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793432 | Composition of covering maps is a covering map if the inverse image is finite.
Let [imath]q: X \to Y[/imath] and [imath]r: Y \to Z[/imath] be covering maps. Let [imath]p=r \circ q[/imath]. Show that if [imath]r^{-1}(z)[/imath] is finite [imath]\forall z[/imath], then [imath]p[/imath] is a covering map. [imath]\textbf{My Attempt:}[/imath] Let [imath]U[/imath] be an arbitrary open set in [imath]Z[/imath]. Then [imath]p^{-1}(U)=q^{-1}(r^{-1}(U))[/imath]. Since [imath]r[/imath] is a covering map, we know that [imath]r^{-1}(U)=\bigcup_{i=1}^{N} V_i[/imath] where [imath]V_i[/imath] is open [imath]\bigcap_{i=1}^{N} V_i = \emptyset[/imath]. Is the following true, [imath]q^{-1}(\bigcup_{i=1}^{N} V_i)=\bigcup_{i=1}^{N} q^{-1}(V_i)[/imath] ? Now since [imath]q[/imath] is covering for each [imath]V_i[/imath] we have that [imath]q^{-1}(V_i)=\bigcup_{\alpha} K_{\alpha}^i[/imath] where the [imath]K_{\alpha}^{i}[/imath] are all disjoint and open. Then, [imath]q^{-1}(\bigcup_{i=1}^{N} V_i)=\bigcup_{i=1}^{N}(\bigcup_{\alpha} K_{\alpha}^i)[/imath]. So we just need to show that [imath]\bigcap_{i=1}^{N}(\bigcup_{\alpha} K_{\alpha}^i)=\emptyset[/imath] which is fairly straightforward. I am not sure if this is the right way to go about this proof as I have not utilized the fact that the number of [imath]V_i[/imath] will be finite. How do I use the fact that [imath]r^{-1}(z)[/imath] is finite? | 308490 | composition of certain covering maps
This problem was posted before, but not the proof (because the asker knowed the answer), only a counterexample without the hypothesis of finite fibres. I want to know how to prove this proposition: Let [imath]q:X\to Y[/imath] and [imath]r:Y\to Z[/imath] be covering maps. Let's suppose that for each [imath]z\in Z[/imath] , the set [imath] r^{-1}(z)[/imath] is finite, then the composition [imath]p = r\circ q[/imath] is also a covering map. Well I have to consider [imath]z\in Z[/imath] and show that there exist a neighborhood that is evenly covered by [imath]p[/imath], I know that there exist a neighborhood [imath]U[/imath] that is evenly covered by [imath]r[/imath], I think that this will be the desired neighborhood. First of all [imath] r^{-1}(U) = \cup_{i=1}^{n} V_i [/imath] (it's easy to see that the unions is finite using the fact that the fibres are finite and this is a local homeomorphism between the [imath]V_i[/imath]) Then [imath] p^{-1}(U)=q^{-1} (r^{-1}(U) ) = q^{-1}(\cup_{i=1}^{n} V_i)=\cup_{i=1}^{n} q^{-1}(V_i) [/imath] With [imath] q^{-1}(V_i) \cong V_i[/imath] under [imath]q[/imath] and so [imath]\cong[/imath] U under [imath]p[/imath] I never used the fact that the preimage is finite, so my proof is obviously not correct, please help me with this |
771588 | When is every matrix in the span of two matrices singular?
Given two square matrices [imath]A, B[/imath], when is [imath]\det(A+tB) = 0[/imath] for all [imath]t\in \mathbb{R}[/imath]? An easy sufficient condition is that [imath]A[/imath] and [imath]B[/imath]'s kernels have nontrivial intersection. Per Henning's comment below, this is not also necessary. Does there exist a nice necessary and sufficient characterization? | 173026 | Matrices whose Linear Combinations are All Singular
I'd like to know if the following problem of elementary linear algebra is already solved / solvable. For two (singular) [imath]n\times n[/imath] matrices [imath]P[/imath] and [imath]Q[/imath], if [imath]\det(\lambda P+\mu Q)=0[/imath] for any [imath]\lambda,\mu\in\mathbb{R}[/imath], what are conditions on [imath]P[/imath] and [imath]Q[/imath]? |
793821 | Expected Value of a Carnival Game Paradox
I was in my Algebra 2 class today, and we're learning basic probability. Talking with my friend, he proposed this carnival game. The game begins with two dollars in the bowl, and every turn you flip a coin. If the result is heads, the carnival doubles the money in the bowl, and if tails, you take the money out of the bowl. The question is, what is the expected value you would get from this game? Starting from the first flip, you have a [imath]1/2[/imath] chance of getting 2 dollars. Multiplying these together, you get 1. Measuring the next flip, you have a [imath]1/4[/imath] change of getting 4 dollars, which multiply to 1 as well. Continuing this, you end up with the infinite sequence [imath]1 + 1 + 1 + 1...[/imath] How is this possible? Would you really statistically pay infinite money to play a game where you have a chance [imath]50[/imath]% chance of getting $2? | 495053 | St. Petersburg Paradox
A fair coin will be tossed until a heads results. You will then be paid [imath]2^{n-1}[/imath] dollars where [imath]n[/imath] equals the number of flips. Now why is the expected pay out infinite? [imath] \sum_{n \geq 1} (\frac{1}{2^n})2^{n-1} = \sum_{n \geq 1} \frac{2^{n-1}}{2^n} [/imath] Why does this give the payout? The payout shouldn't be infinite, it should be a potential infinity. For example, if I flip the coin once and it is a tails, then I will receive two dollars for certain. On the second flip there is a [imath]25[/imath] percent chance that I will receive a minimum of [imath]4[/imath] dollars. but a [imath]75[/imath] percent chance that I will receive only two dollars. And so on as the chances of receiving more money decreases and eventually approaches [imath]0[/imath]. Why then is it suggested that an investor pay any amount of money for such an opportunity? Source: A Brief History of the Paradox, Philosophy and Labyrinths of the Mind Obviously this appears false by intuition, but I'd like to know the argument for the paradox; why one would pay a trillion dollars to enter such a game. The argument just seems far too fallacious to me. How many times would one have to play? After almost twenty thousand plays, we are still making less than 8 dollars. IF every twenty thousand plays we increased our winnings by 8 dollars, we would have to play hundreds of billions of times just to start to have an average earnings equal to one trillion... we'd run out of time in the universe it seems... Edit I am curious, however, if the overall slope of the simulation data would remain consistent, if so why? If not, why? What accounts for the vertical jumps we see occurring less and less often? Why do they occur at all? |
794171 | Solution Verification: [imath]\sum_{n = 1}^{\infty} a_n \lt \infty \implies \sum_{n = 1}^{\infty} \dfrac{\sqrt{a_n}}{n} \lt \infty[/imath] if [imath]a_n \ge 0[/imath]
I would appreciate it if someone can verify my answer to the following and maybe point me towards alternate solutions. Thanks in advance. I'm not entirely convinced with the use of the Comparison Test here. Prove the convergence of [imath] \sum_{n = 1}^{\infty} a_n[/imath] implies the convergence of [imath]\sum_{n = 1}^{\infty} \dfrac{\sqrt{a_n}}{n}[/imath] if [imath]a_n \ge 0[/imath] for each [imath]n \in \Bbb N[/imath]. Solution: If [imath]a_n \ge \frac 1 n[/imath] for infinitely many [imath]n \in \Bbb N[/imath] it follows that [imath]\sum a_n[/imath] diverges leading to a contradiction. Then there is [imath]m \in \Bbb N[/imath] such that [imath]n \ge m \implies a_n \lt \frac 1 n[/imath]. Therefore, [imath] n \ge m \implies \dfrac{\sqrt{a_n}}{n} \lt \dfrac{\sqrt{\frac{1}n{}}}{n} = \dfrac{1}{n^{\frac 3 2}} [/imath] Since [imath]\sum \dfrac{1}{n^{\frac 3 2}} [/imath] converges so does [imath]\sum \dfrac{\sqrt{a_n}}{n}[/imath] by the Comparison Test. | 659180 | Convergence of [imath]\frac{\sqrt{a_{n}}}{n}[/imath]
Can anyone help me with the following question. If [imath]a_{n} \geq 0[/imath] and [imath]\sum a_{n}[/imath] converges then how to prove [imath]\sum \frac{\sqrt{a_{n}}}{n}[/imath] converges. Any idea where to start. My idea was to try using comparison test since [imath]\sqrt{a_{n}} \leq a_{n}[/imath] but it appears that it wouldn't work if [imath]0 \leq a_{n} <1[/imath]. |
64848 | How does [imath](12\cdots n)[/imath] and [imath](ab)[/imath] generate [imath]S_n[/imath]?
I know that [imath]S_n[/imath] is generated by a number of things, like all transpositions, all transpositions of form [imath](1a)[/imath], the transpositions [imath](12),(23),(34),\cdots(n-1n)[/imath], and just the two elements [imath](123\cdots n),(12)[/imath]. Suppose [imath]n[/imath] is prime. If you just have [imath](123\cdots n)[/imath] and some arbitrary transposition [imath](ab)[/imath], how does this also generate [imath]S_n[/imath]? Can you somehow get to [imath](12)[/imath] or reduce it to some other previous case? | 2313186 | Prove that [imath]Sym(p)[/imath] is generated by any transposition and any p-cycle
Let [imath]p[/imath] be a prime. Prove that [imath]Sym(p)[/imath] is generated by any transposition and any p-cycle. Without loss of generality, we may assume the p-cycle to be [imath](1\space 2\space \dots\space p)[/imath] just by 'relabelling' the entries with a bijection in any p-cycle. Again, without loss of generality, we actually want to prove that [imath]Sym(p)=<(1\space k), (1\space 2\space \dots\space p)>[/imath], where [imath]k[/imath] can take any value from [imath]2[/imath] to [imath]p[/imath], because even if the first entry of the transposition is not '[imath]1[/imath]', say, it's '[imath]2[/imath]', then since [imath](1\space 2\space \dots\space p)=(2\space 3\space \dots\space p\space 1)[/imath], we can again 'relabel' the entries with a bijection. Since we already know that [imath]Sym(p)=<(1\space 2), (1\space 2\space \dots\space p)>[/imath]. Why do [imath](1\space k)[/imath] and [imath](1\space 2\space \dots\space p)[/imath] generate [imath](1\space 2)[/imath] for arbitrary [imath]k[/imath]? |
69955 | How Convergence in two different norms relates to Equivalence of Norms
Question: Why was this marked as a duplicate? The referenced question asked nearly a year later than this question. In fact I'm not even sure that they are identical at all. I was working on a homework problem which stated: Let [imath]X[/imath] be a normed linear space with two norms [imath]\|\cdot\|_1[/imath] and [imath]\|\cdot\|_2[/imath]. Assume that [imath]\|x_n - x^1\|_1\to 0[/imath] for some [imath]x^1\in X[/imath] if and only if [imath]\|x_n - x^2\|_2\to 0[/imath] for some [imath]x^2\in X[/imath]. Show that [imath]\|\cdot\|_1[/imath] and [imath]\|\cdot\|_2[/imath] are equivalent. At first I errantly misinterpreted the question as [imath]x_n\to x[/imath] in [imath]\|\cdot\|_1[/imath] if and only if [imath]x_n\to x[/imath] in [imath]\|\cdot\|_2[/imath]. UPDATE: I found a way to go directly from the convergence condition to equivalence of norms. So I don't need to prove that the sequences converge to the same limit in this way (since it now follows from equivalence of norms). I won't erase anything though, as the answer from Andres below may be helpful to someone else. Now that I realize my mistake I'm trying to fix the proof. Since the conclusion I'm trying to prove (that the norms are equivalent) implies that the limits in either norm must agree, I know that must also be true, even though it's not given in the question. After failing to prove this extra condition, I am getting suspicious that it might be needed as a hypothesis after all. Can anyone confirm or refute this? My attempt: Taking [imath]\alpha = \dfrac12\min\left(\|x^1 - x^2\|_1, \|x^1 - x^2\|_2\right)[/imath], I let [imath]B_{1} = \{x\in X : ||x - x^1||_{1} < \alpha\}[/imath] and let [imath]B_{2} = \{x\in X : ||x - x^{2}||_{2} < \alpha\}[/imath]. Now suppose for a contradiction that there is some [imath]x\in B_{1}\cap B_{2}[/imath]. Then [imath]||x - x^{1}||_{1} < \alpha[/imath]. And [imath]||x - x^{2}||_{2} < \alpha[/imath]. I want to use this to prove [imath]||x^{1} - x^{2}||[/imath] (in either norm) is less than [imath]2\alpha[/imath] (which would give me a contradiction). But I keep ending up with these floating [imath]||x - x^{2}||_{1}[/imath] or [imath]||x - x^{1}||_{2}[/imath] terms that I can't do anything with: [imath] \begin{eqnarray*} ||x^{1} - x^{2}||_{1} &\leq& ||x^{1} - x||_{1} + ||x - x^{2}||_{1}\\ &<& \alpha + ||x - x^{2}||_{1} \end{eqnarray*} [/imath] and [imath] \begin{eqnarray*} ||x^{1} - x^{2}||_{2} &\leq& ||x^{1} - x||_{2} + ||x - x^{2}||_{2}\\ &<& ||x^{1} - x||_{2} + \alpha \end{eqnarray*} [/imath] | 159810 | When are two norms equivalent on a Banach space?
I'm working on an exercise from functional analysis. Let [imath]E[/imath] be a vector space and [imath]\|\cdot\|_1[/imath] and [imath]\|\cdot\|_2[/imath] be two complete norms on [imath]E[/imath]. Now suppose that [imath]E[/imath] satisfies the following property: [imath]\bullet[/imath] if [imath](x_n)[/imath] is a sequence in [imath]E[/imath] and [imath]x,y\in E[/imath] such that [imath]\|x_n-x\|_1\to 0[/imath] and [imath]\|x_n-y\|_2\to 0[/imath], then [imath]x=y[/imath]. Now we want to show that the norms [imath]\|\cdot\|_1[/imath] and [imath]\|\cdot\|_2[/imath] are equivalent. My idea is as follows: If for any [imath]n>0[/imath], there is an element [imath]x_n\in E[/imath] such that [imath]\|x_n\|_1>n\|x_n\|_2[/imath]. Then consider [imath](\frac{x_n}{\|x_n\|_1})_{n\geq 1}[/imath]. Clearly, [imath](\frac{x_n}{\|x_n\|_1})_{n\geq 1}[/imath] converges to [imath]0[/imath]. However, I cann't get a contradicition from this. Maybe my idea is wrong. In fact, I even don't konw how to show that a Cauchy sequence in norm [imath]\|\cdot\|_1[/imath] is also a Cauchy sequence in norm [imath]\|\cdot\|_2[/imath]. Anyone can give me some hints or a counter example? Thank you very much. |
794592 | Show that if sup{∑|f(a)|}<∞, then {a∈A:f(a) is not zer0} is countable.
Let f:A→R and suppose that sup{∑a∈F|f(a)|:F is finite subset of A}<∞ then {a∈A:f(a)is not zero} is countable. and show that Define these element of { a ∈ A : f(a) is not 0} a1, a2, a3, ... then ∑(n=1 to infinity)|f(an)| is determined, Regardless of the method of arrange Its question may similar to Show that if [imath]\sup\big\{\sum\lvert\, f(a)\rvert\big\} < \infty[/imath], then [imath]\{ a \in A : f(a) > 0\}[/imath] is countable. | 792738 | Show that if [imath]\sup\big\{\sum\lvert\, f(a)\rvert\big\} < \infty[/imath], then [imath]\{ a \in A : f(a) > 0\}[/imath] is countable.
Let [imath]f:A \to \mathbb R[/imath] and suppose that [imath] \sup\Big\{\sum_{a\in F}\lvert\, f(a)\rvert : F\text{ is finite subset of }A\Big\} < \infty [/imath] then the set [imath]\{ a \in A : f(a) > 0\}[/imath] is countable. My try: Define these element of [imath]{ a ∈ A : f(a) > 0}[/imath], then [imath]∑_{n=1}^\infty\lvert\, f(a_n)\rvert[/imath] is determined, regardless of the method of arrangement. |
791845 | How and in what context are polynomials considered equal?
There's two notions of equivalent polynomials floating around, one saying that [imath]f = g[/imath] iff they're equivalent as maps, and the other saying [imath]f = g[/imath] iff they're equal on each coefficient when written in standard form. I'm interested in polynomials over a finite field, irreducible polynomials and factoring so what type of equivalence should I use? For instance if we take map equivalence, then there are only a finite number of polynomials. And that makes a huge difference! Please explain when it's okay to use what. | 390260 | Do we really need polynomials (In contrast to polynomial functions)?
In the following I'm going to call a polynomial expression an element of a suitable algebraic structure (for example a ring, since it has an addition and a multiplication) that has the form [imath]a_{n}x^{n}+\cdots+a_{1}x+a_{0}[/imath], where [imath]x[/imath] is some fixed element of said structure, a polynomial function a function of the form [imath]x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}[/imath] (where the same algebraic considerations as above apply) a polynomial an element of the form [imath]a_{n}X^{n}+\cdots+a_{1}X+a_{0}[/imath] with indeterminates [imath]X[/imath] (these can be formalized, if we are in a ring, with strings of [imath]0[/imath]s and a [imath]1[/imath]). Note that when we are very rigorous/formal, polynomial expressions and polynomials are something different (although in daily life we often use them synonymously). Polynomial functions and expressions are also different from each other although in this case the relationship is a closer one, since every polynomials expression can be interpreted as a polynomial function evaluated at a certain point (thus "polynomial functions" are something more general than "polynomial expressions"). My question is: Why do we use polynomials ? It seems to me that every piece of mathematics I have encountered so far, one could replace every occurrence of polynomials with polynomial expressions/functions without any major change in the rest of the proof/theorem/definition etc. The only reasons that I can see to use polynomials are the following two:[imath] [/imath] 1. After one makes the idea precise that one can plug "suitable" elements into polynomials (which may lie a ring containing the ring in which the coefficients live in), one can save time in certain setting, by handling the "plugging of suitable elements into polynomials" more elegantly: For example in case of the theorem of Cayley-Hamilton, which in its "polynomial function" version would look like: Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix over [imath]K[/imath], whose characteristic polynomial (function) is [imath]x\mapsto a_{n}x^{n}+\cdots+a_{1}x+a_{0}[/imath]. Then [imath] a_{n}A^{n}+\cdots+a_{1}A+a_{0}I=0. [/imath] whereas the "polynomial" version looks more elegant: Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix over [imath]K[/imath], whose characteristic polynomial is [imath]p_{A}\in K\left[X\right][/imath]. Then [imath] p_{A}\left(A\right)=0. [/imath] 2. The only thing that polynomials can "do", but algebraic expressions/functions can't, is to be different, when the algebraic expressions/functions are the same (i.e. there's a theorem that tells us that the mapping of polynomials to polynomials expressions/functions isn't injective, if the field is finite). Maybe this small difference makes a big enough difference to consider polynomials after all, but as I said, I haven't encountered any situation in which this difference could manifest itself. (I'm guessing that maybe cryptography or higher number theory really needs polynomials and not just polynomial expressions/functions. Since I don't know anything about these subjects, I would be very happy with an example of a theorem (whose content isn't merely technical as it is the case with the theorem above) involving polynomials, where these absolutely cannot be replaced by polynomial expressions/functions. Conversely I would also be happy with an authoritative statement from someone knowledgeable that indeed we could dispense of polynomials, if we wanted to.) |
794092 | Squaring the plane with consecutive integer squares. And a related arrangement.
Q1. I was fiddling around with squaring-the-square type algebraic maths, and came up with a family of arrangements of [imath]n^2 [/imath]squares, with sides [imath]1,2\ldots n^2[/imath] (n odd). Which seems like it would work with ANY odd [imath]n[/imath]. It's so simple, surely it's well-known, but I haven't seen it in my (brief) web travels. I have a page with pics here of the [imath]n=7,9,11[/imath] versions, and description of how to construct them. It seems the same method could square the plane, well, fill greater than any specified area, no matter how huge, at least. Which is what 'infinite' means, practically, isn't it? Anyway, it seems, if somehow not known (which it must be, surely - if anyone has links etc to where it's discussed I would be very grateful) then it's another way of squaring the plane. Q2. Each of these arrangements can be extended, but the ones I've tried (5 or 6) have a little gap to the south-east, (i.e. where the squares don't fit neatly together) but otherwise can be extended forever. Is there an n for which there is no gap? There are more than 1 of each sized square in this arrangement, but still, it would be a nice tessellation with integer squares. Here's a picture of the 7x7 version extended, and detail of the centre. | 794389 | Squaring the plane, with consecutive integer squares. And a related arrangement
Q1. I was fiddling around with squaring-the-square type algebraic maths, and came up with a family of arrangements of [imath]n^2[/imath] squares, with sides [imath]1, 2, 3\ldots n^2[/imath] ([imath]n[/imath] odd). Which seems like it would work with ANY odd [imath]n[/imath]. It's so simple, surely it's well-known, but I haven't seen it in my (brief) web travels. I have a page with pics here of the [imath]n=7,9,11[/imath] versions, and description of how to construct them; [imath]n=11[/imath] is: It seems the same method could square the plane, well, fill greater than any specified area, no matter how huge, at least. Which is what 'infinite' means, practically, isn't it? Anyway, it seems, if somehow not known (which it must be, surely - if anyone has links etc to where it's discussed I would be very grateful) then it's another way of squaring the plane. Q2. Each of these arrangements can be extended, but the ones I've tried (5 or 6) have a little gap to the south-east, (i.e. where the squares don't fit neatly together) but otherwise can be extended forever. Is there an [imath]n[/imath] for which there is no gap? There are more than 1 of each sized square in this arrangement, but still, it would be a nice tessellation with integer squares. Here's a picture of the 7x7 version extended, and detail of the centre. Thanks for any answers or help. P.s. This was too long for the comments section: I wrote to Jim Henle the other day asking about this method, he hadn't seen it, thought it was nice, but not plane-filling in the way his method is. He wrote in part: You are sort of "squaring the plane," but not in the sense that we did it. You are squaring larger and larger areas of the plane, but you don't square the whole thing. ... There are many meanings to "infinite". Aristotle distinguished between the "potential infinite" (more and more, without bound) and the "actual infinite" (all the numbers, all at once). Your procedure is the first sort, and ours is the second sort. Which is what I had thought. But the more I think about it.. the less clear the difference seems. Well, e.g. 'there are an infinite number of primes' means: there is no highest one; any number you can say, there's a higher prime. That's how that is defined, spoken about, to my (non-mathematician's) understanding. Similarly, there are an infinite number of these [imath]n\times n[/imath] square groups, there's no biggest one, any area you name, there's a bigger one. The Henle's method consists in adding more squares, ideally forever, but practically, you stop at some point and say 'and so on forever'. The procedure requires an infinite number of steps. I can't quite see how this series of [imath]n\times n[/imath] squares is so different. You have to start drawing again with each new [imath]n[/imath], sure, but I can't see that matters so much - there are an infinite number of arrangements, and there's the same 'and so on forever'.. i.e. "there is, strictly speaking, no such thing as an infinite summation or process." Nov 2015. [I can't add comments to questions below, not sure why.] Ross M, sorry about the delay! It seems you might be confused between the 2 parts, my fault for combining them in one question. (I still haven't heard anything much about either 2 from anyone.) The first part is the basic [imath]n^2[/imath] arrangements of squares. (The second part takes just one of these and tries to extend it outwards, wonders about the possibility and mathematics of there being no gaps, and has many more than 1 of each sized square) I still don't quite see why 'having to rearrange each step' makes a huge difference to anything. Imagine I had a method of going from [imath]n^2[/imath] to [imath](n+1)^2[/imath] squares by adding more around the edges. Then, according to what people seem to be saying, I 'could tile the whole thing'? The way I've done it has exactly the same area as that would be, just it has to be redrawn. I don't see how that affects whether it 'tiles the plane' or not, or anything else. If someone could explain that to me, I would be very grateful. |
794883 | Discriminant of a splitting field
Let [imath]f(x)\in\mathbf Q[x][/imath] and let [imath]K[/imath] be the splitting field of [imath]f(x)[/imath]. How is the discriminant of [imath]K[/imath] (as a number field) related to the discriminant of [imath]f(x)[/imath]? Are they divisible by the same primes? Does one divide the other? | 8312 | Discriminant of a monic irreducible integer polynomial vs. discriminant of its splitting field
Let [imath]f\in\mathbb{Z}[x][/imath] be monic and irreducible, let [imath]K=[/imath] splitting field of [imath]f[/imath] over [imath]\mathbb{Q}[/imath]. What can we say about the relationship between [imath]disc(f)[/imath] and [imath]\Delta_K[/imath]? I seem to remember that one differs from the other by a multiple of a square, but I don't know which is which. On a more philosophical note: why are these quantities related at all? Is there an explanation for why they can be different, i.e. some information that one keeps track of that the other doesn't? |
784605 | An inequality concerning the sides of convex quadrilateral
[imath]ABCD[/imath] is a convex quadrilateral such that [imath] \angle ABC\ge120^{o} , \angle BCD\ge120^{o}[/imath] , then is it true that [imath]AC+BD>AB+BC+CD[/imath] ? | 489147 | In a convex Quadrilateral ABCD, [imath]\measuredangle ABC = \measuredangle BCD = 120^{\circ}[/imath].Prove that: [imath] AC + BD \ge AB + BC + CD[/imath]
In a convex Quadrilateral ABCD, [imath]\measuredangle ABC = \measuredangle BCD = 120^{\circ}[/imath].Prove that: [imath] AC + BD \ge AB + BC + CD[/imath] My attempt Tried to use cosine formula twice' i.e ([imath]\triangle ABC[/imath] and [imath]\triangle BCD[/imath]) and tried to make and inequality.But couldn't prove. |
795600 | Combinatorial proof of [imath]k\binom{n}{k} = n\binom{n-1}{k-1}[/imath]
I'm trying to prove this combinatorially. [imath]k\binom{n}{k} = n\binom{n-1}{k-1}[/imath] I know the first step is to relate a question to the equation. My question was if you have [imath]n[/imath] friends how many ways can you choose [imath]k[/imath] of them. I know this isn't correct because that would be the question if the left side wasn't being multiplied by k. Can anyone help me figure out what multiplying [imath]{n \choose k}[/imath] by [imath]k[/imath] means? | 299598 | Combinatorial argument for the identity [imath]k\binom{n}{k} = n\binom{n-1}{k-1}[/imath]
I am looking for the combinatorial argument for the identity: \begin{equation} k\binom{n}{k} = n\binom{n-1}{k-1} \end{equation} This is easy to show algebraically as: \begin{equation} \binom{n}{k} = \dfrac{n(n-1)(n-2)(n-k+1)}{k(k-1)!} \end{equation} What is the combinatorial argument? What are some general ideas to get started? Here is a clarification of 2. From what I have seen so far, proving (combinatorially) an identity with an addition sign usually implies that we need to partition a set (this makes sense because of the addition rule and provides a nice visual). On the contrary, the previous observation leads me to believe that multiplication in identities can be resolved with the multiplication principle, but what is the "visual/interpretation" for this? Could someone provide such an interpretation for the example identity given above? |
791617 | Repository of functions which do not have elementary integrals
If there is some function and I suspect that the primitive function cannot be expressed using elementary functions, I would like to have some argument that there indeed is no such expression. One possibility is that I learn about theories that are used to prove thinks like this, as discussed, for example, here, here, here. But if I can take for granted that some functions are "famously" non-integrable in terms of elementary functions, for example [imath]\int e^{x^2}\,\mathrm{d}x[/imath], then I can use this fact when discussing integrals of other functions. (The argument would go like: If it was possible to integrate this function, then using the following steps, we would also be able to integrate [imath]e^{x^2}[/imath]. Hence, the given function does not have elementary integral.) (If I may make a comparison, this is somewhat similar to situation with set-theoretical results: Wise people, who have knowledge of forcing and other advanced stuff, have proven many results which are undecidable in ZFC. If I obtain some of these statements as a consequence of some conjecture, I get that this conjecture is also undecidable.) So I wonder whether there is some book or website which gives some large list of integrals, which are known to be non-elementary. (And thus can be used in proofs that other functions do not have elementary integrals.) Ideally, it would be good if there were also references to the proofs that the functions in the list do not have elementary integrals. | 680478 | List of functions not integrable in elementary terms
When teaching integration to beginning calculus students I always tell them that some integrals are "impossible" (with a bit of expansion on what that actually means). However I must admit that the examples I give mostly come from "folklore" or guesswork. Can anyone point me to a list (not a complete list of course!) of fairly simple elementary functions whose antiderivatives are not elementary? I'm thinking of things like [imath]\exp(x^2)[/imath] which is the standard example, [imath]\sin(\exp(-x))[/imath] perhaps, things like this, not hugely complicated formulae. |
795832 | existence of special functions
Whether there exists a function [imath]f(x,y)[/imath] defined on [imath][0,1]\times(0,1][/imath] satisfies the following conditions: for any [imath]x\in(0,1][/imath], [imath]f(x,y)[/imath] is decreasing with respect to [imath]y[/imath] and [imath]\lim_{y\rightarrow0}f(x,y)=\log x[/imath]. | 784681 | existence of a special function
Whether there exists a function [imath]f(x,y)[/imath] defined on [imath][0,1]\times(0,1][/imath] satisfies the following conditions: for any [imath]x\in(0,1][/imath], [imath]f(x,y)[/imath] is decreasing with respect to [imath]y[/imath] and [imath]\lim_{y\rightarrow0}f(x,y)=\log x[/imath]. |
795944 | P.d.f. of [imath]XY[/imath], where [imath]X, Y[/imath] are independent uniformly distributed over [imath][0,1][/imath]
I tried to change the variables: Let [imath]U=XY[/imath] and [imath]V=Y[/imath]; so then the Jacobian is [imath]1/v[/imath]. So joint pdf [imath]g(u,v) = f(x,y)\cdot (1/v) = 1/v[/imath] Would you then integrate over [imath]v[/imath] from [imath]0[/imath] to [imath]1[/imath] to get the p.d.f., of just [imath]U[/imath]? However this gives infinity. | 792554 | Find the pdf of [imath]\prod_{i=1}^n X_i[/imath], where [imath]X_is[/imath] are independent uniform [0,1] random variables.
How do I find the pdf of [imath]\prod_{i=1}^n X_i[/imath], where [imath]X_is[/imath] are independent uniform [0,1] random variables. I know X~U[0,1], -ln(x) is exponential(1). I also know the sum of two or more independent exponential random variable is gamma. For [imath]Y = \sum_{i=1}^n -ln(x_i)[/imath], which is a gamma(n, 1), I found the pdf for Y is [imath]\int_0^\infty \frac{1}{\Gamma (n)} y^{n-1} e^{-y} dy[/imath]. Let [imath]Z = e^Y[/imath] I am trying to the pdf for Z, what I found is [imath]\int_1^\infty \frac{1}{\Gamma (n)} ln(z)^{n-1} \frac{1}{z^2} dz[/imath], which does not look right to me. Could someone check it? |
798538 | Prove that [imath]n=a^2+b^2-c^2[/imath].
For any natural number [imath]n[/imath] prove that there exist natural numbers [imath]a,b,c[/imath] for which [imath]n=a^2+b^2-c^2.[/imath] I think we can have a proof by induction because : If [imath]n=0[/imath] we choose [imath]a=b=c=0[/imath] or [imath]a=0,b=c[/imath]. If [imath]n=1[/imath] we choose [imath]a=1,b=c[/imath]. Now we need to prove the relationship holds for [imath]n+1[/imath] but I can't prove that. | 444865 | An identity which applies to all of the natural numbers
Prove that any natural number n can be written as [imath]n=a^2+b^2-c^2[/imath] where [imath]a,b,c[/imath] are also natural. |
799798 | How can I prove that the open interval [imath](c,d)[/imath] is not of measure zero?
I am currently trying to prove this theorem. I know how to do the closed interval case, it is a contradiction proof that uses the heine-borel theorem. Does anyone know how I can do the open interval case? I suspect it uses contradiction too, but dont know where to obtain the contradiction from (in closed case it was from the heine-borel theorem). Thanks! | 799756 | Why does an open interval NOT have measure zero?
I am currently working on a proof that requires me to show that an open ball [imath]B_{\epsilon}(x)[/imath] has nonzero measure. I currently have the following proof in my book: "The closed interval [imath][a,b][/imath] is not of measure zero." Hence, if I take the contrapositive, does it follow that "if an interval doesn't have measure zero, the interval is open"? Is there a way for me to prove that the open interval on the line [imath]\mathbb{R}[/imath] is of measure not zero? Thanks! |
126122 | Intersections of all subgroups is a nontrivial subgroup, so every element has finite order.
I need help to prove this result: "Let [imath]G[/imath] be a group such that the intersection of all its subgroups other than [imath]\{1\}[/imath] is a subgroup different from [imath]\{1\}[/imath]. Then all its elements have a finite order". I know I must think of an element [imath]g[/imath] of infinite order. That will imply that every subgroup has infinite order (because it will contain an element like [imath]g^k[/imath]). After this, I don't know which step I can take. Can someone help? | 2753270 | Every element of a group has finite order
Suppose that [imath]G[/imath] is a group, and that the intersection over the non-trivial subgroups of [imath]G[/imath] is non-trivial, i.e., that [imath]\bigcap_{H<G\atop H\ne \{e\}} H\neq\{e\}.[/imath] I need to demonstrate that every element of [imath]G[/imath] has finite order. I tried showing that if the intersection is trivial, an element of infinite order would be there, but couldn't get anywhere. |
702692 | Prove that [imath]H[/imath] is a abelian subgroup of odd order
Question is: Let [imath]G[/imath] be a group of order 2n. Suppose half of the element of G are of order 2 and the other half form a subgroup [imath]H[/imath] of order n . Prove that [imath]H[/imath] is of odd order and is an abelian subgroup of [imath]G[/imath] What could i see is.. if we prove that order of every element of [imath]H[/imath] is odd , then order of [imath]H[/imath] is odd . Also i am unable to use the fact that half of the element of [imath]G[/imath] are of order 2. Please help me to clear this. Thank You. | 1186429 | .Show that [imath]H[/imath] is abelian
Let [imath]G[/imath] be a group of [imath]2n[/imath] .Let H be a subgroup of [imath]G[/imath] consisting of only those elements of [imath]G[/imath] which are of order [imath]\neq 2[/imath].Suppose [imath]o(H)=n[/imath] Show that [imath]n[/imath] is odd and [imath]H[/imath] is abelian. Pairing the elements [imath](g,g^{-1})[/imath] of [imath]H[/imath] since [imath]g\neq g^{-1}\forall g\in H[/imath].Also [imath]H[/imath] has identity [imath]e[/imath].Thus [imath]o(H)=n=[/imath] odd How to show [imath]H[/imath] is abelian. |
800923 | Is every sub-lattice of [imath]\mathcal P(X)[/imath] isomorphic to a sub-lattice of [imath]\mathcal P(X')[/imath] containing a singleton set?
Let [imath]X[/imath] be a finite set. [imath]\mathcal{P}(X)[/imath] denotes the set of all subsets of [imath]X[/imath]. Let [imath]\Gamma[/imath] be a sub-lattice of [imath]\mathcal{P}(X)[/imath], i.e. [imath]\Gamma[/imath] is a collection of subsets of [imath]X[/imath] closed under union and intersection. Suppose [imath]\Gamma[/imath] non-empty and [imath]\Gamma[/imath] not made only of the empty set. Set [imath]n:=n(\Gamma):=\min\{m\geq1\,|\, \Gamma\text{ contains at least one set of cardinality }m\}[/imath] and suppose [imath]n\geq2[/imath]. I would like to find a lattice [imath]\Gamma'[/imath] isomorphic to [imath]\Gamma[/imath], but with [imath]n(\Gamma')=1[/imath]. With this purpose I thought to define a map [imath]\Phi\!:\Gamma\to\Phi(\Gamma)=:\Gamma'[/imath] doing the following operations: from each set [imath]A\in\Gamma[/imath] with cardinality [imath]|A|=n[/imath], choose [imath]n-1[/imath] elements and delete them; (Notice that all these sets [imath]A[/imath]'s are pairwise disjoint, by the minimality of [imath]n[/imath] and since [imath]\Gamma[/imath] is closed under intersection) call [imath]Y[/imath] the set of all deleted elements; from each set [imath]B\in\Gamma[/imath] with cardinality [imath]|B|>n[/imath], delete the elements of [imath]Y[/imath]. Clearly in this way one obtains [imath]\Gamma'\subseteq\mathcal{P}(X')[/imath], where [imath]X':=X\smallsetminus Y[/imath]. By construction [imath]n(\Gamma')=1[/imath]. It is trivial that [imath]\Phi[/imath] is surjective. It is also easy to check that [imath]\Phi[/imath] is an homomorphism of lattices (i.e. [imath]\Phi(A\cup B)=\Phi(A)\cup\Phi(B)\,[/imath], [imath]\,\Phi(A\cap B)=\Phi(A)\cap\Phi(B)\,[/imath]): it suffices to write [imath]\Phi(A)=A\smallsetminus Y[/imath] and apply elementary facts of set theory. Therefore to conclude it remains only to prove that [imath]\Phi[/imath] is injective. I think it is true because I don't manage to build any counter-example. Can you help me to prove it? | 800200 | Is every sub-lattice of [imath]\mathcal P(X)[/imath] isomorphic to a sub-lattice of [imath]\mathcal P(X')[/imath] containing singleton sets?
Let [imath]X[/imath] be a finite set. [imath]\mathcal{P}(X)[/imath] denotes the set of all subsets of [imath]X[/imath]. Let [imath]\Gamma[/imath] be a sub-lattice of [imath]\mathcal{P}(X)[/imath], i.e. [imath]\Gamma[/imath] is a collection of subsets of [imath]X[/imath] closed under union and intersection. Suppose [imath]\Gamma[/imath] non-empty and [imath]\Gamma[/imath] not made only of the empty set. Set [imath]n:=n(\Gamma):=\min\{m\geq1\,|\, \Gamma\text{ contains at least one set of cardinality }m\}[/imath] and suppose [imath]n\geq2[/imath]. I would like to find a lattice [imath]\Gamma'[/imath] isomorphic to [imath]\Gamma[/imath], but with [imath]n(\Gamma')=1[/imath]. With this purpose I thought to define a map [imath]\Phi\!:\Gamma\to\Phi(\Gamma)=:\Gamma'[/imath] doing the following operations: from each set [imath]A\in\Gamma[/imath] with cardinality [imath]|A|=n[/imath], choose [imath]n-1[/imath] elements and delete them; (Notice that all these sets [imath]A[/imath]'s are pairwise disjoint, by the minimality of [imath]n[/imath] and since [imath]\Gamma[/imath] is closed under intersection) call [imath]Y[/imath] the set of all deleted elements; from each set [imath]B\in\Gamma[/imath] with cardinality [imath]|B|>n[/imath], delete the elements of [imath]Y[/imath]. Clearly in this way one obtains [imath]\Gamma'\subseteq\mathcal{P}(X')[/imath], where [imath]X':=X\smallsetminus Y[/imath], and [imath]n(\Gamma')=1[/imath]. My questions are: is [imath]\Gamma'[/imath] a sub-lattice of [imath]\mathcal P(X')[/imath] ? is the map [imath]\Phi[/imath] an isomorphism of lattices (i.e. [imath]\Phi[/imath] bijection, [imath]\Phi(A\cup B)=\Phi(A)\cup\Phi(B)\,[/imath], [imath]\,\Phi(A\cap B)=\Phi(A)\cap\Phi(B)\,[/imath]) ? Edit. It is trivial that [imath]\Phi[/imath] is surjective (by definition of [imath]\Gamma'[/imath]). It seems also clear that [imath]\Phi[/imath] is an homomorphism of lattices: it suffices to write [imath]\Phi(A)=A\smallsetminus Y[/imath] and apply elementary facts of set theory. Therefore the only thing that remains to prove it that [imath]\Phi[/imath] is injective. |
263919 | [imath]\Delta \subset \Phi[/imath] is a base in a root system imples [imath]\Delta^\vee \subset \Phi^\vee[/imath] is a base in a root system
(the notation here is compatible with J.E. Humphrey's "Introduction to Lie Algebras and Representation Theory") Let [imath]\Phi \subset E[/imath] be a root system. Let [imath]\Delta \subset \Phi[/imath] be a base. I already know that [imath]\Phi^\vee \subset E[/imath] is a root system. I'd like to show that [imath]\Delta^\vee \subset \Phi^\vee[/imath] is a base. (the notation [imath]\alpha^\vee[/imath] means [imath]\frac{2\alpha}{(\alpha,\alpha)}[/imath]) I tried 2 approaches and was stuck with about the same problem. Here's one attempt: I'll denote [imath]\Delta=\Delta(\gamma)[/imath] for some regular [imath]\gamma \in E[/imath]. I'd like to show that [imath]\Delta(\gamma)^\vee=\Delta(\gamma^\vee)[/imath]. I can easily show that [imath]\Phi^+(\gamma)^\vee=(\Phi^\vee)^+(\gamma^\vee)[/imath]. But then, how do I show that the indecomposable elements in [imath](\Phi^\vee)^+(\gamma^\vee)[/imath] are exactly the duals of the indecomposable elements in [imath]\Phi^+(\gamma)[/imath]? (If this is true, than it goes the other way around too, so I'll try to show that the dual of a decomposable element is decomposable). Denote [imath]\alpha=\beta_1+\beta_2[/imath] for some [imath]\alpha,\beta_1,\beta_2 \in \Phi^+(\gamma)[/imath]. I'd like to show that [imath]\alpha^\vee[/imath] is decomposable too (in [imath](\Phi^+(\gamma))^\vee[/imath]). It is not true that [imath]\alpha^\vee=\beta_1^\vee+\beta_2^\vee[/imath] as I've seen by checking 2-dimensional examples. That's where I'm stuck. (the other attempt was to go straight from the definition of a base, and there I had a problem with proving the integrality of the coefficients) Another idea was to use the correspondence between Weyl chambers and bases. Both [imath]\Phi[/imath] and [imath]\Phi^\vee[/imath] define the same Weyl chambers. This gives a correspondence between the bases of [imath]\Phi[/imath] and the bases of [imath]\Phi^\vee[/imath]. The correspondence is probably given by taking the dual of a base, but that's what I'm stuck at showing. | 724491 | How to prove that [imath]B^\vee[/imath] is a base for coroots?
Let [imath]\Phi[/imath] be a root system in a real inner product space [imath]E[/imath]. Define [imath]\alpha^\vee = \frac{2\alpha}{(\alpha, \alpha)}[/imath]. Then the set [imath]\Phi^\vee = \{\alpha^\vee: \alpha \in \Phi \}[/imath] is also a root system. Let [imath]B[/imath] be a base for the root system [imath]\Phi[/imath], ie. [imath]B[/imath] is a basis for [imath]E[/imath] and each [imath]\alpha \in \Phi[/imath] can be written as [imath]\alpha = \sum_{\beta \in B} {k_\beta} \beta[/imath] such that [imath]k_\beta[/imath] are integers of same sign. Question: How to prove that [imath]B^\vee = \{ \alpha^\vee: \alpha \in B\}[/imath] is a base for the root system [imath]\Phi^\vee[/imath]? It is easy to see that [imath]B^\vee[/imath] is a basis for [imath]E[/imath]. For the other property, here is what I have so far. Let [imath]\alpha = \sum_{\beta \in B} {k_\beta} \beta[/imath] be a root. Then [imath]\alpha^\vee = \sum_{\beta \in B} \frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)} \beta^\vee[/imath] so to prove the claim, it is necessary to prove that [imath]\frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)}[/imath] is an integer. I don't see how that follows. In the case where [imath](\alpha, \beta) = 0[/imath] the proportion [imath]\frac{(\beta, \beta)}{(\alpha, \alpha)}[/imath] could be anything, so I am a bit confused. |
5219 | How do I prove this sum is not an integer
Assume that [imath]k,n\in\mathbb{Z}^+[/imath]. Prove that the sum \begin{equation*} \dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+\ldots +\dfrac{1}{k+n-1}+\dfrac{1}{k+n} \end{equation*} is not an integer. The case [imath]k=0[/imath] is proved in this question: "Is there an elementary proof that [imath]\sum_{k=1}^{n}\frac{1}{k}[/imath] is never an integer?" | 1086641 | Can the difference of two Harmonic numbers be an integer?
It is well known that the Harmonic numbers [imath]H_n=\sum_{k=1}^n \frac{1}{k}[/imath] are never integers for [imath]n>1[/imath]. Can the difference of two Harmonic numbers [imath]H_n-H_m=\sum_{k=m+1}^n \frac{1}{k}[/imath] be an integer if [imath]n>m>1[/imath]? |
801453 | Why is this relation anti-symmetric?
Why is the following relation anti-symmetric? {(1,2), (2,3), (3,4), (1,4), (1,3), (2,4)} From my understanding, it is anti-symmetric if: [imath] (a, b) \in R, (b, a) \in R, b=a [/imath] Must hold. Which is it this case not true. I would rather say it is asymmetric since there is no symmetric nor anti symmetric relation at all. | 255683 | Antisymmetric Relations
Given a set [imath]\{1,2,3,4\}[/imath], how is the following relation [imath]R[/imath] antisymmetric? [imath]R = \{(1, 2), (2, 3), (3, 4)\}[/imath] Note: Antisymmetric is the idea that if [imath](a,b)[/imath] is in [imath]R[/imath] and [imath](b,a)[/imath] is in [imath]R[/imath], then [imath]a = b[/imath]. In my textbook it says the above is antisymmetric which isn't the case as whenever [imath](a,b)[/imath] is in [imath]R[/imath], [imath](b,a)[/imath] is not. |
802447 | Why is [imath]S_{\ast}\left(X,A\right)[/imath] free?
Why is [imath]S_{\ast}\left(X,A\right)[/imath] free? it is the quotient of two free groups [imath]S_{\ast}\left(X\right)[/imath] & [imath]S_{\ast}\left(A\right)[/imath] | 773607 | Are quotients of chain groups [imath]C_n(X)/C_n(A)[/imath] still free?
Suppose you have a topological space [imath]X[/imath] and a subspace [imath]A[/imath]. Their chain complexes are made up of free abelian groups [imath]C_n(X)[/imath] and [imath]C_n(A)[/imath] are the free abelian groups on the [imath]n[/imath]-simplexes on [imath]X[/imath] and [imath]A[/imath], respectively. If we take the quotient complex we get another chain [imath] \cdots\to C_2(X)/C_2(A)\to C_1(X)/C_1(A)\to C_0(X)/C_0(A)\to\cdots [/imath] of abelian groups. Are these quotients still free abelian groups though? I know in general that the quotient of a free abelian need not be free, but is it any different in this case however? |
775387 | How come if [imath]\ i\ [/imath] not of the following form, then [imath]12i + 5[/imath] must be prime?
I know if [imath]\ i\ [/imath] of the following form [imath]\ 3x^2 + (6y-3)x - y\ [/imath] or [imath]\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}[/imath], then [imath]\ 12i + 5\ [/imath] must be composite number, because: [imath]12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x + 1)[/imath] How come if [imath]\ i\ [/imath] not of the following form [imath]\ 3x^2 + (6y-3)x - y\ [/imath] and [imath]\ 3x^2 + (6y-3)x + y - 1\ [/imath], then [imath]12i + 5[/imath] must be prime? [imath]\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}[/imath]. For example: [imath]\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\ [/imath] ,when [imath]\ x = y = 1\ [/imath], as proved before,[imath]\ 5*12+5\ [/imath] must be composite number; [imath]\ 0, \ 1. \ 2.\ 3.\ 4\ [/imath] can't of the following form [imath]\ 3x^2 + (6y-3)x - y\ [/imath] and [imath]\ 3x^2 + (6y-3)x + y - 1\ [/imath], [imath]\ 12*0 + 5 = 5\ [/imath], [imath]\ 5 [/imath] is prime,[imath]\ 12*1 + 5 = 17\ [/imath], [imath]\ 17 [/imath] is prime, and so on, sure this is true and there's a long proof, but too long to see. | 775648 | How come if [imath]\ i\ [/imath] not of the following form, then [imath]12i + 5[/imath] must be prime?
I know if [imath]\ i\ [/imath] of the following form [imath]\ 3x^2 + (6y-3)x - y\ [/imath] or [imath]\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}[/imath], then [imath]\ 12i + 5\ [/imath] must be composite number, because: [imath]12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x + 1)[/imath] How come if [imath]\ i\ [/imath] not of the following form [imath]\ 3x^2 + (6y-3)x - y\ [/imath] and [imath]\ 3x^2 + (6y-3)x + y - 1\ [/imath], then [imath]12i + 5[/imath] must be prime? [imath]\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}[/imath]. For example: [imath]\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\ [/imath] ,when [imath]\ x = y = 1\ [/imath], as proved before,[imath]\ 5*12+5\ [/imath] must be composite number; [imath]\ 0, \ 1. \ 2.\ 3.\ 4\ [/imath] can't of the following form [imath]\ 3x^2 + (6y-3)x - y\ [/imath] and [imath]\ 3x^2 + (6y-3)x + y - 1\ [/imath], [imath]\ 12*0 + 5 = 5\ [/imath], [imath]\ 5 [/imath] is prime,[imath]\ 12*1 + 5 = 17\ [/imath], [imath]\ 17 [/imath] is prime, and so on, |
5818 | Proving [imath]{n \choose p} \equiv \Bigl[\frac{n}{p}\Bigr] \ (\text{mod} \ p)[/imath]
This is an exercise from Apostol, which i have been struggling for a while. Given a prime [imath]p[/imath], how does one show that [imath]{n \choose p} \equiv \biggl[\frac{n}{p}\biggr] \ (\text{mod} \ p)[/imath] Note that [imath]\Bigl[\frac{n}{p}\Bigr][/imath] denotes the integral part of [imath]\frac{n}{p}[/imath]. I would also like to know as to how does one try to solve this problem. Well, what we need is to show is whenever one divides [imath]{n \choose p}[/imath] by a prime [imath]p[/imath] the remainder is the integral part of [imath]\frac{n}{p}[/imath]. Now, [imath]{ n \choose p} = \frac{n!}{p! \cdot (n-p)!}[/imath] Now [imath]n![/imath] can be written as [imath]n!= n \cdot (n-1) \cdot (n-2) \cdots (n-p) \cdots 2 \cdot 1[/imath] But i am really struggling in getting the integral part. | 806831 | Prove the following :
Prove the following : [imath] {{n}\choose{7}}-\left \lfloor{\frac{n}{7}}\right \rfloor [/imath] is divisible by 7. |
802732 | Need help with a definite integral [imath]\int_0^{\infty} \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}\,dx[/imath]
Evaluate: [imath]\int_0^{\infty} \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}\,dx[/imath] I am not sure where to start or what should be the best approach towards this problem. I tried the substitution [imath]2^x-1=t^2[/imath] but that seems to make things more worse. Using this substitution, I got: [imath]\int_0^{\infty} \frac{1}{\ln^2 2}\frac{\ln\left(\frac{1+t^2}{2}\right)}{(1+t^2)\ln t}\,dt[/imath] I don't see how to proceed after this. :( Any help is appreciated. Thanks! | 520657 | A conjectured closed form of [imath]\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx[/imath]
Consider the following integral: [imath]\mathcal{I}=\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx.[/imath] I tried to evaluate [imath]\mathcal{I}[/imath] in a closed form (both manually and using Mathematica), but without success. However, if WolframAlpha is provided with a numerical approximation [imath]\,\mathcal{I}\approx 3.2694067500684...[/imath], it returns a possible closed form: [imath]\mathcal{I}\stackrel?=\frac\pi{2\,\ln^2 2}.[/imath] Further numeric caclulations show that this value is correct up to at least [imath]10^3[/imath] decimal digits. So, I conjecture that this is the exact value of [imath]\mathcal{I}[/imath]. Question: Is this conjecture correct? |
81805 | How to show that this binomial sum satisfies the Fibonacci relation?
The binomial sum [imath]s_n=\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}+\cdots[/imath] satisfies the Fibonacci relation. I failed to prove that [imath]\binom{n-k+1}{k}=\binom{n-k}{k}+\binom{n-k-1}{k}[/imath]... Any hints or suggestions? | 1079825 | Prove that [imath]F_n={n-1 \choose 0 }+{n-2 \choose 1 }+{n-3 \choose 2 }+\ldots[/imath] where [imath]F(n)[/imath] is the [imath]n[/imath]-th fibonacci number
If [imath]F_n[/imath] is the [imath]n[/imath]-th fibonacci number, then prove that, [imath]F_n={n-1 \choose 0 }+{n-2 \choose 1 }+{n-3 \choose 2 }+\ldots[/imath] I tried the idea of using Pascal's triangle, but it seems to need some adjustment, so I am stuck, please help. Any combinatorial argument would be of greater help than rigorous proof. Thank you. |
802788 | Finding an area of a propeller using double integration? Attempted, please help! :(
The boundary of the internal hole is given by [imath]r = a + b\cos(4α)[/imath] where [imath]a > b >0[/imath]. The external boundary of the propeller is given by [imath]r = c + d\cos(3α)[/imath] where [imath]c - d > a + b[/imath] and [imath]d > 0[/imath]. 1/Calculate the total area of the propeller as shown by the shaded region 2/calculate the average temperature of the propeller, given by [imath]T (r, α) = C/r[/imath], where [imath]C[/imath] is a constant. 3/Evaluate the average temperature numerically for a=3, b=1, c=12, d=5 and C=250 where distance is in cm and temperature is Celsius I got an idea but don't know if it's right and when I ask my friend none of them know how to :( Are you suppose to integrate [imath]r = c + d\cos(3α)[/imath] then minus the hole [imath]r = a + b\cos(4α)[/imath] ? [imath]{1\over2}\int_0^{2\pi}\int_0^{p+q\cos(n\phi)} r^2\>dr\ d\phi\>,\qquad {1\over2}\int_0^{2\pi}\int_0^{p+q\cos(n\phi)}{C\over r}\> r^2\>dr\ d\phi\ .[/imath] since the area of the propeller in polar coordinate is: [imath]{\rm d(area)}={1\over2} r^2{\rm d}(r,\phi)\ .[/imath] If someone help me solve it, it would make my day because I have no idea how to do it! :( I'll thank you if you could solve it! Pretty please. | 800179 | Area of a weird ellipse shape.
A propeller has the shape shown below. The boundary of the internal hole is given by [imath]r = a + b\cos(4α)[/imath] where [imath]a > b >0[/imath]. The external boundary of the propeller is given by [imath]r = c + d\cos(3α)[/imath] where [imath]c - d > a + b[/imath] and [imath]d > 0[/imath]. Calculate the total area of the propeller as shown by the shaded region and calculate the average temperature of the propeller, given by [imath]T (r, α) = C/r[/imath], where [imath]C[/imath] is a constant. Have absolutely no idea how to solve this. |
794495 | A polynomial [imath]\ f(x)[/imath] has integer coefficients such that [imath]\ f(0)[/imath] and [imath]\ f(1)[/imath] are both odd numbers. Prove that [imath]\ f(x) = 0[/imath] has no integer solutions.
Let there be a polynomial [imath]\ f(x)[/imath]. It has integer coefficients such that [imath]\ f(0)[/imath] and [imath]\ f(1)[/imath] are both odd numbers. Prove that [imath]\ f(x) = 0[/imath] has no integer solutions. | 182593 | Test for an Integer Solution
This came up an a training piece for the Putnam Competition and also in Ireland and Rosen. The question posed was basically: Let [imath]p(x)[/imath] be a polynomial with integer coefficients satisfying that [imath]p(0)[/imath] and [imath]p(1)[/imath] are odd. Show that [imath]p[/imath] has no integer zeros. I&R give an example: [imath]p(x) = x^2 - 117x + 31[/imath] and show (no problem) that for any [imath]n[/imath] whether even or odd, [imath]p(n)[/imath] will be odd. And claim that this shows [imath]p(n)[/imath] will never be [imath]0[/imath]. I can see, e.g., that [imath]x^2 + 2x + 1[/imath] will be odd substituting an even [imath]n[/imath] and even for an odd [imath]n[/imath]. But would appreciate help in understanding the underlying math and what is happening here. Also, as a second part, can a general statement about the existence of an integer solution be made if [imath]n[/imath], even and odd, generates an even and an odd as in the last example. I can see that if you look at these equations (mod [imath]2[/imath]), you can distinguish whether there is an integer solution. And I would guess this is intimately connected with the question. Thanks as always. |
802721 | prove that using uniform bounded theorem
Let [imath]y=(\eta_j),\eta_j\in \mathbb C[/imath], be such that [imath]\sum \xi_j\eta_j[/imath] converges for every [imath]x=(\xi_j)\in c_0[/imath] where [imath]c_0\subset l^\infty[/imath] is the subspace of all complex sequences converging to zero. Show that [imath]\sum |\eta_j| < \infty[/imath] using theorem ( Let (Tn) be a sequence of bounded linear operators Tn:X->Y from a Banach space X into a normed space Y such that ‖T(n)x‖ is bounded for every x∈X,say ,Absolutely (T(n)x) less than or equal cx, where cx is real number. Then the sequence of the norms ‖T(n)‖is bounded that is , there is a c such that Absolutely (T(n)) less than or equal c) This question in book Introductory Functional Analysis with Applications page 255 question (10) | 794798 | show that the element in [imath]\ell^1[/imath].
If [imath]x=(x_n)[/imath] is a sequence of complex numbers such that the series [imath]\sum x_ny_n[/imath] is convergent for all [imath]y=(y_n)\in{c_0}[/imath]. Then prove that [imath]x\in{\ell^1}[/imath]. Can anyone tell me what is the meaning of [imath]c_0[/imath] in this question and how to proceed on proving it. |
804898 | Does right continuity imply only countably many discontinuities?
Does right continuity imply only countably many discontinuities? That is, if [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] is right continuous then does it only have countable many discontinuities? Thanks | 65941 | If [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is a left continuous function can the set of discontinuous points of [imath]f[/imath] have positive Lebesgue measure?
If [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is a left continuous function can the set of discontinuous points of [imath]f[/imath] have positive Lebesgue measure? I wondered this today, but made little progress. Thank you. |
804965 | confusion about proof on Banach limits
I am reading a example in John N McDonald's intro to analysis and I am very confused about one line. Let [imath]l_r^{\infty}[/imath] denote that real linear space consisting of all bounded sequences of real numbers and set [imath]E=\{x\in l_r^{\infty} \ : \ \lim_{n\rightarrow \infty}x_n \ \text{ exists}\}[/imath]. Consider the linear functional [imath]L_0[/imath] on E defined by [imath]L_0(x) = \lim_{n\rightarrow \infty}x_n.[/imath]. We will use the Hahn-banach theorem to extend [imath]L_0[/imath] to all [imath]l_r^{\infty}[/imath]. Define [imath]p:l_r^{\infty}\rightarrow \mathbb{R}[/imath] by [imath]p(x) = \limsup_{n\rightarrow \infty}n^{-1}\sum_{k=1}^nx_k[/imath]. Then [imath]p[/imath] satisfies the hypothosis of the Hahn-Banach theorem and, according to part (b) of Excersise 2.35 [imath]p(x) = L_0(x)[/imath] for all [imath]x\in E[/imath] It follows that there is a linear functional [imath]L[/imath] on [imath]l_r^{\infty}[/imath] that agrees with [imath]L_0[/imath] on [imath]E[/imath] and satisfies [imath]L(x)\le p(x)[/imath] for all [imath]x\in l_r^\infty[/imath]. The functional [imath]L[/imath] shares with [imath]L_0[/imath] the following properties: [imath]\color{blue}{\liminf_{n\rightarrow \infty}x_n\le L \le \limsup_{n\rightarrow \infty}x_n}[/imath] and [imath]\color{blue}{L(x) = L(x^{(1)} \text{ where } x_n^{(1)}=x_{n+1}}[/imath] for each n. I am having trouble understanding the stuff in blue. I know if I can show [imath]L(x)\le \limsup_{n\rightarrow \infty}x_n[/imath] then [imath]L(-x)\le \limsup_{n\rightarrow \infty}-x_n=-\liminf_{n\rightarrow \infty}x_n[/imath] and [imath]L(-x) = -L(x)[/imath] so [imath]\liminf_{n\rightarrow \infty}x_n \le L(x)[/imath] but I am not how they got [imath]L(x)\le \limsup_{n\rightarrow \infty}x_n[/imath], and I have no idea how they got the other blue part... any help would be great. | 345260 | A question about Banach limits
I'm trying to prove that there exists a multiplicative linear functional in [imath]\ell_\infty^*[/imath] that extends the limit funcional that is defined in [imath]c[/imath] (i.e., im looking for a linear functional [imath]f \colon \ell_\infty \to \mathbb K[/imath] such that [imath]f( (x_n * y_n) ) = f (x_n) f(y_n)[/imath], for every [imath](x_n), (y_n) \in \ell_\infty[/imath], and such that [imath]f((x_n)) = \lim x_n[/imath] , if [imath](x_n)[/imath] converges). I found a lot of references saying that it exists but I can't find a detailed proof. The usual Hahn-Banach approach doesn't work because I get a shift invariant functional, and thats inconsistent with multiplicavity. The references I found suggest that using ultrafilters to define the limit should work. I found this interesting proof of the reciprocal: every multiplicative linear functional in [imath]\ell_\infty^*[/imath] is a limit along an ultrafilter: Every multiplicative linear functional on [imath]\ell^{\infty}[/imath] is the limit along an ultrafilter. it assumes [imath]\mathbb K = \mathbb R[/imath], but I could adapt so I think (assuming I made no mistakes) it works for [imath]\mathbb C[/imath] as well. In sum, I'm looking for a proof of the result here http://planetmath.org/BasicPropertiesOfALimitAlongAFilter but it should work for [imath]\mathbb C[/imath] and the [imath](x_n) \mapsto \mathcal F -\lim (x_n)[/imath] functional should be continuous. Is there a good reference for this? or is it just trivial? thanks |
803954 | Surely You're Joking, Mr. Feynman! [imath]\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx[/imath]
Prove the following \begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}\end{equation} I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>‿◠)✌ Original question: And of course, for the sadist with a background in differential equations, I invite you to try your luck with the last integral of the group. \begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx\end{equation} Source: Integration: The Feynman Way | 691798 | Differentiation wrt parameter [imath]\int_0^\infty \sin^2(x)\cdot(x^2(x^2+1))^{-1}dx[/imath]
Use differentiation with respect to parameter obtaining a differential equation to solve [imath] \int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)}dx [/imath] No complex variables, only this approach. Interesting integral and it should have a nice ODE. I have not found the right way yet. we have singularities at [imath]x=\pm i[/imath]. |
805506 | [imath]\delta[/imath]-neighbourhood of a set
I have seen in a book about [imath]\delta[/imath]- neighbourhood of a set. What could exactly that mean ? It is not defined there. Is it union of [imath]\delta[/imath] neighbourhood of all the elements of that set ? | 248834 | What exactly is a [imath]\delta[/imath] neighborhood
The set of all point [imath]x[/imath] such that [imath]|x-a| < \delta[/imath] is called a [imath]\delta[/imath] neighborhood of the point [imath]a[/imath]. The set of all points [imath]x[/imath] such that [imath]0<|x-a|<\delta[/imath] in which [imath]x=a[/imath] is excluded, is called a deleted [imath]\delta[/imath] neighborhood of [imath]a[/imath] or an open ball radius [imath]\delta[/imath] about [imath]a[/imath]. I don't understand this definition (and because of that also the definition of a limit point): Is [imath]\delta[/imath] just a random positive integer? What exactly is the use of a [imath]\delta[/imath] neighborhood, I don't see how it could be meaningful at all. |
20282 | Fundamental group of [imath]S^2[/imath] with north and south pole identified
Consider the quotient space obtained by identifying the north and south pole of [imath]S^2[/imath]. I think the fundamental group should be infinite cyclic, but I do not know how to prove this. If it is infinite cyclic, would this and [imath]S^1[/imath] be an example of two spaces which have isomorphic fundamental groups but are not of the same homotopy type? | 1486331 | The fundamental group of [imath]\mathbb S^2[/imath] attached with a diameter?
What is the fundamental group of [imath]\mathbb S^2[/imath] attached with a diameter? And what is the fundamental group of a hemisphere attached with a diameter? I guess for the latter one, we can deformation retract it to a circle by compressing around the diameter. So the fundamental group should be [imath]\mathbb Z[/imath]. And for the first one we can combine this result with Van Kampen's theorem, and the answer should also be [imath]\mathbb Z[/imath]. Am I right? And how can I prove the case for hemisphere more rigorously? |
795138 | Find the distribution of [imath]X_1^2 + X_2^2[/imath]?
Let [imath]X_1[/imath] and [imath]X_2[/imath] are independent [imath]N(0, \sigma^2)[/imath] which means (mean = 0, variance = [imath]\sigma^2[/imath]) random variables. What is the distribution of [imath]X_1^2 + X_2^2[/imath]? My approach is that [imath]X_1\sim N(0, \sigma^2)[/imath] and [imath]X_2\sim N(0, \sigma^2)[/imath]. Transforming [imath]X_1[/imath] and [imath]X_2[/imath] into standard normal, [imath]X_1/\sigma\sim N(0, 1)[/imath] and [imath]X_2/\sigma\sim N(0, 1)[/imath]. Then [imath]X_1^2/\sigma[/imath] and [imath]X_2^2/\sigma[/imath] have chi-squared distribution with 1 degree of freedom. Then I found the moment-generating function for [imath]X_1^2[/imath] and [imath]X_2^2[/imath];[imath]m_{X_1^2} = (1-2t)^{-1/2}[/imath] and [imath]m_{X_2^2} = (1-2t)^{-1/2}[/imath] So the moment generating function for [imath]X_1^2 + X_2^2[/imath] is [imath]m_{X_1^2}(t) m_{X_2^2}(t) = (1-2t)^{-2/2}[/imath] So [imath]X_1^2 + X_2^2[/imath] has a chi-squared distribution with 2 degrees of freedom. My question can I treat [imath]X_1^2/\sigma[/imath] + [imath]X_2^2/\sigma[/imath] as [imath]X_1^2[/imath] + [imath]X_2^2[/imath] like I did above? | 792556 | Let [imath]X_1[/imath] and [imath]X_2[/imath] are independent [imath]N(0, \sigma^2)[/imath] random variables. What is the distribution of [imath]X_1^2 + X_2^2[/imath]?
Let [imath]X_1[/imath] and [imath]X_2[/imath] are independent [imath]N(0, \sigma^2)[/imath] which means (mean = 0, variance = [imath]\sigma^2[/imath]) random variables. What is the distribution of [imath]X_1^2 + X_2^2[/imath]? My approach is that [imath]X_1\sim N(0, \sigma^2)[/imath] and [imath]X_2\sim N(0, \sigma^2)[/imath] Then [imath]X_1^2[/imath] and [imath]X_2^2[/imath] have chi-squared distribution with 1 degree of freedom. (I am not sure the degree of freedom and not sure how to show it as well(please help on this)) Then I found the moment-generating function for [imath]X_1^2[/imath] and [imath]X_2^2[/imath];[imath]m_{X_1^2} = (1-2t)^{-1/2}[/imath] and [imath]m_{X_2^2} = (1-2t)^{-1/2}[/imath] So the moment generating function for [imath]X_1^2 + X_2^2[/imath] is [imath]m_{X_1^2}(t) m_{X_2^2}(t) = (1-2t)^{-2/2}[/imath] So [imath]X_1^2 + X_2^2[/imath] has a chi-squared distribution with 2 degrees of freedom. Is this correct? |
805512 | Pigeon Hole Principle in Unit Disk
Let [imath]n[/imath] be a natural number such that [imath]n \ge 2[/imath] and given complex number [imath]z_1, z_2, \ldots, z_n[/imath] that is contained in an open unit disk centered at origin. Prove that there exists [imath]\epsilon_l = \pm 1[/imath] with [imath]l = 1, 2, \ldots, n[/imath] such that \begin{align*} \left|\sum_{l=1}^n \epsilon_l z_l\right|\le\sqrt{3} \end{align*} | 437430 | For fixed [imath]z_i[/imath]s inside the unit disc, can we always choose [imath]a_i[/imath]s such that [imath]\left|\sum_{i=1}^n a_iz_i\right|<\sqrt3[/imath]?
Let [imath]z_1,z_2,\ldots,z_n[/imath] be complex number such that [imath]|z_i|<1[/imath] for all [imath]i=1,2,\ldots,n[/imath]. Show that we can choose [imath]a_i \in\{-1,1\}[/imath], [imath]i=1,2,\ldots,n[/imath] such that [imath]\left|\sum_{i=1}^n a_iz_i\right|<\sqrt3.[/imath] |
137794 | Probability of given onto Function
Let [imath]X = \{1, 2, 3, \ldots , 25\}[/imath]. If a student selects a function randomly from the set of all functions from [imath]X[/imath] onto [imath]X[/imath], then what is the probability that the selected function maps prime numbers to prime numbers? | 805580 | what is the probability that the selected function maps prime numbers to prime numbers?
Let [imath]X = {1, 2, 3, . . . , 25}[/imath]. If a student selects a function randomly from the set of all functions from X onto X, then what is the probability that the selected function maps prime numbers to prime numbers ? No idea how to proceed, please guide. |
806690 | Differences of two real-valued periodic function is zero in limit
Given [imath]f, g \colon \mathbb{R} \rightarrow \mathbb{R}[/imath] be a periodic function. If [imath]\lim\limits_{x \rightarrow \infty} \left(f(x) - g(x)\right) = 0[/imath], prove that [imath]f = g[/imath]. | 566219 | The difference between two periodic functions converges to zero, is this two functions identical?
If [imath]f(x)[/imath] and [imath]g(x)[/imath] are two periodic functions, that is, [imath]f(x+T_1)=f(x)[/imath] and [imath]g(x+T_2)=g(x)[/imath] for every [imath]$x \in \Bbb R$[/imath]. Now that [imath]$\displaystyle\lim_{x\to\infty}(f(x)-g(x))=0$[/imath]. Conjecture: [imath]f(x) \equiv g(x)[/imath]. |
806715 | Why does manipulating the expression help?
Consider the limit [imath]\lim_{x\rightarrow 1}\frac {x^2-1}{x-1} [/imath] Plugging in [imath]1 [/imath] for [imath] x [/imath] will yield [imath]\frac00 [/imath], which is no good. However, if we simplify the expression, we get: [imath]\lim_{x\rightarrow 1}\frac {x^2-1}{x-1} = \lim_{x\rightarrow 1}(x+1)[/imath] Now, plugging in [imath]1 [/imath] yields the answer [imath]2 [/imath]. My question is: why on earth should simplifying the expression help us find the limit? The two expressions above are absolutely equivalent. I.e., if I plug in, say [imath]10 [/imath] in [imath](x+3)^2 [/imath] and [imath]x^2+6x+9 [/imath], I get the same answer - [imath]169 [/imath]. | 462199 | Why does factoring eliminate a hole in the limit?
[imath]\lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)[/imath] I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the [imath]x^2-25[/imath] into [imath](x-5)(x+5)[/imath], but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the [imath](x-5)[/imath], which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure." I really want to understand this in abstract algebra terms, please elaborate. Thank you very much. |
806861 | Eigenvalues of Matrix with 1s everywhere but diagonal
I'm not sure if this type of matrix has a name, but I feel as if there's a trick to finding the eigenvalues that i'm missing: [imath] a \in R [/imath] [imath] M = \begin{bmatrix} 1 + a & 1 & 1 \\[0.3em] 1 & 1 + a & 1 \\[0.3em] 1 & 1 & 1+ a \end{bmatrix}[/imath] | 806160 | Find the eigenvalues and eigenvectors with zeroes on the diagonal and ones everywhere else.
I have been working on this problem for a couple hours and am completely stuck. Any help would be greatly appreciated. Let [imath]A[/imath] be the [imath]n \times n[/imath] matrix which has zeros on the main diagonal and ones everywhere else. Find the eigenvalues and eigenvectors of [imath]A[/imath]. |
806847 | Parametric Equation of a 3D Helix Tube Surface?
I am trying to parameterize the surface of a 3D helix tube like below. As known, a parametric curve only requires one parameter [imath]t[/imath] [imath]x=R\sin(t)[/imath], [imath]y=R\cos(t)[/imath], [imath]z=p t[/imath], where [imath]R[/imath] is the radius, and [imath]p[/imath] represents the pitch. However for a surface, I think I need at least two parameters [imath]t_1[/imath] and [imath]t_2[/imath] to represent the 3D closed surface of a 3D helix tube. Here, I need [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath] in the parametric equations to be "decoupled". i.e., I am expecting something like [imath]x=f_1(t_1, t_2)[/imath], [imath]y=f_2(t_1, t_2)[/imath], and [imath]z=f_3(t_1, t_2)[/imath]. What are the expressions? | 461547 | what's the equation of helix surface?
I know for the helix, the equation can be written: [imath]x=R\cos(t)[/imath] [imath]y=R\sin(t)[/imath] [imath]z=ht[/imath] this is the helix curve, and there are two parameters: outer radius [imath]R[/imath] and the pitch length [imath]2\pi h[/imath]. However, I would like to generate the 3D helix with another minor radius [imath]r[/imath]. This is not the helix curve, but a 3D object something like spring. I don't know exactly the name of such structure, but when I search helix equation, they usually give the equations for helix curve but not for the 3D helix object (spring). Does anyone know the equation of such object? Thank you so much for any help and suggestions. ps. the shape looks like the following way: |
806713 | Find [imath]\sum\frac{a(n)}{n(n+1)}[/imath], where [imath]a(n)[/imath] --- number of 1's in binary expansion of n.
Let [imath]a(n)[/imath] is a number of 1's in binary expansion of n, find the sum [imath] \sum\limits_{n=1}^{\infty}\frac{a(n)}{n(n+1)}. [/imath] | 432250 | How does one [easily] calculate [imath]\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}[/imath]?
How does one [easily] calculate [imath]\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}[/imath], where [imath]\mathrm{pop}(n)[/imath] counts the number of bits '1' in the binary representation of [imath]n[/imath]? Is there any trick to calculate the sum? From what I already have, it definitely converges. |
807219 | Looking for irrational Numbers Proof
[imath]a,b,c,~[/imath]and [imath]d[/imath] are rational numbers. [imath]b>0[/imath] and [imath]d>0[/imath] the [imath]\sqrt{b}[/imath] and the [imath]\sqrt{d}[/imath] are both irrational. if [imath]a+\sqrt{b}=c+\sqrt{d}[/imath] show that [imath]a = c[/imath] and [imath]b = d[/imath]. I know that a=c and b=d intuitively, but I'm not sure how to prove it. | 737190 | Let a, b, c, d be rational numbers...
Let [imath]a, b, c, d[/imath] be rational numbers, where [imath]\sqrt{b}[/imath] and [imath]\sqrt{d}[/imath] exist and are irrational. If [imath]a + \sqrt{b} = c + \sqrt{d}[/imath], prove that [imath]a=c[/imath] and [imath]b=d[/imath]. |
798751 | rank([imath]A[/imath])=rank([imath]A^T[/imath])
Is there an elementary explanation of why the row-rank of a matrix equals its column-rank (without using adjoint maps, resp. lots of technical computations)? What is the geometric intuition behind this? | 2315 | Is the rank of a matrix the same of its transpose? If yes, how can I prove it?
I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view: "The rank of a matrix A is the number of non-zero rows in the reduced row-echelon form of A". The lecturer then explained that if the matrix [imath]A[/imath] has size [imath]m * n[/imath], then [imath]rank(A) \leq m[/imath] and [imath]rank(A) \leq n[/imath]. The way I had been taught about rank was that it was the smallest of the number of rows bringing new information the number of columns bringing new information. I don't see how that would change if we transposed the matrix, so I said in the lecture: "then the rank of a matrix is the same of its transpose, right?" And the lecturer said: "oh, not so fast! Hang on, I have to think about it". As the class has about 100 students and the lecturer was just substituting for the "normal" lecturer, he was probably a bit nervous, so he just went on with the lecture. I have tested "my theory" with one matrix and it works, but even if I tried with 100 matrices and it worked, I wouldn't have proven that it always works because there might be a case where it doesn't. So my question is first whether I am right, that is, whether the rank of a matrix is the same as the rank of its transpose, and second, if that is true, how can I prove it? Thanks :) |
807391 | Listing equivalence relations class for [imath]x\sim y \Leftrightarrow x^2=y^2[/imath]
Can someone help me on the right track for my proof for the statement below. I started and got stuck but I need help. Please guide me to answer what this statement requires and how to word it out correctly. For all [imath]x,y\in\mathbb R[/imath] define that [imath]x\equiv y[/imath] if [imath]x^{2}=y^{2}.[/imath] Then [imath]\equiv[/imath] is an equivalence relation on [imath]\mathbb R[/imath], there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. True. Let [imath]p=x,y,z\in\mathbb{R}[/imath]. For [imath]x\equiv y[/imath] if [imath]x^{2}\equiv y^{2}[/imath] on [imath]\mathbb{R}.[/imath] To be an equivalence relation on [imath]\mathbb{R}[/imath] we need to show that: [imath]\text{i).}\ \forall x\in\mathbb{R}; x\equiv x\ \text{such that}\ x^{2}=x^{2}[/imath] [imath]\text{ii).}\ \forall x,y\in\mathbb{R}\ \text{if}\ x\equiv y\ \text{then}\ y\equiv x\ \text{such that}\ x^{2}=y^{2}\ \text{but}\ y^{2}=x^{2}[/imath] [imath]\text{iii).}\ \forall x,y,z\in\mathbb{R}\ \text{if}\ x\equiv y\ \text{and}\ y\equiv z, \text{then}\ x\equiv z\ \text{such that}\ x^{2}=y^{2}\ \text{and}\ y^{2}=z^{2}\Rightarrow x^{2}=z^{2}[/imath] Therefore the equivalence class of [imath][0,0]=\{p=(x,y,z\in\mathbb{R}|p\equiv(0,0)\}=\{(0,0)\}[/imath] then [imath]p\equiv(0,0)\Leftrightarrow x^{2}=y^{2}[/imath] which is equal to [imath]0^{2}=0^{2}=0[/imath] if and only if [imath]x=y=0.[/imath] Comment: I am stuck here I dont know where to continue on. Please help..thank you | 807358 | For all [imath]x,y\in\mathbb{R}[/imath] define that [imath]x\equiv y[/imath] if [imath]x^{2}=y^{2} [/imath]; prove [imath]\equiv[/imath] is an equivalence relation.
For all [imath]x,y\in\mathbb{R}[/imath] define that [imath]x\equiv y[/imath] if [imath]x^{2}=y^{2}[/imath] . Then [imath]\equiv[/imath] is an equivalence relation on [imath]\mathbb{R}[/imath] , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Solution True. To show that \equiv is reflexive we need to show that \forall x\in\mathbb{R} :x=x. Let x\in\mathbb{R} , x\equiv\mbox{\ensuremath{x}} if x^{2}=x^{2}, which is obvious. [x] ={y\in\mathbb{R} |x\equiv y} =[0]={y\in\mathbb{R}|0\equiv y}={0} . Hence y^{2} =0^{2} =0 which implies that y=0 . |
807092 | Proof of Drinker paradox
I searched all over the internet but didn't find a formal proof for this paradox, so here is my attempt: [imath]\exists x[P(x)\implies \forall yP(y)][/imath] Let [imath]x=x_0[/imath]. Thus [imath]P(x_0)[/imath] is given. Let [imath]y[/imath] be arbitrary. So we have to prove [imath]P(y)[/imath]. [imath]y[/imath] being arbitarary means that either [imath]y=x_0[/imath] or [imath]y\ne x_0[/imath]. So now we prove [imath]P(x_0) \lor P(z)[/imath] where [imath]z[/imath] is arbitrary and [imath]z\ne x_0[/imath]. [imath]P(x_0)[/imath] is given. So the paradox is true. [imath]\square[/imath] Is this correct ? If not what is wrong ? | 412387 | Why is this true? [imath](\exists x)(P(x) \Rightarrow (\forall y) P(y))[/imath]
Why is this true? [imath](\exists x)(P(x) \Rightarrow (\forall y) P(y))[/imath] |
807900 | Find a set A of rational numbers that is isomorphic to sup{[imath]\omega[/imath], [imath]\omega^\omega[/imath], [imath]\omega^{\omega^\omega}[/imath], [imath]\cdots[/imath]}
Find a set A of rational numbers such that (A, [imath]\le_Q[/imath]) is isomorphic to sup{[imath]\omega[/imath], [imath]\omega^\omega[/imath], [imath]\omega^{\omega^\omega}[/imath], [imath]\cdots[/imath]}. This is a question that is listed in "Introduction to Set Theory" by Hrbacek and Jech on page 123. Here, [imath]\omega[/imath] represents for the ordinal corresponding to the set of natural numbers. The prior question was to find such a set that is isomorphic to [imath]\omega^\omega[/imath], and I've constructed A as {[imath]\prod {{1} \over {p_i^{a_i}}} [/imath]} there, where [imath]p_i[/imath] represents for the [imath]i[/imath]th prime number and [imath]a_i[/imath] being its exponents. Honestly, I've tried my best but couldn't construct such a set corresponding to sup{[imath]\omega[/imath], [imath]\omega^\omega[/imath], [imath]\omega^{\omega^\omega}[/imath], [imath]\cdots[/imath]}. And now I have another question: what is the biggest ordinal number that is isomorphic to (A, [imath]\le_Q[/imath]) where [imath]\le_Q[/imath] denotes the usual ordering in rational number? According to Well Ordering Principle, we know that Q can be well-ordered, then (Q, [imath]<_W[/imath]) is isomorphic to a unique ordinal, but clearly [imath]\le_Q[/imath] isn't a well ordering. Is there no limit in this procedure? | 165151 | Question of an isomorphism of [imath]\epsilon_ 0[/imath] and a subset of the rationals.
I don't know if this question is appropriated for this site. Anyway, I'm searching for an isomorphism of order [imath]f:K \longrightarrow \epsilon_o [/imath], such that [imath](K, \leq)[/imath] is a subset(proper or not) of [imath](\mathbb{Q}, \leq)[/imath] and [imath]\epsilon_o = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, ... \}[/imath]. Actually, I'm not finding neither an isomorphism between [imath]K[/imath] and [imath]\omega^{\omega}[/imath]. Thanks in advanced. |
808022 | Does [imath]f(x)\,dx[/imath] denote multiplication of [imath]f(x)[/imath] by [imath]dx[/imath]?
In the integral form [imath]\int \! f(x) \, \mathrm{d}x[/imath] does [imath]f(x)\,\mathrm{d}x[/imath] can be seen as a multiplication of [imath]f(x)[/imath] and [imath]\mathrm{d}x[/imath]? | 200393 | What is [imath]dx[/imath] in integration?
When I was at school and learning integration in maths class at A Level my teacher wrote things like this on the board. [imath]\int f(x)\, dx[/imath] When he came to explain the meaning of the [imath]dx[/imath], he told us "think of it as a full stop". For whatever reason I did not raise my hand and question him about it. But I have always shaken my head at such a poor explanation for putting a [imath]dx[/imath] at the end of integration equations such as these. To this day I do not know the purpose of the [imath]dx[/imath]. Can someone explain this to me without resorting to grammatical metaphors? |
808200 | Can existence of aleph one be proved without the power set axiom?
Cantor's construction of [imath]\aleph_1[/imath] is to notice that all ordinals constructed after [imath]\omega[/imath] are countable, take the union of all countable ordinals, and then show that this union can not be countable. However, in ZFC the union can only be taken over a set. I can see how to relate countable ordinals to elements of [imath]2^{\aleph_0}[/imath] proving that their collection is a set by specification, but that requires the power set axiom. The "slicker" modern way, declaring [imath]\aleph_1[/imath] to be the least uncountable cardinal, already presupposes the existence of uncountable cardinals. That's even worse since in addition to power set one needs choice to prove that [imath]2^{\aleph_0}[/imath] is an aleph, and then it is greater by the diagonal argument. It wouldn't surprise me that some voodoo with collection or replacemnt schemas does the trick, but then my question is what does it translate to in "naive" terms of Cantor? What reason do we have for the collection of countable ordinals to be a set (without invoking formulas in the first order logic and definable functions)? If power set isn't required why can't Cantor's argument be reproduced to create inaccessible cardinals by noticing that all cardinals produced after [imath]\aleph_0[/imath] are, by construction, accessible? | 4778 | Uncountable ordinals without power set axiom
Assume [imath]M[/imath] is a set, in which all axioms of [imath]ZF - P + (V=L)[/imath] hold. Does then [imath]M[/imath] believe that there exists an uncountable ordinal? I mean, why should the class of all countable ordinal numbers be a set? |
808457 | A logarithmic integral question: [imath]\int_{0}^{1} \frac{\ln^2{x}\ln^2{(1-x)}} {1+x}dx[/imath]
[imath]\int_{0}^{1} \frac{\ln^2{x}\ln^2{(1-x)}} {1+x}dx[/imath] Can the integral be calculated in terms of well known constants? | 805298 | How to find [imath]\int_{0}^{1}\dfrac{\ln^2{x}\ln^2{(1-x)}}{2-x}dx[/imath]
How to find [imath] I=\int_{0}^{1}{\ln^{2}\left(x\right)\ln^{2}\left(1 - x\right) \over 2 - x} \,{\rm d}x [/imath] My idea: Let [imath]x=1-t[/imath], then [imath] I =\int_{0}^{1}{\ln^{2}\left(1 - x\right)\ln^{2}\left(x\right) \over 1 + x}\,{\rm d}x [/imath] I can't proceed any further. I can't remember that Math SE has this similar problem already posted. this integral problem is from This Euler Sums : [imath]\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right) = - \frac{{23}}{{1440}}{\pi ^4}\ln 2 + \frac{1}{{18}}{\pi ^2}{\ln ^3}2 - \frac{1}{{18}}{\ln ^5}2 - 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 8L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{{12}}{\pi ^2}\zeta \left( 3 \right) + \frac{{27}}{4}\zeta \left( 5 \right)[/imath] when I deal this series,then we must solve this integral . Thank you. |
808473 | odd/even binomial coefficient identity
For all n\geq1 : [imath]\left(\begin{matrix}2n\\ 0 \end{matrix}\right) +\left(\begin{matrix}2n\\ 2 \end{matrix}\right) +\left(\begin{matrix}2n\\ 4 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n \end{matrix}\right) [/imath] is equal to [imath]\left(\begin{matrix}2n\\ 1 \end{matrix}\right) +\left(\begin{matrix}2n\\ 3 \end{matrix}\right) +\left(\begin{matrix}2n\\ 5 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k+1 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n-1 \end{matrix}\right) .[/imath] MY SOLUTION Binomial Formula : (a+b)^{2n} = \sum_{k=0}^{2n} \left(\begin{matrix}2n\\ k \end{matrix}\right) a^{k} b^{n-k} let a=1 , let b=-1 , let n=2 (1+(-1))^{2(n)} = \left(\begin{matrix}2n\\ 0 \end{matrix}\right) -\left(\begin{matrix}2n\\ 1 \end{matrix}\right) +...\mbox{-}\left(\begin{matrix}2n\\ 2n-1 \end{matrix}\right) + \left(\begin{matrix}2n\\ 2n \end{matrix}\right) (1+(-1))^{2(4)} = \left(\begin{matrix}2(2)\\ 0 \end{matrix}\right)-\left(\begin{matrix}2(2)\\ 1 \end{matrix}\right)+\left(\begin{matrix}2(2)\\ 02 \end{matrix}\right)-\left(\begin{matrix}2(2)\\ 2(2)-1 \end{matrix}\right)+ \left(\begin{matrix}2(2)\\ 2(2) \end{matrix}\right) 0=1-4+6-4+1 0=0 Thus for all n\geq1 : \left(\begin{matrix}2n\\ 0 \end{matrix}\right) +\left(\begin{matrix}2n\\ 2 \end{matrix}\right) +\left(\begin{matrix}2n\\ 4 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n \end{matrix}\right) is equal to \left(\begin{matrix}2n\\ 1 \end{matrix}\right) +\left(\begin{matrix}2n\\ 3 \end{matrix}\right) +\left(\begin{matrix}2n\\ 5 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2k+1 \end{matrix}\right) + \ldots +\left(\begin{matrix}2n\\ 2n-1 \end{matrix}\right) . Can anyone please give feedback and tell me if my solution is correct. | 807423 | Binomial Expression
Please give me feedback on my answer to this question. Question: For all [imath] n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} [/imath] is equal to [imath] \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} [/imath]. Answer: False, since; Let [imath]a=1, b=-1[/imath] and [imath]n=1[/imath] ,[imath]k=0[/imath] , using Binomial expression. [imath](a+b)^n= \sum \binom{n}{k} a^{k}b^{n-k}[/imath] Then [imath](1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}[/imath] Thus :For all [imath]n\geq1: \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}[/imath] . |
808659 | Direct sums, Span,Union,Independence
Given: K and T are sub-sets of Linear Space V and also [imath]Sp(K) + Sp(T)[/imath] is a direct sum, and [imath]Sp(K \cup T) = Sp(K) \oplus Sp(T)[/imath] I need to do show that [imath] K \cup T[/imath] isn't necessary independent. I said that: [imath] V = R^2, [/imath] [imath]K = \{(1,0),(0,1)\}[/imath] ,[imath] T = \{(1,1)\}[/imath] From here I can say that [imath]Sp(K) + Sp(T)[/imath] is a direct sum because their sum is V and their intersection is zero. and also, [imath]Sp(K \cup T) = Sp(K) \oplus Sp(T)[/imath] because their sum is [imath]Sp(K \cup T)[/imath] and their intersection is zero. From here we see that [imath]K \cup T = \{(1,0),(0,1),(1,1)\}[/imath] but [imath](1,1)[/imath] is dependent by [imath](1,0) [/imath]and [imath](0,1)[/imath] which makes it not necessary independent. The question is, did I prove it right, maybe my sub-sets are wrong and I need other ones? Or just adding [imath](0,0)[/imath] to K and T is enough to show that there isn't necessary independence(which is a another solution)? | 808624 | Base and Spans, linear space
Given, K and T are sub-sets of linear space V. How can I prove that if [imath] V = Sp(K) + Sp(T)[/imath] then [imath] K \cup T[/imath] is NOT a base of V. I proved that it IS a base of V but the answer says that it's not, can you give me an example of 2 sub-sets K,T that will make their union not a base of V? Here how I tried: [imath] V = R^2 [/imath] , [imath]K = \{(0,1)\} [/imath],[imath] T = \{(1,0)\}[/imath] then [imath]V = Sp(K) + Sp(T) = Sp\{(K \cup T)\} = Sp\{(1,0),(0,1)\} = R^2[/imath] , then [imath]K \cup T = \{(1,0),(0,1)\} = R^2[/imath] Which makes the union of K,T a base of V, but I need your help to find 2 sub-sets that will make it wrong. |
808727 | Prove that [imath]\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} \ldots + \frac{1}{2n} > \frac{13}{24}[/imath] for every integer [imath]n>1[/imath]
Prove that for any positive integer [imath]n>1[/imath] [imath] \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} \ldots + \frac{1}{2n} > \frac{13}{24} [/imath] I can prove that the series is greater than [imath]\frac{12}{24}[/imath] however i can't prove that it is greater than [imath]\frac{13}{24}[/imath]. | 508664 | Proving [imath]\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}[/imath] for [imath]n>1,n\in\Bbb N[/imath] by Induction
Proving [imath]\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}[/imath] for [imath]n>1,n\in\Bbb N[/imath] To solve it I used induction but it is leading me nowhere my attempt was as follows: Lets assume the inequality is true for [imath]n = k[/imath] then we need to prove that it is true for [imath]k+1[/imath] so we need to prove [imath]\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} > 13/24[/imath] I don't know where to go from here please help. |
808881 | Any counterexample for inverse limit functor not to be right exact
We know that inverse limit is a "left" exact functor on the category of modules in the sense that whenever [imath]r:(A_i,α_j^i )→(B_i,β_j^i )[/imath] and [imath]s:(B_i,β_j^i )→(C_i,γ_j^i )[/imath] are transformations of inverse systems over an index set [imath]I[/imath] between modules, and [imath]0→A_i→B_i →C_i→0[/imath] is exact for each [imath]i∈I[/imath], where the first function is [imath]r_i[/imath] and the second is [imath]s_i[/imath], then there is an induced (left) exact sequence [imath]0→\varprojlim A_i→\varprojlim B_i→\varprojlim C_i[/imath] . I search for a counterexample to the latter to be not "right" exact. Thanks very much in advance! | 808155 | Do pullbacks commute with filtered limits in this sense?
Let [imath]A_n, B_n, C_n[/imath] be directed systems in some abelian category. Denote by [imath]A \times_C B[/imath] the fibre product of [imath]A[/imath] and [imath]B[/imath] over [imath]C[/imath]. Is it true that [imath](\varprojlim A_n) \times_{\varprojlim C_n} (\varprojlim B_n) = \varprojlim (A_n \times_{C_n} B_n)[/imath], if [imath]\varprojlim (A_n \rightarrow C_n)= (A \rightarrow C)[/imath] and similarly for [imath]B[/imath]? I know that "morally" limits preserve limits, but I don't know the exact statement I need here (i.e. limits in which categories?) Edit: Analogously, what can we say after replacing filtered limits with filtered colimits? Edit2: I think in the case of filtered colimits, I just checked the universal property of a colimit. In case I didn't make a mistake, the statement holds. The other case (filtered limits), and arbitrary colimits/limits are not as important for my purposes, but if you have a nice statement, please feel free to post it. |
809397 | Feedback on my answer for [imath]X^n + Y^n = Z^n [/imath]
The equation [imath]X^n + Y^n = Z^n [/imath] , where [imath]n \ge 3[/imath] is a natural number, has no solutions at all where X; Y;Z are integers. solution: the above is a false statement counter example: let: n=3 ,x=0 y=0 and z=1 since [imath]0^3 + 0^3 + 1^3=1^3[/imath] . | 807657 | The equation [imath]X^{n} + Y^{n} = Z^{n}[/imath] , where [imath] n \geq 3[/imath] is a natural number, has no solutions at all where [imath]X,Y,Z[/imath] are intergers.
The equation [imath]X^{n} + Y^{n} = Z^{n}[/imath] , where [imath]n \geq 3[/imath] is a natural number, has no solutions at all where [imath]X,Y,Z[/imath] are integers. My solution: False. Because if we let [imath]X=0 ,Y=0[/imath], then [imath]0^{3}+0^{3}=0^{3}[/imath] . Thus there is a solution for the equation [imath]X^{n} + Y^{n} = Z^{n}[/imath], where [imath]X,Y,Z[/imath] are integers. Can anyone please give feedback on my answer and state whether it is correct on not. |
809419 | feedback on my solution regarding eqivalence relations.
For all [imath]x, y \in \mathbb{R}[/imath] define that [imath]x \equiv y[/imath] if [imath]x^2 = y^2[/imath]. Then [imath]\equiv[/imath] is an equivalence relation on [imath]\mathbb{R}[/imath], there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: True. Reflectivity: To show that [imath]\mathbb{R}[/imath] is reflexive we need to show that [imath]\forall x \in \mathbb{R} : x \equiv x[/imath]. Let [imath]x \in \mathbb{R}[/imath], [imath]x \equiv x[/imath] if [imath]x^2 = x^2[/imath], which is obvious. Symmetry: Let [imath]x,y \in \mathbb{R}[/imath] with [imath]x \equiv y[/imath], so we have [imath]x^2=y^2[/imath] and trivially we get [imath]y^2=x^2[/imath], hence [imath]y \equiv x.[/imath] Transitivity: Let [imath]x,y,z \in \mathbb{R}[/imath] with [imath]x \equiv y[/imath] and [imath]y \equiv z[/imath]. We now show that [imath]x \equiv z[/imath]: Since [imath]x^2=y^2[/imath] and [imath]y^2=z^2[/imath] we get by transitivity of equality [imath]x^2=z^2[/imath] and therefore [imath]x \equiv z.[/imath] Is the answer to the question correct? And please tell me how I can prove that there are infinitely many equivalence classes one of them consists of one element and the rest consist of two elements. If it's not correct then please provide the correct answer. | 808591 | For all [imath]x,y∈\Bbb{R}[/imath] define that [imath] x\equiv y[/imath] if[imath] x^2=y^2[/imath]
For all [imath]x,y\in\Bbb{R}[/imath] define that [imath]x\equiv y[/imath] if [imath]x^2=y^2[/imath] . Then [imath]\equiv[/imath] is an equivalence relation on [imath]\Bbb{R}[/imath] , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Solution True. To show that [imath]\equiv[/imath] is reflexive we need to show that [imath]\forall x\in\mathbb{R} :x\equiv x[/imath]. Let [imath]x\in\mathbb{R} , x\equiv x \mbox{ if } x^{2}=x^{2}[/imath], which is obvious. [imath][x] ={y\in\mathbb{R} |x\equiv y} =[0]={y\in\mathbb{R}|0\equiv y}={0}[/imath]. Hence [imath]y^{2} =0^{2} =0[/imath] which implies that [imath]y=0[/imath]. Can anyone please give me correct answer to this question. |
802567 | Consider The Differential Equation [imath]\frac{dy}{dx} = 3y^{\frac{2}{3}}, y(0)=0[/imath]
Consider: [imath]\frac{\mathrm{d}y}{\mathrm{d}x} = 3y^{2/3}, \quad y(0)=0[/imath] does exist a solution to this initial value problem? how many solutions if exist? If the answer for the first question is affirmative, and a solution or several (> 2) solutions (if your answer to the second question is `not unique'). I think I need to use existence and uniquness theorem but im not sure, so any tips/solutions? | 734773 | [imath]\frac{dy}{dx} = 3y^{2/3}[/imath] general solution?
What's the general solution of [imath]\frac{dy}{dx} = 3y^{2/3}[/imath] ? Im pretty sure this is a separable equation, but I'm not sure how to go forward? Just multiply by [imath]dx[/imath] and [imath]\frac{1}{3y^{2/3}}[/imath] well then I get [imath]y^{1/3} =x+ C[/imath], correct? |
809773 | If [imath](ab)^{3}=a^{3}b^{3}[/imath] and [imath](ab)^{5}=a^{5}b^{5}[/imath] for all [imath]a,b \in G[/imath] then [imath]G[/imath] is Abelian?
How does [imath](ab)^{3}=a^{3}b^{3}[/imath] and [imath](ab)^{5}=a^{5}b^{5}[/imath] for all [imath]a,b \in G[/imath] implies that [imath]G[/imath] is abelian? I know that the first criteria alone isn't enough because there's a counterexample. What could we say if in those equation instead 3 and 5 we had two other relatively prime numbers? | 737786 | Let [imath]G[/imath] be a group, where [imath](ab)^3=a^3b^3[/imath] and [imath](ab)^5=a^5b^5[/imath]. Prove that [imath]G [/imath] is an abelian group.
Let G be a group, where [imath](ab)^3=(a^3)(b^3)[/imath] and [imath](ab)^5=(a^5)(b^5)[/imath]. Prove that [imath]G[/imath] is abelian group. Thank you in advance. Any help is appreciated. |
810213 | Complex integration around a singularity
I am trying to integrate the function [imath]f(z)=[/imath][imath]\frac{5}{z^2}[/imath] from -3 to 3 and I am supposed to develop a closed region that avoids the origin and use the analyticity of the function in this region to integrate the function in a way that is simpler than parametrizing along some path. Any suggestions? | 809569 | Contour integral of analytic function with singularity
I am supposed to integrate [imath]f(z)=[/imath][imath]\frac{5}{z}[/imath] from -3 t0 3 but I am having trouble understanding how to do this. I've done the integration the "hard" way by using parametrizations but now I need to use the fact that [imath]f(z)[/imath] is analytic everywhere except at [imath]z=0[/imath] and a different method to perform the integration but I am stumped. |
810447 | CHECK: Let a and b be relatively prime integers. Show that [imath]\gcd(a^2+b^2,a+b)=[/imath]1 or 2
Let a and b be relatively prime integers. Show that [imath]\gcd(a^2+b^2,a+b)=[/imath]1 or 2 Proof: [imath]s|a^2+b^2[/imath] and [imath]s|a+b[/imath] implies [imath]s|a^2+b^2[/imath] and [imath]s|(a+b)^2=a^2+b^2+2ab[/imath] implies [imath]s|a^2+b^2-(a+b)^2=2ab[/imath] implies [imath]s|2a[/imath] and [imath]s|2b[/imath] implies [imath]s|\gcd(2a,2b)=2gcd(a,b)=2*1[/imath] Hence [imath]\gcd(a^2+b^2,a+b)=1 or 2[/imath] | 606395 | If [imath]a,b \in\mathbb N[/imath] and [imath]\gcd(a,b)=1[/imath], prove that [imath]\gcd(a+b;a^2+b^2)= 1[/imath] or [imath]2[/imath].
If [imath]a,b \in\mathbb N[/imath] and [imath]\gcd(a,b)=1[/imath], prove that [imath]\gcd(a+b,a^2+b^2)[/imath] is always equal to either 1 or 2, where [imath]\gcd[/imath] is the greatest common divisor. I haven't really ever solved a problem like this before, so I'd like to get some help. Thanks. |
810537 | How to calculate the integration [imath]\int_{0}^{\pi}\frac{dx}{(2-\cos{x})^2}[/imath]
Given that [imath] \int_{0}^{\pi}\frac{dx}{2-\cos{x}}=\frac{\pi}{\sqrt{a^2-1}} [/imath] How to calculate the integral [imath] \int_{0}^{\pi}\frac{dx}{(2-\cos{x})^2} [/imath] | 810087 | Evaluating the integral with trigonometric integrand
While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you. [imath]\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.[/imath] |
47927 | Motivation for Napier's Logarithms
In the wikipedia article on logarithms, I am clueless about the approach and motivation for the following computations done by Napier (and the mysterious appearance of Euler's number) in this section. Please help me understand: By repeated subtractions Napier calculated [imath]10^7(1 − 10^{−7})^L[/imath] for L ranging from 1 to 100. The result for [imath]L=100[/imath] is approximately [imath]0.99999 = 1 − 10^{-5}[/imath]. Napier then calculated the products of these numbers with [imath]10^7(1 − 10^{−5})^L[/imath] for L from 1 to 50, and did similarly with [imath]0.9995 ≈ (1 − 10^{−5})^{20}[/imath] and [imath]0.99 ≈ 0.995^{20}[/imath]. These computations, which occupied 20 years, allowed him to give, for any number ''N'' from 5 to 10 million, the number ''L'' that solves the equation [imath]N=10^7 {(1-10^{-7})}^L[/imath] In modern notation, the relation to natural logarithms is: [imath]L = \log_{(1-10^{-7})} \!\left( \frac{N}{10^7} \right) \approx 10^7 \log_{ \frac{1}{e}} \!\left( \frac{N}{10^7} \right) = -10^7 \log_e \!\left( \frac{N}{10^7} \right)[/imath] where the very close approximation corresponds to the observation that [imath]{(1-10^{-7})}^{10^7} \approx \frac{1}{e}. [/imath] I understand the computations, but I don't know what's going on. | 2848686 | Motivation for logarithmic function
What was John Napier's motivation to introduce the logarithmic function? It neither makes the calculations easier nor accurate (at least for me). How could people think of its importance even before Calculus came into play and showed its connection with the exponential function or, area under the graph [imath]\dfrac{1}{x}[/imath] ? |
810857 | Subadditivity of outer measure on real numbers
In the field of measure theory, the outer measure is defined on the set of real numbers as: [imath]m^{*}(A) = \inf\left\{ \sum_{n=1}^{\infty} l(E_n): \space A \subset \bigcup_{n=1}^{\infty} E_n \right\} [/imath] NOTE - [imath]l(E_n)[/imath] is the length of the interval [imath]E_n[/imath] I want to show that the real number outer measure is subadditive, so [imath]m^{*}\left( \bigcup_{n=1}^{\infty} S_n \right) \le \sum_{n=1}^{\infty} m^{*}(S_n)[/imath] As a first step in the proof, all the measure theory textbooks I have consulted start by stating that we assume that [imath]m^*(S_n)[/imath] is finite for all [imath]n[/imath] (because the result is trivially true otherwise), which makes sense to me. Next, they assert that for any [imath]S_n[/imath], we can find an open cover [imath]\space S_n \subset \bigcup_{n=1}^{\infty} E_n[/imath] such that [imath] \sum_{j=1}^{\infty} l(E_n) \le \mu^*(S_n)+\frac{\epsilon}{2^n}[/imath] Why does such a cover necessarily exist? Isn't it possible that the only covers are countably infinite on the real numbers? Also, what explains the epsilon inequality - how do we know that a countable cover necessarily meets that criterion? | 810745 | Outer measure is countably subadditive
Working on the proof that outer measure is countably subadditive in Royden For a set [imath]A \subset X[/imath] we define the outer measure: [imath]\mu^{*}(A) = \inf\left\{ \sum_{n=1}^{\infty} \tau(T_n): \space T_n \in \mathcal{T}, \space A \subset \bigcup_{n=1}^{\infty} T_n \right\} [/imath] We want to prove that [imath]\mu^{*}[/imath] has the property of subadditivity: for a sequence [imath]\{ A_n\}_{n=1}^{\infty}[/imath] the following is true: [imath]\mu^{*}\left( \bigcup_{n=1}^{\infty} A_n \right) \le \sum_{n=1}^{\infty} \mu^{*}(A_n)[/imath] [imath]\mathbf{Proof}[/imath]: The first step is that there exists [imath](n,j)[/imath] such that [imath]A_n \subset \bigcup_{j=1}^{\infty} T_{(n,j)}[/imath] and [imath] \sum_{j=1}^{\infty} \tau(T_{(n,j)}) \le \mu^*(A_n)+\frac{\epsilon}{2^n}[/imath] ~~~~~~~~~~~~~~~~~~~~~~~~~~ I'm confused as to how we know that such an open cover [imath]T_{(n,j)}[/imath] exists, especially subject to the condition that [imath] \sum_{j=1}^{\infty} \tau(T_{(n,j)}) \le \mu^*(A_n)+\frac{\epsilon}{2^n}[/imath] How would such an open cover be constructed? If anyone could help me develop some intuition for this step, that would be great |
811160 | [imath]AB-BA=A[/imath] implies [imath]A[/imath] is singular and [imath]A[/imath] is nilpotent.
Let [imath]A[/imath] and [imath]B[/imath] be two real [imath]n\times n[/imath] matrices such that [imath]AB-BA=A[/imath] Prove that [imath]A[/imath] is not invertible and that [imath]A[/imath] is nilpotent. My attempt is the following. It holds that [imath]AB=(B+I)A[/imath] If [imath]A[/imath] were invertible, [imath]B=A^{-1}(B+I)A[/imath]. Taking trace on both sides yields [imath]tr(B)=tr(B)+n[/imath], hence [imath]n=0[/imath], which is a contradiction. I can't prove that [imath]A[/imath] is nilpotent though. | 724294 | If [imath]A=AB-BA[/imath], is [imath]A[/imath] nilpotent?
Let matrix [imath]A_{n\times n}[/imath], be such that there exists a matrix [imath]B[/imath] for which [imath]AB-BA=A[/imath] Prove or disprove that there exists [imath]m\in \mathbb N^{+}[/imath]such [imath]A^m=0,[/imath] I know [imath]tr(A)=tr(AB)-tr(BA)=0[/imath] then I can't.Thank you |
811515 | Ten people standing in circle.
I'm trying to solve the question: Ten people are standing are standing in a circle. Find the number of ways to choose [imath]3[/imath] people such that no [imath]2[/imath] of them are consecutive. I have got the answer to be [imath]{10\choose1}{6\choose2}[/imath], but I'm not sure if this is right. Please tell me if my answer is correct! If it's not, then where did I go wrong? | 568045 | combinatorics circular arrangement problem
If [imath]n[/imath] distinct things are arranged in a circle, then what are the number of ways selecting three of these things so that no two of them are next to each other? |
811806 | Proof that frobenius norm is a norm
It's pretty basic and I'm sure I'm missing something dumb here, but I'd like to know why [imath]||A+B||_F \leq ||A||_F+||B||_F[/imath] The way I understand it, [imath]||A+B||^2_F=tr((A+B)^T(A+B))=tr((A^T+B^T)(A+B))=tr(A^TA+A^TB+B^TA+B^TB)[/imath] Now using the property the trace is linear: [imath]tr(A^TA+A^TB+B^TA+B^TB) = tr(A^TA)+tr(B^TB)+tr(A^TB+B^TA)=||A||^2_F+||B||^2_F+tr(A^TB+B^TA)[/imath] Now if we were to prove that [imath]2||A||_F||B||_F \geq tr(A^TB+B^TA)[/imath] that would solve the question. But I don't see how that's trivial, and generally multiplying [imath]\sum[/imath]s together is something I avoid like the plague. Is this indeed the way? would someone help me with this last step? | 564873 | Frobenius Norm Triangle Inequality
How can I go about proving the triangle inequality holds for the Frobenius norm? I worked through [imath]\|A+B\|_F \le \|A\|_F + \|B\|_F[/imath] and was not able to make it work =/. |
762671 | Show that [imath]B(X)[/imath] is semisimple for a Banach space [imath]X[/imath]
Show that [imath]B(X)[/imath] is a semisimple Banach algebra, where [imath]X[/imath] is a Banach space. That is, to show that rad [imath]B(X)=\{0\}[/imath], or equivalently, to show [imath]\sigma(AT)={0} \, \forall T\in B(X)\Rightarrow A=0[/imath]. I have been trying to find a simple proof of the following without using the more general result that for a Banach space [imath]X[/imath], [imath]\sigma(AT)=\sigma(BT) \, \forall T\in B(X)\Rightarrow A=B[/imath]. | 544638 | equality of two operators...
Please help me with the following problem( give some hints or references): Let [imath]X[/imath] be a Banach space and [imath]B(X)[/imath] be the algebra of bounded linear operators on [imath]X[/imath]. Suppose that [imath]A[/imath] and [imath]B[/imath] are two operators in [imath]B(X)[/imath] such that for every [imath]T \in B(X)[/imath] we have [imath]\sigma(AT)=\sigma(BT)[/imath]. Show that [imath]A=B[/imath]. Here [imath]\sigma(A)[/imath] denotes the spectrum of [imath]A[/imath], i.e., the the set of all complex numbers [imath]\lambda[/imath] such that [imath]\lambda I − A[/imath] has no inverse in [imath]B(X)[/imath]. Thanks in advance. |
812449 | isomorphism between group and product of kernel by image
If [imath]\phi[/imath] is a morphism between groups [imath]G[/imath] and [imath]H[/imath], is [imath]G[/imath] isomorphic to [imath]ker(\phi)\times im(\phi)[/imath] ? Why ? Thanks. | 397274 | When does the isomorphism [imath]G\simeq ker(\phi)\times im(\phi)[/imath]? hold?
Suppose you have a group isomorphism given by the first isomorphism theorem: [imath]G/ker(\phi) \simeq im(\phi)[/imath] What can we say about the group [imath]ker(\phi)\times im(\phi)[/imath]? In particular, when does the following hold: [imath]G\simeq ker(\phi)\times im(\phi)[/imath]? I ask this question because i want to prove that [imath]GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{>0}[/imath], with [imath]GL_n^+(\mathbb{R})[/imath] the group of matrices with positive determinant. I proved that [imath]SL_n(\mathbb{R})[/imath] is a normal subgroup and that [imath]GL_n^+(\mathbb{R})/ SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{>0}[/imath], using the surjective homomorphism [imath]det(M)[/imath]. I tried something with semidirect products but I got stuck. |
812922 | What is the probability of covering all [imath]6[/imath] faces of a die after rolling it [imath]14[/imath] times or less?
In order to answer this question, I want to sum up the probabilities for [imath]1 \leq N \leq 14[/imath]: [imath]P(N<6)=0[/imath] [imath]P(N=6)=\displaystyle\frac{6!}{6^6}[/imath] [imath]P(N>6)=\displaystyle\frac{?}{6^N}[/imath] Can you please advise on how to work out the third bullet? Any suggestion for an alternative solution will also be appreciated. Thanks | 669685 | What is the probability of rolling [imath]n[/imath] dice until each side appears at least once?
We roll a die until each side appears at least once, then we stop. What is the probability of rolling exactly [imath]n[/imath] dice? I guess the answer is [imath]6-6\left(\dfrac{5}{6}\right)^n\;,[/imath] but this may be wrong. In trying to solve this problem, I read this paper, but I couldn't solve the problem. |
813194 | Are [imath]1[/imath] and [imath]3[/imath] the only numbers of the form [imath]2^n-1[/imath] that exist in the Fibonacci sequence?
Are [imath]1[/imath] and [imath]3[/imath] the only numbers of the form [imath]2^n-1[/imath] that exist in the Fibonacci sequence? Then, if they are not the only ones, are there infinite cases? I have tried finding another example using a program, but could not find any, thus my conclusion. I think there might be a good proof in the fact that all Fibonacci numbers [imath]Fn[/imath] create perfect squares when [imath]5{Fn}^2+4[/imath] or [imath]5(Fn)^2-4[/imath]. | 16655 | Prove: the intersection of Fibonacci sequence and Mersenne sequence is just [imath]\{1,3\}[/imath]
[imath]\frac{{{\varphi ^n} - {{(1 - \varphi )}^n}}}{{\sqrt 5 }} = {2^m} - 1 .[/imath] Here [imath]\varphi = \frac{{1 + \sqrt 5 }}{2}[/imath] . This integer equation has no solution for [imath]n>3[/imath] and [imath]m>2[/imath]. How to prove? |
813613 | Suppose [imath]X[/imath] has two topologies [imath](X,T),(X,T')[/imath] and both are compact and hausdorffs. Suppose [imath]T \subset T'[/imath]. Prove that [imath]T=T'[/imath]
Suppose [imath]X[/imath] has two topologies [imath](X,T),(X,T')[/imath] and both are compact and hausdorffs. Suppose [imath]T \subset T'[/imath]. Prove that [imath]T=T'[/imath] Because of the compactness of both topologies and the fact that one is an open subset of the other, I can clearly see that we can use the same finite open coverings to cover these topologies. But how do i use the hausdorff definition in this question? Kees Til | 800452 | Assume that [imath](\text{X}, T)[/imath] is compact and Hausdorff. Prove that a comparable but different topological space [imath](\text{X},T')[/imath] is not.
Say that a topological space is CH if it is both compact and Hausdorff. Let [imath]T[/imath] and [imath]T'[/imath] be two topologies on the same set X that are comparable but different, i.e., [imath]T[/imath] is either strictly smaller or strictly larger than [imath]T'[/imath]. Assume that [imath](\text{X}; T )[/imath] is CH. Prove that [imath](\text{X}; T')[/imath] is not CH. I just can seem to get started, so I was wondering if anybody could give me a hint... If [imath]T' \subsetneqq T[/imath], then the identity [imath]id: (\text{X},T) \rightarrow (\text{X}, T')[/imath] is continuous and surjective. This implies that [imath]id(\text{X},T) = (\text{X},T')[/imath] is also compact. So in order to show that [imath](\text{X},T')[/imath] is not CH, I need to show that it is not Hausdorff, right? But I'm stuck... |
813714 | Sheaf of ideals
Let [imath](X, \mathcal{O}_X)[/imath] be a scheme and and [imath]f \in \mathcal{O}_X(X)[/imath]. We define the sheaf of ideals by the assignment [imath] U \longmapsto f|_U \mathcal{O}_X(U) [/imath] and denote this sheaf by [imath]f\mathcal{O}_X[/imath] (one need to check the gluing condition which is easy when [imath]f[/imath] is not a zero divisor). I need to show that [imath]\mathcal{F} = f\mathcal{O}_X[/imath] is finitely-generated and quasi-coherent sheaf. By "finitely-generated" I suppose they mean that for any open subset [imath]U[/imath] the ideal [imath]f|_U \mathcal{O}_X(U)[/imath] is finitely-generated as an [imath]\mathcal{O}_X(U)[/imath]-module. I claim it is so since [imath]f|_U \mathcal{O}_X(U)[/imath] is a principal ideal and hence finitely-generated as an ideal. I need the reasoning that this implies that [imath]f|_U \mathcal{O}_X(U)[/imath] is also finitely-generated as a module. For quasi-coherence I need to construct an exact sequence [imath] \mathcal{O}_X^{(J)}|_U \longrightarrow \mathcal{O}_X^{(I)}|_U \longrightarrow \mathcal{F}|_U \longrightarrow 0 [/imath] where [imath]J[/imath] and [imath]I[/imath] are indexing sets and [imath]\mathcal{O}_X^{(I)}[/imath] is the direct sum of it. Intuitively, this sequence says that [imath]\mathcal{F}(U)[/imath] is a free [imath]\mathcal{O}_X(U)[/imath] modulo some relations. An attempt to construct this sequence is to take for [imath]\mathcal{O}_X^{(J)}(U)[/imath] all the relations of the form [imath]g \sim g + f \cdot h[/imath] for any [imath]g,h \in \mathcal{O}_X^{(I)}[/imath] but how to represent them as a free (sub)module of [imath]\mathcal{O}_X^{(J)}(U)[/imath]? Another approach is to use the following theorem [Liu, page 170, Theorem 1.7]: Theorem 1.7. The sheaf [imath]\mathcal{F}[/imath] is quasi-coherent if and only if for any open affine subset [imath]U[/imath] the sheaf [imath]\tilde{\mathcal{F}}[/imath] induced by [imath]\mathcal{F}(U)[/imath] is canonically isomorphic to [imath]\mathcal{F}|_U[/imath]. That is, [imath]\widetilde{\mathcal{F}(U)} \cong \mathcal{F}|_U[/imath]. The "tilde" represents the following construction: let [imath]M[/imath] be an [imath]\mathcal{O}_X(X)[/imath]-module where [imath]X[/imath] is affine scheme. For every [imath]x \in X[/imath] take the stalk of a sheaf [imath]\tilde{M}[/imath] at [imath]x[/imath] to be [imath] \tilde{M}_x := M_x[/imath] where right hand side is a localization ([imath]x[/imath] is a prime ideal as [imath]X[/imath] is affine). I think that [imath]\mathcal{F}_x = (f \mathcal{O}_X)_x = f_x \mathcal{O}_{X,x} = f_x \mathcal{O}_X(U)_x[/imath] but what is [imath]f_x[/imath] (the element of [imath]f \in \mathcal{O}_X(X)[/imath] localized at [imath]x[/imath]). I think it isjust the image of [imath]f[/imath] at [imath](\mathcal{O}_X(X))_x[/imath], that is [imath]f \mapsto \frac{f}{1}[/imath] so [imath]f_x[/imath] is the same as [imath]f[/imath]. These are my attempts. | 308215 | An exercise in Liu regarding a sheaf of ideals (Chapter II 3.4)
I'm fairly certain there is both a typo and an omission in this exercise. It reads "Let [imath]X[/imath] be a scheme and [imath]f \in \mathcal{O}_X(X)[/imath]. Show that [imath]U \mapsto f|_U \mathcal{O}_X(U)[/imath] for every affine open subset [imath]U[/imath] defines a sheaf of ideals on [imath]X[/imath]. We denote this sheaf of ideals by [imath]f \mathcal{O}_X[/imath]..." Surely "affine" above should be omitted, but my real question is are we missing any hypotheses on [imath]f[/imath]? It is clear that [imath]f \mathcal{O}_X[/imath] defines a presheaf satisfying the uniqueness condition, i.e., if an global section restricts to zero everywhere on an open over, then the element is identically [imath]0[/imath], as [imath]\mathcal{O}_X[/imath] is a sheaf to begin with. However, in proving that is satisfies the criterion regarding the glueing of local sections, do we not need the hypothesis that [imath]f|_U[/imath] is not a zero-divisor for all [imath]U[/imath]? This is the only condition in which I'm able to get at the result. Am I missing anything? EDIT: Please see Martin Brandenburg's answer. I was mistaken. (But it was not a regrettable mistake, as the point which I missed deserves an extra line or two of qualification.) |
813925 | Why is [imath]\sum\limits_{k=0}^{n}(-1)^k\binom{n}{k}^2=(-1)^{n/2}\binom{n}{n/2}[/imath] if [imath]n[/imath] is even?
Why is [imath]\displaystyle\sum\limits_{k=0}^{n}(-1)^k\binom{n}{k}^2=(-1)^{n/2}\binom{n}{n/2}[/imath] if [imath]n[/imath] is even ? The case if [imath]n[/imath] is odd, is clear, since [imath]\displaystyle(-1)^k\binom{n}{k}+(-1)^{n-k}\binom{n}{n-k}=0[/imath] (we have [imath](n+1)/2[/imath] such pairs.) but if [imath]n[/imath] is even we have no symmetry, I tried to consider the odd and even terms separately but with no success. Do you have an idea ? Thanks in advance. | 2338011 | finding the value of [imath]\sum_k (-1)^k \binom nk^2[/imath]
I am trying to find the value of: [imath]\sum_k (-1)^k {n \choose k}^2[/imath] The hint is that [imath](1-x^2)^n = (1-x)^n(1+x)^n[/imath] I tried solving it using convolution but it didn't quite work. |
813803 | Prove [imath]\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1[/imath].
Given [imath]a,b,c,d\ge 0[/imath] and [imath]a^2+b^2+c^2+d^2=4[/imath] show the following holds : [imath]\displaystyle\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1[/imath] Now I tried to fully expand but that becomes too ugly. I want a non-expand proof of this fact. For some motivation I solved a three variable version which I assumed to be true and it came out to be true! That one is : [imath]a,b,c\ge 0[/imath] and [imath]a^2+b^2+c^2=3[/imath] show that : [imath]\displaystyle \frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\le 1[/imath]. The second one is easier and I DO NOT NEED a solution of the second one. Please help me out on the first one because it is similar but not solvable by a similar method. | 624846 | How prove this inequality [imath]\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1[/imath]
Question: let [imath]a,b,c,d\ge 0[/imath],such [imath]a^2+b^2+c^2+d^2=4[/imath] show that [imath]\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1[/imath] My try: By Cauchy-Schwarz inequality,we have [imath]\sum_{cyc}\dfrac{a}{b+3}\le\sqrt{(\sum_{cyc}a^2)\left(\sum_{cyc}\dfrac{1}{(b+3)^2}\right)}[/imath] then we have only prove this [imath]\sum_{cyc}\dfrac{1}{(a+3)^2}\le \dfrac{1}{4}?[/imath] This is not true,in fact, we have [imath]\sum_{cyc}\dfrac{1}{(a+3)^2}\ge \dfrac{1}{4}?[/imath] because we have [imath]\dfrac{1}{(a+3)^2}\ge\dfrac{5-a^2}{64}[/imath] this is true because [imath]\Longleftrightarrow \dfrac{(a-1)^2(a^2+8a+19)}{64(a+3)^2}\ge 0[/imath] so [imath]\sum_{cyc}\dfrac{1}{(a+3)^2}\ge\sum_{cyc}\dfrac{5-a^2}{64}=\dfrac{1}{4}[/imath] can see:http://www.wolframalpha.com/input/?i=1%2F%28a%2B3%29%5E2-%285-a%5E2%29%2F64 This methods is from: can see:Prove this equality [imath]\frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\le\frac{1}{2}[/imath] By the way I have see this three variable inequality let [imath]a,b,c[/imath] be non-negative numbers such that [imath]a^2+b^2+c^2=3[/imath] show that [imath]\dfrac{a}{b+2}+\dfrac{b}{c+2}+\dfrac{c}{a+2}\le 1[/imath] proof: By expanding,the inequality becomes [imath]ab^2+bc^2+ca^2\le abc+2[/imath] without loss of generality,assume that [imath]\min(a,b,c)\le b\max(a,b,c)[/imath] then \begin{align*} 2-ab^2-bc^2-ca^2+abc&=2-ab^2-b(3-a^2-b^2)-ca^2+abc\\ &=(b^3-3b+2)-a(b^2-ab+ca-bc)\\ &=(b-1)^2(b+2)-a(b-a)(b-c)\ge 0 \end{align*} Equality occurs for [imath](a,b,c)=(1,1,1)[/imath] and also for [imath](a,b,c)=(0,1,\sqrt{2})[/imath] or any cyclic permutation. so my Four-inequality variable inequality,can use this methods [imath]\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1[/imath] [imath]\Longleftrightarrow a^2cd+3a^2c+3a^2d+9a^2+ab^2d+3ab^2+abc^2-abcd+3ac^2+9ac+3b^2d+9b^2+3bc^2+bcd^2+3bd^2+9bd+9c^2+3cd^2+9d^2-81\le 0[/imath] [imath]\Longleftrightarrow a^2cd+3a^2c+3a^2d+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45[/imath] [imath]\Longleftrightarrow a^2(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45[/imath] [imath]\Longleftrightarrow (4-b^2-c^2-d^2)(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45[/imath] Then I can't Thank you very much |
814443 | Multivariable Calc Proof
Suppose that [imath]F(x,y,z)=0[/imath] is an equation so that any variable can be solved in terms of the other two. Show that [imath]\displaystyle \frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z}\cdot \frac{\partial z}{\partial x}= -1.[/imath] I understand what the question means and I am able to see how it works for a linear equation but assuming I can't set up a general linear equation (I don't think it would be sufficient enough) how would I do this proof. | 392770 | Proof on showing if F(x,y,z)=0 then product of partial derivatives (evaluated at an assigned coordinate) is -1
The task is as follows: Given: [imath]F(x,y,z) = 0[/imath] Goal: Show [imath]\frac{\partial z}{\partial y}|_x \frac{\partial y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1[/imath] Here is my work so far: (1) Differentiate with respect to y, I get: [imath]0 + F_2 + F_3 \frac{\partial z}{\partial y} = 0[/imath] So [imath] F_3 \frac{\partial z}{\partial y} = - F_2[/imath] (2) Differentiate with respect to x, I get: [imath]F_1 + F_2 \frac{\partial y}{\partial x} + 0 = 0[/imath] So [imath] F_2 \frac{\partial y}{\partial x} = - F_1[/imath] (3) Differentiate with respect to z, I get: [imath]F_1 \frac{\partial x}{\partial z} + 0 + F_3 = 0[/imath] So [imath] F_1 \frac{\partial x}{\partial z} = - F_3[/imath] (4) After some manipulations with the [imath]F_i[/imath], I get to the conclusion that [imath]\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z} = -1[/imath], so when evaluated with x, z, y respectively, conclusion is still true My questions are: (1) Is my proof correct? (2) For example, when I differentiate with respect to y, I "let" [imath]F_1[/imath] be 0 and find partials for other coordinates. I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing. Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too. Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y. But is my thought ok? Would someone please help me on this question? Thank you very much ^_^ |
814076 | Diophantine equation: [imath]2 a^2 + 2 b^2 = c^2 + d^2[/imath]
I am looking for integer solutions of the following equation: [imath] 2a^{2} + 2b^{2} - c^{2} - d^{2} = 0 [/imath] Preferentially the solutions should obey [imath]a+b+c+d=0[/imath]. By inspection I found the solutions: [imath](a,b,c,d)=(1,0,1,1)[/imath], and [imath](a,b,c,d)=(1,1,2,0)[/imath]. Additional solutions can be generated by changing the signs of [imath]a,b,c,d[/imath], or by scaling by an integer, or by swapping [imath]a[/imath] and [imath]b[/imath], and [imath]c[/imath] and [imath]d[/imath]. As a theoretical physicist I am rarely working with diophantine equations, and hence I am wondering what other solutions exist that I have not taken into account. | 814045 | Solutions to the diophantine equation: [imath]2a^2 + 2b^2- c^2- d^2 = 0[/imath]
As suggested on Mathoverflow (https://mathoverflow.net/questions/168536/solutions-to-the-diophantine-equation-2a2-2b2-c2-d2-0) I am transfering this question to math-stackexchange: I am looking for integer solutions of the following algebraic equations: [imath] 2a^{2} + 2b^{2} - c^{2} - d^{2} = 0 [/imath] Preferentially the solutions should obey [imath]a+b+c+d=0[/imath]. By inspection I found the following solutions: [imath](a,b,c,d)=(1,0,1,1)[/imath] , [imath](a,b,c,d)=(0,1,1,1)[/imath], [imath](a,b,c,d)=(1,1,-2,0)[/imath] and [imath](a,b,c,d)=(1,1,0,-2)[/imath]. Additional solutions can be generated by swapping the sign of [imath]a,b,c,d[/imath] or by scaling the solutions that I have given by an integer. As a theoretical physicist I am rarely working with these diophantine equations and hence I am wondering what other solutions of the equation exist that I have not taken into account. I am looking forward to your responses. |
814682 | Continuous is sequentially continuous
Is it possible to prove [imath]f: \mathbb{R} \to \mathbb{R}[/imath] is continuous everywhere iff it is sequentially continuous everywhere without the axiom of choice? | 126010 | Continuity and the Axiom of Choice
In my introductory Analysis course, we learned two definitions of continuity. [imath](1)[/imath] A function [imath]f:E \to \mathbb{C}[/imath] is continuous at [imath]a[/imath] if every sequence [imath](z_n) \in E[/imath] such that [imath]z_n \to a[/imath] satisfies [imath]f(z_n) \to f(a)[/imath]. [imath](2)[/imath] A function [imath]f:E \to \mathbb{C}[/imath] is continuous at [imath]a[/imath] if [imath]\forall \varepsilon>0, \exists \delta >0:\forall z \in E, |z-a|<\delta \implies |f(z)-f(a)|<\varepsilon[/imath]. The implication [imath](2)\implies(1)[/imath] is trivial (though I will happily post a proof if there is sufficient interest). The proof of the implication [imath](1)\implies(2)[/imath] is worth remarking on, though. Proof that [imath](1)\implies(2)[/imath]: Suppose on the contrary that [imath]\exists \varepsilon>0:\forall \delta>0, \exists z \in E:\left (|z-a|<\delta \; \mathrm{and} \; |f(z)-f(a)|\ge \varepsilon\right )[/imath]. Let [imath]A_n[/imath] be the set [imath]\{z\in E:|z-a|<\frac{1}{n} \; \mathrm{ and }\; |f(z)-f(a)|\ge\varepsilon\}[/imath]. Now use the Axiom of Choice to construct a sequence [imath](z_n)[/imath] with [imath]z_n \in A_n \; \forall n \in \mathbb{N}[/imath]. But now [imath]a-\frac{1}{n}<z_n<a+\frac{1}{n}\; \forall n \in \mathbb{N}[/imath] so [imath]z_n \to a[/imath]. So [imath]f(z_n) \to f(a)[/imath]. But [imath]|f(z_n)-f(a)|\ge\varepsilon\; \forall n \in \mathbb{N}[/imath], which is a contradiction. You will have noticed that the above proof uses the Axiom of Choice (the lecturer didn't explicitly spell out the dependence, but it's definitely there). My question is: is it possible to prove that [imath](1) \implies (2)[/imath] without using the Axiom of Choice. I strongly suspect that it isn't. In that case, can anyone prove that we have to use the Axiom of Choice? I can think of three ways to do this: (A) Show that [imath]\left( (1) \implies (2)\right)\implies \mathrm{AC}[/imath]. I suspect that this statement is untrue. This is definitely untrue, as Arthur points out, because I only used the axiom of countable choice, which is strictly weaker than AC. (B) Show that [imath](1)\implies (2)[/imath] is equivalent to some other statement known to require the Axiom of Choice (the obvious example being the well-ordering of the real numbers). (C) Construct or show the existence of a model of ZF in which there exist sequences which satisfy [imath](1)[/imath] but not [imath](2)[/imath]. Of course, if anyone can think of another way, I would be very interested to hear about it. One final note - I am aware that very many theorems in Analysis use the Axiom of Choice in one way or another, and that this is just one example of such a theorem. If there exists a model of ZF like the one described in (C), is the study of Analysis in that model interesting? |
812629 | How to show that [imath]P_n(x)[/imath] have n distinct roots
Let [imath]P_n : \mathbb R \to \mathbb R [/imath] , [imath]n\in \mathbb N[/imath] be defined by [imath]P_n(x)=\frac{\displaystyle1}{\displaystyle2^n n!}\frac{\displaystyle d^n}{\displaystyle dx^n}[(x^2-1)^n][/imath] I need to show that [imath]P_n(x)[/imath] has exactly [imath]n[/imath] distinct roots in [imath](-1,1)[/imath] I think it is important to use induction and Rolle's theorem. To use induction, [imath]P_1(x)=x[/imath] and it is trivial that the root is [imath]x=0[/imath] in [imath](-1,1)[/imath] , but the inductivity is problem. I assumed that [imath]P_n(x)[/imath] has [imath]n[/imath] distinct roots, and yet I can't advance further. | 527140 | Proof the Legendre polynomial [imath]P_n[/imath] has [imath]n[/imath] distinct real zeros
I need a proof to show that the inequality [imath]m < n[/imath] leads to a contradiction and [imath]P_n[/imath] has [imath]n[/imath] distinct real roots, all of which lie in the open interval [imath](-1, 1)[/imath]. |
276583 | Dirac Delta Function of a Function
I'm trying to show that [imath]\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|}[/imath] Where [imath]a_{i}[/imath] are the roots of the function [imath]f(x)[/imath]. I've tried to proceed by using a dummy function [imath]g(x)[/imath] and carrying out: [imath]\int_{-\infty}^{\infty}dx\,\delta\big(f(x)\big)g(x)[/imath] Then making the coordinate substitution [imath]u[/imath] = [imath]f(x)[/imath] and integrating over [imath]u[/imath]. This seems to be on the right track, but I'm unsure where the absolute value comes in in the denominator, and also why it becomes a sum. [imath]\int_{-\infty}^{\infty}\frac{du}{\frac{df}{dx}}\delta(u)g\big(f^{-1}(u)\big) = \frac{g\big(f^{-1}(0)\big)}{\frac{df}{dx}\big(f^{-1}(0)\big)}[/imath] Can any one shed some light? Wikipedia just states the formula and doesn't actually show where it comes from. | 1533086 | Equivalence between Dirac delta of a function to a usual Dirac delta
Let [imath]f_1(s,\tau)=\delta(e^{(\tau+s)} \sinh \tau )[/imath]. This should be equal to [imath]0[/imath] everywhere unless [imath]\tau=0[/imath], but I think there should be some constant multiplying the delta, i.e. it should be equivalent to [imath]f_2(s,\tau)=c(\tau,s)\delta(\tau)[/imath] where [imath]c(\tau,s)[/imath] is a constant. How can I manipulate [imath]f_1[/imath] to resemble [imath]f_2[/imath]? Thanks. |
767805 | Banach subspace of [imath]C(X)[/imath]
Let [imath]X[/imath] be compact and suppose that [imath]Y[/imath] is a Banach subspace of [imath]C(X)[/imath]. If E is a closed subset of [imath]X[/imath] such that for every [imath]g\in C(E)[/imath] there is an [imath]f\in Y[/imath] with [imath]f_{|_{E}}=g[/imath]. Show that there is a constant [imath]c>0[/imath] such that for each [imath]g\in C(E)[/imath] there is an [imath]f\in Y[/imath] with [imath]f_{|_{E}}=g[/imath] and [imath]||f||\leq c||g||[/imath]. For this I think about using the inverse mapping theorem for the functional [imath]\phi:Y\to C(E)[/imath] but it's not injective. Please help me. | 240023 | Is the differentiation operator an open mapping?
Consider the vector spaces [imath]C([0, 1])[/imath] and [imath]C^1([0, 1])[/imath] with norm, [imath]\begin{align*} \displaystyle \|f\|_\infty=\sup_{x\in [0, 1]}|f(x)|, \end{align*}[/imath] and let [imath]T:C^1([0, 1])\to C([0, 1])[/imath] the operator given by, [imath]\begin{align*} \displaystyle Tf=f'. \end{align*}[/imath] Is [imath]T[/imath] defined this way an open mapping? If it's, how to prove it? |
104440 | Finite-dimensional subspaces of normed vector spaces are direct summands
Here is a problem in functional analysis from Folland's book: If [imath]\mathcal{M}[/imath] is a finite-dimensional subspace of a normed vector space [imath]\mathcal{X}[/imath], then there is a closed subspace [imath]\mathcal{N}[/imath] such that [imath]\mathcal{M}\cap \mathcal{N} = 0[/imath] and [imath]\mathcal{M}+\mathcal{N} = \mathcal{X}[/imath]. I tried the following approach: I am trying to define a projection map [imath]\pi_{\mathcal{M}}[/imath] from [imath]\mathcal{X}[/imath] to [imath]\mathcal{M}[/imath], which would be continuous and hence taking the inverse of any closed set would give a closed set in [imath]\mathcal{X}[/imath]. I am confused about what the projection map would be. Please suggest some approach. | 2289233 | Are finite-dimensional normed spaces injective?
A Banach space [imath]X[/imath] is said to be injective if every linear continuous [imath]T:Y\to X[/imath], with [imath]Y\subseteq Z[/imath] Banach spaces, has a linear continuous extension [imath]T^*:Z\to X[/imath] with the same operator-norm. My question is: is every finite-dimensional normed space [imath]X[/imath] injective? All examples I know are infinite-dimensional and besides they're not so simple. Any hint? Thank you. |
815571 | Integral [imath]\int^{ \pi /2}_{0} \ln (\sin x)\ dx[/imath]
[imath]\int^{ \pi /2}_{0} \ln (\sin x)\ dx[/imath] The answer is [imath]- \frac{\pi}{2} \ln 2[/imath]. I have changed it into [imath]\frac{1}{2} \int^{1}_{0} t d \ln t^{2}[/imath] But I didn't get the answer with it. | 37829 | Computing the integral of [imath]\log(\sin x)[/imath]
How to compute the following integral? [imath]\int\log(\sin x)\,dx[/imath] Motivation: Since [imath]\log(\sin x)'=\cot x[/imath], the antiderivative [imath]\int\log(\sin x)\,dx[/imath] has the nice property [imath]F''(x)=\cot x[/imath]. Can we find [imath]F[/imath] explicitly? Failing that, can we find the definite integral over one of intervals where [imath]\log (\sin x)[/imath] is defined? |
815710 | Calculate integral using the method of parameter derivation or integration
[imath]F( \alpha ) = \int^{ \pi }_{ 0} \ln ( 1 - 2 \alpha \cos x + \alpha^{2} ) dx[/imath] Should I derive the inner function? But I can't process the derived outcome. | 650513 | Computing [imath]\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx[/imath]
For [imath]a\ge 0[/imath] let's define [imath]I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.[/imath] Find explicit formula for [imath]I(a)[/imath]. My attempt: Let [imath]\begin{align*} f_n(x) &= \frac{\ln\left(1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\ &=\frac{\ln\left(\displaystyle\frac{1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\ &=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}. \end{align*}[/imath] Now it is easy to see that [imath]f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}[/imath] as [imath]n \to \infty[/imath]. [imath]|f_n(x)|\le \frac{2a+2}{(1-a)^2}[/imath] RHS is integrable so [imath]\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)[/imath]. But [imath]\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.[/imath] Consider [imath]\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.[/imath] So [imath]I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).[/imath] It looks too easy, is there any crucial lack? [imath](*)[/imath] — we have to check [imath]a=1[/imath] here by hand and actually consider [imath][0,1), (1,\infty)[/imath] but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof. |
815925 | Does [imath] \sum_{n=1}^{\infty} \left(\frac{\sin(n)}{n}\left ( \sum_{m=1}^{n}\frac{1}{m} \right )\right) [/imath] converge?
I am trying to determine whether the [imath] \sum_{n=1}^{\infty} \left(\frac{\sin(n)}{n}\left ( \sum_{m=1}^{n}\frac{1}{m} \right ) \right) [/imath] converges or not. I have tried the popular tests, but all were inconclusive (so perhaps I am doing it wrong...). I would appreciate any help and explanation on how I should solve it. Thanks. | 812455 | Does the sum [imath]\sum _{n=1}^{\infty }\left(\frac{\sin\left(n\right)}{n}\sum _{k=1}^n\left(\frac{1}{k}\right)\:\right)[/imath] absolutely converge?
I have the following sum: [imath]\sum _{n=1}^{\infty }\left(\frac{\sin\left(n\right)}{n}\sum _{k=1}^n\left(\frac{1}{k}\right)\:\right)[/imath] And I need to determine if it absolutely converges, conditionally converges or diverges. I know it converges conditionally with the help of the Dirichlet test, but I'm having a hard time determining if it converges absolutely. Intuitively, I have a feeling it does not, because [imath]\left|\frac{\sin\left(n\right)}{n}\right|[/imath] looks to be partial to the harmonic sum, which diverges. |
816060 | Show that if [imath]\sum_{n=1}^{\infty}a_n[/imath] converges abs. and [imath](b_n)_{n \in \mathbb{N}}[/imath] is bounded , then [imath]\sum_{n=1}^{\infty} a_nb_n[/imath] converges abs.
My attempt: Since [imath](b_n)_{n \in \mathbb{N}}[/imath] is bounded it follows that there exists a bounded sequence [imath](c_n)_{n \in \mathbb{N}}: 0 \leq |b_n| \leq |c_n|, \forall n \in \mathbb{N}[/imath] Therefore it follows that: [imath]\sum_{n=1}^{\infty} a_nb_n \leq \sum_{n=1}^{\infty} a_nc_n[/imath] Since [imath]\sum_{n=1}^{\infty}a_n[/imath] converges absolutely [imath]\Rightarrow[/imath] [imath]\sum_{n=1}^{\infty}a_n[/imath] converges [imath]\Rightarrow (a_n)_{n \in \mathbb{N}} [/imath] is a null sequence [imath]\Rightarrow \exists N \in \mathbb{N}: \forall n \geq N: |a_n| \lt \frac{\epsilon}{|c|p}[/imath], where [imath]c:= max\{c_1,\dots,c_n,\dots\}, p \in \mathbb{N}[/imath] Then it follows that: [imath]\forall n \geq N: |a_nc_n|=|a_n||c_n| \lt |a_n||c| \lt \frac{\epsilon}{|c|p}|c|=\frac{\epsilon}{p}[/imath] [imath]\Rightarrow \forall n \geq N: |a_{n+1}c_{n+1}+\dots+a_{n+p}c_{n+p}|\leq |a_{n+1}c_{n+1}|+ \dots +|a_{n+p}c_{n+p}| \lt p\frac{\epsilon}{p}=\epsilon[/imath] So after Cauchy's convergence test, it follows that [imath]\sum_{n=1}^{\infty} a_nc_n[/imath] is a convergent majorant of [imath]\sum_{n=1}^{\infty} a_nb_n[/imath] and therefore [imath]\sum_{n=1}^{\infty} a_nb_n[/imath] converges absolutely. Was this proof correct? And a second part of that question is: If [imath]\sum_{n=1}^{\infty}[/imath] converges, but not absolutely, does [imath]\sum_{n=1}^{\infty} a_nb_n[/imath] converge? For my proof I didn't really needed [imath]\sum_{n=1}^{\infty}[/imath] to converge absolutely, I just used the fact that it converges when it converges absolutely and since I proved that [imath]\sum_{n=1}^{\infty} a_nb_n[/imath] converges absolutely, it converges as well. Is that right? | 85287 | [imath]b_n[/imath] bounded, [imath]\sum a_n[/imath] converges absolutely, then [imath]\sum a_nb_n[/imath] also
a) Prove that if [imath]\sum a_n[/imath] converges absolutely and [imath]b_n[/imath] is a bounded sequence, then also [imath]\sum a_nb_n[/imath] converges absolutely. I wanted to use the comparison test to show it's true, but I think I got something mixed up. Here's what I have so far. [imath]b_n[/imath] is bounded [imath]\Rightarrow \exists c > 0: |b_k| < c, \forall k \in \mathbb N[/imath] [imath]\sum a_nb_n < \sum a_nc < c\sum a_n[/imath]. I'm very tempted to use the comparison test here and say, as [imath]\sum a_n[/imath] converges absolutetly, then so does [imath]c\sum a_n[/imath], but for that I needed the inverted relation, right? I would need [imath]\sum a_n > c\sum a_n[/imath], which is not true. However isn't it obvious that something absolutely convergent multiplied by a constant is also absolutetly convergent? Is there a mathematical way to write this? Thanks a lot in advance. b) Refute with a counter example: if [imath]\sum a_n[/imath] converges and [imath]b_n[/imath] is a bounded sequence, then also [imath]\sum a_nb_n[/imath] converges. Is this even possible? I mean, it has to be, but it doesn't make sense to me. If something converges, it means it's bounded, right? And I thought, well, since bounded + bounded = bounded, then bounded*bounded would also get me something bounded again. Anyway, my idea would be to use for [imath]a_n[/imath] an alternating series that ist only convergent, for example [imath](-1)^n \frac 1 {n^2}[/imath]. But something tells me I'm trying to prove [imath]a_n b_n[/imath] is not absolutely convergent. Thanks a lot in advance guys! |
816250 | Sum of all the positive integers problem
The staff of Numberphile has shown that the sum of all the integers from [imath]0[/imath] to [imath]\infty[/imath] is [imath]-\frac1{12}[/imath]. Recently I was looking for the sum of all the (positive) integers from [imath]0[/imath] to [imath]n[/imath] and I found that: [imath]\sum_{i=0}^n i=\frac{n(n+1)}{2}[/imath] So I decided to take the limit: [imath]\lim_{n\to \infty}\frac{n(n+1)}{2}[/imath] but that tends towards [imath]\infty[/imath] when I expected that to be [imath]-\frac1{12}[/imath]! Where did I got wrong? (the result is also confirmed by Wolfram Alpha) | 1749689 | Whats wrong with this proof? (infinite sequences)
So I just watched a video where they explained that the sum of all natural numbers is [imath]-1/12[/imath]. However, there was an interesting comment: S1 = 1 + 2 + 3 + 4 + 5 ... = - 1/12 S1 - S1 = 1 + 2 + 3 + 4 + 5 + 6 ... - 1 - 2 - 3 - 4 - 5 ... = 1 + 1 + 1 + 1 + 1 + 1 ... Since S1 - S1 = - 1/12 - (- 1/12) = 0 It follows that 1 + 1 + 1 + 1 + 1 .... = 0 Let's name this sequence S2: S2 = 1 + 1 + 1 + 1 + 1 ... = 0 Now let's subtract it from itself: S2 - S2 = 1 + 1 + 1 + 1 + 1 ... - 1 - 1 - 1 - 1 .... = 1 Given that S2 equals 0, we can also write this as: 0 - 0 = 1 For those who are wondering about the video: https://www.youtube.com/watch?v=w-I6XTVZXww It does the same trick of "shifting". With infinity and stuff things get tricky. What is wrong with this proof? I am a math student, so involved answers with limits is fine. |
816273 | How does [imath]\arccos[/imath] actually work?
[imath]\arccos\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)=θ[/imath] I regard the above as witchcraft. How would I work this out if I didn't have a calculator? Once I know how to workout arcos without a calculator, will I know how to work out arcsine and arctangent? Is it simply a case of swapping around the the adjacent, opposite and hypotenuse? | 816264 | How is arccos derived?
Stupid question but I need to understand: If [imath] \cos = \dfrac{\text{adjacent}}{\text{hypotenuse}} [/imath] What is arcosine? [imath] \text{adjacent} \cdot \text{hypotenuse}[/imath]? Is this the same for arcsine and arctangent? |
816585 | 10 non-increasing or non-decreasing sequence from 101 random numbers
In [imath]101[/imath] random integer numbers [imath]a[i],i=0, \cdots,100[/imath], prove that we can always find [imath]10[/imath] non-increasing or non-decreasing sequence. A sequence is a sequence of numbers is an array of numbers picking up from the original array with increasing index. | 738906 | Collection of numbers always in increasing or decreasing order
Anyone have any ideas on this question? I think you have to use the pigeon hole principle..but I am not sure about that? The numbers [imath]1,2,3,\ldots,101[/imath] are written down in a row in some order. Is it always possible to cross out [imath]90[/imath] numbers in a way such that all [imath]11[/imath] numbers left will stay either in increasing or decreasing order? |
816722 | Limit of a strange sequence [imath]a_{n} = a_{[n/2]} + a_{[n/3]} + a_{[n/6]}[/imath]
Recently I came across this problem : If [imath]a_{0} = 1[/imath] and [imath]a_{n} = a_{[n/2]} + a_{[n/3]} + a_{[n/6]}[/imath] where [imath][x][/imath] denotes the greatest integer not exceeding [imath]x[/imath] then show that [imath]\lim_{n \to \infty}\dfrac{a_{n}}{n} = \frac{12}{\log 432}[/imath] The recurrence relation is very strange and seems to be multiplicative (has terms like [imath]a_{[n/2]}[/imath]) instead of being additive (having terms like [imath]a_{n - 1}[/imath] etc) and I wonder how such a recurrence can be solved. Any suggestions are welcome! | 487957 | How prove this nice limit [imath]\lim\limits_{n\to\infty}\frac{a_{n}}{n}=\frac{12}{\log{432}}[/imath]
Nice problem: Let [imath]a_{0}=1[/imath] and [imath]a_{n}=a_{\left\lfloor n/2\right\rfloor}+a_{\left\lfloor n/3 \right\rfloor}+a_{\left\lfloor n/6\right\rfloor}.[/imath] Show that [imath]\lim_{n\to\infty}\dfrac{a_{n}}{n}=\dfrac{12}{\log{432}},[/imath] where [imath]\lfloor x \rfloor[/imath] is the largest integer not greater than [imath]x[/imath]. It is said this problem was created by Paul Erdős, and I can't find this problem's solution, does anyone have any nice methods? Thank you. |
816799 | How many eulerian circuits are in [imath]K_{2,n}[/imath] ?
Hello I have been asked to find a formula of how many different Euler circuits or as it also called eulerian cycles are in [imath]K_{2,n}[/imath]. when [imath]n[/imath] is an even number ofcourse otherwise there won't be any eulerian circuits. for example in [imath]K_{2,2}[/imath] you have vertices [imath]a[/imath] [imath]b[/imath] [imath]c[/imath] [imath]d[/imath] so there is only one cycle possible, in other words [imath]adbca[/imath] is the same as [imath]cbdac[/imath]. thanks from advance :) | 814021 | Count Eulerian cycles in [imath]K_{n,2}[/imath]
So I need to count how many Eulerian cycles exist in the complete graph [imath]K_{n,2}[/imath]. Two cycles are considered identical if: 1) The order of the edges is identical in both cycles. For example: The cycles [imath]cabc, bcab, abca[/imath] are identical. Or 2) If we get the first cycle by reversing the order of the edges in the second cycle. For example: [imath]abca[/imath] and [imath]acba[/imath] are the same cycle. Note: The order of the edges is what matters, not the vertices. Also, of course if [imath]n[/imath] is odd, there's no such cycle. So I tried this in different ways and arrived at different solutions: I can think of this as all the permutations of the side with [imath]n[/imath] vertices, removing all the reverse permutations, so that's [imath]\frac {n!}{2}[/imath]. Or, I can think of it as permutations in a cycle, which means [imath]\frac {(n-1)!}{2}[/imath]. Then I manually counted the Eulerian cycles in [imath]K_{2,4}[/imath] and I got 6, which works well with [imath](n-1)![/imath]. So either I counted incorrectly, or [imath](n-1)![/imath] is the answer, but I don't understand why. Any help would be appreciated. |
815619 | Prove that [imath]x_1x_2x_3\cdots x_n>y_1y_2y_3\cdots y_m[/imath] with some given conditions
[imath]x_1, x_2, x_3,.\cdots ,x_n[/imath] and [imath]y_1, y_2, y_3,\ldots,y_m[/imath] are two series of positive integers, such that [imath]m\not=n[/imath]. Given that [imath]1<x_1< x_2<x_3<\cdots<x_n<y_1< y_2< y_3<\cdots<y_m[/imath] and [imath]x_1+ x_2+ x_3+\cdots+x_n > y_1+ y_2+ y_3+\cdots+y_m[/imath], prove that [imath]x_1x_2x_3\cdots x_n>y_1y_2y_3\cdots y_m[/imath]. I have tried to prove it for conservatives. I have no idea how to deal with this type of inequality properly. | 790980 | Prove that: [imath]x_1\cdot x_2\cdots x_n>y_1\cdot y_2\cdots y_m[/imath].
For two positive integer sequences [imath]x_1,x_2,\ldots,x_n[/imath] and [imath]y_1,y_2,\ldots,y_m[/imath] satisfying [imath]x_i\neq x_j\quad \text{and}\quad y_i\neq y_j\quad \forall i,j, i \ne j[/imath] [imath]1<x_1<x_2<\cdots<x_n<y_1<\cdots<y_m.[/imath] [imath]x_1+x_2+\cdots+x_n>y_1+\cdots+y_m.[/imath] Prove that: [imath]x_1\cdot x_2\cdots x_n>y_1\cdot y_2\cdots y_m[/imath]. (from internet) I don't have an idea for this problem. Thanks for your help. |
817269 | Evaluating [imath]\int_{\mathbb R^n}e^{-\langle Ax,x \rangle}dx[/imath] where [imath]A[/imath] is symmetric and positive definite
I was asked the following question, and while I think I made little progress, I'd like a push in the right direction. Let [imath]A[/imath] be [imath]n[/imath]x[imath]n[/imath] positive definite symmetric matrix with real entries. then for all [imath]x \in \mathbb R^n[/imath], evaluate the integral: [imath]\int_{\mathbb R^n}e^{-\langle Ax,x \rangle}dx[/imath] Hint: Start by assuming [imath]A[/imath] is diagonal and remember that any symmetric matrix is also diagonlizable. What I did: I think the hint is good. Assume [imath]A[/imath] is diagonal with the entries [imath]\lambda_1 ,\lambda_2 ,...\lambda_n[/imath], then [imath]\langle Ax,x \rangle=\sum_{i=1}^{n} \lambda_ix_i^2[/imath] So now the integral becomes [imath]\int_{\mathbb R^n}e^{\sum_{i=1}^{n}\lambda_ix_i^2}dx[/imath] But what now? can we split integral and integrate by [imath]x_i[/imath] each time? | 653159 | If [imath]A[/imath] is positive definite, then [imath]\int_{\mathbb{R}^n}\mathrm{e}^{-\langle Ax,x\rangle}\text{d}x=\left|\det\left({\pi}^{-1}A\right)\right|^{-1/2}[/imath]
Let [imath]A[/imath] be a positive definite real [imath]n\times n[/imath] matrix. How can I prove that [imath] \int_{\mathbb{R}^n}\mathrm{e}^{-\langle Ax,x\rangle}\text{d}x=\left|\,\det\left(\pi^{-1}{A}\right)\right|^{-1/2}=\pi^{n/2}\lvert\,\det A\rvert^{-1/2}\!, [/imath] where [imath]\langle\cdot,\cdot\rangle[/imath] denotes the inner product in [imath]\mathbb R^n[/imath], i.e. [imath]\langle x,y\rangle =x^Ty[/imath]. |
817453 | Conformal mapping vertical strip [imath]\{-1 < Re(z) < 1\}[/imath] onto open unit disk [imath]D[/imath] in [imath]\mathbb{C}[/imath].
As the title states, I need to find a conformal mapping from the vertical strip [imath]\{-1 < Re(z) < 1\}[/imath] onto open unit disk [imath]D[/imath]. I'm not sure how to do this.. so do I first have to find a few points? For example, do I want the line [imath]Re(z) = 1[/imath] to map to (0,0) and then [imath]Re(z) = -1[/imath] to map to the [imath]|z| = 1[/imath]? Thanks! | 393799 | Conformal Map from Vertical Strip to Unit Disc
I haven't found a similar question on here, though I suspect the question may be rather well-covered. I want to find a conformal map from the vertical strip [imath]\{z:-1<Re(z)<1\}[/imath] onto the unit disc. Under the exponential map the region is taken to the annulus with radii [imath]e[/imath] and [imath]e^{-1}[/imath], but I'm not sure how useful this will be. Can anyone advise on what the conformal map may be? Thanks. |
397837 | lebesgue measure of difference of two sets
Suppose [imath]A,B[/imath] are in [imath]\mathbb R^n[/imath], [imath]A[/imath] lebesgue measurable and [imath]|(A \setminus B)\cup(B\setminus A)| = 0[/imath]. Show [imath]B[/imath] must be lebesgue measurable as well and [imath]|A| = |B|[/imath] (where [imath]|\cdot |[/imath] is the lebesgue outer measure). | 817654 | Lebesgue measurability of nearly identical sets
If [imath]A[/imath] is a Lebesgue measurable set such that [imath]m^*(A\triangle B)=0[/imath], then [imath]B[/imath] is Lebesgue measurable. I tried the problem as follows: [imath]m^*(A\cup B)=m^*((A\triangle B)\cup(A\cap B))\leq m^*(A\triangle B)+m^*(A\cap B)=m^*(A\cap B)[/imath] Thus [imath]m^*(A\cap B)=m^*(A\cup B)[/imath] and so [imath]m^*(A)=m^*(B)[/imath]. But I got stuck at this point. Please give any hint! |
31785 | How many sides does a circle have?
My son is in 2nd grade. His math teacher gave the class a quiz, and one question was this: If a triangle has 3 sides, and a rectangle has 4 sides, how many sides does a circle have? My first reaction was "0" or "undefined". But my son wrote "[imath]\infty[/imath]" which I think is a reasonable answer. However, it was marked wrong with the comment, "the answer is 1". Is there an accepted correct answer in geometry? edit: I ran into this teacher recently and mentioned this quiz problem. She said she thought my son had written "8." She didn't know that a sideways "8" means infinity. | 2894077 | Is this proof of [imath]0 = \infty[/imath] just a mathematical joke?
Is this "proof" just a mathematical joke or might there be some deeper truth in it, eventhough the theorem is obviously false? Definition: Let a regular [imath]n[/imath]-gon be a geometric figure [imath]f^d_n[/imath] that is uniquely defined by one integer-valued parameter [imath]n[/imath] (the figure's number of corners) one real-valued parameter [imath]d \neq 0[/imath] (the figure's diameter) (Let's assume [imath]d=1[/imath] and drop the parameter in the following.) has an area [imath]\neq 0[/imath] Lemma 1: A regular [imath]n[/imath]-gon [imath]f_n[/imath] is a regular [imath]m[/imath]-gon [imath]f_m[/imath] iff [imath]n=m[/imath]. Theorem: [imath]0 = \infty[/imath] Proof: A circle is a regular [imath]0[/imath]-gon [imath]f_0[/imath] because it's defined by its diameter and its number of corners, which is 0. On the other side, a circle is a regular [imath]\infty[/imath]-gon [imath]f_\infty[/imath] being defined by its diameter and its number of corners, which is [imath]\lim_{n\rightarrow \infty} n = \infty[/imath]. From Lemma 1 it follows that [imath]0 = \infty[/imath]. [imath]\blacksquare[/imath] Lemma 2: There are no regular [imath]1[/imath]- and [imath]2[/imath]-gons. My questions are: Which are the obvious errors in this proof? Is there some deeper truth in this proof? If it's just a mathematical joke: Who told it first? If it contains some deeper truth: Who told it first? |
817501 | Prove that integrable implies bounded
If function [imath]f[/imath] is integrable at [imath][a,b][/imath] prove that [imath]f[/imath] is bounded at [imath][a,b][/imath]. Could you explain without using rieaman integrable? I have did so far : if [imath]f[/imath] is integrable then it is continuous at [imath][a,b].[/imath] let [imath]a\in I=[a,b][/imath] so for any [imath]\epsilon>0[/imath] there is [imath]\delta[/imath] such as [imath]|x-a|\leq\delta\Rightarrow|f(x)-f(a)|\leq\epsilon[/imath] assume [imath]|t-a|\leq\delta[/imath] then [imath]-\epsilon\leq f(t)-f(a)\leq \epsilon[/imath] so [imath]\int_a^x-\epsilon\leq \int_a^xf(t)-f(a)\leq \int_a^x \epsilon[/imath] [imath]\int_a^x|f(t)-f(a)|\leq\epsilon|x-a| [/imath] | 610054 | If a function [imath]f(x)[/imath] is Riemann integrable on [imath][a,b][/imath], is [imath]f(x)[/imath] bounded on [imath][a,b][/imath]?
Most statements regarding Riemann integrals (at least the ones that I have encountered) begin with the statement "for [imath]f(x)[/imath] bounded on [imath][a,b][/imath]." I am wondering if Riemann integrability implies boundedness. I think that this has to be the case, but I am not sure. If Riemann integrability does imply boundedness, are improper integrals considered Riemann integrals? I would think that improper integrals wouldn't be Riemann integrals since improper integrals are allowed to be equal to [imath]+\infty[/imath] or [imath]-\infty[/imath]. Or are improper integrals that are not equal to [imath]\infty[/imath] considered Riemann integrals? I am confused. |
818410 | conditional expectation of the Brownian motion
[imath](B_t)[/imath] is a Brownian motion and i assume that [imath]s<t<u[/imath] we have [imath]E[B_t |\sigma(B_s,B_u)] = G(B_s,B_u)[/imath] Does anyone knows the explicit expression of [imath]G[/imath] ? (the calculus is easy but fastidious...). Thank you. | 805820 | conditional expectation of brownian motion
Let [imath](B_t)_{t\geq 0}[/imath] be a standard Brownian motion in [imath]\mathbb{R}^d[/imath]. It is intuitive that, for fixed [imath]s<t<u[/imath] [imath]\mathbb{E}[B_t\mid \sigma(B_s,B_u)]=B_s+\frac{t-s}{u-s}(B_u-B_s).[/imath] However, I cannot think of a way to show this rigorously. If first attempted to take [imath]A\in\sigma(B_s,B_u)[/imath] and show that [imath]\mathbb{E}[1_A B_t]=\mathbb{E}[1_A(B_s+\frac{t-s}{u-s}(B_u-B_s))][/imath]. But I cannot manage to show this equality. I'd be very thankful for any ideas and suggestions on how to tackle this problem. |
766197 | Ring Properties
If [imath](S,.,+)[/imath] is a ring with the property that [imath]a^2 = a[/imath] for all a an element of [imath]S[/imath], which of the following must be true, given: I [imath]a + a = 0[/imath] for all [imath]a\in{ S}[/imath]. II [imath](a + b)^2 = a^2 + b^2 [/imath] for all [imath]a, b \in{ S}.[/imath] III S is commutative. A. III only B. I and II only C. I and III only D. II, and III only E. I, II, and III Help please I know that it is not only III so A. is out. | 71617 | Why is a ring [imath]R[/imath] with the property that [imath]r=r^2[/imath] for each [imath]r\in R[/imath] so special?
The question is motivated by the following multiple-choice problem: If [imath]R[/imath] is a ring with the property that [imath]r=r^2[/imath] for each [imath]r\in R[/imath], which of the following must be true? I. [imath]r+r=0[/imath] for each [imath]r\in R[/imath]. II. [imath](r+t)^2=r^2+t^2[/imath] for each [imath]r,t\in R[/imath]. III. [imath]R[/imath] is commutative. Here are my questions: What theorems do I need to solve the problem above? Why is a ring with the property that [imath]r=r^2[/imath] for each [imath]r\in R[/imath] so special? Is there a name for such rings? |
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