [imath]G[/imath] such that [imath]x = x^{-1}[/imath], then [imath]a_{1}a_{2} \cdot \cdot \cdot a_{n} = e[/imath]. Since the only element in [imath]G[/imath] that is an inverse of itself is the identity element [imath]e[/imath], for every other element [imath]k[/imath], it must have an inverse [imath]a_{k}^{-1} = a_{j}[/imath] where [imath]k \neq j[/imath]. Thus [imath]a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = e[/imath]. If there is exactly one [imath]x \neq e[/imath] in [imath]G[/imath] such that [imath]x = x^{-1}[/imath], then [imath]a_{1}a_{2} \cdot \cdot \cdot a_{n} = x[/imath]. This is stating that [imath]x[/imath] is not the identity element but is its own inverse. Then every other element [imath]p[/imath] must also have an inverse [imath]a_{p}^{-1} = a_{w}[/imath] where [imath]p \neq w[/imath]. Similarly to the first question, a rearrangement can be done: [imath]a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot xx^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = xx^{-1} = e[/imath]. And this is where I am stuck since I proved another statement. Any comments would be appreciated for both problems. |
2708947 | How do I prove that the product of two closed sets is closed?
Given [imath]A \in \mathbb{R^q}[/imath] and [imath]B \in \mathbb{R^p}[/imath] two closed sets, then [imath]A \times B[/imath] is a closed set in [imath]\mathbb{R^m} = \mathbb{R^q} \times \mathbb{R^p}[/imath] How do I prove the previous result by using sequences? | 479777 | Is product of two closed sets closed?
If [imath]A[/imath] is closed in [imath]X[/imath] and [imath]B[/imath] is closed in [imath]Y[/imath], does it follow that [imath]A \times B[/imath] is closed in [imath]X \times Y[/imath] ? |
2709067 | Proving that continuous function [imath]f:[0,2] \to \mathbb{R}[/imath] such as [imath]f(0)=f(2)[/imath], [imath]\exists x \in [0,1]:f(x)=f(x+1)[/imath]
I am having trouble formalizing the answer to this question. Any help would be appreciated. Is it true that for every continuous function [imath]f:[0,2] \to \mathbb{R}[/imath] such as [imath]f(0)=f(2)[/imath], [imath]\exists x \in [0,1]:f(x)=f(x+1)[/imath]? If the function is periodic of degree one this seems true to me. I tried proving by contradiction, supposing that [imath]\forall x \in [0,1], f(x) \neq f(x+1)[/imath], but I couldn't get any further. | 2307896 | Proof that continuous function on [imath] [0, 2][/imath] has [imath]f(x) = f(x-1)[/imath] for some [imath] x[/imath]
Proof that if continuous function on [imath][0, 2][/imath] and [imath]f(0) = f(2)[/imath], then for some [imath]x: f(x) = f(x - 1)[/imath] |
2709416 | Finding the sum of the infinite series [imath]\sum_{k=2}^{\infty} \frac 1 {k^2-1}[/imath]
Edit: Thanks for the help everyone. Turns out I was just making an arithmetic error. Whoops! Question answered I've got this infinite series which starts at k=2 [imath]\sum_{k=2}^{\infty} \frac 1 {k^2-1}[/imath] https://www.wolframalpha.com/input/?i=infinite+sum+1%2F(k%5E2-1) I thought this series might be telescoping, so I decomposed it to: 1/2(k-1) - 1/2(k+1). Then I wrote out partial sums and found that after k=4, terms begin to cancel. Then I summed up the parts that don't cancel and got the sum = 7/12. However according to wolfram, the sum is 3/4. I don't understand what I'm missing. If following my work in text form is confusing, I will gladly write it out and upload a picture. | 1624203 | Hint on computing the series [imath]\sum_{n=2}^\infty \frac{1}{n^2-1}[/imath].
I'm supposed to determine whether this sum diverges or converges and if it converges then find its value: [imath] \sum_{n=2}^\infty \frac{1}{n^2-1}. [/imath] Using the comparison test I eventually showed that this converges. But I can't figure out how to show what this sum converges to. The only sums we actually found values for in my notes are geometric series which this clearly isn't. I saw that I could use partial fraction decomposition to represent the terms as [imath]\frac{1/2}{n-1}- \frac{1/2}{n+1} [/imath] but that just gets me [imath]\infty - \infty[/imath], so this isn't the way to do it. I'm not sure how to find the value of this sum. I don't need the full solution but a hint would be appreciated. Thanks. :) |
2708633 | Finite [imath]2[/imath]-group with derived subgroup of order 8
Does there exist a finite non-abelian [imath]2[/imath]-group [imath]G[/imath] such that [imath]G^{\prime}\cong D_8[/imath] or [imath]G^{\prime}\cong Q_8[/imath]? By an easy inspection with GAP, I could not find any example! Any answer or comment will be greatly appreciated! | 384081 | [imath]D_8[/imath] as a derived subgroup
Every undergraduate student knows that there are (exactly) two non-abelian groups of order 8: Dihedral ([imath]D_8[/imath]) and Quaternion ([imath]Q_8[/imath]). The group [imath]Q_8[/imath] has many interesting properties; simple of them are: it has unique element of order [imath]2[/imath], all subgroups are normal, etc. A property, which is strange and not obvious, the group [imath]Q_8[/imath] has, is that there is no group [imath]G[/imath] such that [imath]G/Z(G)[/imath] is isomorphic to [imath]Q_8[/imath]. On the other hand, [imath]D_8[/imath] has no such property; there are (infinitely many) groups [imath]G[/imath] such that [imath]G/Z(G)[/imath] is isomorphic to [imath]D_8[/imath]; example - [imath]D_{16}[/imath]. The question, I would like to ask, is related to a property of [imath]D_8[/imath], which is not shared by [imath]Q_8[/imath]. There are (infinitely many) groups whose derived subgroup is [imath]Q_8[/imath]; example- [imath]SL_2(3)[/imath]. For groups [imath]G[/imath] of order <100, one can check on GAP that [imath]D_8[/imath] is not a derived subgroup of any group. The natural question is then: Question 1: Does there exist a finite/infinite group whose derived subgroup is equal to [imath]D_8[/imath]? Question 2: Does there exist a finite/infinite group whose derived subgroup is equal to (general) dihedral group [imath]D_{2n}[/imath] of order [imath]2n[/imath] ([imath]n>2[/imath])? |
2709088 | Extrema Question
[imath]f(x,y)=x^2+x\cdot y+x+y^2[/imath] on the set [imath]S=\{(x,y) : x^2+y^2 \leq 9\}[/imath]. I found the critical point on the interior of [imath]S[/imath] to be [imath]\left(\frac{-2}{3},\frac{1}{3}\right)[/imath]. I'm not sure how to parametrize the boundary of [imath]S[/imath] in terms of theta using sine and cosine and write [imath]f[/imath] in terms of theta on the boundary by putting restrictions on theta. I also am not sure how to find the extrema of [imath]f[/imath] on [imath]S[/imath]. | 2708313 | Parametrization and Extrema for Multivariable Function
We are given the function: [imath]f(x,y)=x^2+xy+x+y^2[/imath] On the set [imath]S=\{(x,y):x^2+y^2 \leq 9\}[/imath] I found the critical point on the interior of [imath]S[/imath] to be [imath](-2/3,1/3)[/imath]. I am asked to parametrize the boundary of [imath]S[/imath] in terms of theta using sine and cosine, and write [imath]f[/imath] in terms of theta on the boundary. (don't forget restrictions on theta). I'm not sure how to do this. After we do that, we are asked to find the extrema of [imath]f[/imath] on [imath]S[/imath]. I am also not sure how to do this. Any help would be immensely appreciated! |
2709372 | The cardinality of [imath]\{f\subseteq\mathbb R^2\mid f\colon\mathbb{R\to R}\text{ is a function}\}[/imath] is strictly greater than the cardinality of [imath]\mathbb R[/imath]
I have a weird problem. Let's define [imath]\mathcal{F}=\{f\subseteq \mathbb{R}^2\mid f:\mathbb{R}\rightarrow\mathbb{R}\text{ is a function}\}[/imath]. I'm trying to prove that there does not exist a surjection [imath]g:\mathbb{R}\twoheadrightarrow \mathcal{F}[/imath]. I'm completely stumped, though. I tried modifying Cantor's diagonal argument, but that didn't lead anywhere because, well, [imath]\mathbb{R}[/imath] is uncountable. Next, I tried doing something with Cantor's theorem, because it was the only thing I could find that proved that a surjection of something wasn't possible. That didn't work either. Is there anything that could point me in the right direction here? | 1690593 | Cardinality of [imath]\mathbb{R}^\mathbb{R}[/imath] by diagonal argument
I want to prove that the set of all real functions [imath]\mathbb{R}^\mathbb{R}[/imath] has a higher cardinality than the real numbers [imath]\mathbb R[/imath], by Cantor's diagonal argument. I'm having difficulties with approaching this problem. What I'm looking for is a hint in the right direction. I've seen an example where it is shown that the power set [imath]2^S[/imath] of a countable set [imath]S[/imath] is uncountable, by the diagonal argument. |
2710035 | If [imath] x + \frac{1}{x}=3 [/imath], what is [imath] x^{3} + \frac{1}{x^{3}} [/imath] and the generalization [imath] x^{n} + \frac{1}{x^{n}} [/imath]?
*This problem is from a math olympiad. It is known that [imath] x + \frac{1}{x} = 3[/imath] What is the value of [imath] x^{3} + \frac{1}{x^{3}} ?[/imath] I know that we can solve for [imath]x[/imath] from [imath] x + 1/x = 3[/imath] first, by [imath] x^{2}-3x+1 = 0[/imath] Which gives [imath] x = \frac{3 \pm \sqrt{5} }{2} [/imath] Then we can relatively tedious find the values of [imath] x^{3} + \frac{1}{x^{3}} [/imath] Are there better ways? Can we generalize a way for [imath] x^{n} + \frac{1}{x^{n}} ? [/imath] Thanks. | 421995 | If [imath]x+{1\over x} = r [/imath] then what is [imath]x^3+{1\over x^3}[/imath]?
If [imath]x+{1\over x} = r [/imath] then what is [imath]x^3+{1\over x^3}[/imath] Options: [imath](a) 3,[/imath] [imath](b) 3r,[/imath] [imath](c)r,[/imath] [imath](d) 0[/imath] |
2709701 | Why is it assuming in both sides that [imath]a^{n}=e[/imath]?
Let [imath]G[/imath] be a cyclic group of order [imath]n[/imath] generated by [imath]a[/imath]. Then [imath]a^{k} = e[/imath] if and only if [imath]n | k[/imath]. Proof: I have a question about the above proof. Why is it assuming in both sides that [imath]a^{n}=e[/imath]? | 595644 | In a finite cyclic group, the generator has the order of the group
Let [imath]G[/imath] be a finite cyclic group of order [imath]n[/imath] generated by an element [imath]g[/imath]. I want to show that the order of [imath]g[/imath] is exactly the order of [imath]G[/imath]. My try: We can write [imath]G=\{g^i\;|\;i\in \mathbb Z\}[/imath]. Since [imath]G[/imath] is finite, we know that there will be identifications [imath]g^i=g^j[/imath] that will make [imath]G[/imath] finite. By the multiplication law on [imath]G[/imath] we must have [imath]g^0=1[/imath]. Also if [imath]i<j[/imath] and if [imath]g^i=g^j[/imath] then [imath]g^{j-i}=1[/imath]. I can see all these but not see why [imath]g^n=1[/imath] and not [imath]g^k=1[/imath] for all [imath]0<k<n[/imath]? Thank your for your help!! |
2710448 | Does the [imath]\int_{-\infty}^\infty e^{{-x}^2} \, dx[/imath] integral converge?
I got this exercise as an optional challenge for students. This integral is supposed to converge, and in fact, the Math Stack community has already done it, but I'm looking for something more detailed, integral there aren't shown with the detail I'd expect, hence the overall understanding of the method is somewhat difficult. I hope I don't botter the community by asking this. Greetings and thank you. | 1187577 | Use integral test to show that Gaussian Integral [imath]\int_{-\infty}^{\infty} e^{-x^2} \ dx[/imath] converges?
There's a question in my book (it's a school-specific textbook) saying: Use the integral test to show [imath]\int_{-\infty}^{\infty} e^{-x^2} \ dx[/imath] (the Gaussian Integral) converges. The integral test says if the integral converges, so does the series. But I have no idea how to represent this as a series, do I need to? I'm not looking to evaluate it - I've read about it online - but how do I use the integral test here? Thanks |
322564 | A projection which has [imath]||E\alpha|| \le ||\alpha||[/imath] is an orthogonal projection
this problem screwed me up for a long time. Please help me solve this. [imath]V[/imath] is an inner product space and [imath]W[/imath] is a finite-dimensional subspace of [imath]V[/imath]. Prove that if [imath]E[/imath] is a projection with range [imath]W[/imath], then [imath]E[/imath] is the orthogonal projection on [imath]W[/imath] iff [imath]||E\alpha|| \le ||\alpha||[/imath] for all [imath]\alpha[/imath] in [imath]V[/imath]. I can prove that if [imath]E[/imath] is an orthogonal projection then [imath]||E\alpha|| \le ||\alpha||[/imath] using property: [imath](E\alpha | \beta) = (\alpha|\beta)[/imath] for all [imath]\alpha[/imath] in [imath]V[/imath] and [imath]\beta[/imath] in W. But the converse seems more difficult. | 97004 | A projection satisfying [imath]\| Px \| \leq \|x\|[/imath] for all [imath]x[/imath] is an orthogonal projection
How to prove that if [imath]V[/imath] is a finite dimensional inner product space and [imath]W[/imath] a subspace of [imath]V[/imath], if [imath]P[/imath] is projection map ([imath]P^2=P[/imath]) having [imath]W[/imath] as its range and is such that [imath]\|Px\| \leq \|x\|[/imath] for all [imath]x \in V[/imath], then [imath]P[/imath] is orthogonal projection of [imath]V[/imath] onto [imath]W[/imath]. |
1834617 | Comparing different definitions of tightness for measures
Let [imath]X[/imath] be a Hausdorff space, [imath]\mathcal{B}(X)[/imath] the Borel [imath]\sigma[/imath]-algebra and [imath]\mu : \mathcal{B}(X) \to [0, \infty][/imath] a measure. Consider the following properties: (1) [imath]\forall A \in \mathcal{B}(X): \mu(A) = \sup \left\{ \mu(K) \mid K \subseteq A, K \text{ compact} \right\}[/imath] (2) [imath]\forall \varepsilon > 0 \, \exists K \subseteq X[/imath] compact: [imath]\mu\left(X \setminus K\right) < \varepsilon[/imath] If [imath]\mu[/imath] satisfies (1) then some authors say that [imath]\mu[/imath] is "tight" or "inner regular" or "inner regular w.r.t. the compact sets". Often those authors that reserve the notion "inner regular" for "inner regular w.r.t. the closed sets" call [imath]\mu[/imath] "tight" if it satisfies (1). Other authors, mostly probabilists, refer to a finite measure (probability measure) [imath]\mu[/imath] as "tight" if it satisfies (2). (2) seems not to be an interesting property for measures [imath]\mu[/imath] that are not totally finite (but at least finite on compact sets). As an example, the Lebesgue measure on [imath]\mathbb{R}[/imath] is "tight" in the sense of (1) but not "tight" in the sense of (2). If [imath]\mu[/imath] is finite then (1) [imath]\Rightarrow[/imath] (2). It is also folklore that if [imath]X[/imath] is metrizable and [imath]\mu[/imath] finite then (1) [imath]\Leftrightarrow[/imath] (2) (see also here). Questions: Can the assumption on [imath]X[/imath] being metrizable be weakened, e.g. perfectly normal or even weaker? Are there interesting applications in which tightness in the sense of property (1) is of interest when [imath]\mu[/imath] is finite, but (2) is not applicable? | 1437644 | Tightness and Inner Regularity
Let [imath]P[/imath] be a probability measure on a Borel [imath]\sigma[/imath]-algebra (on some metric space, [imath]\Omega[/imath]). It is called tight if for every [imath]\epsilon >0[/imath], there exists a compact [imath]K[/imath] such that [imath]P(X \in K) \geq 1 - \epsilon[/imath]. It is called inner regular if [imath]P(A) = \sup \{P(K) | K \subset A, K \text{ compact}\}[/imath] for every [imath]A[/imath] in the Borel [imath]\sigma[/imath]-algebra. It is clear that inner regularity implies tightness if we choose [imath]A=\Omega[/imath]. For an arbitrary set [imath]A[/imath], using tightness we can obtain a compact [imath]K(\epsilon)[/imath] such that [imath]P(A\cap K(\epsilon)) \geq P(A) - \epsilon[/imath]. If [imath]A[/imath] is closed then [imath]K \cap A[/imath] is compact and hence we get inner regularity for closed sets [imath]A[/imath]. How can I prove this for general [imath]A[/imath]? Thanks. |
2710914 | Prove that the series with terms [imath]a_n[/imath] diverges.
where the series has the nth term given by [imath]a_n = \frac {1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}[/imath] I managed to show that [imath]a_n = \frac {(2n)!} {4^n (n!)^2}[/imath] But it doesn't really help.. I'd like to compare it with something since the ratio test doesn't work. | 1887601 | How would you show that the series [imath]\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}[/imath] diverges?
How would you show that the series [imath]\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}[/imath] diverges? Wolfram Alpha says it diverges "by comparison", but I'd like to know to what you would compare it? I've tried some basic things to no avail: The ratio test is inconclusive, comparing with something "smaller" which I know diverges is proving difficult. If possible, I'd like to show this without using anything fancy (such as Stirling's approximation). Can you do this with a "basic" comparison? |
2711300 | A problem from abstract algebra of finite group?
A and B are to nonempty subset of a group G with [imath]|A|+|B| > |G|[/imath] then G = AB. I am a bit confused of the order of a subset of a group. Generally a subset of a group is not a subgroup but when it occurs [imath]|A|+|B| > |G|[/imath], I want to figure out the relation between A and B and if A will have the inverse of some elements of A if [imath]|A| > \frac{|G|}{2}[/imath]? | 1034240 | Prove either [imath]G=ST[/imath] or |[imath]G|\geq|S|+|T|[/imath]
Let G be a finite group, and let S and T be (not necessarily distinct) nonempty subsets. prove that either [imath]G=ST[/imath] or |[imath]G|\geq|S|+|T|[/imath] That's my thougt, I am thinking suppose [imath]G[/imath] does not equal to [imath]ST[/imath], then there exist a [imath]h[/imath] in [imath]G[/imath], but [imath]h[/imath] [imath]\notin ST[/imath]. Thus, I suppose T={[imath]x_1,x_2......x_N[/imath]}and [imath]S=\{{t_1,t_2,.......,t_m,x_{k+1},......,x_{N}}\}[/imath], where [imath]x_k+1,...x_N[/imath] are the common elements between two sets S and T. So, I want to construct two totally different sets A and B so that [imath]|A|+ |B| > |G|[/imath], firstlt,if [imath]1[/imath] is not in [imath]S[/imath] and [imath]T[/imath], then suppose T=A, and then suppoese B={[imath]h,t_1,...,t_m, hx_{k+1}^{-1},...,hx_N^{-1}[/imath]}. I think if we can prove B[imath]\neq[/imath]A(orT) and B[imath]\neq[/imath]S, then I can know the question will be proved since [imath]|A|+|B|>|G|[/imath], but I only can prove B[imath]\neq[/imath]S, I don't know how to prove B[imath]\neq[/imath]T. Maybe my thoughts are wrong, can someone tell me how to solve this question? I think my methods are not correct, I hope someone can help me solve this question. |
1806836 | How to deduce the usual definition of Quasi-coherent Module over a scheme from the general definition over ringed Spaces
Quasi-Coherent Modules over a Ringed Space : Let [imath](X,\mathcal O_X)[/imath] be a ringed space. A sheaf of modules [imath]F[/imath] over [imath](X,\mathcal O_X)[/imath] is called quasi-coherent if for every point [imath]x\in X[/imath] [imath]\exists U\subset X[/imath] s.t [imath]{F_{|}}_{U}[/imath] is the cokernel of a map [imath]\oplus_{i\in I}\mathcal O_U\rightarrow\oplus_{j\in J}\mathcal O_U[/imath] Quasi-Coherent Modules over a Scheme: Let [imath](X,\mathcal O_X)[/imath] be a Scheme. A sheaf of modules [imath]F[/imath] over [imath](X,\mathcal O_X)[/imath] is called quasi-coherent if X can be covered by open affine subsets [imath]U_i=Spec(A_i)[/imath], such that for each [imath]i[/imath] there is an [imath]A_i[/imath] module [imath]M_i[/imath] with [imath]{F_{|}}_{U_i}=\tilde{M_i}[/imath] Question: How can one deduce the definition of quasi-coherent sheaf of module over a scheme from the definition of quasi-coherent sheaf of modules over a ringed space? | 1599597 | Why are Grothendieck's and Hartshorne's definitions of quasi-coherence equivalent?
Hartshorne's Algebraic Geometry defines an [imath]\mathcal O_X[/imath]-module [imath]\mathscr F[/imath] to be quasi-coherent if there is an open affine cover [imath](U_i=\operatorname{Spec} A_i)_{i\in\mathcal I}[/imath] of [imath]X[/imath] such that each [imath]\mathscr F\rvert_{U_i}[/imath] is isomorphic to the sheaf of modules [imath]\widetilde{M_i}[/imath] associated to some [imath]A_i[/imath]-module [imath]M_i[/imath]. It is later proved that this is equivalent to the statement that for any open affine [imath]U=\operatorname{Spec} A\subseteq X[/imath], there exists an [imath]A[/imath]-module [imath]M[/imath] with [imath]\mathscr F\rvert_U\cong\widetilde M[/imath]. On the other hand, Grothendieck's Éléments de Géométrie Algébrique defines [imath]\mathscr F[/imath] to be quasi-coherent if every point [imath]x\in X[/imath] admits an open neighborhood [imath]U\subseteq X[/imath] such that [imath]\mathscr F\rvert_U[/imath] is isomorphic to the cokernel of a morphism of free [imath]\mathcal O_U[/imath]-modules. Now I wonder how those definitions are equal. I tried proving it by myself, but both directions seem quite nontrivial to me... Any hints? |
2711051 | How to show the canonical embedding [imath]c_0\rightarrow c_0^{**}[/imath] is not surjective?
I want to show [imath]c_0[/imath] is not reflexive by showing the canonical embedding [imath]T:c_0\rightarrow c_0^{**}[/imath] is not surjective. And I know [imath]c_0[/imath] is a closed subspace of [imath]\ell_\infty[/imath] and [imath]\ell_\infty[/imath] is isometrically isomorphic to [imath]c_0^{**}[/imath]. But I couldn't find [imath]x^{**}\in c_0^{**}[/imath] such that [imath]T(x)\neq x^{**}[/imath] ([imath]x\in c_0[/imath]). Can anyone help me to show the canonical embedding [imath]T:c_0\rightarrow c_0^{**}[/imath] is not surjective? Thank you! | 1027212 | The space [imath]c_0[/imath] of sequences that converge to 0 is not reflexive
Consider the Banach space [imath]X[/imath] of null sequence whose elements are complex sequence which converges to [imath]0[/imath]. In addition the norm is defined as [imath]\|(a_1, \dots, a_n)\| := \sup_n |a_n|.[/imath] Show this space is NOT reflexive. Recall that the dual of [imath]X[/imath] is [imath]l_1[/imath]. Also use the following fact. If [imath]X[/imath] is reflexive space and [imath](x_n)[/imath] is a sequence in [imath]X[/imath] and for all [imath]\psi \in X^*[/imath], the sequence [imath]\psi(x_n)[/imath] has a limit in [imath]\mathbb C[/imath], then [imath]x_n[/imath] converges weakly to some [imath]x \in X[/imath]. Suppose that [imath]X[/imath] is reflexive. Consider the sequence [imath]u_n = (1, \dots, 1, 0, \dots) \in X,[/imath] where first [imath]n[/imath] entries are [imath]1[/imath]. Then for any [imath]y \in l_1[/imath], we have [imath]y(u_n) := \sum_n y_n\times u_n = y_1+\dots+y_n \to \sum_n y_n < \infty,[/imath] since [imath]y[/imath] is absolutely convergent. Now it is only left to show that [imath]u_n[/imath] does not converges weakly to some [imath]u \in X[/imath] which I guess is [imath](1, 1, \dots)[/imath]. That is, for all [imath]y \in l_1[/imath], [imath]y(u_n) \nrightarrow y(u).[/imath] However, this seems correct to me. Where did I make mistake, please? Thank you! |
2711775 | Prove [imath]\cos(\alpha − \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta[/imath] using vectors
Suppose that [imath]v[/imath] and [imath]w[/imath] are unit vectors. If the angle between [imath]v[/imath] and [imath]\hat{i}[/imath] (the unit vector in the positive [imath]x[/imath] direction) is [imath]\alpha[/imath] and that between [imath]w[/imath] and [imath]\hat{i}[/imath] is [imath]\beta[/imath], prove that [imath]\cos(\alpha − \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta[/imath] | 900700 | Use vectors to prove [imath]\cos(\beta - \alpha) = \cos \alpha \cos\beta + \sin\alpha \sin\beta[/imath]
If two vectors [imath]a[/imath] and [imath]b[/imath] make angle [imath]\alpha[/imath] and [imath]\beta[/imath] with the [imath]x[/imath]-axis, prove, using vectors, that: [imath]\cos(\beta - \alpha) = \cos \alpha \cos\beta + \sin\alpha \sin\beta[/imath] I tried taking components of each vector along the plane, but that didn't get me anywhere. I tried taking component of vector [imath]b[/imath] along vector [imath]a[/imath], but I also got stuck there. Can anyone point me in the right direction? |
471826 | Orthogonal projection: more linear algebra questions
I'm working on proving the following: If [imath]P:V\to V[/imath] is a linear operator and [imath]P^2=P[/imath] and [imath]\|Pv\|\le\|v\|[/imath] for all [imath]v \in V[/imath] then [imath]P[/imath] is an orthogonal projection. My thoughts on this are that a linear map with the following properties is an orhtogonal projection: [imath]null(P) [/imath] is the orthogonal complement of its range, [imath]P^2 = P[/imath] and [imath]\|Pv\|\le\|v\|[/imath]. Therefore it should be enough to show that the range of [imath]P[/imath] is orthogonal to its null space. How to do this? Let [imath]n \in null(P)[/imath] and [imath]u \in range(P) = U[/imath]. Want to show: [imath]\langle u,n \rangle = 0[/imath]. Am I doing this right? And how can I show [imath]\langle u,n \rangle = 0[/imath]? | 21668 | Orthogonal Projection
Seems like I still don't get it, I think I am missing something important. Let [imath]V[/imath] be an [imath]n[/imath] dimensional inner product space ([imath]n \geq 1[/imath]), and [imath]T\colon\mathbf{V}\to\mathbf{V}[/imath] be a linear transformation such that: [imath]T^2 = T[/imath] [imath]||T(a)|| \leq ||a||[/imath] for every vector [imath]a[/imath] in [imath]\mathbf{V}[/imath]; Prove that a subspace [imath]U \subseteq V[/imath] exist, such that [imath]T[/imath] is the orthogonal projection on [imath]U[/imath]. Now, I know these things: The fact hat [imath]T^2 = T[/imath] guarantees that [imath]T[/imath] is indeed a projection, so I need to prove that T is an orthogonal projection (I guess this is where [imath]||T(a)|| \leq ||a||[/imath] kicks in). To do this I can prove that: For every [imath]v[/imath] in [imath]ImT^{\perp}[/imath], [imath]T(v) = 0[/imath] Alternatively, I can prove that for every [imath]v[/imath] in [imath]ImT[/imath] and [imath]u[/imath] in [imath]KerT[/imath], [imath](v,u)=0[/imath]. [imath]T[/imath] is self-adjoint (according to Wikipedia) The matrix [imath]A = [T]_{E}[/imath] when [imath]E[/imath] is an orthonormal basis, is hermitian (this is equivalent to the previous point). What else? I've been thinking about it for quite some time now, and I'm pretty sure there is something big I'm missing, again. I just don't know how to use the data to prove any of these things. Thanks! |
1862343 | (Non-)Canonicity of using zeta function to assign values to divergent series
This article http://blogs.scientificamerican.com/roots-of-unity/does-123-really-equal-112/ got me thinking about the "identity" [imath]1 + 2 + 3 + \cdots = -1/12,[/imath] and I wanted to convince myself there was nothing particularly unique about this identity or the Riemann zeta construction. More precisely, this identity only really makes sense if you think of an integer [imath]n[/imath] as being the specialization at [imath]z=-1[/imath] of the function [imath]n^{-z}[/imath]. So here's a question: For any complex number [imath]c[/imath], does there exist a domain [imath]\Omega \subset \mathbb{C}[/imath], and analytic functions [imath]F(n, s)_{n\in\mathbb{N}}[/imath] and [imath]f(s)[/imath] on [imath]\Omega[/imath], such that the following hold i. [imath]F(n,0) = n[/imath] ii. [imath]\sum_{n=1}^\infty F(n,s) = f(s)[/imath] on [imath]\Omega[/imath] in some reasonable sense (maybe converges uniformly on compact subsets of [imath]\Omega[/imath]?) iii. [imath]f[/imath] can be extended holomorphically to some domain containing both [imath]\Omega[/imath] and [imath]0[/imath] such that [imath]f(0) = c[/imath]. So shifting the Euler series and Riemann zeta would be such a construction for [imath]c=-1/12[/imath]. As the question stands, I feel that the answer is almost certainly yes, although to be fair the functions [imath]n^{-s}[/imath] have a lot more structure than "holomorphic functions on some domain". So a follow-up question would be: are there "natural" additional constraints for which the answer to this question is No? I apologize that this is kind of open-ended, but the goal is to convince myself that there is nothing particularly canonical about [imath]-1/12[/imath] (or to hear an explanation of why it is canonical). | 1854642 | Can different choices of regulator assign different values to the same divergent series?
Physicists often assign a finite value to a divergent series [imath]\sum_{n=0}^\infty a_n[/imath] via the following regularization scheme: they find a sequence of analytic functions [imath]f_n(z)[/imath] such that [imath]f_n(0) = a_n[/imath] and [imath]g(z) := \sum_{n=0}^\infty f_n(z)[/imath] converges for [imath]z[/imath] in some open set [imath]U[/imath] (which does not contain 0, or else [imath]\sum_{n=0}^\infty a_n[/imath] would converge), then analytically continue [imath]g(z)[/imath] to [imath]z=0[/imath] and assign [imath]\sum_{n=0}^\infty a_n[/imath] the value [imath]g(0)[/imath]. Does this prescription always yield a unique finite answer, or do there exist two different sets of regularization functions [imath]f_n(z)[/imath] and [imath]h_n(z)[/imath] that agree at [imath]z=0[/imath], such that applying the analytic continuation procedure above to [imath]f_n(z)[/imath] and to [imath]h_n(z)[/imath] yields two different, finite values? |
2712187 | Prove that: [imath]e^{-\frac{|t|}{2}}= \int_{\mathbb R} \frac{e^{ixt}}{1+x^2} dx.[/imath]
I need help for prove that [imath]e^{-\frac{|t|}{2}}= \int_{\mathbb R} \frac{e^{ixt}}{1+x^2} dx, \quad \forall t\in \mathbb R.[/imath] Thank you in advance | 846961 | Cauchy distribution characteristic function
I know that it's easy to calculate integral [imath]\displaystyle\int_{-\infty}^\infty \frac{e^{itx}}{\pi(1+x^2)} \, dx[/imath] using residue theorem. Is there any other way to calculate this integral (for someone who don't know how to use residue theorem)? |
2711971 | Find the tangent planes to [imath]f(x,y) = 2+x^2 + y^2[/imath] that contain the [imath]x[/imath] axis
I am asked to find the tangent planes to [imath]f(x,y) = 2+x^2 + y^2[/imath] that contain the [imath]x[/imath] axis. Knowing that the [imath]x[/imath] axis has a direction vector [imath]\left< 1,0,0 \right>[/imath]. That said, the planes must contain that vector. I'm not sure how I would use the plane approximation [imath]L = f(x_0, y_0) + 2x (x-x_0) + 2y(y-y_0)[/imath] to get to the answer. The answer from the textbook is \begin{align} ax + 2\sqrt{2} - cz + d &=0\\ ex - 2\sqrt{2} -gz + h &= 0 \end{align} (so finding [imath]a[/imath], [imath]c[/imath], [imath]d[/imath], [imath]e[/imath], [imath]g[/imath] and [imath]h[/imath]) Any guidance is highly appreciated | 1401716 | Tangent planes to [imath]2+x^2+y^2[/imath] and that contains the [imath]x[/imath] axis
I need to find the tangent planes to [imath]f(x,y) = 2+x^2+y^2[/imath] and that contains the [imath]x[/imath] axis, so that's what I did: [imath]z = z_0 + \frac{\partial f(x_0,y_0)}{\partial x}(x-x_0)+\frac{\partial f(x_0,y_0)}{\partial y}(y-y_0) \implies \\ z = 2 + x_0^2 + y_0^2 + 2x_0(x-x_0) + 2y_0(y-y_0) \implies \\ 2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0[/imath] So since the plane must contain the [imath]x[/imath] axis, its normal vector must have the form [imath](0,y,z)[/imath]. The normal vector fot the plane I found is: [imath](2x_0, 2y_0, -1)[/imath] so [imath]x_0 = 0, y_0 = y_0[/imath] our plane has the form: [imath]2y_0y -z -y_0^2+2 = 0[/imath] but when I plot this graph for some values of [imath]y_0[/imath] I only get 1 tangent plane at [imath]y_0\approx 1.5[/imath] |
2712353 | Bridge Between Series and Set Theory
I was wondering if it was possible to express infinite divergent sums using set theory (I do apologize if some of the notation is improper) I wanted to consider [imath]\sum_{n=1}^{\infty}n[/imath] i.e. the sum of all natural numbers. I thought you might be able to express this by adding up the cardinality of sets with n elements. So [imath]card(\{1\}) +card(\{2,3\})+card(\{4,5,6\})+...[/imath] and so on. This then would equal [imath]card(\{1\} \bigcup\{2,3\}\bigcup\{4,5,6\}\bigcup...)=\mathbb{N}[/imath] Hence [imath]\sum_{n=1}^{\infty}n=\aleph_0[/imath] Is my math correct and can this be generalized to compute other infinite sums, even those which involve terms with decimals? | 844818 | Is the sum of all natural numbers countable?
I do not even know if the question makes sense. The point is rather simply. If I have the sum of all natural numbers, [imath]\sum_{n\in \mathbb{N}}n[/imath] this is clearly "equal to infinity". But since almost a century ago, we know that there are (a lot of) different "infinities". So, is this sum equal to something countable or something bigger? I tried to look for references, but couldn't find anything and, since I am not an expert in Logic, Set Theory or Foundations of Mathematics, I thought that it would be good to ask here. Thanks in advance, Davide PS: This question is about sum of cardinals. It is not about Calculus. |
2712648 | Proving an expression is an integer (number theory)
Suppose that [imath]r[/imath] is a real number such that [imath]r + \frac{1}{r}[/imath] is an integer. Prove that [imath]r^{2017} + \frac{1}{r^{2017}}[/imath] is an integer. I thought of using prove by induction, but I don't think it really suits in this case. I've also thought of using modulos in some kind of way but not sure how. Also, can we assume that both [imath]r^{2017}[/imath] and [imath]\frac{1}{r^{2017}}[/imath] are both integers, since adding them make up an integer? | 1041134 | If [imath]a+1/a[/imath] is an integer, then so is [imath]a^t+1/a^t[/imath] for [imath]t\in\mathbb N[/imath]
I need to show if [imath]a[/imath] is in [imath]\mathbb{R}[/imath] but not equal to [imath]0[/imath], and [imath]a+\dfrac{1}{a}[/imath] is integer, [imath]a^t+\dfrac{1}{a^t}[/imath] is also an integer for all [imath]t\in\mathbb N[/imath]. Can you provide me some hints please? |
2712333 | Proof that a [imath]k[/imath]-regular bipartite graph has a [imath]1[/imath]-factor
So I'm trying to proof that a [imath]k[/imath]-regular bipartite graph has a [imath]1[/imath]-factor, for [imath]k>0[/imath]. Proof: Let [imath]G=(V,E)[/imath] be a [imath]k[/imath]-regular bipartite graph such that [imath]E\subset\{e=\{a,b\}\in E\::a\in U, b\in W\}[/imath] for certain [imath]U,W\subset V[/imath]. Then certainly [imath]|U|=|W|:=n[/imath] and thus [imath]|V|=2n[/imath] since [imath]U\cap W=\varnothing[/imath]. Because [imath]G[/imath] is [imath]k[/imath]-regular, [imath]G[/imath] is [imath]k[/imath]-edge colourable. Because [imath]|E|=\frac12\sum_{v\in V}d(v)=\frac122nk=nk[/imath], every colour is used on exactly [imath]n[/imath] edges, since in that case the colour is used on [imath]2n[/imath] different vertices, and it cannot be used on any vertex twice. Choose a colour [imath]i[/imath] and let [imath]E_i=\{e\in E \ | \ k(e)=i\}[/imath], then [imath]G_i=(V,E_i)[/imath] is a [imath]1[/imath]-factor. Done. Question: Is this proof correct, and if not how do I proof the statement? And second: obviously not every [imath]k[/imath]-regular graph has a [imath]1[/imath]-factor, so if my proof works, which part of my proof does not extend to any [imath]k[/imath]-regular graph? Is it the part where I conclude that we have an even amount of edges? And if so, does a [imath]k[/imath]-regular graph on an even amount of vertices contain a [imath]1[/imath]-factor? Edit: My proof does not seem exactly the same to me as the linked one; also I don't quite understand the notation and much of the terminology in the linked post, I have only followed an 80-hour course on graph theory, covering an introduction, colouring and directed graphs/edges/menger's theorem and the like. I'm looking for some feedback on my proof; is it correct? | 856201 | Show that a finite regular bipartite graph has a perfect matching
Some preliminaries: A matching in a bipartite graph with vertex set [imath]X \cup Y[/imath] is a subset [imath]E_1[/imath] of the edge set such that no vertex is incident with more than one edge in [imath]E_1[/imath] A complete matching in a bipartite graph is a matching such that every vertex in [imath]X[/imath] is incident with an edge in [imath]E_1[/imath]. A perfect matching in a graph [imath]G[/imath] (not necessarily bipartite) is a matching such that each vertex of [imath]G[/imath] is incident with one edge of the matching. Theorem: A necessary and sufficient condition for there to be a complete matching from [imath]X[/imath] to [imath]Y[/imath] in [imath]G[/imath] is that [imath]|Γ(A)|≥|A|[/imath] for every [imath]A ⊆ X[/imath]. Here [imath]\Gamma(A)[/imath] is used to denote the set of all neighbors of vertices in [imath]A[/imath]. Here's my proof. Can someone please verify it or improve it? I feel that it is a bit hand-waving and lacks much-needed rigor. Show that a finite regular bipartite graph has a perfect matching Let [imath]X[/imath] and [imath]Y[/imath] be the (disjoint) vertex sets of the bipartite graph. Let [imath]A \subseteq X[/imath]. Then, there are [imath]d|A|[/imath] edges incident with a vertex in [imath]A[/imath]. But then, [imath]|\Gamma(A)| \geq |A|[/imath]. If that were not the case, then there would exist a vertex [imath]v \in \Gamma(A)[/imath] with [imath]\operatorname{deg}(v) > d[/imath]. So, there exists a complete matching between the vertex sets [imath]X[/imath] and [imath]Y[/imath] of the bipartite graph. Note that this matching is perfect, since [imath]|X| = |Y|[/imath]. |
2713079 | Sum of squares in a general field
Can we prove that for any field [imath]K[/imath], the following two are equivalent? i) for any [imath]a,b,c\in K[/imath], [imath](a,b,c)\neq(0,0,0)\implies a^2+b^2+c^2\neq0[/imath] ii) for any [imath]a,b,c,d\in K[/imath], [imath](a,b,c,d)\neq(0,0,0,0)\implies a^2+b^2+c^2+d^2\neq0[/imath] What I have done: since [imath]-1[/imath] is a sum of two squares in any [imath]\mathbb F_p[/imath] with [imath]p[/imath] a prime number. Such [imath]K[/imath] must be of characteristic [imath]0[/imath]. | 133883 | Is it true in an arbitrary field that [imath]-1[/imath] is a sum of two squares iff it is a sum of three squares?
Here's a statement from Lam's First Course in Noncommutative Rings. (Paraphrased) Let [imath]k[/imath] be a field. Then the following conditions are equivalent. [imath](\forall a,b,c,d\in k)\;\;(a,b,c,d)\neq 0\implies a^2+b^2+c^2+d^2\neq 0;\tag 1[/imath] [imath]-1\text{ is not a sum of two squares.}\tag2[/imath] I can't prove this fact. If I replace [imath](2)[/imath] with [imath]-1\text{ is not a sum of three squares,}\tag3[/imath] then the equivalence between [imath](1)[/imath] and [imath](3)[/imath] is easy to prove. Indeed, suppose [imath](1)[/imath] is false. Then we may assume without loss of generality that there are [imath]a,b,c,d\in k[/imath] with [imath]a\neq 0[/imath] such that [imath]a^2+b^2+c^2+d^2=0.[/imath] But then [imath]-1=\left(\frac ba\right)^2+\left(\frac ca\right)^2+\left(\frac da\right)^2,[/imath] and so [imath](3)[/imath] is false. Conversely, if [imath](3)[/imath] is false, then there exist [imath]x,y,z\in k[/imath] such that [imath]-1=x^2+y^2+z^2[/imath] and therefore [imath]0=x^2+y^2+z^2+1^2,[/imath] which means that [imath](1)[/imath] is false. I can also prove [imath](1)\implies (2).[/imath] Suppose [imath](2)[/imath] is false. Then [imath]-1=x^2+y^2,[/imath] whence [imath]0=x^2+y^2+1^2+0^2,[/imath] which implies that [imath](1)[/imath] is false. But I can't prove [imath](2)\implies (1)[/imath] and I'm starting to suspect that it may actually be false. But I can't find a counterexample either. It would require finding a field in which [imath]-1[/imath] is a sum of three squares but not a sum of two squares. But in finite fields, every element is a sum of two squares, as proven here. In the subfields of [imath]\mathbb R[/imath] on the other hand, [imath]-1[/imath] is not a sum of three squares. And in [imath]\mathbb C,[/imath] it is a sum of two squares: [imath]-1=0^2+i^2.[/imath] In general, an example cannot be algebraically closed, which leaves me with the fields of rational functions as the only other fields I know. Also, the example cannot have characteristic [imath]2[/imath] as then [imath]-1=1=1^2+0^2.[/imath] Edit: The theorem Gerry Myerson cites in his answer is proven here. A Wikipedia article on the Stufe is here. Both links have been given by joriki in the comments. |
2713714 | How to prove that [imath]X^{T}X[/imath] is singular if [imath]X[/imath] is singular.
Let's suppose [imath]X[/imath] is [imath]n \times m[/imath] not full rank matrix (ie any column vector in [imath]X[/imath] can be expressed as linear combination of others column vectors). If [imath]B = X^TX[/imath] why [imath]B[/imath] is singular? The question is inspired from the statistics, wherein, if vector of independent variable is linearly dependent, then it is assumed that [imath]X^TX[/imath] is also dependent (or singular). It is difficult to grasp if [imath]X[/imath] is dependent then how [imath]X^TX[/imath] is dependent? | 526797 | If [imath]X[/imath] is a singular matrix, then is [imath]X'X[/imath] singular?
If [imath]X[/imath] is a singular matrix, then is [imath]X'X[/imath] singular? And why? (' means transpose) Thanks in advance! |
2713431 | If the series [imath]\sum_{n=1}^{\infty} na_{n} [/imath] converges, then [imath]\sum_{n=1}^{\infty} na_{n+1} [/imath] also converges?
I was tasked with proving or disproving the following statement: If the series [imath]\sum_{n=1}^{\infty} na_{n} [/imath] converges, then [imath]\sum_{n=1}^{\infty} na_{n+1} [/imath] also converges. I tried to disprove this using [imath]\sum_{n=1}^\infty\frac{1}{n}[/imath] and the fact it diverges, and turning it to [imath]\sum_{n=1}^\infty\frac{1}{n^3} \cdot n[/imath], but that didn't work. Intuitively the statement doesn't sound right because it is too specific. | 588009 | If [imath]\sum\limits_{n=1}^\infty na_n[/imath] converges , does [imath]\sum\limits_{n=1}^\infty na_{n+1}[/imath] converge?
I ask for some help with this question: Prove or provide counter example: If [imath]\sum\limits_{n=1}^\infty na_n[/imath] converges then [imath]\sum\limits_{n=1}^\infty na_{n+1}[/imath] also converges. I tries this way: If [imath]\sum\limits_{n=1}^\infty na_n[/imath] converges then [imath]na_n \to 0[/imath], therefore [imath]a_n \to 0[/imath]. There are 3 possible cases: 1) If [imath]a_n >0 [/imath] and [imath]a_n[/imath] is monotonic decreasing sequence then [imath]na_{n+1}<na_n[/imath] and [imath]\sum_{n=1}^\infty na_{n+1}[/imath] converges by Comparison Test. 2) If [imath]a_n >0 [/imath] and [imath]a_n[/imath] is not monotonic decreasing sequence : it is not possible that [imath]a_{n+1}>a_n[/imath] because in this case [imath]a_n \to \infty[/imath], therefore it must be [imath]a_{n+1} \le a_n[/imath] and [imath]\sum_{n=1}^\infty na_{n+1}[/imath] converges by Comparison Test. 3) If [imath]a_n[/imath] is sign-alternating series. There I have a problem to find a solution. Thanks. |
2430101 | Torsion subgroup of [imath]\mathbb{Z}\times\mathbb{Z}_n[/imath]
Fix some [imath]n\in{\mathbb{Z}}[/imath] with [imath]n>1[/imath]. Find the torsion subgroup of [imath]\mathbb{Z}\times\mathbb{Z}_n[/imath]. Show that the set of elements of infinite order together with the identity is not a subgroup. I've seen a solution of this where [imath]0\times\mathbb{Z}_n[/imath] is the torsion subgroup, but there was no explanation. I don't understand where this comes from or what it means, and I'm not even sure what the group operation is. | 198160 | Subgroups of [imath]\mathbb Z \times(\mathbb Z/n\mathbb Z)[/imath]
So I am dealing with a problem from Dummit (specifically 2.1.7) and am having some issues. The part of the problem in question is: Prove the set of elements of the direct product [imath]\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})[/imath] of infinite order together with the identity is not a subgroup of [imath]\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})[/imath]. I am assuming this question is referring to addition as the binary operation for both sets. I know the proof basically involves showing that the product is not closed; however, do not all the elements of the integers have infinite order (except the identity)? In that case, the elements of infinite order together with the identity would form all of [imath]\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})[/imath] would it not? Thanks |
2714126 | expectation of maximum of iid random variables from normal distribution.
1) How to find expectation of max of random variables , i.e : [imath]\mathbb{E}[max(x_1,x_2,\dots,x_n)][/imath] where [imath]x[/imath] are IID random variables from [imath]\mathcal{N}(\mu,\sigma^2)[/imath]. I know that CDF is [imath]F(x)^n[/imath] and PDF is [imath]nF(x)^{n-1}f(x)[/imath]. I have also seen simplifications for uniform and exponential distributions but not for Gaussian distribution. 2) In general, How to solve [imath]\int_{x=0}^{\infty} [nx F(x)^{n-1}f(x)] dx[/imath] | 89030 | Expectation of the maximum of gaussian random variables
Is there an exact or good approximate expression for the expectation, variance or other moments of the maximum of [imath]n[/imath] independent, identically distributed gaussian random variables where [imath]n[/imath] is large? If [imath]F[/imath] is the cumulative distribution function for a standard gaussian and [imath]f[/imath] is the probability density function, then the CDF for the maximum is (from the study of order statistics) given by [imath]F_{\rm max}(x) = F(x)^n[/imath] and the PDF is [imath]f_{\rm max}(x) = n F(x)^{n-1} f(x)[/imath] so it's certainly possible to write down integrals which evaluate to the expectation and other moments, but it's not pretty. My intuition tells me that the expectation of the maximum would be proportional to [imath]\log n[/imath], although I don't see how to go about proving this. |
2714822 | Prove that if a prime [imath]p[/imath] divides [imath]a^2 + b^2[/imath] then it must divide both a and b.
I proved this by taking cases for different remainders and using modular arithmetic, when the prime is [imath]3[/imath].However this method won't give a general proof for all primes.I was eager to get a proof for all primes [imath]p[/imath], but i couldn't get one.I have seen in Pythagorean triples that if both [imath]a[/imath] and [imath]b[/imath] are not divisible by [imath]p[/imath], [imath]c=a^2 + b^2[/imath] is also not, but how do i prove it? Proving by contradiction may help...maybe induction. Also we can try proving for all numbers. of the form [imath]6n+1[/imath] and [imath]6n-1[/imath] so that it becomes evident for all primes. The question changed and now is about proving that for all primes equivalent to [imath]3 \bmod 4.[/imath] | 105034 | show if [imath]n=4k+3[/imath] is a prime and [imath]{a^2+b^2} \equiv 0 \pmod n[/imath] , then [imath]a \equiv b \equiv 0 \pmod n[/imath]
[imath]n = 4k + 3 [/imath] We start by letting [imath]a \not\equiv 0\pmod n[/imath] [imath]\Rightarrow[/imath] [imath]a \equiv k\pmod n[/imath] . [imath]\Rightarrow[/imath] [imath]a^{4k+2} \equiv 1\pmod n[/imath] Now, I know that the contradiction will arrive from the fact that if we can show [imath]a^2 \equiv 1 \pmod n [/imath] then we can say [imath]b^2 \equiv -1 \pmod n [/imath] then it is not possible since solution exists only for [imath]n=4k_2+1 [/imath] so [imath]a \equiv b\equiv 0 \pmod n [/imath] So from the fact that [imath]a^{2^{2k+1}} \equiv 1 \pmod n[/imath] I have to conclude something. |
2715210 | Why can reversing the order of quantifiers in a statement change its truth value?
The statement [imath]\forall x \in \mathcal{R} \exists n \in \mathcal{N} (n>x)[/imath] Is true, but [imath]\exists n \in \mathcal{N} \forall x \in \mathcal{R} (n>x)[/imath] Is false. Why does reversing the order of the quantifiers cause the first statement to become false? | 20248 | Does the order of quantification matter?
I'm working with the universal and the existential quantifiers and I was wondering is there an order of operations that applies to them? What is the difference between [imath](\forall x)P(x)(\exists x)Q(x)[/imath] And [imath](\exists x)Q(x)(\forall x)P(x)[/imath] Lets say we have this relationship (a) [imath][(\forall x)P(x) \vee (\exists x)Q(x)] \implies (\forall x)[P(x) \vee Q(x) ][/imath] |
2715104 | Proving [imath]Gal(K/\mathbb{Q})[/imath] is [imath]D_4[/imath].
Let [imath]a = \sqrt{2+i}[/imath] and [imath]K[/imath] is the splitting field of minimal polynomial of [imath]a[/imath] over [imath]\mathbb{Q}[/imath]. Prove that [imath]Gal(K/\mathbb{Q})[/imath] is [imath]D_4[/imath]. I find the minimal polynomial of [imath]a[/imath] is [imath]p(x)=x^4-4x^2+5[/imath] and its roots are [imath]\sqrt{2+i},-\sqrt{2+i},\sqrt{2-i},-\sqrt{2-i}[/imath]. Let [imath]b=\sqrt{2-i}[/imath]. So the splitting field of [imath]p[/imath] is [imath]K=\mathbb{Q}(a,b)[/imath]. Also by rational root theorem, [imath]p[/imath] is irreducible over [imath]\mathbb{Q}[/imath]. Thus [imath][\mathbb{Q}(a):\mathbb{Q}]=4[/imath]. Also [imath]b\not\in\mathbb{Q}(a)[/imath], and minimal polynomial of [imath]b[/imath] over [imath]\mathbb{Q}(a)[/imath] is [imath]x^2-2+i[/imath]. Thus [imath][\mathbb{Q}(a,b):\mathbb{Q}]=[\mathbb{Q}(a,b):\mathbb{Q}(a)]\cdot[\mathbb{Q}(a):\mathbb{Q}]=2\cdot 4=8.[/imath] Now since [imath]K[/imath] is the splitting field over [imath]\mathbb{Q}[/imath] of a separable polynomial, [imath]K/\mathbb{Q}[/imath] is Galois. Hence [imath]|Gal(K/\mathbb{Q})|=8[/imath]. Hence the 8 automorphisms are [imath]a\to \begin{cases}a\\-a\\b\\-b\end{cases}\quad\text{and}\quad b \to \begin{cases}b\\-b\end{cases}.[/imath] From this how to conclude [imath]Gal(K/\mathbb{Q})[/imath] is isomorphic to [imath]D_4[/imath]? Thanks | 1231921 | Galois group of [imath]x^4-2[/imath]
I am trying to explicitly compute the Galois group of [imath]x^4-2[/imath] over [imath]\mathbb{Q}[/imath]. I found that the resolvent polynomial is reducible and the order of the Galois group is [imath]8[/imath] using the splitting field [imath]K=\mathbb{Q}(2^{1/4}, i)[/imath]. Hence I need to find 8 automorphisms. Thus do I just map elements of the same order to each other that fix the base field [imath]\mathbb{Q}[/imath] [imath]2^{1/4}[/imath] to [imath]i2^{1/4}[/imath] |
2709895 | Prove that a certain element of the commutator subgroup of [imath]F_4[/imath] is not a commutator
Let [imath]F_4[/imath] be the free group on the letters [imath]a,b,c,d[/imath]. I would like to prove that the element [imath][a,b][c,d] = aba^{-1}b^{-1}cdc^{-1}d^{-1}[/imath] is not equal to [imath][x,y] = xyx^{-1}y^{-1}[/imath] for any elements [imath]x,y \in F_4[/imath]. | 59816 | Why is the set of commutators not a subgroup?
I was surprised to see that one talks about the subgroup generated by the commutators, because I thought the commutators would form a subgroup. Some research told me that it's because commutators are not necessarily closed under product (books by Rotman and Mac Lane popped up in a google search telling me). However, I couldn't find an actual example of this. What is one? The books on google books made it seem like an actual example is hard to explain. Wikipedia did mention that the product [imath][a,b][c,d][/imath] on the free group on [imath]a,b,c,d[/imath] is an example. But why? I know this product is [imath]aba^{-1}b^{-1}cdc^{-1}d^{-1}[/imath], but why is that not a commutator in this group? |
2715098 | Does the zero product property hold in vector spaces?
Suppose [imath]V[/imath] is a vector space over a field [imath]F[/imath]. Let [imath]v \in V\setminus \{0\}[/imath] and [imath]\lambda \in F[/imath]. Does [imath]\lambda v= 0[/imath] imply [imath]\lambda = 0[/imath]? | 2279978 | Does [imath]\lambda[/imath] [imath]\mathbf u[/imath] [imath]= 0[/imath] imply [imath]\lambda = 0[/imath] or [imath]\mathbf u[/imath] [imath]= 0[/imath]?
Does [imath]\lambda \mathbf u = 0[/imath] imply [imath]\lambda = 0 [/imath] or [imath]\mathbf u= 0[/imath] where [imath]\mathbf u[/imath] is in a vector space [imath]V[/imath] over a field [imath]F[/imath]. The case for the vector equalling zero is trivial due to the fact field axioms include a multiplicative inverse, so how could I show lambda is 0? |
2714969 | Prove if A is a set finite and B is infinite then [imath]|A|+|B|=|B|[/imath]
Prove if [imath]A[/imath] is a finite set and [imath]B[/imath] is infinite then [imath]|A|+|B|=|B|[/imath] My work I was thinking in solve this exercise by contradiction. Suppose [imath]|A|+|B|\not = |B|[/imath] then [imath]|A|+|B|=|A|[/imath] Here im a little stuck. Can someone help me? | 1412470 | How to prove that if [imath]A[/imath] is infinite and [imath]B[/imath] is finite, then [imath]|A\cup B|=|A|[/imath]?
I'm studying logic and unfortunately, I'm a newbie at this, so I don't see the stuff everyone sees at the moment. I want to solve following exercise, but get nowhere: Let [imath]A[/imath] be an infinite set and let [imath]B[/imath] be a finite set. Use the AC to prove that [imath]|A| +|B| = |A|[/imath]. I've proved this one though: If [imath]X[/imath] is an infinite set, there is an injective function [imath]\mathbb{N} \rightarrow X[/imath], hence [imath]\omega \leq |X|[/imath]. But I don't see how I could use this. What's the logic here? |
2716285 | Prove [imath]\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})[/imath] is divergent.
Evaluate if the following series is convergent or divergent: [imath]\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})[/imath]. I tried to evaluate the divergence, applying the Weierstrass comparison test: [imath]\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})>\sum\limits_{n=1}^\infty \ln({\frac 1 {{n}}})[/imath]. Since the function is decreasing then as [imath]1+\frac{1}{n}[/imath] is closer to [imath]0[/imath] then the inequality follows. Obviously [imath]\sum\limits_{n=1}^\infty \ln({\frac 1 {{n}}})[/imath] diverges since [imath]\lim_{n\to\infty}\ln({\frac 1 {{n}}})=-\infty[/imath]. Questions: 1) Is my answer right? If not why? 2) What other kind of approach do you propose? Thanks in advance! | 1876350 | Ways of showing [imath]\sum_\limits{n=1}^{\infty}\ln(1+1/n)[/imath] to be divergent
Show that the following sum is divergent [imath]\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)[/imath] I thought to do this using Taylor series using the fact that [imath] \ln\left(1+\frac1n\right)=\frac1n+O\left(\frac1{n^2}\right) [/imath] Which then makes it clear that [imath] \sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)\sim \sum_{n=1}^{\infty}\frac1n\longrightarrow \infty [/imath] But I feel like I overcomplicated the problem and would be interested to see some other solutions. Also, would taylor series be the way you would see that this diverges if you were not told? |
2715580 | If last two digits of [imath] 7^{7^{7}}=k[/imath], find [imath]k-38[/imath]?
I have managed to get the answer as [imath]07[/imath] using modular arithmetic but answer seems wrong to me because the answer after evaluation of [imath]k-38[/imath] is [imath]5[/imath]. Please help. | 390685 | The last 2 digits of [imath]7^{7^{7^7}}[/imath]
What is the calculation way to find out the last [imath]2[/imath] digits of [imath]7^{7^{7^7}}[/imath]? WolframAlpha shows [imath]...43[/imath]. |
2717172 | Will Fermat's last theorem hold true on the Complex Plane?
Fermat's Last Theorem In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation [imath]a^n + b^n = c^n[/imath] for any integer value of [imath]n[/imath] greater than two. Now the question is will Fermat's last theorem hold true if we extend the question to the complex plane. Ie when [imath]a[/imath], [imath]b[/imath] or [imath]c[/imath] can be complex numbers. Why or Why not and is there any prove to it? | 466705 | Solutions in complex number field, instead of [imath]\mathbb{N}[/imath], to Fermat's Last Theorem
As far as I know, we are searching solutions in set of positive integers for [imath]x^n + y^n = z^n[/imath] for [imath]n > 2[/imath]. There are many proofs are stated that Fermat is true and there is no solutions for this equations. Now my question is, can we have solutions in complex numbers field? If yes, how to find them. Thanks in advance. |
2713888 | Alternative to Standard Deviation as an Average Spread
I have heard on numerous occasions that the Standard Deviation of a data set is meant to represent "the average spread of the data around the mean", and if that's the case why don't we use this formula instead which yields a much more precise representation of the average spread around the mean: [imath]\frac{\sum_{i=1}^n \Bigl(+\sqrt{(x_i-\overline{x})^2}\Bigl)}{n}[/imath] where [imath]n[/imath] is the total number of data values , and the [imath]+\sqrt{}[/imath] symbolises taking explicitly the positive square root of [imath](x_i-\overline{x})^2[/imath] Take the data set [imath](1,2,2,2,5,5,7,8)[/imath] for example. My formula produces the value 2.25, whereas the traditional Standard Deviation formula, [imath]s=\sqrt{\frac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1}}[/imath] produces the value [imath]2.619[/imath] to 3.d.p. I am aware that simplifying the expression inside of the sum using laws of indices results in [imath]\Bigl(+(x_i-\overline{x})\Bigl)[/imath] which would result in the sum being equal to zero. See my proof of this result here. My question is, is the reason I just stated, why this forumla is not used in place of the Standard Deviation, or can this formula be interpreted in the way I intend it to, and if so why is it not used? | 1200211 | Standard Deviation. Why do we take the square root of the entire equation?
Please forgive my lack of maths knowledge, It is my understanding that: Standard Deviation is the average distance from the mean in a data set of numbers. Therefore it stands to reason that working out the standard deviation of the data set [imath]x_i = \{1,2,3,4,5\}[/imath] would involve the following. First working out the mean [imath]\mu(x_i) = 3[/imath] and Then working out the sum of the distance from the average [imath]\sum{|x_i-\mu|} = 6[/imath] then we do [imath]\frac{\sum{|x_i-\mu|}}{N} = \frac{6}{5} = 1.2[/imath] This means that, according to my method/thinking, 1.2 is the standard deviation. However when using the formula [imath]\sqrt{\frac{\sum{(x_i-\mu)}^2}{N}}[/imath] I get [imath]1.414[/imath] Can someone explain why I'm wrong in layman terms. Thankyou |
2717123 | Bisection Method - Number of steps required proof
I am currently reading [imath]\textit{Scientific Computing: An Introductory Survey}[/imath] by Michael Heath. In Section 5.2.1 when talking about the Bisection Method it is said that the number of iterations [imath]n[/imath] required to achieve a tolerance [imath]tol[/imath] is [imath] n = \left\lceil log_{2}\left( \frac{b-a}{tol} \right)\right\rceil [/imath] where the starting interval is [imath][ a, b][/imath]. [imath]\textbf{I would like to prove this claim but can't figure out where to start}[/imath]. I know I need to prove that [imath]n - 1< log_{2}\left( \frac{b-a}{tol} \right) \leq n [/imath] and I also know that the length of the interval after [imath]k[/imath] iterations is [imath]\frac{b-a}{2^{k}}[/imath] but I don't know where to go from there. Could anyone please give me a hint? | 646857 | Bisection Method number of steps for convergence
For the bisection method, determine the formula [imath]n \geq \frac{\log(b_0-a_0)-\log(\epsilon)}{\log(2)}-1[/imath] for the number of steps to guarantee that [imath]|{r-c_n}| < \epsilon.[/imath] This is one of the questions from my numerical analysis class. I understand how to use the bisection method, but I don't understand the analysis of the method or how to start deriving an answer. |
1378626 | Let [imath]T[/imath] be the set of all positive integer divisors of [imath]2004^{100}[/imath]. Size of largest subset [imath]S[/imath] of [imath]T[/imath] such that no element in [imath]S[/imath] divides another?
I am getting an answer slightly over [imath]100^2[/imath]. Is this right (working below), or is there a better way of selecting elements of S? The following question appeared on the 2004 Canada National Olympiad: Let [imath]T[/imath] be the set of all positive integer divisors of [imath]2004^{100}[/imath]. What is the largest possible number of elements of a subset [imath]S[/imath] of [imath]T[/imath] such that no element in [imath]S[/imath] divides any other element in [imath]S[/imath]? Factorising, [imath]2004 = 2^2\cdot3\cdot167[/imath]. So any element of [imath]T[/imath] is of the form [imath]2^p 3^q 167^r\text{ with }0\le{p}\le200,0\le{q}\le100,0\le{r}\le100 \tag{1}[/imath] We start by excluding from [imath]S[/imath] the element mapping to [imath]p=q=r=0[/imath] because [imath]1[/imath] divides any natural number. [imath]\text{The key to the method is to choose elements so that the sum }p+q+r\text{ is always the same. Then distinct elements cannot be multiples.}\\[/imath] [imath]\text{For }p=0:\text{ choose all }q,r\text{ s.t. }q+r=200:2^{0}3^{100}167^{100}[/imath] [imath]\text{For }p=1:\text{ choose all }q,r\text{ s.t. }q+r=199:2^{1}3^{100}167^{99},2^{1}3^{99}167^{100}\\{\cdots}[/imath] [imath]\text{For }p=100:\text{ choose all }q,r\text{ s.t. }q+r=100:2^{100}3^{100}167^{0},\dots,2^{100}3^{0}167^{100}[/imath] [imath]\text{For }p=101:\text{ choose all }q,r\text{ s.t. }q+r=99:2^{101}3^{99}167^{0},\dots,2^{101}3^{0}167^{99}\\{\cdots}[/imath] [imath]\text{For }p=200:\text{ choose all }q,r\text{ s.t. }q+r=0:2^{200}3^{0}167^{0}\\\text{So }\\ \max|S| = \sum\limits_{k=1}^{101}{k}+\sum\limits_{k=1}^{100}{k}=\frac{1}{2}101\cdot102+\frac{1}{2}100\cdot101=101^2=10201\\[/imath] | 1116160 | A set of integers whose elements all divide [imath]2015^{200}[/imath] but do not divide each other
Let [imath]S[/imath] be a set of natural numbers,such that each element divides [imath]2015^{200}[/imath] but for no two elements [imath]a[/imath] and [imath]b[/imath], [imath]a|b[/imath]. Find the maximum number of elements in [imath]S[/imath] . [imath]2015^{200}=(5\cdot 13\cdot 31)^{200}[/imath] Hence each element is of the form [imath]5^p13^q31^r[/imath] where [imath]p,q,r[/imath] lie between [imath]0[/imath] and [imath]200[/imath], inclusive.The given condition also says that for no two elements [imath]5^p13^q31^r[/imath] and [imath]5^a13^b31^c[/imath],[imath]p\le a,q\le b,r\le c[/imath]. I have tried to guess the solution by fixing the value of [imath]a[/imath] and then letting [imath]b[/imath] and [imath]c[/imath] run through the range of values that they are allowed to take.But I keep getting confused again and again. I will appreciate any sort of comment,hint or answer. EDIT: As has been noted in the comments below,a trivial upper bound is [imath]201^2[/imath]. But is this achievable?Can we show that it's not?Because then we may gain some insight. |
2717861 | Is factor ring field?
There is defined Z[i] = {a+bi|a,b [imath]\in[/imath] Z} with standard operations of addition and multiplications complex number. Question is, if factor ring Z[i]/(1-i) is field. How could I prove it? Do you have any hint? | 751418 | Is [imath]\langle1-i\rangle[/imath] a maximal ideal in [imath]\mathbb{Z}[i][/imath]?
I'm trying to show that [imath]\langle1-i\rangle[/imath] is a maximal ideal of [imath]\mathbb{Z}[i][/imath]. I started by assuming there is some ideal [imath]A[/imath] that properly contains [imath]\langle1-i\rangle[/imath], and then I want to show that [imath]1 \in A[/imath], and then I will have that [imath]A= \mathbb{Z}[i] [/imath], so [imath]\langle1-i\rangle[/imath] is maximal. Is this the right approach? Thanks Second Attempt: Because [imath]\forall a+bi \in \mathbb{Z}[i][/imath], we have [imath]a+bi = (a+b) -b(1-i)[/imath], then we have [imath]a+bi - (a+b) = -b(1-i) \in \langle1-i\rangle[/imath], which implies [imath]a+bi +\langle1-i\rangle = a+b +\langle1-i\rangle[/imath]. So every coset is of the form [imath]c +\langle1-i\rangle[/imath], where [imath]c\in \mathbb{R}[/imath]. Furthermore, [imath]-i(1-i)^2=2 \in \langle1-i\rangle[/imath]. Write [imath]c = 2q + r[/imath], where [imath]q,r\in \mathbb{R}[/imath] and [imath]r=0,1[/imath]. Then because [imath]2 \in \langle1-i\rangle[/imath], we have [imath]2q\in \langle1-i\rangle[/imath] and therefore [imath]2q= 2q+r-r = c-r\in \langle1-i\rangle[/imath], so [imath]c+\langle1-i\rangle = r+\langle1-i\rangle[/imath], and every coset is of the form [imath]0 + \langle1-i\rangle[/imath] or [imath]1 +\langle1-i\rangle[/imath]. Thus, [imath]1 +\langle1-i\rangle[/imath] is the only nonzero element, and is invertible with itself as its inverse. Thus, [imath]\mathbb{Z}[i]/\langle1-i\rangle[/imath] is a field with two elements, and by a theorem, [imath]\langle1-i\rangle[/imath] must be maximal. Is this correct? |
2717725 | Show that any upper bounded subset of [imath]\mathbb{Z}[/imath] has a unique maximal element
Show that any upper bounded subset of [imath]\mathbb{Z}[/imath] has a unique maximal element This is what I've managed to do so far. Let [imath]X \subseteq \mathbb{Z}[/imath] be a nonempty upper bounded nonempty subset of [imath]\mathbb{Z}[/imath]. Since [imath]\mathbb{Z} \subseteq \mathbb{R}[/imath] we have that [imath]X[/imath] is upper bounded in [imath]\mathbb{R}[/imath] hence [imath]\sup X[/imath] exists. Let [imath]\alpha = \sup X[/imath]. We claim [imath]\alpha = \max X[/imath]. This is where I get stuck, ultimately I have to show that [imath]\alpha \in X[/imath], and I'm not exactly sure how to go about doing that. | 915420 | Every non-empty subset of the integers which is bounded above has a largest element.
I was reading a proof about every non-empty subset of the integers which is bounded above has a largest element, but i have troubles in one step. Here is the proof: Since [imath]S[/imath] is a non-empty subset of [imath]\mathbb{Z}[/imath](hence [imath]\mathbb{R}[/imath]) which is bounded above, by the supremum property [imath]sup S[/imath] exists. Let [imath]w=supS[/imath], then we have to show that [imath]w \in S[/imath], so we suppose [imath]w \notin S[/imath], then we have [imath]w-1<w[/imath], then exists [imath]m\in S[/imath] such that [imath]w-1<m<w[/imath](here is my problem, why can we take [imath]m\in S[/imath] such that happens? we are in [imath]\mathbb{Z}[/imath], so that does not happens). Then he takes [imath]n\in S[/imath] such that [imath]w-1<m<n<w[/imath]. The inequality [imath]n<w[/imath] implies that [imath]-w<-n[/imath], and [imath]w-1<m[/imath], if we add up both inequalities we have [imath]-1<m-n[/imath] implies [imath]n-m<1[/imath], then [imath]0<n-m<1[/imath] which is a contradiction because there are no integers between [imath]0[/imath] and [imath]1[/imath] Can anybody explains me if that step is right?? How can I prove this if there is a problem with that step? Thanks! |
546586 | Determining the order of the kernel and image
Let [imath]G[/imath] be a finite group. Let [imath]G'[/imath] be a group and let [imath]\phi : G \to G'[/imath] be a homomorphism. Let [imath]K \leq G[/imath] be the kernel of [imath]\phi[/imath] and [imath]I \leq G'[/imath] be the image of [imath]\phi[/imath]. (a) Find a formula that relates the number of elements in [imath]G[/imath], [imath]K[/imath], and [imath]I[/imath]. (b) Suppose [imath]H \leq G[/imath], Find a formula for the number of elements of [imath]\phi(H)[/imath]. I'm not sure if (a) is some sort of application of the Main Homomorphism Theorem or not. I'm hoping to get some hints to sort of push me in the right direction. | 547276 | Application of Main Homomorphism Theorem
This is related to this question. I just didn't want a prolonged discussion in the comments. Let [imath]\phi: G \to G'[/imath] be a homomorphism. Let [imath]G[/imath] be a finite group. Let [imath]K \leq G[/imath] be the kernel of [imath]\phi[/imath]. Let [imath]I \leq G'[/imath] be the image of [imath]\phi[/imath]. Let [imath]H' \leq G'[/imath]. Find a formula relating the order of [imath]\phi^{-1}(H')[/imath] in terms of [imath]H', I, K[/imath]. Attempt at a solution: [imath]|\phi^{-1}(H')|=|H'\cap I|\cdot |K \cap \phi^{-1}(H')|[/imath]. My justification for this is as follows. [imath]\phi^{-1}(H')/(K \cap \phi^{-1}(H'))\cong H' \cap I[/imath] |
2718409 | Prove that [imath]\mathbb{Z}_5[/imath] is a field
I have looked all over but I can't really find a clear example of this... If [imath]\mathbb{Z}_5[/imath] is set [imath]\{0,1,2,3,4\}[/imath] prove that it is a field... I understand that from the table we can see that the set is commutative, associative, and it has an identity. I'm not sure how to show the inverse [imath]-a[/imath] is in [imath]F[/imath], but by the table we can see that for take [imath]2[/imath] is in [imath]F[/imath] then if the inverse exists [imath]-2[/imath] is in [imath]F[/imath], and [imath]2 + (-2) = 0[/imath] but in the table we can see that [imath]2 + 3 = 0[/imath] so then [imath]-2 = 3[/imath]? But I am just confused how that is possible, I know maybe it has something to do with mod but not sure how to tie it all together table | 1941945 | Show that if [imath]p[/imath] is prime then [imath]\Bbb Z_p[/imath] is a field
Check my proof please. Let [imath]\Bbb Z_p:=\Bbb Z/p\Bbb Z[/imath] be the quoting ring modulus [imath]p[/imath]. I want to prove that if [imath]p[/imath] is prime then [imath]\Bbb Z_p[/imath] is a field. Known facts about the quotient rings of the kind [imath]\Bbb Z_n[/imath] (I dont list all the characteristics of a ring, just some ones): They are commutative rings with unity. [imath][0]=[n][/imath] is the identity of addition or zero. Then [imath][n]+[a]=[a][/imath] and [imath][n]\cdot[a]=[n][/imath] for all [imath][a]\in\Bbb Z_n[/imath]. [imath][1][/imath] is the multiplicative identity or unity. Then [imath][1]\cdot[a]=[a][/imath] for all [imath][a]\in\Bbb Z_n[/imath]. The addition is defined as [imath][a]+[b]=[a+b][/imath], and the multiplication is defined as [imath][a]\cdot[b]=[ab][/imath]. [imath][a]=[b][/imath] means that exists [imath]z\in\Bbb Z[/imath] such that [imath]a=b+nz[/imath]. [imath]|\Bbb Z_p|=p[/imath]. A ring [imath]\Bbb Z_p[/imath] dont have zero divisors, i.e. doesnt exist [imath][a],[b]\in\Bbb Z_p[/imath] distinct of [imath][p][/imath] such that [imath][a]\cdot[b]=[p][/imath]. Proof by contradiction: suppose that exists such [imath][a],[b][/imath] distinct of [imath][p][/imath] that are zero divisors. Then exist some [imath]z_j\in\Bbb Z[/imath] such that: [imath](a+pz_1)(b+pz_2)=pz_3\iff ab+pz_4=pz_3\iff ab=pz_5[/imath] but this is a contradiction with the assumption that [imath]a,b\neq p[/imath] and [imath]p[/imath] prime, so dont exist divisors of zero on any [imath]\Bbb Z_p[/imath].[imath]\Box[/imath] For any [imath][a]\in\Bbb Z_p[/imath] distinct of [imath][p][/imath] exist [imath][b][/imath] such that [imath][a][b]=[1][/imath]. Because [imath]\Bbb Z_p[/imath] dont have zero divisors and is finite for some [imath][a]\neq[p][/imath] we have that if [imath][b]\neq[c][/imath] then [imath][a][b]\neq[a][c][/imath]. Proof by contradiction: if [imath][a][b]=[a][c][/imath] in the above conditions then [imath][a][c]=[ac]=[ab]=[a][b]\iff ac+pz_1=ab+pz_2\iff a(c-b)=pz_3[/imath] for some [imath]z_j\in\Bbb Z[/imath]. But [imath]pz_3\in[p][/imath], then [imath][a][c-b]=[p][/imath] but because there are no zero divisors in [imath]\Bbb Z_p[/imath] then or [imath][a]=[p][/imath] or [imath][c-b]=[p][/imath] what contradicts the above conditions.[imath]\Box[/imath] Because [imath][a][b]\neq[a][c][/imath] for [imath][a]\neq[p][/imath] and [imath][b]\neq[c][/imath] then for any [imath][a]\neq[p][/imath] we have that [imath][a]\cdot\Bbb Z_p=\Bbb Z_p\implies \exists [b]\in\Bbb Z_p:[a][b]=[1][/imath] In other words: every element of [imath]\Bbb Z_p[/imath] but [imath][p][/imath] have a multiplicative inverse as stated above.[imath]\Box[/imath] Because [imath]\Bbb Z_p[/imath] is a commutative ring with [imath][1]\neq[0][/imath] and have multiplicative inverse for all their elements but [imath][p]=[0][/imath] then [imath]\Bbb Z_p[/imath] is a field.[imath]\Box[/imath] |
2192550 | If U is a T-invariant subspace of V is there necessarily another T-invariant subspace[imath]-W S.T. U\oplus W = V?[/imath]
Suppose [imath]V[/imath] is a vector space over [imath]\Bbb F, T:V\to V[/imath] is a linear operator and [imath]U[/imath] is a T-invariant subspace of V. is there necessarily another T-invariant subspace[imath]-\,W\,\, S.T. \,U\oplus W = V?[/imath] I suspect this is in fact not true but cannot come up with a counter example. | 258502 | Do [imath]T[/imath]-invariant subspaces necessarily have a [imath]T[/imath]-invariant complement?
Suppose [imath]T[/imath] is a linear operator on some vector space [imath]V[/imath], and suppose [imath]U[/imath] is a [imath]T[/imath]-invariant subspace of [imath]V[/imath]. Does there necessarily exist a complement (a subspace [imath]U^c[/imath] such that [imath]V=U\oplus U^c[/imath]) in [imath]V[/imath] which is also [imath]T[/imath]-invariant? I'm curious because I'm wondering if, given such [imath]U[/imath], it is always possible to decompose the linear operator [imath]T[/imath] into the sum of its restrictions onto [imath]U[/imath] and [imath]U^c[/imath], but I don't know if such a [imath]T[/imath]-invariant [imath]U^c[/imath] exists. |
2719275 | Change derivative and limit order
I want to find the derivative of a function [imath]f[/imath]. If I manage to find a function [imath]g[/imath], whose derivative I know for all [imath]n[/imath] such that [imath]f(x)=\lim_{n\rightarrow \infty} g(x,n),[/imath] is it true that [imath]\frac{\partial f}{\partial x}(x)=\lim_{n\rightarrow \infty}\left( \frac{\partial g}{\partial x}g(x,n)\right)?[/imath] I'm asking this because I want to know [imath]\frac{\partial \theta}{\partial x_i}(||\vec{x}||),[/imath] where [imath]\theta[/imath] is the Heaviside step function. | 409178 | Can I exchange limit and differentiation for a sequence of smooth functions?
Let [imath](f_n)_{n\in \mathbb N}[/imath] be a sequence of smooth functions converging to some [imath]f[/imath]. Under what circumstances can I exchange limit and derivative?, i.e. [imath]\lim_{n\rightarrow \infty} \frac{\partial f_n(x)}{\partial x} = \frac{\partial f(x)}{\partial x}[/imath] |
2719200 | If the Sums of all [imath]k[/imath]-th Powers are Equal then the Terms are Equal
Consider the following problem: Let [imath]a_1, \ldots, a_n[/imath] and [imath]b_1, \ldots, b_n[/imath] be real numbers such that [imath]a_1^k+\cdots a_n^k= b_1^k+ \cdots+b_n^k[/imath] for all [imath]1\leq k\leq n[/imath]. Then there is a bijection [imath]\varphi:\{1, 2, \ldots, n\}\to \{1, 2, \ldots, n\}[/imath] such that [imath]a_i=b_{\varphi(i)}[/imath] for all [imath]i[/imath]. This can be proved using Newton's identities, which gives that the polynomials [imath]p(x)=(x-a_1)\cdots(x-a_n)[/imath] and [imath]q(x)=(x-b_1)\cdots(x-b_n)[/imath] are identical. My question is, can this be said about countable sequences. More precisely, is the following true? Let [imath]a_1, a_2, a_3, \ldots[/imath] and [imath]b_1, b_2, b_3,\ldots[/imath] be real numbers such that [imath]a_1^k+a_2^k+a_3^k+\cdots = b_1^k+ b_2^k+b_3^k+ \cdots[/imath] are equal for all [imath]k\geq 1[/imath] (Here we insist that both sums are convergent for each [imath]k[/imath]). Then there is a bijection [imath]\varphi:\mathbb N \to \mathbb N[/imath] such that [imath]a_i=b_{\varphi(i)}[/imath] for all [imath]i[/imath]. | 2675228 | If for any [imath]k[/imath], [imath]\sum\limits_{n=0}^\infty a_n^k=\sum\limits_{n=0}^\infty b_n^k[/imath], then[imath](a_n)=(b_{σ(n)}),\ σ \in{\mathfrak S}_{\mathbb N}[/imath]
Let [imath](a_n)_{n≥0}[/imath] and [imath](b_n)_{n≥0}[/imath] be two sequences of a nomed algebra such that [imath]\sum{\| a_n\|}[/imath] and [imath]\sum{\| b_n\|}[/imath] converge, and$ [imath]$\forall n, \ a_n, b_n \neq 0[/imath] Show that $(\forall k \in \mathbb N^*, \ \sum_{n=0}^{\infty}{a_n}^k = \sum_{n=0}^{\infty}{b_n}^k) \implies \exists\ \sigma \in{\mathfrak S}_{\mathbb N} \ ,\ (a_n) = (b_{\sigma(n)})[imath][/imath] The vector-space should be a finite-dimensional \mathbb R$-vector-space. I do not see how to deal with the hypothesis. If you have ideas/hints... It would be relevant to deal with complex sequences even if it is less general. But I don't think one can generalize using real and complex cases. |
2719714 | How to prove the following equalities
Show that for [imath]|z|<1[/imath], one has [imath]\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+...+\frac{z^{2^n}}{1-z^{2^{n+1}}}+...=\frac{z}{1-z}, \frac{z}{1+z}+\frac{2z^2}{1+z^2}+...+\frac{2^kz^{2^k}}{1+z^{2^k}}+...=\frac{z}{1-z}.[/imath] Justify any change in the order of summation. [Hint: Use the dyadic expansion of an integer and the fact that [imath]2^{k+1}-1=1+2+2^2+...+2^k[/imath]] I can get some feelings of the equalities, but I do not know how to prove it using the hint. For example, the first equality: [imath]z+z^2+z^3+z^4+z^5+z^6+z^7+...=(z+z^3+z^5+z^7+...)+(z^2+z^6+z^{10}+...)+(z^4+z^{12}+...)+...+(z^{2^n}+z^{2^n+2^{n+1}}+z^{2^n+2^{n+2}}+...)+...=z(1+z^2+z^4+z^6+...)+z^2(1+z^4+z^8+...)+...[/imath] so, [imath]\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+...+\frac{z^{2^n}}{1-z^{2^{n+1}}}+...=\frac{z}{1-z}[/imath]. How can I get a strict proof of the above equalities? | 1628569 | Prove that [imath] \sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z} [/imath]
Show that for [imath]|z|<1[/imath], one have [imath] \frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots+\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots = \frac{z}{1-z} [/imath] and [imath] \frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots+\frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots = \frac{z}{1-z} [/imath] Use the dyadic expansion of an integer and the fact that [imath]2^{k+1}-1=1+2+2^2+\cdots+2^k[/imath]. |
2720440 | On the cardinality of quotient set
I have read here that a quotient set may have a cardinality strictly greater of the starting set in ZF. I would to know if we add to ZF the axiom "for every set [imath]X[/imath] and equivalence relation [imath]\thicksim[/imath] on [imath]X[/imath], we have [imath]|X/\thicksim|\leq|X|[/imath]". I was wondering what theory we obtain, certainly it is strictly stronger than ZF and weaker than ZFC, but is strictly weaker than ZFC or is equivalent? I was looking to proof that is equivalent considering a nonempty family [imath]\{X_i|i\in I\}[/imath] of nonempty set and i considered the relation on the product [imath]X=\prod\{X_i|i\in I\}[/imath] fixed a factor [imath]X_j[/imath], and let [imath]p_j[/imath] the projection, [imath](x_i)\thicksim(y_i)[/imath] iff [imath]x_j=y_j[/imath], my idea was that the quotient [imath]X/\thicksim[/imath] is substantially [imath]X_j[/imath] (in particular they are in bijection) but now i think this doesn't work. | 2158581 | Is denying "Paradoxical Partitioning" equivalent to accepting the Axiom of Choice?
In this accepted answer (to this question here, from a couple of years ago) it has been noted: One more point to make about the paradoxical decomposition to more parts than elements [...] currently we do not know of any model of ZF+[imath]\lnot[/imath]AC where such decomposition does not exist. Namely, as far as we know, in all models where choice fails there is some set which can be partitioned into more parts than elements. (Just for clarification: From the context of the above references the comparison "more parts than" is obviously meant in terms of the cardinality of the suitable partition being strictly larger than the cardinality of the suitable initial set itself.) Now, I find the negation or denial of such "paradoxial partitioning" interesting; i.e. the statement (proposition, "[imath]\lnot[/imath]PP"): "Each partition of each given set has cardinality less than, or at most equal to, the cardinality of the given set." And I like to further explore how this suggested "proposition [imath]\lnot[/imath]PP" (being considered along with ZF, of course) relates to "the standard set theory including Axiom of Choice", ZFC. Therefore My questions: Are there any models of ZFC known (or could there be any such models, in principle) in which there is some set which can be partitioned into more parts than elements ? And: Are there any models of ZF known (or could there be any such models, in principle) in which there is no set which can be partitioned into more parts than elements, and which is not also a model of ZFC ? (And just for reference: Is there a conventional or concise way of expressing the suggested "proposition [imath]\lnot[/imath]PP" in terms of standard notation, such as used in the sources linked above? Has it perhaps been discussed already, by some other name? ...) |
2720527 | Use mathematical Induction to prove For all [imath]n ≥ 6[/imath] we have [imath]4n^2 + 1 < 3 · 2^n[/imath]
Im stuck on how to start this question. | 2719636 | Solving an inequality induction
For all: [imath] n \ge 6, 4n^2+1 < 3*2^n [/imath] p(n): [imath] 4n^2+1 < 3*2^n [/imath] My work: Basis Step: P(6) 145 < 192 is true Induction Step: Let [imath] n\ge 6 [/imath], Assume [imath] 4(n+1)^2 +1 < 3*2^{n+1}[/imath] What should I do next?? |
2720617 | If [imath]g \circ f[/imath] is injective and [imath]f[/imath] is surjective, then [imath]g[/imath] is injective.
Let [imath]f : A \to B[/imath] and [imath]g : B \to C[/imath] be functions. If [imath]g \circ f[/imath] is injective and [imath]f[/imath] is surjective, then [imath]g[/imath] is injective. This question that correctly proves this theorem, but my proof seems to be different, so I'd like clarification on whether it's correct. My Proof We want to show that, for any [imath]b_1, b_2 \in B[/imath], [imath]g(b_1) = g(b_2)[/imath] iff [imath]b_1 = b_2[/imath]. Let [imath]g(f(a_1)) = g(f(a_2))[/imath], where [imath]a_1, a_2 \in A[/imath]. [imath]\therefore f(a_1) = b_1[/imath] and [imath]f(a_2) = b_2[/imath], where [imath]b_1, b_2 \in B[/imath]. (Since [imath]f[/imath] is surjective.) [imath]\therefore g(b_1) = g(b_2)[/imath], where [imath]b_1 = b_2[/imath]. (Since [imath]g \circ f[/imath] is injective.) [imath]\therefore[/imath] [imath]g[/imath] is injective. [imath]\ \ \ Q.E.D.[/imath] I would greatly appreciate it if people could please review my proof and elaborate on anything I've done incorrectly. | 606257 | If [imath]g\circ f[/imath] is injective and [imath]f[/imath] is surjective then [imath]g[/imath] is injective
Let [imath]f:A\rightarrow B[/imath] and [imath]g:B\rightarrow C[/imath] be functions, prove that if [imath]g\circ f[/imath] is injective and [imath]f[/imath] is surjective then [imath]g[/imath] is injective. I need advise or correction if something is incorrect with my proof. Thank you beforehand. We must show that [imath]g[/imath] is injective, i.e for [imath]x,y\in B, g(x)=g(y)\implies x=y[/imath] Let [imath]x,y\in B[/imath] such that [imath]g(x)=g(y)[/imath]. Because [imath]f[/imath] is surjective there exists [imath]a,b \in A[/imath] such that [imath]f(a)=x[/imath] and [imath]f(b)=y[/imath] [imath]\implies g(f(a))=g(f(b))[/imath] [imath]\implies g\circ f(a)=g\circ f(b)[/imath] [imath]\implies a=b[/imath] (by injectivity of [imath]g\circ f[/imath]) [imath]\implies f(a)=f(b)[/imath] [imath]\implies x=y[/imath] Would appreciate any correction in proof writing also! |
2720156 | Image of [imath]\exp(z)[/imath] under condition
I want to find the image of the function that defined as [imath]\exp(x+iy)=e^x(\cos(y)+i\sin(y))[/imath] for [imath]z\in L[/imath] where [imath]L=\left\{z\in \mathbb C \mid a\mathrm{Re}(z)+b\mathrm{Im}(z)+c=0 \right\}[/imath] and [imath]a^2+b^2\ne 0[/imath]. I tried to get to an expression for [imath]z[/imath] uing the identities [imath]\mathrm{Re}(z)=\dfrac{z+\bar z}{2}[/imath] and [imath]\mathrm{Im}(z)=\dfrac{z-\bar z}{2i}[/imath] but I got to a dead end. How should I approach this problem? | 2717761 | Image of [imath]L=\{z\in\Bbb C \mid a\operatorname{Re}(z)+b\operatorname{Im}(z)+c=0; a^2+b^2\neq0; a,b,c\in \Bbb R \}[/imath] under the function [imath]e^z[/imath]
I'm trying to find out what is the image of the group [imath]L=\{z\in\Bbb C\mid a\operatorname{Re}(z)+b \operatorname{Im}(z)+c=0\,; a^2+b^2\neq0\,; a,b,c\in \Bbb R \}[/imath] under the function [imath]e^z[/imath] in the Imaginary-real plane. In other words, find [imath]\{e^z\mid z\in L\}[/imath]. Correct me if something is not good in the question. I understand it should be something like a spiral, but I don't really understand how to show it. |
2721254 | How to get a closed form for [imath]\sum m f(n,m)[/imath] where [imath]f(n,m)[/imath] is number of weak compositions?
First, consider how many [imath]n[/imath]-dimensional vector of non-negative integers [imath](x_1,x_2,\cdots,x_n)[/imath] are there whose sum of all entries satisfies [imath]x_1+x_2+\cdots+x_n=m[/imath]? For example for [imath]n=2,m=2[/imath], there are [imath](2,0),(1,1),(0,2)[/imath], so [imath]f(2,2)=3[/imath]. For [imath]n=3,m=2[/imath] there are [imath](2,0,0),(0,2,0),(0,0,2),(1,1,0),(0,1,1),(1,0,1)[/imath], so [imath]f(3,2)=6[/imath]. How about for general [imath]m,n[/imath]? What I know is [imath]f(n,m)\le[/imath] n+m choose n by How many [imath]k-[/imath]dimensional non-negative integer arrays [imath](x_1,\cdots,x_k)[/imath] satisfies [imath]x_1+x_2+\cdots+x_k\le n[/imath] -- Then, I would like a closed form solution for [imath]\sum_{m=0}^Mf(n,m)[/imath] | 507413 | Proof for number of weak compositions
I'm looking for an alternative to the following (possibly standard) proof for the number of weak compositions: The number of [imath]k[/imath]-compositions of [imath]n+k[/imath] corresponds to a weak one by subtracting 1 from each "bin". Thus we have [imath]\binom{n+k-1}{k-1}[/imath]. While I kind of like this proof and always felt it made sense lately I've been left wanting something a little more convincing. Does anyone have a different proof of this result? |
2721100 | Counting how many functions exist given a relation
Let [imath]A = \{1, 2, 3, 4\}[/imath], and let [imath]F[/imath] be the set of all functions from [imath]A[/imath] to [imath]A[/imath]. Let [imath]S[/imath] be the relation on [imath]F[/imath] defined by: [imath]\forall f, g \in F[/imath], [imath]fSg \iff f(i) = g(i)[/imath] for some [imath]i \in A[/imath] Let [imath]h: A \to A[/imath] be the function defined by [imath]h(x) = 1[/imath] for all [imath]x \in A[/imath] How many functions [imath]g \in F[/imath] are there so that [imath]gSh[/imath]? So for whatever [imath]x \in A[/imath], [imath]h(x) = 1[/imath]. So I want to find how many functions [imath]g \in F[/imath] exist so that [imath]gSh[/imath]. So I made a sort of "recipe" for counting how many functions [imath]g[/imath] exist. Since in order for [imath]gSh[/imath], [imath]g(i) = 1[/imath] for any [imath]i \in A[/imath], since [imath]g(i) = h(i)[/imath] if [imath]gSh[/imath]. Step 1: Map [imath]1 \to 1[/imath] Step 2: Map [imath]2 \to 1[/imath] Step 3: Map [imath]3 \to 1[/imath] Step 4: Map [imath]4 \to 1[/imath] So this is just equal to [imath]1 * 1 * 1 * 1[/imath]. So my assumption is that there exists only [imath]1[/imath] function [imath]g[/imath] such that [imath]gSh[/imath]. Is this correct? I think I might be misunderstanding the wording, and just want to make sure. Thanks in advance. | 2715308 | How many functions are possible to create in this example?
Let [imath]A = \{ 1,2,3,4 \}[/imath] Let [imath]F[/imath] be a set of all functions from [imath]A \to A[/imath]. Let [imath]S[/imath] be a relation defined by : [imath]\forall f,g \in F[/imath] [imath]fSg \iff f(i) = g(i)[/imath] for some [imath]i \in A[/imath] Let [imath]h: A \to A[/imath] be the function [imath]h(x) = 1 [/imath] for all [imath]x \in A[/imath]. How many functions [imath]g \in F[/imath] are there so that [imath]gSh[/imath] ? My solution : [imath]gSh[/imath] means that [imath]g(i) = h(i)[/imath] for some [imath]i \in A[/imath]. So [imath]g(i) = 1[/imath]. So number 1 always needs to connect to some x - 4 choices. Then we have [imath]3[/imath] left-over numbers that can connect to [imath]4[/imath] numbers each. So solution is [imath]4 \cdot 4 \cdot 4 \cdot 4[/imath]. Is this correct at all? Thanks in advance !! :) |
2721424 | Defiintion of a.s. convergence: sequence of R.V.s [imath]z_1, z_2, ...[/imath] s.t. [imath]P(z_i=1)=P(z_i=-1)=\frac 1 2[/imath]
Let [imath]z_1, z_2, z_3,[/imath] ...be a sequence of independent random variables s.t. [imath]P(z_i=1)=P(z_i=-1)=\frac 1 2[/imath]. Does this sequence converge almost surely? I am still bothered by the definition of almost sure convergence. If we define a RV Z([imath]\omega_1[/imath]=negative sign)=-1, Z([imath]\omega_2[/imath]=positive sign)=1 with sample space {[imath]\omega_1[/imath], [imath]\omega_2[/imath]} doesnt this sequence then converge to Z? Since P(lim n→∞ Zi(w))=Z(w)=1? But the answer should be no, so I am confused bout this defintion (apology in advnace as I am new to this.) I think i know how to use second bernoulli to prove it since sum of probabilities which is 1/2 will be infinite meaning that the sequence does not converge. But would some one please explain the above definition? | 2719967 | Sequence of R.V.s [imath]z_1, z_2, ...[/imath] s.t. [imath]P(z_i=1)=P(z_i=-1)=\frac 1 2[/imath] converge almost surely?
Let [imath]z_1, z_2, z_3,[/imath] ...be a sequence of independent random variables s.t. [imath]P(z_i=1)=P(z_i=-1)=\frac 1 2[/imath]. Does this sequence converge almost surely? My intuition is telling me that the sequence will approach the random variable Z? i.e. [imath]z_n->Z[/imath] as [imath]n->\infty[/imath] and Z is just another RV s.t. [imath]P(Z=1)=P(Z=-1)=\frac 1 2[/imath]. Here is my confusion, if a sequence converges almost surely, then [imath]z_i[/imath] should be different for each term s.t. it approaches Z in the end; but in this sequence [imath]z-i[/imath] is the same for every term. So I dont udnersntand. |
2721702 | If [imath]A = \int_Xf\ d\mu[/imath], then [imath]\sqrt{1+A^2} \ \le \ \int_x\sqrt{x+f^2}\ d\mu \ \le \ 1+A [/imath]
Suppose [imath](X,M,\mu)[/imath] is a measue space with [imath]\mu(X)=1[/imath] and [imath]f:X\rightarrow[0,\infty)[/imath] is measureble. Let [imath]A = \int_Xf\ d\mu[/imath] Show that [imath] \sqrt{1+A^2} \ \le \ \int_X\sqrt{x+f^2}\ d\mu \ \le \ 1+A [/imath] The first equality seems obvious but I don't know how to officially prove and I'm completely stuck with the second equality. Any help would be appreciated! | 2718968 | An integral inequality (Rudin [imath]L^p[/imath]-spaces 3.12)
I've been learning about [imath]L^p[/imath]-spaces and came across this problem in Rudin's Real and Complex Analysis: Suppose [imath]\mu(\Omega)=1[/imath] and [imath]h:\Omega \to [0,\infty][/imath] is measurable. If [imath]A= \int_\Omega {h d\mu}[/imath], prove that: [imath] \sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu \leq 1+A [/imath] I've already shown that [imath]\sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu[/imath] using Jensen's inequality applied to the convex function [imath]\varphi(x)=\sqrt{1+x^2}[/imath]. I can't seem to show the latter inequality. Any help or hints would be appreciated. Thank you in advance! |
2721323 | RSA explanation
Let's say that we have the following RSA setup where [imath]m = pq[/imath], where [imath]p[/imath] and [imath]q[/imath] are distinct primes. In addition, let [imath]d, e \in \mathbb{N}[/imath] such that [imath]d = e^{-1}[/imath](mod [imath]\phi(m)).[/imath] If [imath]M[/imath] stands for a plain text, we let [imath]C \equiv M^e[/imath] (mod [imath]m[/imath]), where [imath]C[/imath] is the cypher text. As long as gcd ([imath]C, m) = 1[/imath], [imath]\implies[/imath] [imath]C^d \equiv M[/imath] (mod [imath]m[/imath]). It was also shown (lecture) that gcd ([imath]C,m) = 1[/imath] exactly when [imath]gcd(M,m) = 1.[/imath] What happens if, in an unfortunate manner, the message I want to send is not relatively prime to [imath]m[/imath]? We know that this never really happens in practice because our choices for [imath]p[/imath] and [imath]q[/imath] are large that the probability of [imath]M[/imath] being divisible by either [imath]p[/imath] or [imath]q[/imath] is roughly about [imath]0[/imath]. It could be that [imath]p[/imath] and [imath]q[/imath] were small such that this situation may arise, but RSA should still work (at least I hope it does). With no assumption on gcd[imath](C,m)[/imath] whatsoever, I want to show that [imath]C^d \equiv M[/imath] (mod [imath]p[/imath]), [imath]C^d \equiv M[/imath] (mod [imath]q[/imath]), [imath]C^d \equiv M[/imath] (mod [imath]m[/imath]). Keep in mind that I know very little about cryptography, but I'm fine working with aspects related to number theory. | 2702037 | Prove the congruence of RSA: [imath]C^d\equiv M\ (\textrm{mod}\ pq)[/imath] still hold even if [imath]gcd(M, pq) > 1[/imath].
Pre-conditions: Let the public key be [imath](n, e)[/imath], and [imath]n=pq[/imath], which [imath]p,q[/imath] are large distinct primes. Let [imath]gcd(e, (p-1)(q-1))=1[/imath], and [imath]d[/imath] is the inverse of [imath]e[/imath] modulo [imath](p-1)(q-1)[/imath]. And let [imath]M[/imath] be the raw message, [imath]M < pq[/imath], and [imath]C[/imath] represents the encrypted message(i.e. [imath]M^e[/imath]). Proof: Since [imath]M < pq[/imath], and [imath]p, q[/imath] are prime, [imath]gcd(M, pq) > 1[/imath] implies that [imath]gcd(M, p)=p[/imath] or [imath]gcd(M,q)=q[/imath], but not both. WLOG, consider the case [imath]gcd(M, p)=p[/imath] and [imath]gcd(M, q)=1[/imath]. Let [imath]ed=1+k(p-1)(q-1)[/imath], then we have: [imath]C^d \equiv (M^e)^d \equiv 0\ (\textrm{mod}\ p) \equiv M\ (\textrm{mod}\ p)[/imath] [imath]C^d \equiv (M^e)^d \equiv M\cdot M^{k(p-1)(q-1)} \equiv M\cdot1^{k(p-1)} \equiv M\ (\textrm{mod}\ q)[/imath] Since [imath]gcd(p,q)=1[/imath], [imath]C^d \equiv M\ (\textrm{mod}\ pq)[/imath] by Chinese Remainder Theorem. [imath]\mathbf{Q.E.D.}[/imath] This is an exercise in my text book Discrete Mathematics and it's Application, Rosen, page 305, section 4.6. And I don't have the solution... |
2715489 | Prove that for all [imath]n\geqslant 1[/imath] we have [imath]F_n<{\left(\frac 74\right)}^n[/imath].
Recall that the Fibonacci sequence is defined by [imath]F_0=0, F_1=1[/imath] and [imath]F_n =F_n−1 +Fn−2[/imath], for [imath]n\geq 2[/imath]. Prove that for all [imath]n\geq 1[/imath] we have [imath]F_n<{\left(\frac 74\right)}^n[/imath]. | 2951228 | Prove [imath]F_{n+1} ≤ (\frac74)^n [/imath], where [imath]F_n[/imath] are Fibonacci numbers
Let [imath]F_n[/imath] be the [imath]n[/imath]-th Fibonacci number, defined recursively by [imath]F_0 = 0[/imath], [imath]F_1 = 1[/imath] and [imath]F_n = F_{n−1} + F_{n−2}[/imath] for [imath]n ≥ 2[/imath]. Prove the following by induction (or strong induction): [imath](a)[/imath] For all [imath]n ≥ 0[/imath], [imath]F_{n+1} ≤ \left(\dfrac74\right)^n[/imath]. [imath](b)[/imath] Let [imath]G_n[/imath] be the number of tilings of a [imath]2 × n[/imath] grid using domino pieces (i.e. [imath]2 × 1[/imath] or [imath]1 × 2[/imath] pieces). Then prove [imath]G_n = F_{n+1}[/imath]. For question [imath](a)[/imath], I've done the proof but the result I kept getting was [imath]\left(\frac74\right)^{k+1}\left(\frac{11}7\right)≤\left(\frac74\right)^{k+1}[/imath] which is wrong. |
2723256 | Error in evaluation of [imath]\displaystyle\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2}[/imath]
Evaluate [imath]\displaystyle\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2}[/imath] Here's my method but that results into an error. \begin{align} \lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2} &=\lim_{x\to 0}\frac{\cos x}{x} - \lim_{x\to 0}\left(\frac{1}{x}\right)\lim_{x\to 0}\left(\frac{\ln(1+x)}{x}\right)\\ &= \frac{\cos x}{x} - \frac{1}{x} \\ &= \frac{\cos x -1}{x}\\ &= 0 \quad \text{(either by L'hôpital or some manipulations)} \end{align} I have used the fact that [imath]\lim_{x\to 0} \frac{\ln(1+x)}{x}=1[/imath] And the answer seems to be [imath]\frac{1}{2}[/imath]. I can do it in different ways (L'hôpital precisely) but please point out my error. | 1783782 | Analyzing limits problem Calculus (tell me where I'm wrong).
I came accross: [imath]\lim_{x \to\ 0}\frac{x\cos x - \log (1 + x)}{x^{2}}[/imath] I tried the following please tell me where I 'm wrong: [imath]\lim_{x \to\ 0}\frac{x\cos x - \log (1 + x)}{x^{2}} [/imath] [imath]\text{(Dividing by }x)[/imath] [imath]=\displaystyle\lim_{x \to\ 0}\dfrac{ \cos x - \dfrac{\log (1 + x)}{x}}{x} [/imath] [imath]=\lim_{x \to\ 0}\frac{\cos x - 1}{x} [/imath] [imath]=\lim_{x \to\ 0}\frac{-2\sin^{2} \dfrac{x}{2}}{x} [/imath] [imath]=\lim_{x \to\ 0}\frac{-2x\sin^{2}\dfrac{x}{2}}{(\dfrac{x}{2})^{2}\times 4} [/imath] [imath]\lim_{x \to\ 0}\dfrac{-x}{2}=0[/imath] But Answer given is [imath]\dfrac{1}{2}[/imath] Please Help. |
2119215 | compact positive operator
let [imath] T \in K ( H ) [/imath] be [imath] T \geqslant 0 [/imath]. What is the best way to show the following statement? There is a compact, positive and unique operator [imath] A[/imath] so that [imath] A^{2} = T [/imath]. ( K ( H ) defined compact operator) | 311310 | compact and self adjoint square root of an operator
Let H a Hilbert space and [imath]T:H\rightarrow H[/imath] a linear bounded, self-adjoint, positive and compact operator. How can i prove that the square root of T, [imath]\ T^{1/2}:H\rightarrow H[/imath] is also compact and self-adjoint; thanks. |
1436320 | Solve [imath]y^3=x^{3}+8x^{2}-6x+8[/imath] for positive integers [imath]x, y [/imath]
How we can solve this equation : [imath]y^3=x^{3}+8x^{2}-6x+8[/imath] for positive integers [imath]x, y [/imath]? I tried to factories the [imath]L. H. S [/imath] but I couldn't complete the solution. How I can solve it? Thanks. | 928674 | Solving [imath]y^3=x^3+8x^2-6x+8[/imath]
Solve for the equation [imath]y^3=x^3+8x^2-6x+8[/imath] for positive integers x and y. My attempt- [imath]y^3=x^3+8x^2-6x+8[/imath] [imath]\implies y^3-x^3=8x^2-6x+8[/imath] [imath]\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8[/imath] Now if we are able to factorise [imath]8x^2-6x+8[/imath] then we can compare LHS with RHS.Am I on the right track?Please help. |
2720878 | A conjectured vanishing sum involving the Thue–Morse sequence
Let's define [imath]t_n[/imath] by the recurrence [imath]t_0 = 1, \quad t_n = (-1)^n \, t_{\lfloor n/2\rfloor}.\tag1[/imath] It is easy to see that [imath]|t_n|=1[/imath], and the signs follow the same pattern as the Thue–Morse sequence: [imath]1,\,-1,\,-1,\,1,\,-1,\,1,\,1,\,-1,\,-1,\,1,\,1,\,-1,\,1,\,-1,\,-1,\,1,\,...\tag2[/imath] (see this question for an example of a non-recursive formula for [imath]t_n[/imath]). Now, define: [imath]\mathcal{S}(m, n)=\sum_{k=0}^{2^n-1}t_k\,k^m, \quad \color{gray}{m,\,n\in\mathbb{Z}^+.}\tag3[/imath] Computing some values of this sum immediately leads to the following conjecture: [imath]\text{For all }\, n>m,\,\,\mathcal{S}(m, n)=0.\tag{$\diamond$}[/imath] (e.g. for [imath]m=2[/imath] and [imath]n=3[/imath] we have [imath]0-1-4+9-16+25+36-49=0[/imath]). How can we prove it? Is it a known result? | 163369 | Polynomials and partitions
There is a question I have based on the fact: If you take a quadratic polynomial with integer coefficients, take the set [imath]\{1,2,3,4,5,6,7,8\}[/imath], make a partition [imath]A=\{1,4,6,7\}[/imath], [imath]B=\{2,3,5,8\}[/imath], and then evaluate the polynomial with the elements of each set you got an equality. If now with a cubic polynomial and the set [imath]\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16\}[/imath], how to pick a suitable partition that accomplish the same fact? I mean is there a general formula for bigger degree? |
2718803 | Injective restriction of [imath]f(x) = \frac{ax-b}{cx-d}[/imath]
For [imath]a,b,c,d \in \mathbb{R}[/imath] and [imath]ad-bc \neq 0[/imath] I have the function [imath]f : \mathbb{R}\setminus{\{d/c\}} \to \mathbb{R}, f(x) = \frac{ax-b}{cx-d}[/imath]. How can I find the largest possible restriction of [imath]f[/imath] such that [imath]f[/imath] is injective on it and the image of this restriction? Thank you for your help. Edit: If I understand correctly, the linked question is concerned with finding a condition for injectivity on the parameters a,b,c,d (the condition they find is [imath]bc-ad \neq 0[/imath]). With mine the values are already given. Also I know that (as a continuous function) it's injective iff it is strictly monotone, which is the case when [imath]f'(x) = \frac{bc-ad}{(cx-d)^2}[/imath] is positive or negative. But I don't understand how this helps me, since the denominator is always positive and the numerator a fixed value for given bc and ad? If this is even the right approach for finding a restriction. | 777571 | [imath]f(x) = \frac {ax+b}{cx+d}[/imath] , bijection [imath]f: \Bbb R \to\Bbb R[/imath]?
For which [imath]a,b,c,d \in \Bbb R[/imath] does [imath]f(x) = \frac {ax+b}{cx+d}[/imath] define a bijection [imath]f: \Bbb R \to \Bbb R[/imath]? I'm guessing I need a system of equations and I know that [imath]cx + d \ne 0[/imath]. Thanks! |
2724075 | On dense sets and isolated points
I need to prove that if [imath]D[/imath] is dense in [imath]X[/imath] and [imath]X[/imath] has no isolated points, then [imath]D[/imath] has no isolated points. I managed to prove that if [imath]A\subset X[/imath] is open, then [imath]A[/imath] has no limit points, but can't seem to make use of that. My attempt was to prove that [imath]X = D'[/imath], since [imath]\overline{D} = X[/imath], so that would imply [imath]D \subset D'[/imath], but it's not working out... Any tips? | 1230296 | If a space [imath]X[/imath] has no isolated points, then nor does any dense subset of [imath]X[/imath].
[imath]X[/imath] is a topological space with no special properties. If a space [imath]X[/imath] has no isolated points, then show the same for any dense subset of [imath]X[/imath]. Thanks a lot. |
2717248 | On an exercise from Royden&Fitzpatrick's Real Analysis Book
Exercise 47 in Section 2.7 of Royden&Fitzpatrick's Real Analysis Book says that "a continuous strictly increasing function that is defined on an interval maps Borel sets to Borel sets". But here do we need the continuity condition really? Because it seems that, without any continuity condition, the result follows directly from two preceding exercises, one of which says that "a strictly increasing function that is defined on an interval has a continuous inverse", and the other asserts that "if [imath]f[/imath] is a continuous function and [imath]B[/imath] is a Borel set, then [imath]f^{-1}(B)[/imath] is a Borel set". I think that the continuity condition is superfluous. Am I right? Or, is there a detail which I cannot see? Thanks in advance for any comment. EDIT: Here I just wanted to know that if the aforementioned exercise in Royden&Fitzpatrick's Real Analysis Book can be stated as "a strictly increasing function that is defined on an interval maps Borel sets to Borel sets". And I see that the answer is "yes". Thanks for the link. | 957572 | Monotone functions and Borel sets
I'm studying measure theory and two question came to my mind: If [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is monotone and [imath]B\subseteq\mathbb{R}[/imath] is borel, is the image [imath]f(B)[/imath] borel? If [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is a monotone function (say, non-decreasing), does there exist a sequence of continuous functions [imath]f_n:\mathbb{R}\to\mathbb{R}[/imath] converging pointwise to [imath]f[/imath]? Here's the motivation for those questions: Let [imath](X,M)[/imath] and [imath](Y,N)[/imath] be measure spaces. It's known that if [imath]\mu[/imath] is a measure on [imath](X,M)[/imath] and [imath]f:X\to Y[/imath] is measure, then we have the pushforward measure [imath]f_*\mu(A)=\mu(f^{-1}(A))[/imath] on [imath](Y,N)[/imath]. What if we were to define a "pullback measure"? Given a function [imath]f:X\to Y[/imath] such that [imath]f(M)\subseteq N[/imath] (i.e. [imath]f[/imath] maps measurable sets to measurable sets) and a measure [imath]\nu[/imath] on [imath]N[/imath], the natural formula would be [imath]f^*\nu(A)=\nu(f(A))[/imath]. So we ask if is there a good amount of functions which map measurable sets to measurable sets in [imath]\mathbb{R}[/imath], and monotone functions seem like good candidates (for strictly monotone, continuous functions, the result is valid and there are several answers on the web). If this were true, maybe we could use some convergence argument to solve the problem above. |
2712748 | How to show that [imath]f[/imath] is Riemann integrable?
Let [imath]f[/imath] be the function on [imath][0, 1][/imath] given by: [imath]f(x) = \left\{\begin{array}{ll} 1 & : x \neq \frac{1}{2}\\\\ 2 & : x = \frac{1}{2}\end{array} \right.[/imath] Prove that [imath]f[/imath] is Riemann integrable and compute [imath]\int_0^1 f(x)\,\mathrm dx[/imath]. Hint: for each [imath]\epsilon > 0[/imath], find a partition [imath]P[/imath] so that [imath]U_P (f) − L_P(f) \leq \epsilon[/imath]. So I understand that I have to show that [imath]U_P (f ) = L_P(f )[/imath] to show that it is integrable, but I have no idea where to start. What partition do I pick and how do I even come about picking it? Also as far as the integral, I figure that it is 1, although I'm not sure if I am right. Since it is a single point ([imath]x = \frac{1}{2} [/imath]) where the function is not uniformly continuous, does it change the area as a whole? | 669237 | Integral with a discontinuity point
Prove that if the real-valued function [imath]f[/imath] on the interval [imath][a,b][/imath] is bounded and is continuous except at a finite number of points then [imath]\int^1_0f(x)dx[/imath] exists. I know that I can break up the interval [imath][a,b][/imath] into subintervals where each subinterval has one discontinuity point. Then if [imath]f[/imath] is integrable on each subinterval then it is integrable on [imath][a,b][/imath]. So if I let [imath]c[/imath] be my point of discontinuity in the subinterval [imath][a_1,b_1]\subset [a,b][/imath], where [imath]c\in[a_1,b_1][/imath]. Then how can I show this subinterval is integrable? |
2724163 | Given the curve [imath]y=2x^3+3x^2−36x[/imath] on the interval [imath]− 1 ≤ x ≤ 4[/imath] , find the absolute maximum?
I have found the first derivative which is [imath]y′=6x^2+6x−36[/imath]. | 2721109 | Given the curve [imath]y = 2 x ^3 + 3 x ^2 − 36 x[/imath] on the interval [imath]− 1 \le x \le4[/imath] , find the absolute minimum?
I know I have to find the first derivative which is [imath]y'=6x^2+6x-36.[/imath] Then I set it to [imath]0.[/imath] Which I got [imath]0 = 6(x^2+x-6)[/imath] and then [imath]x=-3[/imath] or [imath]x=2.[/imath] |
2724313 | [imath]PSL_2(\mathbb{Z})[/imath] is isomorphic to [imath]\mathbb{Z}_2 \ast \mathbb{Z}_3[/imath]
I'm reading Serre's Trees and in Example 1.5.3 he claims that [imath]PSL_2(\mathbb{Z})[/imath] is isomorphic to [imath]\mathbb{Z}_2 \ast \mathbb{Z}_3[/imath]. [imath]\mathbb{Z}_2 \ast \mathbb{Z}_3 = \langle x, y | x^2 = y^3 = 1 \rangle[/imath]. But does [imath]PSL_2(\mathbb{Z})[/imath] even have elements of order [imath]2[/imath] or [imath]3[/imath] to begin with? Previously, I thought since all elements of [imath]PSL_2(\mathbb{Z})[/imath] is in the form of [imath] \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}. [/imath] Hence [imath]\mathbb{Z}[/imath] injects in [imath]PSL_2(\mathbb{Z})[/imath]. Let [imath]\phi: \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \mapsto n[/imath]. [imath]\phi[/imath] is a homomorphism and in fact bijective. Hence [imath]\mathbb{Z} \cong PSL_2(\mathbb{Z})[/imath]. Thus [imath]PSL_2(\mathbb{Z})[/imath] cannot have any [imath]2[/imath]-torsion or [imath]3[/imath]-torsion. I must be missing something. Any help is appeciated! | 1197537 | Prove that [imath]PSL(2,\mathbb{Z})[/imath] is free product of [imath]C_2[/imath] and [imath]C_3[/imath]
Prove that [imath]PSL(2,\mathbb{Z})=C_2 \star C_3[/imath]. Now [imath]C_2 \star C_3=\langle a,b\ |\ a^2, b^3 \rangle[/imath] i.e. the free product. But how do I show that presentation of [imath]PSL(2,\mathbb{Z})[/imath] is this? |
2723626 | The dual cone of [imath]\{(x, t) \,|\, ||x||_1 \le t\}[/imath]
In the course slides for the Stanford Convex Optimization course by Boyd it states that if [imath]K=\{(x, t) \,|\, ||x||_1 \le t\}[/imath] then its dual cone is [imath]K^* =\{(x, t) \,|\, ||x||_\infty \le t\}[/imath] Why is that? If [imath]x = (-t, 0)[/imath] and [imath]y = (t, 0)[/imath] then [imath]x \in K[/imath] and [imath]y \in K^*[/imath] but [imath]x^ty = -t^2[/imath], which is negative, contrary to the definition of a dual cone. | 320038 | Dual cone of a [imath]L^1[/imath] norm cone?
I am listening to convex optimization lectures and I hear that dual cone of a [imath]L^1[/imath] norm cone is a [imath]L^{\infty}[/imath] norm cone. Can anybody please explain how? I understand that every point in the dual cone must have an non-negative inner product with any point in its corresponding cone. How does that bring a diamond shaped [imath]L^1[/imath] norm to a square shaped [imath]L^{\infty}[/imath] norm? |
2724055 | For coprime integers [imath]p,q>0[/imath] and [imath]u,v[/imath] real numbers algebraic over [imath]\mathbb{Q}[/imath] of degrees [imath]p[/imath] & [imath]q[/imath], prove [imath][\mathbb{Q}(u,v) : \mathbb{Q}] = pq[/imath]
I know that [imath]\mathbb{Q}(u,v)[/imath] extends [imath]\mathbb{Q}(u)[/imath] which extends [imath]\mathbb{Q}[/imath] so I was thinking I could use that [imath][\mathbb{Q}(u,v):\mathbb{Q}] = [\mathbb{Q}(u,v):\mathbb{Q}(u)][\mathbb{Q}(u):\mathbb{Q}][/imath] and calculate each of the components on the RHS, but I'm not quite sure how to do that. I suspect that [imath][\mathbb{Q}(u,v):\mathbb{Q}(u)] = q[/imath] and [imath][\mathbb{Q}(u):\mathbb{Q}] = p[/imath] | 1612311 | Product of degree of two field extensions of prime degree
Let [imath]L/K[/imath] be a field extension. Let be [imath]\alpha, \beta \in \mathbb{C}[/imath], such that [imath][\mathbb{Q}(\alpha):\mathbb{Q}] = p[/imath], and [imath][\mathbb{Q}(\beta):\mathbb{Q}] = q[/imath], for some prime numbers [imath]p[/imath] and [imath]q[/imath]. Assume [imath]p \neq q[/imath]. Prove that: [imath]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\mathbb{Q}(\alpha,\beta):Q] = [\mathbb{Q}(\alpha):\mathbb{Q}] \cdot [\mathbb{Q}(\beta):\mathbb{Q}][/imath] I have no idea how to proceed on this, besides proving that the left hand side is [imath]pq[/imath]. |
2725267 | How to show that a set with injections into and from [imath]\mathbb{N}[/imath] is uncountable?
Let's say we have two injective functions [imath]u : \mathbb{N} \to A[/imath] and [imath]v : A \to \mathbb{N}[/imath]. How can I demonstrate that [imath]A[/imath] is uncountable ? If there is an injective function [imath]u : \mathbb{N} \to A[/imath] and [imath]v : A \to \mathbb{N}[/imath], consequently there is a bijection between [imath]\mathbb{N}[/imath] and [imath]A[/imath]. But what can I do from that? Thanks! | 2543551 | Proving a set of functions is uncountable
I am trying to prove that [imath]D(\mathbb{N}[/imath])={ [imath]f\in \mathbb{N}^\mathbb{N}[/imath] | f is a bijection such that [imath]f(n)\neq n[/imath] for all [imath]n\in\mathbb{N}[/imath]} is uncountable. So, I was thinking of showing [imath]D(\mathbb{N})[/imath] ~ [imath]P(\mathbb{N})[/imath] using Cantor Bernstein Theorem. For the one direction, is it found to say since [imath]D(\mathbb{N})\subset \mathbb{N}^\mathbb{N}[/imath] ~[imath]P(\mathbb{N})[/imath] then there is an injection from [imath]D(\mathbb{N})[/imath] to [imath]P(\mathbb{N})[/imath]? I am not too sure how to go about the other direction. Should I show equinumerous to functions from [imath]\mathbb{N}^\mathbb{N}[/imath] instead of using the power set? |
2724881 | Does any series diverge more slowly than the harmonic series?
Sorry if this is too simple of a question. Let [imath]S_n = \sum \limits _{k=0} ^{n} \dfrac{1}{k}[/imath]. For another sequence [imath]\{x_k\}[/imath] define [imath]S'_n = \sum \limits _{k=0} ^{n} x_k[/imath]. Is it possible that [imath]S'_n[/imath] diverges and [imath] lim _{n \rightarrow \infty} \dfrac{S'_n}{S_n} = 0[/imath]? | 452053 | Is there a slowest rate of divergence of a series?
[imath]f(n)=\sum_{i=1}^n\frac{1}{i}[/imath] diverges slower than [imath]g(n)=\sum_{i=1}^n\frac{1}{\sqrt{i}}[/imath] , by which I mean [imath]\lim_{n\rightarrow \infty}(g(n)-f(n))=\infty[/imath]. Similarly, [imath]\ln(n)[/imath] diverges as fast as [imath]f(n)[/imath], as [imath]\lim_{n \rightarrow \infty}(f(n)-\ln(n))=\gamma[/imath], so they 'diverge at the same speed'. I think there are an infinite number of 'speeds of divergence' (for example, [imath]\sum_{i=1}^n\frac{1}{i^k}[/imath] diverge at different rates for different [imath]k<1[/imath]). However, is there a slowest speed of divergence? That is, does there exist a divergent series, [imath]s(n)[/imath], such that for any other divergent series [imath]S(n)[/imath], the limit [imath]\lim_{n \rightarrow \infty}(S(n)-s(n))=\infty[/imath] or [imath]=k[/imath]? If so, are there an infinite number of these slowest series? |
2722269 | Proving a sequence by induction
Trying to prove by induction in a non-math course and I feel like I'm getting the steps but I'm just stuck on the math. [imath]G_1 = 1[/imath] [imath]G_2 = 1[/imath] [imath]G_n = 2G_{n−1} + 3G_{n−2}, \quad n \geq 3[/imath] Using mathematical induction, prove that for every [imath]n ≥ 1[/imath], [imath]G_n ≤ 3^n[/imath]. I used [imath]G_1[/imath] as my base case, since it’s given. My induction hypothesis is to assume [imath]G_n \leq 3^n[/imath] for some arbitrary [imath]n[/imath]. Thus I should prove [imath]G_{n+1}\leq 3^{n+1}[/imath]. Using the given formula I simplified down to [imath]2G_n + 3G_{n-1} ≤ 3^n3[/imath]. I feel like I'm pretty much there, but I just don't see how to solidify the proof from here. Any hints would be greatly appreciated! | 927549 | Proof by induction that [imath]a_0 = 1, a_1 = 1, a_n=2a_{n-1} + 3a_{n-2}[/imath] satisfies [imath]a_n = \frac12 (3^n) + \frac12 (-1)^n[/imath]
The question: The terms of a sequence are given recursively as \begin{cases} a_0 = 1,\\ a_1 = 1 \\a_n=2a_{n-1} + 3a_{n-2} \quad\text{ for } n \geq 2 \end{cases} prove by mathematical induction [imath]a_n = \frac12(3^n) +\frac12(-1)^n[/imath] for all [imath]n \geq 0[/imath] I've started by proving that the formula works e.g [imath]a_0= 1\implies a_0 = \frac12(3^0)+\frac12(-1)^0 = 1 \\a_1=1\implies a_1 = \frac12(3^1) + \frac12(-1)^{1} = 1[/imath] Therefore [imath]p(1)[/imath] and [imath]p(2)[/imath] are true Now I have to do the hard part.. the inductive step suppose [imath]p(1), p(2),\ldots, p(k)[/imath], that is [imath]a_{k+1} = 2a_{k} + 3a_{k-1}[/imath] Im not sure where to really go from here. |
1251058 | Solving the functional equation [imath]f(x^2+f(y))=(f(x))^2+y[/imath]
Find all [imath] f : R\rightarrow R [/imath] such that [imath]f(x^2+f(y))=(f(x))^2+y, \forall\text{ x,y}\in R[/imath] Thanks in advance! | 1426480 | How to solve the functional equation [imath]f\left(x^2+f(y)\right)=y+f(x)^2[/imath]
How to solve the following functional equation: Find all [imath]f:\mathbb{R}\to\mathbb{R} [/imath] such that: [imath] f\left(x^2+f(y)\right)=y+f(x)^2 [/imath] Holds for every [imath]x,y\in\mathbb{R}[/imath]. A friend gave it to me, probably its an olympiad question. I started with the equation [imath]f(f(y))=y+f(0)^2[/imath] which seems to be quite helpful, but I couldn't do it effectively. How to solve it properly? |
2724895 | If [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath] is holomorphic and [imath]|f(z)| \geq 1 \forall z \in \mathbb{C}[/imath], show [imath]f[/imath] is constant
If [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath] is holomorphic and [imath]|f(z)| \geq 1 \space \space \forall z \in \mathbb{C}[/imath], show that [imath]f[/imath] is constant I have attempted to prove this, but I am not sure if my proof is correct. I tried to follow the proof of Liouville's Theorem to do this: Since [imath]f[/imath] is holomorphic, by Taylor's Thm, we have that [imath]\forall z \in \mathbb{C}[/imath] and [imath]R>0[/imath], [imath]f'(z) = \frac{1}{2\pi i} \oint_{|w-z|=R} \frac{f(w)}{(w-z)^2} dw[/imath] Now using the estimation lemma, [imath]\left|f'(z) \right|\leq \frac{1}{2\pi} \oint_{|w-z|=R} \frac{|f(w)|}{R^2} |dw|[/imath] Then [imath]\left|f'(z) \right|\leq -\frac{1}{2\pi R^2} \oint_{|w-z|=R} -\space|f(w)| |dw|[/imath] [imath]\leq-\frac{1}{2\pi R^2} \oint_{|w-z|=R} 1 |dw| = -\frac{1}{R}[/imath] Letting [imath]R\rightarrow \infty[/imath], we have [imath]f'(z) = 0 \space \forall z \in \mathbb{C}[/imath], i.e. [imath]f[/imath] is constant. Is this correct? | 576724 | If [imath]f[/imath] is entire and [imath]|f|\geq 1[/imath], then show [imath]f[/imath] is constant.
I know I'm going to use Liouville's Theorem, but my main question is why is [imath]1/|f(z)|[/imath] entire as well if [imath]f[/imath] is entire? Is this just a basic property: if [imath]f[/imath] is entire, then [imath]1/f[/imath] is entire? Thanks for the help. |
1395893 | Mathematical Olympiad Problem
Let [imath]\Bbb{R}[/imath] be the set of real numbers. Determine all functions [imath]f:\Bbb{R}\longrightarrow \Bbb{R}[/imath] satisfying the equation [imath]f(x+f(x+y))+f(xy) = x + f(x+y)+yf(x)[/imath] for all real numbers [imath]x[/imath] and [imath]y[/imath]. | 1882449 | Determine all functions satisfying [imath]f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)[/imath]
Let [imath]\Bbb R [/imath] be the set of real numbers .Determine all functions [imath]f:\Bbb R \rightarrow \Bbb R [/imath] satisfying the equation [imath]f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)[/imath] for all real numbers [imath]x[/imath] and [imath]y[/imath]. My work [imath]f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)[/imath] Setting [imath]y=0[/imath] [imath]f(x + f(x)) + f(0) = x + f(x )[/imath] [imath]\implies f(x + f(x)) = x + f(x ) -f(0)[/imath] [imath]\implies f(t) = t -f(0)[/imath] Setting [imath]x=0[/imath] [imath]f(f(y)) + f(0) = 0 + f(y) +yf(0)[/imath] [imath]f(f(y)) = f(y) +f(0)(y-1)[/imath] Setting [imath]y=1[/imath] [imath]f(x + f(x + 1)) = x + f(x + 1)[/imath] |
455740 | How to solve [imath]1+2^{a}+2^{2a+1}=b^2[/imath]
I want to find all integer solutions for [imath]1+2^{a}+2^{2a+1}=b^2[/imath]. How could I do this? | 808509 | Solve the following equation in positive integers [imath]x[/imath] and [imath]y[/imath]
What are the solutions in positive integers of the equation: [imath]{1+2^x+2^{2x+1}=y^2}[/imath] I tried to factorize the equation but it didn't help much. Clearly [imath]y [/imath] is an odd integer. Substituting [imath]y =2n+1[/imath], we get [imath]2^x+2^{2x+1}=(2n)\cdot{(2n+2)}[/imath] [imath]\Rightarrow (2^{x-2})\cdot(1+2^{x+1})=(n)\cdot(n+1)[/imath] Which is the product of 2 consecutive integers. Does it help? I don't know. |
2725791 | Show that [imath]f(x)=\alpha x[/imath] for some [imath]\alpha \in R[/imath]
Suppose [imath]f(x+y)=f(x)+f(y)[/imath] for some [imath]f: R \rightarrow R[/imath] such that [imath]f[/imath] is continuous at [imath]0[/imath].Show that [imath]f(x)=\alpha x[/imath] for some [imath]\alpha \in R[/imath]. Hint: show that [imath]f(nx) = nf(x)[/imath], then show [imath]f[/imath] is continuous on [imath]R[/imath]. Then show that [imath]f(x)/x = f(1)[/imath] for all rational [imath]x[/imath]. How can I show [imath]f(nx)=nf(x)[/imath]?? I am stuck with the first hint, so I can't show any attempt. Thank you in advance! | 1734170 | If [imath]f[/imath] is continuous and [imath]f(x+y) = f(x)+f(y)[/imath], then [imath]f(x) = cx[/imath] for all [imath]x \in \mathbb{R}[/imath]
Assume [imath]f[/imath] is a function over [imath]\mathbb{R}[/imath] satisfying [imath]f(x+y) = f(x)+f(y)[/imath] for all [imath]x,y \in \mathbb{R}[/imath]. Show that if [imath]f[/imath] is continuous, then [imath]f(x) = cx[/imath] for all [imath]x \in \mathbb{R}.[/imath] I find it hard to use the definition of continuity to derive anything from the equation. I think we could write the definition of the limit in terms to get [imath]\forall \epsilon,\exists \delta \quad 0<|x-a|<\delta \quad \implies \quad |f(x)-L| < \epsilon.[/imath] I am not sure where to go from here. |
2726324 | [imath](A+, \cdot)[/imath] finite ring. If [imath](ab-1)b = 0[/imath], then [imath]b(ab-1) = 0[/imath]
Let [imath](A,+,\cdot)[/imath] be a finite ring, and [imath]a,b \in A[/imath] such that [imath](ab-1)b = 0[/imath]. Prove that [imath]b(ab-1) = 0[/imath]. I already have one solution for this problem and it involves using the fact that if [imath](A,+,\cdot)[/imath] is finite, then [imath]\exists k,m \in \mathbb{N}^*[/imath] such that [imath]b^k = b^m[/imath] (we need to show that [imath]bab = b[/imath], so if [imath]b = b^m[/imath], [imath]bab = bab^m = bab^2b^{m-2} = bbb^{m-2} = b^m = b[/imath], because the hypothesis means that [imath]ab^2 = b[/imath]; the solution involves using the hypothesis in proving that [imath]\exists m\in \mathbb{N}, m>1[/imath] such that [imath]b = b^m[/imath]). I am looking for a different solution, because this one seems rather "forced". I have solved other problems like this (for instance: [imath](M, \cdot)[/imath] a finite monoid, then if [imath]ab=1 \implies ba=1[/imath]) by assuming that the conclusion is false, then defining a function on [imath]A[/imath] (in the monoid example, [imath]f:M \to M, f(x) = xa[/imath] ) that is injective or surjective, and because [imath]A[/imath] is a finite set, then it is bijective, and thus I achieve a contradiction. This kind of solutions seem more natural, at least for me, because in cases like this, when I have a relation between two elements in a ring, I try and "play" with them and achieve other relations involving them that will be useful in proving the problem. I've tried constructing such a function for this problem, but I didn't succeed. It's harder on this particular problem because I can't prove that functions such as [imath]f(x) = x(ax-1), f(x) = bx, f(x) = bxb[/imath] (or others that are involving the hypothesis, but are using only the second operation of the ring) are injective or surjective, because multiplying by [imath]b[/imath] or [imath]a[/imath] and using the hypothesis, I get that [imath]0=0[/imath], which doesn't prove anything. Can this problem be solved using this idea, or any other idea, without using the fact that in a finite ring every element [imath]x \in A[/imath] has two powers, [imath]c,d \in \mathbb{N}^*[/imath] such that [imath]x^c = x^d[/imath] ? Any help (tips, solution) is appreciated. | 1156657 | A finite pseudo-ring such that [imath]ab^2=b[/imath] for some [imath](a,b)\in A[/imath]
I have this exercise which I found very difficult: [imath](A,+,.)[/imath] is a finite pseudo-ring (a structure satisfying all the axioms of a ring except for the existence of a multiplicative identity), and [imath](a,b)\in A^2[/imath] such that [imath]ab^2=b[/imath], prove that [imath]bab=b[/imath]. I used every possible combination of powers of [imath]a[/imath] and [imath]b[/imath] but didn't find anything, the idea I was using is the construction of a sequence [imath]a^nb^m[/imath] for example and there for there exist [imath]n,m,n',m'[/imath] such that [imath]a^nb^m=a^{n'}b^{m'}[/imath] and I use the given equality to reduce it, but it' s very difficult. Can someone help me find the right sequence or the right method to solve this , thanks you |
2726588 | Evaluate the limit [imath]\lim_\limits{x\to-\infty} (1+\frac{1}{x})^{x^2}[/imath]
I need to evaluate the limit [imath] \lim_{x\to-\infty} \left(1+\frac{1}{x}\right)^{x^2}[/imath] I can substitute in [imath]x := -x[/imath] to show that this is the same as [imath] \lim_{x\to\infty} \left(1-\frac{1}{x}\right)^{x^2}[/imath] However, I don't know where to proceed from here. I can rewrite the limit as [imath] \lim_{x\to\infty} \left(\left(1-\frac{1}{x}\right)^x\right)^x [/imath] but this does not help as Algebra of Limits Product only applies to finitely many limits. I also tried substituting [imath]x := 1/x[/imath] to get [imath] \lim_{x\to0^+} (1-x)^{\frac{1}{x^2}} [/imath] but this does not help me either. I expect intuitively the answer to be [imath]\infty[/imath] as the larger exponent means the limit is "growing faster" than [imath]\lim\limits_{x\to\infty} (1+\frac{1}{x})^{x} = e[/imath] but I do not know how to show this. | 415090 | Computing limit of [imath](1+1/n)^{n^2}[/imath]
How can I compute the limit [imath]\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n^2}[/imath]? Of course [imath]\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n} = e[/imath], and then [imath]\left(1+\frac{1}{n}\right)^{n^2} = \left(\left(1+\frac{1}{n}\right)^n\right)^n[/imath]. Since the term inside converges to [imath]e[/imath], the whole thing is [imath]e^n[/imath], and can I conclude that the limit is infinity as [imath]n\rightarrow\infty[/imath]? |
468224 | Inequality [imath](1+a_2)^{2}(1+a_3)^{3}... (1+a_n)^{n}>n^{n}[/imath]
I need help proving this inequality. Let [imath]a_2,a_3,...,a_n[/imath] be positive non zero real numbers such that their product is [imath]1[/imath] Prove that [imath](1+a_2)^{2}(1+a_3)^{3}... (1+a_n)^{n}>n^{n}[/imath] I tried Bernoulli that yields a LHS greater than [imath]n! [/imath] I also tried a trig [imath]tan^{2}[/imath] substitution without success... Thanks for helping. | 173644 | Proving [imath](1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n[/imath] for positive reals [imath]a_2,\ldots, a_n[/imath] whose product is [imath]1[/imath]
Let [imath]n \ge3[/imath] be an integer, and let [imath]a_{2},a_{3}, ... ,a_{n}[/imath] be positive real numbers such that [imath]a_{2} a_{3}\cdots a_{n}=1.[/imath] Prove that: [imath](1+a_{2})^{2}(1+a_{3})^{3}\cdots(1+a_{n})^{n}\ge n^n[/imath] This is the 2nd problem of the 53rd IMO and seems pretty interesting. How would we solve that? |
2727442 | Why is being a free variable a necessary condition
In the following formula [imath]\alpha \rightarrow \forall x \alpha[/imath], why is it important that [imath]x[/imath] does not occur free in [imath]\alpha[/imath]? Here are my thoughts: I would like to find a countermodel and formula [imath]\alpha[/imath] where [imath]x[/imath] is free in [imath]\alpha[/imath] and [imath]\alpha \rightarrow \forall x \alpha[/imath] holds. If we take [imath]\alpha[/imath] to be the formula [imath]x = 5[/imath] in some model whose domain is [imath]\mathbb{N}[/imath], we see that neither [imath]\alpha[/imath] nor [imath]\forall x \alpha [/imath] is satisfied which is not what we want. I cannot see how to meaningfully satisfy [imath]\alpha[/imath] while not satisfying [imath]\forall x \alpha[/imath]. Any hints? | 688384 | Why is [imath]\alpha \rightarrow \forall x(\alpha)[/imath] not generally correct in first-order logic?
Why is [imath]\alpha \rightarrow \forall x(\alpha)[/imath] not generally correct in first-order logic? i.e., when there are free occurrences of [imath]x[/imath] in [imath]\alpha[/imath], and, on the same point, why is the formula scheme correct when there are no free occurrences of [imath]x[/imath] in [imath]\alpha[/imath]? |
2727626 | Why in every vector space are the vectors [imath](-1)\vec{u}=-\vec{u}[/imath]?
I am curious why [imath](-1)\vec{u}=-\vec{u}[/imath] for every vector space. Since vector spaces can be defined differently, with different scalar multiplication rules, why do we have that= [imath](-1)\vec{u}=-\vec{u}[/imath] is true for all vector spaces automatically? I'm missing a subtlety here! Thanks in advance! | 952737 | Proof that [imath]-v = (-1)*v[/imath]
I need to prove that for every Vector Space this is valid: [imath] -v = (-1)*v [/imath] -v = inverse element of addition -1 a real number [imath]*[/imath] the multiplication by real number of the Vector Space My teacher said that [imath]-v[/imath] is just a notation for the inverse element of addition. I'd like to prove that [imath]-v = (-1)*v[/imath]. So I came up with the following solution and I'd like to know if it's correct: \begin{align} v + (-1)*v = u \\ (1)*v + (-1)*v = u \\ (1-1)*v = u \\ 0*v = u \\ o = u\\ \end{align} Since [imath]v + -v = o[/imath] [imath]v + (-1)*v = o = v + -v[/imath] adding -v to both sides v + (-1)*v -v = v + -v + -v o + (-1)*v = o + -v (-1)*v = -v Did I commit any mistakes? Did I make any assumptions that may not be valid for EVERY Vector Space? Edit: As S. Sheng said I have not proved that (0)*v = o. I'll try to prove that and come back later with a proof of that. I also haven't proved that (1)*v = v Oh my... I'm starting to think this is beyond my abilities.. |
2727598 | Number theory question involving Wilson's theorem. How to solve??
Given p is a prime number greater than 2, and [imath] 1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{p-1} = \frac{N}{p-1}[/imath] how do I show, [imath] p | N [/imath] ??? The previous part of this question had me factor [imath] x^{p-1} -1[/imath] mod [imath]p[/imath]. Which I think is just plainly [imath](x-1) ... (x-(p-1))[/imath] | 2722409 | Proof that [imath]\sum\limits_i \frac{(p-1)!}i[/imath] is divisible by [imath]p[/imath]
Problem: Let [imath]p[/imath] be a prime. Consider [imath]\sum\limits_{i = 1}^{p-1} \frac1i = \frac K{(p-1)!}[/imath]. Rearranging, we have [imath]K = \sum\limits_{i = 1}^{p-1} \frac{(p-1)!}i[/imath]. Prove that [imath]p \mid K[/imath]. Hint: consider the factorization [imath]x^{p-1} - 1 \equiv (x-1) ...(x - (p-1)) \pmod{p}[/imath]. Attempt: I am finding applying the hint difficult. Sure, the RHS of the congruence relation contains [imath](p-1)![/imath], but I can't see how the factorization would help me prove the expression for [imath]K[/imath]. Any help is appreciated. |
2727799 | Show that EFGH is a parallelogram
[imath]ABCD[/imath] is a parallelogram and [imath]O[/imath] is any point. The parallelogram [imath]OAEB,OBFC,OCGD,ODHA[/imath] are completed. show that [imath]EFGH[/imath] is a parallelogram. | 617863 | Prove, in this figure, that [imath]EFGH[/imath] is a parallelogram
In the following figure, [imath]ABCD[/imath] is a parallelogram, and [imath]O[/imath] is any point. Parallelograms [imath]OAEB, OBFC, OCGD, ODHA[/imath] are completed. Prove that [imath]EFGH[/imath] is a parallelogram. We can obtain a fairly trivial proof using affine geometry. As [imath]OAEB, OBFC, OCGD, ODHA[/imath] are parallelograms, [imath]\vec{A} + \vec{B} = \vec{O} + \vec{E} \implies \vec{A} + \vec{B} - \vec{O} = \vec{E} \tag1[/imath] [imath]\vec{C} + \vec{B} - \vec{O} = \vec{F}\tag2[/imath] [imath]\vec{D} + \vec{C} - \vec{O} = \vec{G} \tag3[/imath] [imath]\vec{D} + \vec{A} - \vec{O} = \vec{H}\tag4[/imath] Adding [imath](1)[/imath] to [imath](3)[/imath] and [imath](2)[/imath] to [imath](4)[/imath], we get, [imath]\vec{E} + \vec{G} = \vec{A} + \vec{B} + \vec{C} + \vec{D} - 2\vec{O} \tag5[/imath] [imath]\vec{F} + \vec{H} = \vec{A} + \vec{B} + \vec{C} + \vec{D} - 2\vec{O} \tag6[/imath] Clearly, [imath](5)[/imath] and [imath](6)[/imath] are equal, therefore, [imath]\vec{E} + \vec{G} = \vec{F} + \vec{H}[/imath] Therefore, [imath]EFGH[/imath] is a parallelogram. Can somebody give an elementary proof using Euclidean geometry? Also, I noticed that in my proof, nowhere did I use the fact that [imath]ABCD[/imath] is a parallelogram, but constructing an example, it was quickly clear that the result stated does not generalize to all quadrilaterals. How come? Is my proof incorrect? |
2728269 | How to prove that this group is abelian?
Let [imath]G[/imath] be a group such that for all [imath]a,b,c\in G[/imath] we have [imath]ab=ca \implies b=c[/imath] How can I show that [imath]G[/imath] is abelian ? I am kind of stuck in the question | 1373359 | In a group [imath]G[/imath], if for all [imath]a,b,c\in G[/imath], [imath]ab=ca\Rightarrow b=c[/imath], then [imath]G[/imath] is abelian
Let [imath]G[/imath] be a group. If for all [imath]a,b,c\in G[/imath], [imath]ab=ca\Rightarrow b=c[/imath], then phow can I prove that [imath]G[/imath] is abelian? |
2022573 | Find all primes [imath]p[/imath] such that [imath]p^3-4p+9=x^2[/imath], where [imath]x[/imath] is a positive integer.
Problem:Find all primes [imath]p[/imath] such that [imath]p^3-4p+9=x^2[/imath], where [imath]x[/imath] is a positive integer. My Attempt: I have made the following observations while trying tot solve this problem: [imath]p=2[/imath] yields [imath]x=3[/imath]. For [imath]p>2[/imath] the following is true: [imath]x^2\equiv 0\pmod 4[/imath] and [imath]p\equiv 3\pmod 4.[/imath] It seems that [imath]p=2,7[/imath] and [imath]11[/imath] are the only primme numbers that satisfy the conditions of this problem. Below are the values of [imath]p^3-4p+9[/imath] evaluated for [imath]1\leq p\leq 100.[/imath] 9 24 114 324 1296 2154 4854 6792 12084 24282 29676 50514 68766 79344 103644 148674 205152 226746 300504 357636 388734 492732 571464 704622 912294 I would like to know how one can prove that [imath]p=2,7[/imath] and [imath]11[/imath] are the only solutions to the above-mentioned equation. | 1580363 | Find all primes [imath]p[/imath] such that [imath] p^3-4p+9 [/imath] is a perfect square.
Find all primes [imath]p[/imath] such that [imath] p^3-4p+9 [/imath] is a perfect square. I tried a few different values for [imath]p[/imath], namely [imath]2,3,5,7,[/imath] and [imath]11[/imath]. The prime [imath]p =2,7,11[/imath] all worked but [imath]p =13[/imath] didn't so it makes me wonder. How can I find all primes such that it is a perfect square? EDIT: it turns out this is problem P25 in post number #63 at http://artofproblemsolving.com/community/c3h1171106p5665470 |
2703854 | X contains all its accumulation points [imath]\iff[/imath] X contains all its closure points.
From (say Apostol Analysis) we find the following two facts: X is closed iff it contains ALL its accumulation points. X is closed iff it contains ALL its boundary points. So, one must conclude that X contains all its accumulation points [imath]\iff[/imath] X contains all its closure points. How to show this fact (without claiming X is closed) ? Thanks in advance. | 226319 | boundary points, isolation points, accumulation points.
For a fixed subset [imath]S[/imath] in [imath]\mathbb{R^n}[/imath] and a point [imath]p[/imath] in [imath]\mathbb{R^n}[/imath] How do you prove that if a point [imath]p[/imath] is a non isolated boundary point of [imath]S[/imath], then it must be an accumulation point in [imath]S[/imath]? Thinking about the solution: I get that a boundary point must either be isolated point or an accumulation point, so if it's not an isolated point it must be an accumulation point, but I'm not sure why it has to be one or the other. I know isolated points are never accumulation points, because accumulation points must have other points in every neighborhood. But I can't quite grasp my head around it or know where to begin in writing a logical proof. |
2728819 | How to prove that [imath]\lim_{n\to \infty}\sin(n!)[/imath] does not exists?
The continued fraction of [imath]\pi[/imath] has not yet be known. I could not see the distribution of the [imath]\{\frac{n!}{2\pi}\}[/imath] where [imath]\{x\}=x-\lfloor x \rfloor[/imath]. Is there any idea how to find two convergent subsequences with different limits? | 8690 | Is there a limit of cos (n!)?
I encountered a problem today to prove that [imath](X_n)[/imath] with [imath]X_n = \cos(n!)[/imath] does not have a limit (when [imath]n[/imath] approaches infinity). I have no idea how to do it formally. Could someone help? The simpler the proof (by that I mean less complex theorems are used) the better. Thanks |
2729051 | How to integrate [imath]\frac{1}{1+x^4}[/imath]
Evaluate [imath]\int \frac{1}{1+x^4}dx[/imath]. I am finding it very difficult to evaluate this integral. I do not know any standard formula for this integral. I cannot think of a suitable substitution. I tried to evaluate the integral by partial fractions but it become a bit messy and integration by parts does not result in a simpler integral. How can you evaluate this integral without using partial fractions? | 1628530 | How to find the antiderivative of this function [imath]\frac{1}{1+x^4}[/imath]?
How to find the antiderivative of the function [imath]f(x)=\frac{1}{1+x^4}[/imath]? My math professor suggested to use the method of partial fractions, but it doesn't seem to work, because the denominator cannot be factored at all. Attempting to integrate with other familiar methods (integration by parts, trigonometric substitution) didn't work out as well. Does anyone have an idea what is the best approach to this problem? |
2488724 | Mathematical Olympiad Treasures Problem 3.27 how many ways to fill a table such that product of entries equal -1
The question is: In how many ways can one fill a [imath]m × n[/imath] table with [imath]±1[/imath] such that the product of the entries in each row and each column equals [imath]−1[/imath]? So far I believe that [imath]m,n[/imath] must have equal parity, but have not made much progress past that. | 100794 | number of table with [imath]1[/imath] and [imath]-1[/imath]
How many [imath]m\times n[/imath] tables are there if they are subject to the following two conditions? In each cell of the table we have [imath]1[/imath] or [imath]-1[/imath]. The product of all the cells in any given row and the product of all cells in any given column equals [imath]-1[/imath]. I already know if Parity of [imath]m[/imath] and [imath]n[/imath] are different there is no such a table. what is the answer in general? |
2729459 | Fubini's Theorem contradiction
Why does the fact that [imath]\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dy \,dx = \frac{\pi}{4}[/imath] and [imath] \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dx \,dy = -\frac{\pi}{4}[/imath] doesn't contradict Fubini's Theorem? Does the tangent function have something to do with that matter? | 85097 | Contradicting Fubini's theorem
I have a function defined as follows: [imath]f(x,y)= \dfrac{x^2-y^2}{\left(x^2+y^2\right)^2}[/imath], if [imath](x,y)\neq (0, 0)[/imath] and [imath]f(x,y)=0[/imath] if [imath](x,y)=(0,0)[/imath]. Now, [imath]\int_0^1\int_0^1 f(x,y)\,\text{d}x~\text{d}y=-\frac{\pi}{4}[/imath] and [imath]\int_0^1\int_0^1 f(x,y)\,\text{d}y~\text{d}x=\frac{\pi}{4}.[/imath] The question I have is this: why does this not contradict Fubini's theorem? thanks. |
2728646 | Why is the integral defined as the limit of the sum [imath]\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x[/imath]?
I am failing to understand why the integral is defined as: [imath]\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x[/imath] instead of: [imath]\int_a^b f(x)dx=\sum_{i=1}^\infty f(x_i^*)\Delta x[/imath] Is the former just popular preference or is there something I am not conceptually understanding here? | 53001 | What's the "limit" in the definition of Riemann integrals?
Consider one of the standard methods used for defining the Riemann integrals: Suppose [imath]\sigma[/imath] denotes any subdivision [imath]a=x_0<x_1<x_2\cdots<x_{n-1}<x_n=b[/imath], and let [imath]x_{i-1}\leq \xi_i\leq x_i[/imath]. Then if [imath]|\sigma|:=\max\{x_i-x_{i-1}|i=1,\cdots,n\},[/imath] which we shall call the norm of the subdivision, we define: [imath]\int_a^bf(x)dx:=\lim_{|\sigma|\to 0}\sum_{i=1}^nf(\xi_i)(x_i-x_{i-1}).[/imath] When one talks about the limit of a function [imath]\lim_{x\to x_0}f(x)[/imath], one has exactly one value [imath]f(x)[/imath] for every [imath]x[/imath]. However, for every [imath]|\sigma|[/imath], the value of the Riemann sum [imath]\sum_{i=1}^nf(\xi_i)(x_i-x_{i-1})[/imath] is not necessarily unique. Using the [imath]\epsilon[/imath]-[imath]\delta[/imath] language, one may restate the definition as follows: Suppose [imath]f:[a,b]\to{\mathbb R}[/imath], [imath]J\in{\mathbb R}[/imath]. If for all [imath]\epsilon>0[/imath], there exists [imath]\delta>0[/imath] such that for any subdivision [imath]\sigma[/imath] and [imath]\{\xi_i\}[/imath] on [imath]\sigma[/imath] (i.e. [imath]x_{i-1}\leq \xi_i\leq x_i[/imath]), [imath]|\sigma|<\delta[/imath] implies [imath]|\sum_{i=1}^nf(\xi_i)\Delta x_i-J|<\epsilon,[/imath] we call [imath]J[/imath] is the Riemann integral of [imath]f[/imath] on [imath][a,b][/imath] and denote [imath]J=\int_a^bf(x)dx.[/imath] Here are my questions: How should I understand this kind of limit? It seems that this is not the "limit of a function" I learned in elementary real analysis. Where does it appear in mathematics besides the definition of Riemann integrals? |
1055302 | [imath]M \times N[/imath] orientable if and only if [imath]M, N[/imath] orientable
For two manifolds [imath]M[/imath] and [imath]N[/imath] I'm trying to prove that [imath]M \times N[/imath] is orientable if and only if [imath]M[/imath] and [imath]N[/imath] are orientable. My attempt so far: [imath]\impliedby)[/imath] Assume [imath]M, N[/imath] are orientable. Then [imath]\widetilde{M}[/imath] and [imath]\widetilde{N}[/imath] the orientation double covers of [imath]M[/imath] and [imath]N[/imath] are disconnected. But [imath]\widetilde{M} \times \widetilde{N}[/imath] is the orientation double cover for [imath]M \times N[/imath]. Since [imath]\widetilde{M}, \widetilde{N}[/imath] are disconnected certainly [imath]\widetilde{M} \times \widetilde{N}[/imath] is disconnected. Therefore, [imath]M \times N[/imath] is orientable. [imath]\implies)[/imath] Proceed by contradiction. Assume [imath]M \times N[/imath] is orientable and WLOG [imath]M[/imath] is nonorientable. Then the orientation double cover [imath]\widetilde{M\times N}[/imath] is disconnected. Now note that [imath]\widetilde{M}[/imath] is connected since [imath]M[/imath] is nonorientable. So [imath]\widetilde{M} \times \widetilde{N}[/imath] is connected since the product of a connected space and a disconnected space is connected. But [imath]\widetilde{M\times N} = \widetilde{M} \times \widetilde{N}[/imath], so [imath]\widetilde{M\times N}[/imath] is connected. But this is a contradiction, therefore it must be that [imath]M[/imath] is orientable. This completes our proof. The validity of this proof hinges on two assumptions that I'm unsure about: 1. [imath]\widetilde{M\times N} = \widetilde{M} \times \widetilde{N}[/imath] 2. The product of a connected space and a disconnected space is connected. Are these two facts true? Is my proof valid? How do I go about proving these two necessary assumptions if they are indeed true? If I'm totally going the wrong way could someone send me in the proper direction? Thanks | 550426 | Product of manifolds & orientability
I'm studying orientability of manifolds currently and I'm having trouble to prove the following: [imath]M\times N[/imath] is orientable iff [imath]M[/imath] and [imath]N[/imath] are orientable. I am able to prove that the product is orientable if components are orientable (chart is [imath]\{(U_\alpha\times V_{\beta},\phi_\alpha\times \psi_\beta):(\alpha,\beta)\in A\times B \}[/imath], and [imath]\det J=\det J_1 \det J_2>0[/imath] by Cauchy-Binet's theorem), but I don't know how to prove the other direction. So why this holds: if [imath]M\times N[/imath] is orientable, then [imath]M[/imath] and [imath]N[/imath] are orientable? Thanks in advance. |
2729539 | Results that seem trivial, but are not
I saw the Cayley-Hamilton theorem Let [imath]\mathbf{A}[/imath] be a [imath]n\times n[/imath]-matrix, and [imath]p(\lambda)=\det(\lambda \mathbf{I}_n-\mathbf{A})[/imath] the characteristic polynomial of [imath]\mathbf{A}[/imath]. Then [imath]p(\mathbf{A})=\mathbf{0}[/imath] and I thought it was trivial since [imath]\det(\mathbf{A}-\mathbf{A})=0[/imath]. I checked with Wikipedia and learned that this didn't work because [imath]\lambda[/imath] is a scalar, and [imath]\det(\mathbf{0})[/imath] is also a scalar, and not a matrix. This brings me to my question: Are there any other results that seem trivial, but aren't? | 2217995 | Simple theorems that are instances of deep mathematics
So, this question asks about how useful computational tricks are to mathematics research, and several people's response was "well, computational tricks are often super cool theorems in disguise." So what "computational tricks" or "easy theorems" or "fun patterns" turn out to be important theorems? The ideal answer to this question would be a topic that can be understood at two different levels that have a great gulf in terms of sophistication between them, although the simplistic example doesn't have to be "trivial." For example, the unique prime factorization theorem is often proven from the division algorithm through Bezut's lemma and the fact that [imath]p\mid ab\implies p\mid a[/imath] or [imath]p\mid b[/imath]. A virtually identical proof allows you to establish that every Euclidean Domain is a unique factorization domain, and the problem as a whole - once properly abstracted - gives rise to the notion of ideals and a significant amount of ring theory. For another example, it's well known that finite dimensional vector spaces are uniquely determined by their base field and their dimension. However, a far more general theorem in Model Theory basically lets you say "given a set of objects that have a dimension-like parameter that are situated in the right manner, every object with finite "dimension" is uniquely determined by its minimal example and the "dimension." I don't actually quite remember the precise statement of this theorem, so if someone wants to explain in detail how vector spaces are a particular example of [imath]k[/imath]-categorical theories for every finite [imath]k[/imath] that would be great. From the comments: In a certain sense I'm interested in the inverse question as this Math Overflow post. Instead of being interested in deep mathematics that produce horribly complicated proofs of simple ideas, I want simple ideas that contain within them, or generalize to, mathematics of startling depth. |
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