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2659719 | Find all integer solutions to the equation [imath]x^2 − x = y^5 − y[/imath].
My professor gave us a worksheet with diophantine equations including this one he claims that it is one of the easier ones but that it has a unique solution. Find all integer solutions to the equation [imath]x^2 − x = y^5 − y[/imath]. Also my professor will give extra credit for all prime number solutions (x and y). | 1242947 | Finding integer solutions of [imath]m^2-n^5 = m - n[/imath]
How to list all integer solutions of [imath]m^2-n^5 = m - n[/imath] Here [imath]m[/imath] and [imath]n[/imath] are some positive integers. Also, I want to know the name of this type equations (if name exist). Regards Rosy |
1246385 | How to prove [imath]\operatorname{Span}(\operatorname{Span}(S)) = \operatorname{Span}(S)[/imath]
Given a Subset [imath]S[/imath] (not necessarily a subspace) of a vector space [imath]V[/imath], [imath]\operatorname{Span}(S)[/imath] indicate the smallest subspace containing [imath]S[/imath]. I need a hint to solve the problem [imath]\operatorname{Span}(\operatorname{Span}(S)) = \operatorname{Span}(S)[/imath] | 1050861 | [imath][S][/imath] means linear span of S. Then [imath][S] = [[S]][/imath]
My definition of span: Suppose a vector space [imath](V,+,\cdot)[/imath], and [imath]S = \{u_1,\cdots,u_n\}[/imath] (and [imath]S[/imath] is a subset of [imath]V[/imath], not a subspace) [imath][S]=:\bigcap_{W\subset V, W\supseteq S} W[/imath] In other words, [imath][S][/imath] is, by definition, the intersection of all [imath]W[/imath], such that [imath]W[/imath] is a subspace of [imath]V[/imath] and [imath]W[/imath] contains [imath]S[/imath]. Well, then I need to prove that [imath][S] = [[S]][/imath] My strategy: [imath][[S]][/imath] means the intersection of all subspaces that contain [imath][S][/imath]. If all the subspaces contains [imath][S][/imath], so does their intersection. Therefore, [imath][S]\subset[[S]][/imath] However, I'm having trouble proving the converse: [imath][[S]]\subset[S][/imath] Any hints? |
2660621 | What is the largest number whose first digits can be calculated?
The way to calculate the first digits of [imath]a^b[/imath] is to compute the [imath]log_{10} a[/imath] and multiply by [imath]b[/imath], and find the fractional part. But if there are too many digits in the integer part of the result, then the digits after the point are inaccessible. By too many digits I mean enough digits to overflow a computer's memory. So this question boils down to, how many digits can be represented in a computer's memory? The largest number I have ever computed the first digits of is [imath]2^{2^{2^{32}}}[/imath] which begins with 315921269337233843004... and has more than [imath]9.34*10^{1292913985}[/imath] digits. It took literally almost all my computer's RAM and 30 hours of my time to calculate. Is there any hope of eventually finding the first digits of, say, [imath]3^{3^{3^{3^3}}}[/imath] at some point? | 1341922 | Which is the largest power of natural number that can be evaluated by computers?
Which is the largest power of natural number that can be evaluated by computers? For example if we take a very large power of 7: [imath]7^{120000000000}[/imath]. Can a computer calculate this number? |
2660327 | Disk on a tile floor
A disk 2 inches in diameter is thrown at random on a tiled floor, where each tile is a square with sides 4 inches in length. Let C be the event that the disk will land entirely on one tile. In order to assign a value to P(C), consider the center of the disk. In what region must the center lie to ensure that the disk lies entirely on one tile? If you draw a picture, it should be clear that the center must lie within a square having sides of length 2 and with its center coincident with the center of a tile. Since the area of this square is 4 and the area of a tile is 16, it makes sense to let [imath]P(C) = \frac{4}{16}.[/imath] I don't really understand the statement for the last sentence "Since the area of this square is 4 and the area of a tile is 16, it makes sense to let [imath]P(C) = \frac{4}{16}.[/imath]" | 2616215 | Hidden variables in probability
I came up with a mental problem in probability and I found out that it's either trivial or more complex than what I thought. The problem is this: Take a coin of diameter [imath]D[/imath] and radius [imath]R[/imath], and toss it. Cover the table with a rectangular tablecloth made of small squares of side [imath]\ell[/imath], with [imath]\ell \geq D[/imath]. What is the probability for the coin to land exactly inside a little square? If the problem has a trivial solution then I expect it to be this: the probability is the area of the coin over the area of the square: [imath]P = \frac{\pi R^2}{\ell^2}[/imath] But here I would a sort of [imath]\theta[/imath] function such that (in case this hypothesis is missing) the probability is zero when [imath]D>\ell[/imath] hence: [imath]P = \frac{\pi R^2}{\ell^2}\theta(\ell - 2R)[/imath] But here many problems arise: first of all my convention demands that [imath]\theta (\ell - 2R) = 1[/imath] for [imath]2R = \ell[/imath], because as improbable it is, there could be a golden toss by which the coin lands exactly perfectly within the square the side of which is identical to the diameter. But now I strongly believe I shall consider the whole number of squares of the tablecloth. But more squares = more probability? I'm stuck on an either trivial or complicated reasoning. Any clarification? |
2659984 | if [imath]2[/imath] hyperbola are conjugates of each other then value of [imath]c[/imath]
If the hyperbola [imath]x^2+3xy+2y^2+2x+3y+2=0[/imath] and [imath]x^2+3xy+2y^2+2x+3y+c=0[/imath] are conjugate of each other . Then [imath]c[/imath] equals solution i try Asymptotes of 1 st hyperbola is [imath]x^2+3xy+2y^2+2x+3y+k=0[/imath] and it represent [imath]2[/imath] pair of lines so its discriminant is zero .so i am getting [imath]k=1[/imath] How to solve it from that point | 182429 | Finding the conjugate hyperbola
Assume we are given a general proper hyperbola [imath] a_{11}x^2 + 2 a_{12} x y + a_{22} y^2 + 2 a_{13}x +2 a_{23}y + a_{33} = 0[/imath] with [imath]D =\det \begin{pmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{pmatrix} < 0[/imath] and [imath] A = \det \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \neq 0[/imath]. Is there an easy way to find the implicit description of the conjugate hyperbola? I could translate and rotate the hyperbola and obtain a canonical representation of the form [imath]\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1[/imath], find the conjugate of the canonical one and the transform back. However, I'm looking for a more straightforward way to achieve this. |
2661145 | Is it formally right to say that [imath]\Bbb N \subset \Bbb Z[/imath] etc?
I was studying the formal construction of numbers from [imath]\mathrm{ZF}[/imath], assuming the elements of the naturals being [imath]\{\}, \{\{\}\}[/imath], etc. Given the naturals, its ring properties and arithmetic, we can define the integers being a set of equivalence classes on [imath]\Bbb N^2[/imath]. This logic is used from the integers to the complex numbers, so the tear always have sets of equivalence classes. My doubt is: since, for example, [imath]\bigcup \Bbb Z = \Bbb N^2[/imath], isn’t it wrong to formally say that [imath]1_{\Bbb N} = 1_{\Bbb Z}[/imath], that is, [imath]1 \in \Bbb Z[/imath]? Since that the only thing we can assume is that there is a bijective map [imath]\Bbb N \mapsto \Bbb Z[/imath] (in fact, they’re both countable). | 1055501 | Why are integers subset of reals?
In most programming languages, integer and real (or float, rational, whatever) types are usually disjoint; 2 is not the same as 2.0 (although most languages do an automatic conversion when necessary). In addition to technical reasons, this separation makes sense -- you use them for quite different purposes. Why did they choose to say [imath]\mathbb{Z} \subset \mathbb{R}[/imath] in math? In other words, why are 2 and 2.0 considered the same? When you are working in [imath]\mathbb{R}[/imath], does it make any difference whether some elements, eg. 2.0, also belong to [imath]\mathbb{Z}[/imath] or not? |
2661168 | Find [imath]\int \sqrt{2+\tan x} dx[/imath]
I am trying to get the answer [imath]\int \sqrt{2+\tan x} dx[/imath] What I did was to write [imath]2+\tan x[/imath] as [imath]\frac{2\cos x+\sin x}{\cos x}[/imath] but then find no way. Is there any simple form or should I proceed some other way ? | 626942 | Calculation of [imath]\int \sqrt{\tan x+2}dx[/imath]
Calculation of [imath]\displaystyle \int\sqrt{\tan x+2}\;dx[/imath] [imath]\bf{My\; solution::}[/imath] Let [imath]\displaystyle \tan x+2 = t^2 [/imath], Then [imath]\displaystyle \sec^2 (x)dx = 2tdt\Rightarrow dx = \frac{2t}{1+\tan^2 x}dt = \frac{2t}{1+(t^2-2)^2}dt[/imath] So Integral convert into [imath]\displaystyle \int \frac{2t^2}{(t^2-2)^2+1^2}dt[/imath] Let [imath]\displaystyle (t^2-2) = u\Rightarrow t^2=u+2\;,[/imath] Then [imath]\displaystyle tdt=\frac{1}{2}du[/imath] Now How can i solve after that please help me Thanks |
1203786 | Does [imath]\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0 [/imath] for all [imath]n[/imath] imply that [imath]\lambda_1= \lambda_2= \dots= \lambda_k = 0 [/imath]?
Suppose [imath]\lambda_1, \lambda_2, \dots, \lambda_k [/imath] are complex numbers that [imath]\forall n \in \mathbb{N}[/imath] satisfy [imath]\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0.[/imath] Can we deduce that [imath]\lambda_1= \lambda_2= \dots= \lambda_k = 0 [/imath]? | 399986 | [imath]\sum_i x_i^n = 0[/imath] for all [imath]n[/imath] implies [imath]x_i = 0[/imath]
Here is a statement that seems prima facie obvious, but when I try to prove it, I am lost. Let [imath]x_1 , x_2 \dots x_k[/imath] be complex numbers satisfying: [imath]x_1 + x_2 \dots + x_k = 0[/imath] [imath]x_1^2 + x_2^2 \dots + x_k^2 = 0[/imath] [imath]x_1^3 + x_2^3 \dots + x_k^3 = 0[/imath] [imath]\dots[/imath] Then [imath]x_1 = x_2 = \dots = x_k = 0[/imath]. The statement seems obvious because we have more than [imath]k[/imath] constraints (constraints that are in some sense, "independent") on [imath]k[/imath] variables, so they should determine the variables uniquely. But my attempts so far of formalizing this intuition have failed. So, how do you prove this statement? Is there a generalization of my intuition? |
2661528 | Prove that [imath] \lim_{n \to \infty}\int_0^{n^2} n\sin (x/n)e^{-x^2} dx = 1/2 [/imath]
Prove that [imath] \lim_{n \to \infty}\int_0^{n^2} n\sin(x/n)e^{-x^2} dx = 1/2.[/imath] I've tried all of the techniques learned in the course so far and none of them seem to work, so if anyone could help me with this it would be much appreciated. Thanks! | 1143964 | Computing [imath]\lim_{n \to \infty} \int_0^{n^2} e^{-x^2}n\sin\frac{x}{n}\,dx[/imath]?
I am trying to compute this integral/limit, I don't feel like I have any good insight... [imath]\lim_{n \to \infty} \int_0^{n^2} e^{-x^2}n\sin\frac{x}{n} \, dx.[/imath] I have tried to make a change of variable to get rid of the [imath]n^2[/imath], I changed to [imath]X=\frac{x}{n^2}[/imath] but got something even worse, I've tried to reach a situation where I could use a convergence theorem for Lebesgue Integrals,... I'm not sure I'm even on the right track! Could you give me a hint on how to start this? Thank you very much! |
2662243 | Evaluate [imath]\sum_{n=1}^\infty \frac1{n(2n-1)}[/imath]
The question is to evaluate [imath]\sum_{n=1}^\infty \frac1{n(2n-1)}[/imath] I have done this: [imath]\sum_1^\infty\frac1{n(2n-1)}=\sum_{n=1}^\infty\frac2{2n-1}-\frac1n=\frac21-\frac11+\frac23-\frac12+\cdots\\=1-\frac12+\frac13-\frac14+\cdots=\log 2[/imath] But Wolfram alpha says the initial sum is [imath]2\log 2[/imath]. I think the wrong step is when I started a new line above - the sum needs to be re-ordered to get it into the form I have got. But the series is not absolutely convergent (I think), so this shouldn't be allowed. Am I right? How can I correct this argument? If it is not possible to correct is, can someone provide an alternative method I could use instead? | 2653938 | Prove that [imath]-2\log(2) = -2 + \sum_{n=1}^{\infty}\frac{1}{n(2n+1)}[/imath]
How to prove that [imath] -2\log(2) = -2 + \sum_{n=1}^{\infty}\frac{1}{n(2n+1)} [/imath] I know that this sum is equal to [imath]\phi(1/2)+\gamma[/imath] where [imath]\phi(x)[/imath] is the digamma function and [imath]\gamma[/imath] is the Euler-Mascheroni constant, but I want to evaluate the sum without knowing it. |
2660361 | How do I interpret Euler's formula?
I don't understand the formula at all: [imath]e^{ix} = \cos(x) + i \sin(x)[/imath] I've tried reading all sorts of webpages and answers on the subject but it's just not clicking with me. I don't understand how we can define things when these are already known quantities. We've already got an exact definition for [imath]e[/imath] as [imath]\lim_{n \to 0} (1+\frac{1}{n})^n[/imath], and we've got [imath]i = \sqrt{-1}[/imath], and we've got [imath]\cos[/imath] and [imath]\sin[/imath] as the [imath]x[/imath] and [imath]y[/imath] coordinates of where right triangles meet the unit circle. So I don't understand how we get from those figures to Euler's formula. Then again I also don't understand why complex numbers are of the form [imath]a + bi[/imath] so that might have something to do with it. I have a hard time separating what identities are "emergent" from the previous mathematics and which pieces are "by definition" and why they are defined that way. I don't understand why we start talking about "rotations" as if it's obvious that's what's happening. I don't even know why [imath]e[/imath] is involved in any of this to begin with. Is [imath]e^{ix}[/imath] simply being written in the slightly rearranged complex form [imath]a + ib[/imath] where [imath]a = \cos(x)[/imath] and [imath]b = \sin(x)[/imath]? Is this to be interpreted as [imath]\cos(x)[/imath] being the [imath]x[/imath]-coordinate and somehow [imath]i \sin(x)[/imath] is the [imath]y[/imath] coordinate? | 291686 | How does [imath]e^{i x}[/imath] produce rotation around the imaginary unit circle?
Euler’s formula states that [imath]e^{i x} = \cos(x) + i \sin(x)[/imath]. I can see from the MacLaurin Expansion that this is indeed true; however, I don’t intuitively understand how raising [imath]e[/imath] to the power of [imath]ix[/imath] produces rotation. Can anyone give me an intuitive understanding? |
2663884 | Suppose [imath]x^2=1[/imath] for all [imath]x \in G[/imath]. Prove that [imath]G[/imath] is abelian.
Suppose that [imath]G[/imath] is a group s.t. [imath]x^2=1[/imath] for all [imath]x \in G[/imath]. Prove that [imath]G[/imath] is abelian. If we assume that [imath]x^2 = 1[/imath] for all [imath] x \in G[/imath], and suppose that [imath]a,b \in G[/imath] and if [imath]x=ab[/imath] we see that [imath]x^2 = abab =aabb = a^2 b^2[/imath]. Simplifying we see that [imath]ab=ba[/imath], and because we know that [imath]x= ab[/imath], we see that [imath]1=1[/imath]. Therefore it is abelian. [imath]\square[/imath] Is this sound? I'm struggling with this stuff because I think it's so simple and I'm concerned I'm overthinking it. I'd like to contest the mods decision that this is an exact duplicate. Though there are similar questions, as has been linked, it is not the same thing. If it is a problem with the abstract algebra tag, then delete it and allow it to be an active thread under other tags. | 598817 | Group with [imath]a=a^{-1}[/imath] for all [imath]a\in G[/imath] is abelian
Let [imath]G[/imath] be a group, and suppose that [imath]a= {a^{-1}}[/imath] for every [imath]a\in G[/imath]. Prove that [imath]G[/imath] is Abelian. I know that I need to prove that [imath]{a^{-1}}[/imath] = a and use right multiplication. I also know that I need to declare a, b [imath]\in[/imath] G and go from there, but I'm confused when I get to ([imath]{a^{-1}}[/imath][imath]{b^{-1}}[/imath]) = ab. |
2664036 | Given [imath]n[/imath], how to construct [imath]n[/imath] points
Given [imath]n[/imath], how to construct [imath]n[/imath] points on a plane s.t. any 3 points are not on a line, and the distance between any 2 points is an integer? Below is how to do it for [imath]n=6[/imath] here ABC is the 3,4,5 triangle. | 85309 | Points at integer distance
How many points can one can place in [imath]\mathbb{R}^n[/imath], with the requirement that no [imath]n+1[/imath] points lie in the same [imath]\mathbb{R}^{n-1}[/imath]-plane, and the euclidean distance between every two points is an integer? |
2663665 | Understanding the Riemann Surface of the [imath]\sqrt{z}[/imath]
I have been trying all the day to understand the Riemann Surface of the [imath]\sqrt z[/imath]. I Can not understand . Can anyone help me to understand how this picture is Riemann Surface of [imath]\sqrt z[/imath]. The other questioner did not ask this question. He already could visualize my question's answer. So I request you not to try to block this question. This is the only source from where I get help. Please read that question.Intuitively understanding Riemann surfaces Explanation in simple words will be highly appreciated. | 995550 | Intuitively understanding Riemann surfaces
I'm looking at the Riemann surface of [imath]f(z) = z^{1/2}[/imath] so the set [imath]\{(z,w) \in \mathbb{C}^2 : w^2 = z \}[/imath]. I understand that the point of the riemann surface is to understand this multi-valued function. Now I visualise this as taking two copies of the complex plane both with a slit in them (negative real axis say) and then I put them on top of each other, flip the top one and then sort of join them up along the slit and that's fine and I can see that that will form some surface where the local coordinates are given by projection onto the z or w axis. But here I am just dealing with the domain of the function [imath]f(z)[/imath]. My difficulty is when I think of the analogous situation with say [imath]\{(x,y) \in \mathbb{R}^2 : y^2 = x\}[/imath] this "surface" is just a line and I can see how given a point on this "line" we have an associated pair [imath](x,y)[/imath]. Now when I look at the Riemann surface described above all I see is the domain of the function [imath]f(z)[/imath] I can't see how each point of the surface is in anyway related to a pair [imath](z,w)[/imath]. I realise this is quite a vague question and I am struggling to put my frustration with this concept into words - I hope this makes sense! I've now added a (terrible) diagram which may help to illustrate my issue: Thanks |
2385348 | Evaluate the integral [imath]\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x dxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}[/imath]
Change the order of integration in: [imath]\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x }{\sqrt{4a^2x^2-(x^2+y^2)^2}}dxdy[/imath] and hence evaluate it. I changed the order of integration and got the limits [imath]y=0[/imath] to [imath]a[/imath] and [imath]x=a-\sqrt{a^2-y^2}[/imath] to [imath]a+\sqrt{a^2-y^2}[/imath]. I then tried changing to polar coordinates, but it is only making the integrand more complex. | 2917017 | Changing order of integration in Double Integral
Change the order of integration & evaluate it. [imath] \int_{0}^{2a} \int_{0}^{\sqrt{2ax-x^2}} \phi'(y) \frac{(x^2+y^2)x}{\sqrt{(4a^2x^2-(x^2+y^2)^2})} dxdy [/imath] I changed the order but could'nt solve it. Is there any 'tricky' variable substitution involved ? I couldn't solve it even with polar co-ordinates due to the presence of [imath] \phi'(y) [/imath] factor in the integrand? Is there a way to circumvent this? |
2663537 | Prove that x and y commute
Suppose G is a group with x and y as elements. Show that [imath](xy)^2 = x^2 y^2[/imath] if and only if x and y commute. My very basic thought is that we expand such that [imath]xxyy = xxyy[/imath], then multiply each side by [imath]x^{-1}[/imath] and [imath]y^{-1}[/imath], such that [imath]x^{-1} y^{-1} xxyy = xxyy x^{-1}[/imath] , and therefore [imath]xy=xy[/imath]. I realize that this looks like a disproportionate amount of work for such a simple step, but that is what past instruction has looked like and that is perhaps why I am confused. Moreover, "if and only if" clauses have always been tricky for me since I took Foundations of Math years ago, but if I remember correctly, the goal here should be to basically do the proof from right to left and then left to right, so to speak. Anyhow, I think that I am overthinking this problem. | 2106896 | Abelian group (Commutative group)
Prove that if in a group [imath](ab)^2= a^2 b^2[/imath] then the group is commutative. I am having a hard time doing this. Here is what I have so far: Proof: [imath]a^2 b^2= a^1 a^1 b^1 b^1[/imath] =[imath]aa^{-1}bb[/imath] =ebb Hence,[imath]aa^{-1}=e[/imath] I am stuck, I do not know if this is the right process in proving this |
2664916 | Divisibility of a Polynomial
Prove that for all [imath]n[/imath], [imath]121\mid n^2+3n+5[/imath]. I thought proving [imath]n^2 +3n+5\pmod{121}\equiv 0[/imath] had no solutions would be a good start, so I completed the square of [imath]n^2+3n+5([/imath], and arrived at [imath](n+62)^2 \pmod{121} \equiv 3718[/imath]. Replacing [imath]n+62[/imath] by [imath]x[/imath], I lastly arrived at [imath]x^2 \equiv 88\pmod{121}[/imath]. Now, Im stuck. Any hints would be appreciated. | 9431 | [imath]n^2 + 3n +5[/imath] is not divisible by [imath]121[/imath]
Question: Show that [imath]n^2 + 3n + 5[/imath] is not divisible by [imath]121[/imath], where [imath]n[/imath] is an integer. |
360888 | Other proofs that subgroups of [imath]A_5[/imath] have order at most 12
How can it be proved that any subgroup of [imath]A_5[/imath] has order at most 12? This is [Herstein, Problem 2.10.15], which also gives the hint that I can assume the result of the previous problem that [imath]A_5[/imath] has no normal subgroups [imath]N \ne (e),A_5[/imath]. This problem appears in an earlier section of the text than the Sylow theorems. There is a proof given at Subgroups of [imath]A_5[/imath] have order at most [imath]12[/imath]?, but it uses the Sylow theorems, and I wonder if a more elementary proof is available. So far, I can prove the following: For [imath]n \ge 3[/imath], the subgroup generated by the 3-cycles is [imath]A_n[/imath]; if a normal subgroup of [imath]A_n[/imath] contains even a single 3-cycle it must be all of [imath]A_n[/imath]; [imath]A_5[/imath] has no normal subgroups [imath]N \ne (e),A_5[/imath]. I showed the latter by repeatedly conjugating a given nontrivial element in [imath]A_5[/imath] by 3-cycles to eventually obtain elements whose product is a 3-cycle. | 2995244 | How do i prove that any subgroup of [imath]A_5[/imath] has order at most 12?
I know this question has been answered Other proofs that subgroups of [imath]A_5[/imath] have order at most 12 But i have difficulty in understanding that proof.The book says we can assume that [imath]A_5[/imath] has no normal subgroup.How to find the proof using this property ? EDIT- (I solved the previous question ( [imath]A_5[/imath] has no normal subgroup ) but i am not able to solve this problem ). I have already mentioned that my question similar to Other proofs that subgroups of [imath]A_5[/imath] have order at most 12 so please don't mark it as duplicate. The book which i am talking about is - Topics In Algerbra by Herstein. (2.10.15) I am thankful if some can explain the same proof ( provided in link ).I have difficulty in understanding the homomorphic part of that answer. |
2665052 | Number of Sylow subgroups (another)
if we look at [imath]S_3[/imath] so we have [imath]|G|=6=3*2[/imath] Now to find the number of Sylows subgroup we look at all the divisors of [imath]6[/imath] which are: [imath]1,2,3,6[/imath] [imath]n_2:[/imath] will be all the number that [imath]2[/imath] does not divide so it is [imath]1,3[/imath] [imath]n_3:[/imath] will be all the number that [imath]3[/imath] does not divide so it is [imath]1,2[/imath] Is that correct? | 566479 | Sylow subgroups of [imath]S_3[/imath] and [imath]S_4[/imath]
Find the three 2-Sylow subgroups of [imath]S_3[/imath] and find a 2-Sylow subgroup and a 3-Sylow subgroup of [imath]S_4.[/imath] I just learned Sylow' theorem at the moment and I don't know how to do these problems. I know a p-Sylow subgroup of a group [imath]G[/imath] is a subgroup of order [imath]p^k[/imath], where [imath]p^k[/imath] divides the order of [imath]G[/imath] but [imath]p^{k+1}[/imath] does not. [imath]|S_3| = 6 = 2\dot\ 3[/imath] and [imath]|S_4| = 24 = 2^3\dot\ 3[/imath]. How can I solve these types of problems? |
2665207 | Suppose [imath]\theta \neq \frac pq * \pi[/imath]. Show [imath]{\{e^{in\theta} : n \ \epsilon \ N}\}[/imath] is dense in [imath]S^1=\{x + iy: x^2 + y^2 = 1\} \subseteq C[/imath]
Essentially, I need to show that e^inx is dense in the complex unit circle for irrational angle x. I'm not sure how to go about this but I think I need to show that between any two points on the unit circle, we can find some e^inx. | 282102 | Prove that the orbit of an iterated rotation of 0 (by (A)(Pi), A irrational) around a circle centered at the origin is dense in the circle.
I think the title of the question says it all. I unfortunately did not seem to conclude anything. Some ideas I had: It is easy to show that (given [imath]T[/imath] is the rotation) [imath]\{T^n(\theta)\}[/imath] is a set of distinct points. Furthermore, given that the circle is a compact metric, it must have a limit point [imath]x[/imath]. By continuity of the rotation function, [imath]T^n(x)[/imath] is a limit as well since taking [imath]T[/imath] of every term yields the same sequence (with only the first term removed). By induction, we have infinitely many distinct limit points [imath]\{T^n(x)\}[/imath]. That's all I could come up with! It was also easy to show that the orbit is infinite. I still don't seem to be able to get close to the required result however. |
2666333 | Analysis proof (compactness)
Let [imath]A[/imath] [imath]\subseteq[/imath] [imath]\mathbb{R}^n[/imath] be compact, and let [imath]F: A \to \mathbb{R}^m[/imath], be a continuous mapping. Prove [imath]F(A)[/imath] is compact. | 111758 | If [imath]A[/imath] is compact, is then [imath]f(A)[/imath] compact?
I just got my exam back, and I still cannot understand this question: Given a continuous function [imath]f:A\subseteq\mathbb{R}\to\mathbb{R}[/imath], show that if [imath]A[/imath] is a compact set, then its image, [imath]f(A)[/imath], is also compact. I know that a set [imath]A\subseteq\mathbb{R}[/imath] is compact if every sequence in [imath]A[/imath] has a subsequence that converges to a limit that is also in [imath]A[/imath], and I know that a function [imath]f[/imath] is continuous on [imath]A[/imath] if for every [imath](x_n)\subseteq A[/imath] such that [imath]x_n\to c\in A[/imath], it follows that [imath]f(x_n)\to f(c)[/imath]. Therefore, all that I need to do is show that for every [imath](y_n)\subseteq f(A)[/imath], there is a subsequence [imath](y_{n_k})[/imath] such that [imath]y_{n_k}\to y\in f(A)[/imath]. Can I then make the assumption that for any sequence [imath](y_n)\subseteq f(A)[/imath], there is a sequence [imath](x_n)\subseteq A[/imath] such that [imath]y_n=f(x_n)[/imath]? If so, I could then continue by stating that since [imath]A[/imath] is compact, there is a subsequence [imath](x_{n_k})[/imath] such that [imath]x_{n_k}\to x\in A[/imath], and since [imath]f[/imath] is continuous, [imath]f(x_{n_k})\to f(x)[/imath]. I believe that this yields the required subsequence [imath](y_{n_k})[/imath] of [imath](y_n)[/imath] such that [imath]y_{n_k}=f(x_{n_k})\to f(x)=y\in f(A)[/imath]. What do you guys think? Is this a sound approach? Thanks in advance. |
2305116 | Inverting the process of composing functions (anti-composition)
The main idea is that if you have some well defined function [imath]f[/imath], is there a function [imath]f^{\frac{1}{2}}[/imath] such that [imath]f^{\frac{1}{2}}\left(f^{\frac{1}{2}}\left(x\right)\right) = f\left(x\right)[/imath]. For lack of a better name, I will call the process of finding [imath]f^{\frac{1}{2}}[/imath] anti-composition. My question is, are there any known techniques to anti-composing function? Is this even a problem that conventional mathematics has looked at? For example, the problem I am working on finding the anti-composition of [imath]f\left(x\right) = x^2+C[/imath]. I know that when [imath]C=0[/imath], [imath]f^{\frac{1}{2}}\left(x\right)=x^\sqrt{2}[/imath], because [imath]\left(x^\sqrt{2}\right)^\sqrt{2}=x^2[/imath]. Due to this rather simple form, I assume that [imath]f^{\frac{1}{2}}\left(x\right)=\left(x-h\right)^n+k[/imath] for some [imath]h[/imath],[imath]k[/imath], and [imath]n[/imath], which may depend on [imath]C[/imath]. However, the composition of this is [imath]f\left(x\right)=\left(\left(x-h\right)^n+k-h\right)^n+k[/imath]. I assume that [imath]n=\sqrt{2}[/imath], so [imath]x^{n^2} = x^2[/imath]. Also, I know that [imath]h=0[/imath], because any change in [imath]h[/imath] will translate the whole function, whereas [imath]x^2+C[/imath] definitely has no x-translation. This means [imath]f\left(x\right)=\left(x^\sqrt{2}+k\right)^\sqrt{2}+k[/imath]. However, in order for this to equal [imath]x^2+C[/imath], [imath]k[/imath] must equal [imath]0[/imath] and [imath]C[/imath] simultaneously, which is not possible unless [imath]C=0[/imath]. That's all I've worked on it so far, but I really have no idea what to try next. I have an idea to use the above properties and use a point to get other points of the function and then manipulate the inputs until it looks like a function, but I have no idea how to generalize it or fill in the intervals between the points produced. My attempt at can be found at https://www.desmos.com/calculator/uhc46vpkif, but I didn't find anything useful, except for the general shape of the function. Any help is appreciated! | 1884366 | Given any function [imath]g(x): \mathbb{R} \rightarrow \mathbb{R}[/imath], find a function [imath]f(x)[/imath] such that [imath]f(f(x)) = g(x)[/imath].
This is a problem I was casually discussing with friends: Given any function [imath]g(x): \mathbb{R} \rightarrow \mathbb{R}[/imath], find a function [imath]f(x): \mathbb{R} \rightarrow \mathbb{R}[/imath] such that [imath]f(f(x)) = g(x)[/imath]. Is it possible to find a solution for [imath]f(x)[/imath] or prove that there isn't a solution for [imath]f(x)[/imath] such that [imath]f(f(x)) = g(x)[/imath] for any given [imath]g(x)[/imath]? For example, if [imath]g(x) = x + 2[/imath], then [imath]f(x)[/imath] could easily be [imath]f(x) = x + 1[/imath]. But if [imath]g(x) = x^2 - 1[/imath], then it's not clear to see what [imath]f[/imath] can satisfy such solution. What approach might be promising? My first gut reaction would be some analytical method but can't think of a way still...Discussions are welcomed! |
1641405 | Given the set [imath]A=\{1,2,\dotsc,14\}[/imath], find all subsets of [imath]7[/imath] elements that sum to a multiple of [imath]7[/imath].
I would appreciate if somebody could help me with the following problem. Given the set [imath]A=\{1,2,\dotsc,14\}[/imath], calculate the number of distinct sets [imath]M \subset A[/imath] such that [imath]|M| = 7[/imath] and such that the sum of the elements of [imath]M[/imath] is a multiple of [imath]7[/imath]. | 1618420 | Find the number of all subsets of [imath]\{1, 2, \ldots,2015\}[/imath] with [imath]n[/imath] elements such that the sum of the elements in the subset is divisible by 5
The problem is as in the question title. Only one addition - [imath]n[/imath] is not divisible by [imath]5[/imath]. I already have a solution involving permutations, but recently I read about generating functions and I was wondering if this problem can be solved with them. A similar problem is the following: Find the number of all subsets of [imath]\{1, 2, \ldots, 2015\}[/imath] and the sum of elements in each subset is divisible by 5. The generating function used is [imath]{((1+x^0)(1+x^1)(1+x^2)(1+x^3)(1+x^4))}^{403}.[/imath] But this function cannot be used for my problem, since we need to count how many elements have been "used" to make the subset. Can anyone help me? |
1177164 | Finding Bijection between subgroups of group [imath]G = NU[/imath].
Let [imath]N \unlhd G[/imath] and [imath]U \le G[/imath] with [imath]G = NU[/imath]. Then there exists an inclusion-preserving bijection from the set of all subgroups [imath]X[/imath] with [imath]U \le X \le G[/imath] on the set of all [imath]U[/imath]-invariant subgroups [imath]Y[/imath] with [imath]U \cap N \le Y \le N[/imath]. A subgroup [imath]V \le G[/imath] is called [imath]U[/imath]-invariant for some subgroup [imath]U \le G[/imath], if [imath]V^g = V[/imath] for each [imath]g \in U[/imath]. It guess the map might be [imath]\varphi(X) = X \cap N[/imath], I could show injectivity by using the so called Dedekind identity (*) by which if [imath]X \cap N = Y \cap N[/imath] and factoring [imath]X = U(X\cap N), Y = U(Y\cap N)[/imath] we have [imath] X = U(X\cap N) = U(Y\cap N) = Y. [/imath] But how to show surjectivity? (*) The Dedekind identity as I know it goes like this, let [imath]G = UV[/imath] for two subgroups [imath]U,V \le G[/imath]. Then every subgroup [imath]H[/imath] with [imath]U \le H \le G[/imath] has a factorisation [imath]H = U(V\cap H)[/imath]. | 552718 | Bijection map from a set of subgroup to another set of subgroup under some condition.
Let [imath]G[/imath] be a finite group. Let [imath]N \trianglelefteq G[/imath] and [imath]U \leq G[/imath] such that [imath]G = NU[/imath]. Then there exists a bijection, preserving inclusion, from the set of subgroups [imath]X[/imath] satisfying [imath]U ≤ X ≤ G[/imath] to the set of [imath]U-[/imath]invariant subgroups [imath]Y[/imath] satisfying [imath]U \cap N ≤ Y ≤ N[/imath]. This is a problem from Kurzweil-Stellmacher book. I dont know how to do it. Thanks for any help. |
2667216 | What's the expected value for a poisoned bottle
There are bottles numbered from 1 to n. Every bottle k has [imath]\frac{1}{k}[/imath] chance of being poisoned. A random bottle is picked (evenly distributed) and is found to be poisoned. What's the expected value for the number of the bottle that was picked? Here's what I had in mind: A - The bottle that was picked is poisoned X - The number of the bottle that was picked We're searching for: [imath]\mathbb{E}[X|A]=\sum_{i=1}^n iP(X=i|A)[/imath] The problem is calculating the probability. I tried using Bayes' theorem but I can't calculate [imath]P(A)[/imath]. Any ideas? | 1063154 | Expected Value of Excellent Wine Age
I Ran into this question and I can't find the right way to approach it. We have [imath]n[/imath] different wine bootles numbered [imath]i=1...n[/imath]. the first is 1 year old, the second is 2 years old ... the [imath]n[/imath]'th bottle is [imath]n[/imath] years old. Each bottle is still good at probability of [imath]1/i[/imath]. We pick out a random bottle and it is good. what is the expected value of the age of the bottle? I'm really not sure what the random varable here is and how to aproach the question. I'd be grateful for a lead. Thanks, Yaron. |
2666941 | Deriving the sin and cos addition formulas using Euler's formula
I am trying to figure out the quick way to remember the addition formulas for [imath]\sin[/imath] and [imath]\cos[/imath] using Euler's formula: [imath]\begin{align} \sin(\alpha + \beta) &= \sin \alpha\;\cos\beta + \cos\alpha\;\sin\beta \\ \cos(\alpha + \beta) &= \cos \alpha\;\cos\beta - \sin\alpha\;\sin\beta \\ \sin(\alpha-\beta) &= \sin\alpha\;\cos\beta - \cos\alpha\;\sin\beta \\ \cos(\alpha-\beta) &= \cos\alpha\;\cos\beta + \sin\alpha\;\sin\beta \end{align}[/imath] I'm now convinced of why it is true that [imath]e^{ix} = \cos(x) + i\sin(x)[/imath] but I don't know how to use this to derive these four rules. | 608109 | Why is [imath]\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha[/imath]
Why is [imath]\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha[/imath]? How can we find the RHS if we don't know what it is? (instead of proving the identity itself) I could find a geometric solution in Wikipedia, but is there any solution that doesnt require drawing something? Edit: I now saw some nice proofs using [imath]e[/imath] and Euler's identity. I would appreciate anything NOT using them too, as a change. Out of curiosity, Is there a proof that says you actually "can't" prove the identity using only other simple 1-variable identities like [imath]\sin^2\alpha+\cos^2\alpha=1[/imath]? |
2668050 | Show that your collection of vectors is in fact linearly independent and spans [imath]M^{m\times n}(\mathbb R)[/imath]?
Determine a basis of [imath]M^{m\times n}(\mathbb R)[/imath].Show that your collection of vectors is in fact linearly independent and spans [imath]M^{m\times n}(\mathbb R)[/imath]. What is the dimension of [imath]M^{m\times n}(\mathbb R)[/imath]? Can someone give me a hint? | 1261057 | Dimension of [imath]m\times n[/imath] matrices
I'm a bit confused on the notion of the dimension of a matrix, say [imath]\mathbb{M}_{mn}[/imath]. I know how this applies to vector spaces but can't quite relate it to matrices. For example take this matrix: [imath] \left[ \begin{array}{ccc} a_{11}&\cdots&a_{1n}\\ \vdots & \vdots & \vdots \\ a_{m1}&\cdots&a_{mn} \end{array} \right] [/imath] Isn't the set of matrices [imath]\mathbb{M}_{mn}[/imath] with exactly one entry [imath]a_{ij}[/imath] set to [imath]1[/imath] on each matrix and [imath]m\times n[/imath] total matrices a basis for [imath]\mathbb{M}_{mn}[/imath]? In the sense that we can take some linear combination of them and add them up to create: [imath] a_{11} \cdot \left[ \begin{array}{cccc} 1&\cdots&\cdots&0\\ \vdots & \vdots & \vdots & \vdots \\ 0&\cdots&\cdots&0 \end{array} \right] + a_{12} \cdot \left[ \begin{array}{cccc} 0 &1 &\cdots&0\\ \vdots & \vdots &\vdots & \vdots \\ 0&0&\cdots&0 \end{array} \right] + \cdots = \left[ \begin{array}{ccc} a_{11}&\cdots&a_{1n}\\ \vdots & \vdots & \vdots \\ a_{m1}&\cdots&a_{mn} \end{array} \right] [/imath] So the dimension of all [imath]\mathbb{M}_{mn}[/imath] is [imath]m\times n[/imath]? |
2667019 | Proving that the unit square [imath]\{(x,y) \in \mathbb{R^{2}} | 0 is open in the metric topology on \mathbb{R^{2}}[/imath]
Proving that the unit square [imath]\{(x,y) \in \mathbb{R^{2}} | 0<x,y<1\}[/imath] is open in the metric topology on [imath]\mathbb{R^{2}}[/imath] What I know so far is that for a Metric space with (X,d) where X is a set and d is the metric, the following properties must hold: i) [imath]d(x,y) = 0[/imath] ii) [imath]d(x,y) = d(y,x)[/imath] iii) [imath]d(x,z)\leq d(x,y) + d(y,x)[/imath] (which is the triangular inequality) I know that the metric topology is the topology on X generated by the basis [imath]\{B_{\epsilon,d}(x): \epsilon > 0, x \in X\}[/imath] I know what all these things mean distinctly but I don't know how I would be able to start a proof like this. Can someone please help? | 2655632 | Show that [imath](a,b)\times (c,d)[/imath] is an open set in [imath]\mathbb{R}^2[/imath] with the Euclidian metric.
I need to show that [imath](a,b)\times (c,d)[/imath] is an open set in [imath]\mathbb{R}^2[/imath] with the Euclidian metric. I know that a set [imath]U[/imath] is open if for [imath]x\in U[/imath] there exists an open ball [imath]B_\epsilon(x)[/imath] such that [imath]B_\epsilon(x)\subset U[/imath] for some [imath]\epsilon >0[/imath]. The euclidian metric in [imath]\mathbb{R}^2[/imath] is given by [imath]d_2(x,y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}[/imath], so I need to show that for [imath]x\in (a,b)\times (c,d)[/imath] there exists [imath]B_\epsilon(x) = \{y\in\mathbb{R}^2:\sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} < \epsilon\}[/imath] such that [imath]B_\epsilon(x)\subset (a,b)\times (c,d)[/imath] for some [imath]\epsilon >0[/imath], but I have no idea how! Question: How do I show that [imath](a,b)\times (c,d)[/imath] is an open set in [imath]\mathbb{R}^2[/imath] with the Euclidian metric? |
2668244 | Unfair 3 Sided Dice
If you roll an unfair 3 sided dice where Side 1 has probability [imath]p_1[/imath] Side 2 has probability [imath]p_2[/imath] Side 3 has probability [imath]p_3[/imath] How can you find a formula that the 8th 3 will be rolled on the 10th roll? I see that this question was asked before, its a good start, but there was no answer on this. | 1172386 | Unfair 3 Sided Die Question
Say you have an unfair 3-sided die. Side 1 will be rolled with probability [imath]P_1.[/imath] Side 2 will be rolled with probability [imath]P_2.[/imath] Side 3 will be rolled with probability [imath]P_3.[/imath] Find a formula for the probability that the 8th 3 will be rolled on the 10th roll. Here is how I started: [imath] \sum_{i=8}^{10}P(T_8 = i) [/imath] Im just having trouble with that expansion? |
2668741 | Question from the British Mathematical Olympiad
Question: A [imath]5×5[/imath] square is divided into 25 unit squares. One of the numbers [imath]1,2,3,4,5[/imath] is inserted into each of the unit squares in such away that each row, each column and each of the two diagonals contains each of the five numbers once and only once. The sum of the numbers in the four squares immediately below the diagonal from top left to bottom right is called the score. Show that it is impossible for the score to be 20. What is the highest possible score? What have I done so far: The only possible way to reach 20 in the row immediately below the diagonal is by adding four 5s, i.e., [imath]5+5+5+5=20[/imath]. It follows that all the squares in those particular four squares immediately below the diagonal must be filled with 5. We show that this is impossible given the statement of the problem. Proof Assume for the sake of contradiction that such a score is possible. We call two squares adjacent if they share any common edge. It it given in the statement of the problem that the diagonals contain all the given integers exactly once. Hence it follows that the diagonal above our problem set of squares must contain exactly one five. It also follows that at least two squares containing 5s each must be adjacent to each other in a row or a column. Hence it is impossible to contain four 5s in the given set of squares. Going back, we conclude that it is impossible for the score to be 20. The part where I am stuck is finding the maximum possible score. I am guessing that all the integers in the four squares have to be distinct but I cannot seem to prove it. Thank you. | 668150 | A 5x5 board has 25 cells.The numbers [imath]\{1,2,3,4,5\}[/imath] are written on every row,every column and the two main diagonals.
A 5x5 board has [imath]25[/imath] cells. The numbers [imath]\{1,2,3,4,5\}[/imath] are written on every row,every column and the two main diagonals without any repetition. If the sum of the numbers of the diagonal below the leading diagonal be represented as [imath]A[/imath]. Show that [imath]A\ne 20[/imath]. Also find the maximum possible value of [imath]A[/imath]. I could prove that [imath]A\ne 20[/imath] by showing that each of the cells of the diagonal below the leading diagonal cannot be [imath]20[/imath] as if such arrangement exists, then the board cannot have [imath]5[/imath] five times, which actually should be by the given condition of the question. For the second part, I can imagine the maxiumum value to be [imath]17[/imath]. For which I proved that [imath]5[/imath] can occur only [imath]2[/imath] times in the four cells and probably [imath]4[/imath] cannot occur twice. But I got no formal proof. Please help! |
2668840 | [imath]a_{n+1}=\frac2{a_n+a_{n−1}}[/imath] . Prove that this sequence has a limit.
The sequense [imath]\{a_n\}[/imath] such that [imath]a_1>0, a_2>0[/imath] and [imath]a_{n+1}=\frac2{a_n+a_{n−1}}[/imath] for [imath]n≥2[/imath]. Prove that this sequence has a limit. I know I need to prove 1) the sequence is monotone; 2) the sequence is bounded. But sequense is not monotone. Any hints? | 2332271 | sequence of positive numbers satisfying [imath]a_{n+1}=\frac{2}{a_n+a_{n-1}}[/imath], prove it converges
Assume that [imath](a_n)[/imath] is a sequence of positive real numbers satisfying [imath]a_{n+1}=\frac{2}{a_n+a_{n-1}}[/imath] for [imath]n=2,3,\dots[/imath]. Prove that [imath](a_n)[/imath] is convergent and find the limit. I have no idea how to prove convergence. I tried to prove that [imath](a_n)[/imath] is bounded and monotone, but both trials failed. If convergence is proven, the limit is easy: if [imath]g=\lim a_n[/imath], then [imath]g\ge 0[/imath] because all [imath]a_n[/imath]'s are positive, the equivalent relation [imath]a_{n+1}(a_n+a_{n-1})=2[/imath] gives [imath]g(g+g)=2[/imath], so finally [imath]g=1[/imath]. |
2668858 | [imath]36 \leq 4(a^3+b^3+c^3+d^3) - (a^4+b^4+c^4+d^4)\leq4 8.[/imath]
Let [imath]a,b,c,d \in \Bbb R[/imath], [imath]a+b+c+d=6[/imath] and [imath]a^2+b^2+c^2+d^2=12[/imath]. Then [imath]36 \leq 4(a^3+b^3+c^3+d^3) - (a^4+b^4+c^4+d^4)\leq4 8.[/imath] I have found only two bounds: [imath]216 \geq a^3+b^3+c^3+d^3[/imath] and [imath]144 \geq a^4+b^4+c^4+d^4[/imath]. How to prove this inequality? | 767189 | [imath] a+b+c+d=6 , a^2+b^2+c^2+d^2=12[/imath] [imath]\implies[/imath] [imath] 36 \leq 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \leq48 [/imath]
Let [imath]a , b, c, d[/imath] be real numbers such that [imath] a+b+c+d=6 \\ a^2+b^2+c^2+d^2=12[/imath] How do we prove that [imath] 36 \space \leq\space 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \space\leq48\space[/imath] ? |
2668965 | Difference between fractions and rational numbers?
What is the difference between fractions and rational numbers? is [imath]\frac{\pi}{1}[/imath] a fraction? | 2348357 | Is [imath]\frac{1}{\sqrt{2}}[/imath] a fraction or not?
Is [imath]\frac{1}{\sqrt{2}}[/imath] a fraction or not? I just came across a Grade 8 student's class work where the teacher has given the above as an example to differentiate fractions from rational numbers. I think this is wrong fundamentally, as [imath]\sqrt2[/imath] is irrational while all fractions would be rational. Please advise. Thanks! |
2669216 | partition of an infinite set with copies
Here is my current concern. Say that a set [imath]E[/imath] has the half-property when there exist sets [imath]F[/imath] and [imath]G[/imath] such that [imath]E[/imath] is the disjoint union of [imath]F[/imath] and [imath]G[/imath], [imath]F[/imath] and [imath]G[/imath] being isomorphic. Trivially, no finite set has the half property, whereas (trivially as well) any countable set or the set of real numbers has the half-property. It seems to me that any infinite set has the half-property, but I feel that this could be related to some axiomatic system (involving the choice? I don't know). If [imath]E[/imath] is (up to isomorphism) a power set, then take [imath]E'[/imath] such that [imath]E=\mathcal P(E')[/imath], a in [imath]E'[/imath] and take [imath]F[/imath] as the subsets containing a and [imath]G[/imath] as the subsets without a; can this be helpful to solve the problem? Thanks in advance for your thoughts! | 1041731 | Prove that if [imath]A[/imath] is an infinite set then [imath]A \times 2[/imath] is equipotent to [imath]A[/imath]
I want to prove that if [imath]A[/imath] in an infinite set, then the cartesian product of [imath]A[/imath] with 2 (the set whose only elements are 0 and 1) is equipotent to [imath]A[/imath]. I'm allowed to use Zorn's Lemma, but I can't use anything about cardinal numbers or cardinal arithmetic (since we haven't sotten to that topic in the course). I read a proof of the fact that if [imath]a[/imath] is an infinite cardinal number, then [imath]a+a=a[/imath], which is something similar to what I want to prove. Any suggestions will be appreciated :) |
2669726 | Is it true that F2n is divided by Fn (Fibonacci)
How can I prove that [imath]F_{2n}[/imath] is divisible by [imath]F_n[/imath] in the Fibonacci sequence? | 2668257 | Prove that for the Fibonacci sequence [imath](F_n)[/imath], [imath]F_n[/imath] divides [imath]F_{2n}[/imath].
I have a seemingly simple question to solve by induction. The question says [imath]F_n[/imath] divides [imath]F_{2n}[/imath] in the Fibonacci sequence. My thoughts on this. First,since the n-th term of the Fibonacci sequence is given as the sum of the two previous terms, simple induction won't suffice. So I should try composite induction. That is, supposing that [imath]F_k[/imath] divides [imath]F_{2k}[/imath] for all [imath]k<n[/imath], I must prove [imath]F_{n}[/imath] divides [imath]F_{2n}[/imath]. Is this approach valid? Doesn't seem to work. If I write [imath]F_{2n}=F_{2n-1}+F_{2n-2}[/imath] then by the inductive hypothesis, [imath]F_{n-1}[/imath] divides [imath]F_{2n-2}[/imath] but this does not help me assert that [imath]F_n[/imath] divides [imath]F_{2n}[/imath]. Am I doing this wrong? |
2668919 | Can two different natural numbers raised to the same irrational power be integers?
Is it possible to find an [imath]\alpha\in\mathbb R[/imath], [imath]\alpha>0[/imath] so that for all [imath]n\in\mathbb N[/imath], [imath]n^\alpha\in\mathbb N[/imath]? A stronger(restricted) problem: Can [imath]2^\alpha[/imath] and [imath]3^\alpha[/imath] be simultaneously integers?? Here [imath]\alpha >0[/imath]. For the second one, I can only reduce to [imath]\cfrac{\log 3}{\log 2} = \cfrac{\log s}{\log q}[/imath] for integer solution [imath](s,q)[/imath], and then no idea..... For the first one, we have excluded all [imath]\alpha\in\mathbb Q[/imath](which is easy). I tried to use large enough [imath]n^\alpha\in\mathbb N[/imath] and to show [imath](n+1)^\alpha[/imath] fails to be an integer but I failed. I think there is more advanced number theoretic technique to be used. | 1643755 | [imath]n^a[/imath] integral for all integer [imath]n[/imath] implies [imath]a[/imath] integral
Let [imath]a>0[/imath] be a real number, such that for all integers [imath]n\geq 1[/imath]: [imath]n^a \in \mathbb N[/imath] Show that [imath]a[/imath] must be an integer. It's not difficult to show this when [imath]a[/imath] is a rational number: [imath]2^\frac{p}{q}[/imath] is irrational when the fraction is in lowest terms and [imath]q \neq 1[/imath]. When [imath]a[/imath] is irrational, then for all [imath]n\geq 1[/imath], there exists [imath]m_n\in \mathbb N^*[/imath], such that: [imath]a = \frac{\log m_n}{\log n}; \quad \text{$m_n$ is not a power of $n$}[/imath] I think considering [imath]n=2,3[/imath] is enough to show a contradiction, but I can't seem to find it. This is what I get: [imath]a = \frac{\log p}{\log 2} = \frac{\log q}{\log 3}[/imath] [imath]p = 2^{\log q/\log 3} [/imath] I think the RHS is irrational when [imath]q[/imath] is not a power of [imath]3[/imath], but I can't prove it. The closest thing I have to a solution is this answer on a similar question. But it uses an unproven conjecture, and I was hoping for a more elementary proof. |
2669456 | The extension of this derivative notation [imath]f^{(0)} = f, f^{(1)} = f', ...[/imath]
The extension of this derivative notation [imath]f^{(0)} = f, f^{(1)} = f', ...[/imath] Just a quick question on this notation, is [imath]f^{(-1)}[/imath] used for antiderivatives? | 1844977 | Could this be proper notation for an antiderivative? Does this notation even exist?
If we define [imath]f(x)[/imath] as some arbitrary function, then we can define [imath]f'(x)[/imath] or [imath]f^{(1)}(x)[/imath] as the first order and [imath]f''(x)[/imath] or [imath]f^{(2)}(x)[/imath] as the second order. My question is: Is there sure thing as a [imath]f^{(-1)}(x)[/imath] notation? Could it be an antiderivative or an integration? And no, I am not talking about [imath]f^{-1}(x)[/imath] that represents an inverse of [imath]f(x)[/imath]. |
2668912 | For [imath]f_n^2 - f_{n-2}^2 = f_{2n-1}[/imath]
We need to provide a visual counting proof for the identity above. At first, we count the number of pairs of [imath]n[/imath] tilings where at least one ends in a square. Then, conditioning on whether the first tiling ends in a square, we need prove the rest. If we consider the R.H.S of the identity, we see that it is a strip with a length of [imath](2n-2)[/imath] that we're tiling. How do we approach the problem then? How should one visualize the diagram, given that we need to find the tiling pairs where at least one ends in a square? N.B what I'm asking for is a visual proof, not by any other means. | 2419622 | How do I prove/disprove that [imath]F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}[/imath], where [imath]F_n[/imath] is the [imath]n[/imath]th Fibonacci number?
I was looking at the Fibonacci sequence: [imath]1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,\ldots, F_n : F_n + F_{n - 1} = F_{n + 1} \land F_0 = 0[/imath] Here I noticed something; [imath]\qquad F_n^2 + F_{n\pm1}^2 = F_{2n\pm1} : n\in \mathbb{N}[/imath] For example; [imath]\ \ F_3^2 + F_{3 + 1}^2 = F_3^2 + F_4^2 = 2^2 + 3^2 = F_{2\times3 + 1} = F_{6 + 1} = F_7 = 13[/imath] [imath]F_3^2 + F_{3 - 1} = F_3^2 + F_2^2 = 2^2 + 1^2 = F_{2\times3 - 1} = F_{6 - 1} = F_5 = 5[/imath] However, no matter how true this can seem, I cannot seem to prove/disprove it. My Attempt: [imath]\because (a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + 2ab : (a,b)\in \mathbb{Z}[/imath] [imath]\implies F_n^2 + F_{n\pm 1}^2 = (F_n + F_{n\pm 1})^2 - 2F_nF_{n\pm 1}[/imath] [imath]\implies F_n^2 + F_{n - 1}^2 = F_{n + 1}^2 - 2F_nF_{n - 1} : F_n + F_{n - 1} = F_{n + 1}[/imath] [imath]\implies F_n^2 + F_{n + 1}^2 = F_{n + 2}^2 - 2F_nF_{n + 1} : F_n + F_{n + 1} = F_{n + 2} \iff F_n + F_{n - 1} = F_{n + 1}[/imath] [imath]\implies F_{2n\pm 1} + 2F_nF_{n\pm 1} = F_{n + 2\lor1}^2 : F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1} \tag{+2 if +, +1 if -}[/imath] [imath]\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1} - F_{n\pm 1}^2}\sqrt{F_{2n\pm 1} - F_n^2}= F_{n + 2\lor1}[/imath] [imath]\implies F_{2n\pm 1} + 2\sqrt{(F_{2n\pm 1} - F_{n\pm 1}^2)(F_{2n\pm 1} - F_n^2)} = F_{n + 2\lor1}^2[/imath] [imath]\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1}(F_{2n\pm 1} - F_n^2 - F_{n\pm 1}^2) - (F_nF_{n\pm 1})^2} = F_{n + 2\lor1}^2[/imath]. [imath]\because F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}[/imath] [imath]\implies F_{2n\pm 1} + 2\sqrt{-(F_nF_{n\pm 1})^2} = F_{n + 2\lor1} \neq F_{2n\pm 1} + 2\sqrt{\big(-(F_nF_{n\pm 1})\big)^2}[/imath] [imath]\therefore F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} : i = \sqrt{-1}[/imath] I have looked over every step and cannot see where I am wrong. The only part that must be wrong is that: [imath]F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} \iff F_n^2 + F_{n\pm1}^2 = F_{2n\pm 1}[/imath] I can seperate [imath]F_{2n\pm 1}[/imath] as [imath]F_n^2 + F_{n\pm 1}^2[/imath] (if true) so I turn the equation into: [imath]F_n^2 + F_{n\pm 1}^2 + 2F_nF_{n\pm 1}i = F_{n + 2\lor1}[/imath] [imath]\therefore (F_n + F_{n\pm 1})\bigg(\frac{F_n}{F_{n\pm 1}} + \frac{F_{n\pm 1}}{F_n} + 2i\bigg) = F_{n + 2\lor1}[/imath] Now I know that the greater the value of [imath]n[/imath], one of the fractions in the above equation (depending on which sign is used in [imath]F_{n\pm 1}[/imath]) will get closer to the golden ratio [imath]φ = \frac{1 + \sqrt{5}}{2}[/imath] but [imath]i = \sqrt{-1}[/imath] is not a "real" number, although [imath]F_n[/imath] is. I then searched for other formulae for finding [imath]F_n[/imath] and came across the one below (being Binet's Formula): [imath]F_n = \bigg\{\frac{φ^n - ψ^n}{φ - ψ} : ψ = φ - \frac{\sqrt{20}}{2} = \frac{1 - \sqrt{5}}{2} \ \land \ φ - ψ = \sqrt{5}\bigg\}[/imath] But I do not know what how to turn that into [imath]F_{n\pm 1}[/imath]. Could you please show me where I went wrong (because I must have gone wrong somewhere), and also if you wish, could you please show me how to turn the above fraction into [imath]F_{n\pm 1}[/imath] (if possible). I only used the [imath](a + b)^2[/imath] expansion rule because my skill level in mathematics is not particularly high compared to most of the users on this site (especially when logging into the MSE and finding fancy symbols like [imath]\nabla[/imath] and [imath]\partial[/imath] and [imath]\sum[/imath] which all look Greek to me, however the only "fancy" thing I know is that [imath]i = i^5 = \cdots = \sqrt{-1}[/imath]) so if you used any other techniques in trying to prove this formula, could you please show me step by step? Much appreciated. Thank you in advance. Edit: I believe this question is a duplicate of this other question [imath]\longrightarrow[/imath] Prove [imath]f_{n+1}^2+f_n^2=f_{2n+1}[/imath] and I can go here to serve as a little hint [imath]\longrightarrow[/imath] Fibonacci Square Indentity |
2670032 | Do there exist any subsets of [imath]\mathbb{R}[/imath] with positive measure but not of size continuum?
As title goes, without assuming continuum hypothesis, is there a subset of [imath]\mathbb{R}[/imath] with positive measure but not being continuum? That is, Does there exist [imath]$A\subseteq \mathbb{R}$[/imath] such that [imath]$\mu(A)>0$[/imath] and [imath]$\aleph_0<|A|<2^{\aleph_0}$[/imath]? I believe that maybe in some system without assuming continuum hypothesis, there exists. And by the regularity of Lebesgue measure, it suffices to deal with compact set. | 1294765 | Let [imath]X\subset \mathbb{R}[/imath] Lebesgue measurable, [imath]|X|<|\mathbb{R}|[/imath], is it true that [imath]X[/imath] is null?
Let [imath]X\subset \mathbb{R}[/imath] Lebesgue measurable, [imath]|X|<2^{\aleph_0}[/imath], is it true that [imath]X[/imath] is null? Of course I am not assuming the Continuum Hypothesis. EDIT: It might be helpful to know that all Borel measurable sets have cardinality either [imath]\aleph_0[/imath] or [imath]2^{\aleph_0}[/imath]. Then a measurable set of cardinality strictly between those two must be Lebesgue but not Borel measurable. |
2668295 | Induction proof that the number of edges of a complete graph with [imath]x[/imath] vertices is always [imath]\frac{x^2-x}{2}[/imath]
I am trying to prove that the number of edges of a complete graph with [imath]x[/imath] vertices is always [imath]\frac{x^2-x}{2}[/imath]. I'm having trouble figuring out this proof by induction. I've only just started doing induction proofs and I'm not entirely sure where to begin. Thank you very much! | 1054448 | number of edges induction proof
Proof by induction that the complete graph [imath]K_{n}[/imath] has [imath]n(n-1)/2[/imath] edges. I know how to do the induction step I'm just a little confused on what the left side of my equation should be. [imath]E = n(n-1)/2[/imath] It's been a while since I've done induction. I just need help determining both sides of the equation. |
2670099 | Proving that [imath]x^3-9[/imath] is irreducible in [imath]\mathbb{F}_{31}[x][/imath]
I am attempting to solve part (b) of problem 4.2 in Chapter 12 of Artin's Algebra textbook. I have already solved part (a). Prove that the following polynomials are irreducible: (a) [imath]x^2+1[/imath] in [imath]\mathbb{F}_7[x][/imath] (b) [imath]x^3-9[/imath] in [imath]\mathbb{F}_{31}[x][/imath] My attempt Assume [imath]x^3-9[/imath] is reducible. It is easy to see that it is not a unit, so it must have a proper divisor. So we can write [imath]x^3-9=a(x)b(x)[/imath] for some [imath]a,b\in \mathbb{F}_{31}[x][/imath] where [imath]a,b[/imath] are not units, hence are not constant. One of them has degree 2 and the other degree 1. WLOG assume [imath]a[/imath] has degree 2 and [imath]b[/imath] has degree 1. Then we can write [imath]a(x)=a_0+a_1x+a_2x^2[/imath] and [imath]b(x)=b_0+b_1x[/imath]. Where [imath]a_2,b_1\neq0.[/imath] We obtain the equations [imath]a_0b_0=-9[/imath] [imath]a_0b_1+a_1b_0=0[/imath] [imath]a_1b_1+a_2b_0=0[/imath] [imath]a_2b_1=1[/imath] Now, I would like to obtain a contradiction by playing with these four equations, but I am not sure how to do this. Please help me. I would prefer a solution which deduces a contradiction from the four equations rather than a solution using some other method. I am aware that this math problem has already been asked on stackexchange, but I do not believe that my question is a duplicate since I desire to approach the problem from a different angle. | 82355 | Is [imath]x^3-9[/imath] irreducible over the integers mod 31?
Is [imath]x^3-9[/imath] irreducible over the integers mod 31? I know that one method is to check whether it has any roots or not, but that turns out to be very tedious in this case. So, is there any other simpler method to find it out? |
2670658 | Prove that for any [imath]a, b, c>0[/imath] we have:
Prove that for any [imath]a,b,c>0[/imath] we have: [imath]\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\frac{(a+b)^2}{b+c}+\frac{(b+c)^2}{c+a}+\frac{(c+a)^2}{a+b}[/imath] Can someone help me? I tried Radon inequality for [imath]p=1[/imath], and I obtained [imath]2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\frac{(a+b)^2}{b+c}+\frac{(b+c)^2}{c+a}+\frac{(c+a)^2}{a+b}[/imath] but that's the far I could go. | 2614390 | I have a inequality, I don't know where to start
Show that for [imath]x,y,z > 0[/imath] the inequality is true: [imath]\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}[/imath] I have tried Holder, but i had no luck. Please give me a hint of how to start. |
2671153 | Show that if [imath]\lim(x_n - x_{n-2}) = 0[/imath] then [imath]\lim(\frac{x_n - x_{n-1}}{n}) = 0[/imath]
I know that [imath]\frac{1}{n}[/imath] converges to zero. So if I can show that [imath]x_n - x_{n-1}[/imath] is bounded then by the squeeze theorem, I would have the desired result. I have tried to show that [imath]x_n - x_{n-1}[/imath] is bounded by seeing [imath](x_n - x_{n-2}[/imath] as a Cauchy sequence but I am not getting anywhere. | 418035 | Show that: [imath]\lim \limits_{n\to\infty}\frac{x_n-x_{n-1}}{n}=0 [/imath]
Here is an exercise: Suppose that [imath]\{x_n\}[/imath] is a sequence such that [imath]\lim \limits_{n\to\infty}(x_n-x_{n-2})=0[/imath]. Show that: [imath]\lim \limits_{n\to\infty}\frac{x_n-x_{n-1}}{n}=0 [/imath] Thanks. |
1839913 | Axiom of Choice: Where does my argument for proving the axiom of choice fail? Help me understand why this is an axiom, and not a theorem.
In terms of purely set theory, the axiom of choice says that for any set [imath]A[/imath], its power set (with empty set removed) has a choice function, i.e. there exists a function [imath]f\colon \mathcal{P}^*(A)\rightarrow A[/imath] such that for any subset [imath]S[/imath] of [imath]A[/imath], [imath]f(S)\in S.[/imath] Is this correct? My question then is about proving this fact, so that we do not need to put it as an axiom. Now as per the research done on this single object- Axiom of Choice, I believe here that there should be some falsity in my argument. I do not find the mistake. For any [imath]S\in \mathcal{P}^*(A)[/imath], since [imath]S\neq \emptyset[/imath], [imath]\exists s\in S[/imath]. Define [imath]f(S)=s[/imath]. Then [imath]f[/imath] is a choice function. This was showing me that the axiom of choice is proved, but then why it had been put as an axiom? For example, in this book, the author asserts that It is a metatheorem of mathematical logic that it is impossible to specify the function that assigns to each non-empty subset of [imath]\mathbb{R}[/imath], an element of itself. There are several notes and books on axiom of choice, but here I am trying to understand through doing some argument for some problem, where problem actually arises. | 2909445 | What's wrong with my argument of choice function from infinite family of nonempty sets?
Let [imath](A_i \mid i \in I)[/imath] be a family of non-empty set. Then [imath]A_i \neq \emptyset[/imath] for all [imath]i \in I[/imath], thus [imath]\exists a_i \in A_i[/imath] for all [imath]i \in I[/imath]. We define [imath]f:I \to \bigcup A_i[/imath] by [imath]f(i)=a_i[/imath]. From reading other posts in MSE, I'm sure that my argument is wrong without Axiom of Choice, but I can not understand why it's wrong. The usual answer is that you can not appeal to Existential Instantiation infinite many times. However, I'm unable to understand this explanation under set theory. Maybe the problem is related to underlying logic, with which I'm not familiar. Please explain me why you can not appeal to Existential Instantiation infinite many times? |
2670509 | Comparing Topologies of Metric Spaces
How can I generally show that two metric spaces induce the same topology, for instance, in my textbook it says metric space [imath]d(x,y[/imath]) and [imath]\sqrt{d(x,y)}[/imath] induce the same topology. How do I show this? Much help would be appreciated. ok so very generally I wrote up some solution, and it goes: Let [imath]d^{'}=\sqrt{d}[/imath], then we want to show that [imath]B_d(x,\epsilon) \subseteq B_{\sqrt{d}}(x,\epsilon)[/imath]. Let [imath]y \in B_d(x,\epsilon)[/imath],but [imath]d^{'}[/imath] is the square root metric of [imath]d[/imath], so we have [imath]d^{'} \le d[/imath], hence [imath]d^{'}(x,y)\le d(x,y)<\epsilon \implies y \in B_{\sqrt{d}}(x,\epsilon)[/imath], therefore [imath]B_d(x,\epsilon) \subseteq B_{\sqrt{d}}(x,\epsilon)[/imath]. is this the correct way? | 1435027 | Showing that two topologies give rise to the same topology
Suppose that for a metric space [imath](X,d)[/imath] we have another metric [imath]p[/imath] forming the metric space [imath](X,p)[/imath]. Both of these will give rise to two topologies [imath]\tau_{1}[/imath] and [imath]\tau_{2}[/imath]. I want to show that [imath]\tau_{1}[/imath] and [imath]\tau_{2}[/imath] give rise to the same topology. How could one do it generally in this case? I was thinking of taking an open set [imath]U_{d}[/imath] in the metric space [imath](X,d)[/imath] and showing that this is open in [imath](X,p)[/imath] and conversely, taking an open set [imath]U_{p}[/imath] in [imath](X,p)[/imath] and showing its open in [imath](X,d)[/imath]. Would this be the idea to prove that the topologies are equivalent? (I'm sure this can be used with more generality for the case that we have two topological spaces rather than two metric spaces). If not, what would be the correct way? Thanks for the help. |
2090656 | Prove that the interior of the boundary of an open or closed set is empty.
Let [imath]X[/imath] be a topological space, [imath]A\subset X[/imath] and [imath]B=Fr(A)[/imath]. I have two questions. How do you prove that if [imath]A[/imath] is open or closed, then [imath]B^\circ=\emptyset[/imath]? Why can this not be proven for a set that is both open and closed? | 1142892 | Prove that the Interior of the Boundary is Empty
Suppose X is a Metric Space Let S [imath]\subset X[/imath] Prove that if S is Closed then, the Interior of the Boundary of S is Empty Totally stuck on how to solve this. |
2671673 | Sum of two matrices equals product; prove that the matrices commute
Let [imath]A,B[/imath] be square real matrices of the same size such that [imath]A+B=AB[/imath]. Show that [imath]AB=BA[/imath]. I have got the result for invertible matrices, but what about non-invertible ones? | 2242629 | If [imath]A+B=AB[/imath] then [imath]AB=BA[/imath]
I was doing the problem [imath] A+B=AB\implies AB=BA. [/imath] [imath]AB=BA[/imath] means they're invertible, but I can't figure out how to show that [imath]A+B=AB[/imath] implies invertibility. |
2671686 | P(Z+) is uncountable (Power set of Positive Integers)
I have been trying to prove that the Power set of Positive Integers is uncountable using : Is [imath]i[/imath] an element of the [imath]i[/imath]-th subset But I can not seem to prove it. | 1427032 | Proving Power Set of [imath]\mathbb N[/imath] is Uncountable
I'm getting hung up on a proof that I remember being fairly easy... Showing that the power set of [imath]\mathbb N[/imath] is uncountable. Supposing it's countable, say [imath]A=\{A_1,...\}[/imath], we choose a set [imath]B[/imath] composed of elements [imath]b_i[/imath], where [imath]b_i\notin A[/imath]. How to we guarantee that [imath]B[/imath] isn't the naturals itself, an element of its own power set? Thanks! |
2670685 | Prove that [imath] E \cap (F + (E \cap G)) = (E \cap F) + (E \cap G)[/imath]
I have three vector subspaces of a vector space [imath]H[/imath] , I need to prove the following : [imath] E \cap (F + (E \cap G)) = (E \cap F) + (E \cap G)[/imath] I've already proved that [imath]E + ( F \cap G) \subset (E+F) \cap (E + G)[/imath] , The first thing I thought of is to prove the double inclusion, i.e : [imath] E \cap (F + (E \cap G)) \subset (E \cap F) + (E \cap G)[/imath] and [imath] (E \cap F) + (E \cap G) \subset E \cap (F + (E \cap G)) [/imath] But I have no idea how to do that . | 1131397 | Associative rule of subspaces and vector spaces: [imath](U+V) \cap W = U + (V \cap W)[/imath]
If [imath]U, V, W[/imath] are subspaces of a vector space [imath]X[/imath] and if [imath]U[/imath] is a subspaces of [imath]W[/imath], then [imath](U+V) \cap W = U + (V \cap W)[/imath] Do we need also that [imath]V[/imath] is a subspace of [imath]W[/imath]? I can't see how to prove this otherwise. This is typed up by my professor so it's definitely possible there's a typo. |
2672126 | Finding value of [imath]\int^{\pi}_{0}\frac{\sin^2(10x)}{\sin^2 (x)}dx[/imath]
Finding value of [imath]\int^{\pi}_{0}\frac{\sin^2(10x)}{\sin^2 (x)}dx[/imath] Try: Using [imath]\displaystyle \sin(10x)=\frac{e^{i(10x)}-e^{-i(10x)}}{2i}[/imath] and [imath]\displaystyle \sin (x)=\frac{e^{i(x)}-e^{-i(x)}}{2i}[/imath] So [imath]\frac{\sin(10x)}{\sin x}=e^{-i(9x)}\cdot \frac{e^{i(20x)}-1}{e^{i(2x)}-1}=e^{-i(9x)}\sum^{9}_{n=1}e^{i(2nx)}[/imath] So [imath]\frac{\sin^2(10x)}{\sin^2(x)}=e^{-i(18x)}\bigg[\sum^{9}_{n=1}e^{i(2nx)}\bigg]^2[/imath] So [imath]\int^{\pi}_{0}\frac{\sin^2(10x)}{\sin^2(x)}=\frac{1}{2}\int^{\pi}_{-\pi}\frac{\sin^2(10x)}{\sin^2(x)}dx= \frac{1}{2}\int^{\pi}_{-\pi}e^{-i(18x)}\bigg[\sum^{9}_{n=1}e^{i(2nx)}\bigg]^2[/imath] Could some help me to proceed further, Thanks | 2008044 | How do I prove [imath]\int_0^\pi\frac{(\sin nx)^2}{(\sin x)^2}dx = n\pi[/imath]?
I need to show that the following integral: [imath]\int_0^\pi\frac{(\sin nx)^2}{(\sin x)^2}dx[/imath] = [imath]n\pi[/imath], for all natural numbers [imath]n[/imath]. What is the method to evaluate the above? I initially thought of the Leibniz rule, but that wouldn't work as the result only holds for positive integers, meaning the integral would not be a continuous function in [imath]n[/imath]. Is there some other way to prove it? |
2672573 | Integral inequality [imath]\int_{0}^{1}f^2\left(x\right)dx\geq\frac{3}{13}[/imath]
Let [imath]f: \left[0,\infty\right)[/imath] [imath]\rightarrow\left[0,\infty\right)[/imath] be a continuous function so that [imath]f(f(x))=x^2, \forall x\in \left[0,\infty\right).[/imath] Show that: [imath]\int_{0}^{1}f^2\left(x\right)dx\geq\frac{3}{13} [/imath] I noticed [imath]f[/imath] is injective, so it is strictly monotone. Also, [imath]f^2(x)=f(x^2), \forall x \geq 0[/imath], but by substituting [imath]x^2[/imath] I'm getting nowhere. Do you have any ideas? | 2539976 | Prove that:[imath]f(f(x)) = x^2 \implies \int_{0}^{1}{(f(x))^2dx} \geq \frac{3}{13}[/imath]
Let [imath]f: [0,\infty) \to [0,\infty)[/imath] be a continuous function such that [imath]f(f(x)) = x^2, \forall x \in [0,\infty)[/imath]. Prove that [imath]\displaystyle{\int_{0}^{1}{(f(x))^2dx} \geq \frac{3}{13}}[/imath]. All I know about this function is that [imath]f[/imath] is bijective, it is strictly increasing*, [imath]f(0) = 0, f(1) = 1, f(x^2) = (f(x))^2, \forall x \in [0, \infty)[/imath] and [imath]f(x) \leq x, \forall x \in [0, 1][/imath]**. With all these, I am not able to show that [imath]\displaystyle{\int_{0}^{1}{(f(x))^2dx} \geq \frac{3}{13}}[/imath]. *Suppose that [imath]f[/imath] is strictly decreasing. Then, [imath]\forall x \in (0,1), x^2 < x \implies f(x^2) > f(x) \iff (f(x))^2 > f(x) \iff f(x) > 1[/imath], which is false because, if we substitute [imath]x[/imath] with [imath]0[/imath] and with [imath]1[/imath] in [imath]f(x^2) = (f(x))^2[/imath] we get that [imath]f(0) \in \{0,1\}[/imath] and [imath]f(1) \in \{0,1\}[/imath]. So [imath]f[/imath] is strictly increasing. **Suppose that there exists [imath]x_0 \in [0, 1][/imath] such that [imath]f(x_0) > x_0[/imath]. Then, [imath]x_0^2 = f(f(x_0)) > f(x_0) > x_0[/imath], which is false. Then [imath]f(x) \leq x, \forall x \in [0,1][/imath]. Edit: I have come up with an idea to use Riemann sums, but I reach a point where I cannot continue. Let [imath]\epsilon < 1[/imath]. Then [imath]f(\epsilon) = x_1[/imath] and [imath]f(x_1) = \epsilon^2[/imath]. And now [imath](f(\epsilon))^2 = f(\epsilon^2) = x_2[/imath] and so on. Now we will use the Riemann sum: We will take the partition [imath]\Delta = (1 > \epsilon > \epsilon ^2 > ... \epsilon ^{2^n} >0 ) [/imath]and the intermediate points will be the left margin of each interval. Then we have: [imath]\displaystyle{\int_{0}^{1}{(f(x))^2 dx}} = \displaystyle{ \lim_{\epsilon \to 1}{\lim_{n \to \infty}{\sum_{k = 0}^{n}{(\epsilon^{2^k} - \epsilon^{2^{k+1}})\epsilon^{2^{k+1}}}}}}[/imath] I do not know how to compute this. |
2672140 | Stability of injective, surjective and bijective functions under composition
Let [imath]f : A \to B, \quad g: B \to C, [/imath] be two functions. Show the following: 1) If [imath]f[/imath] and [imath]g[/imath] are surjective then [imath]g \circ f[/imath] is surjective 2) If [imath]f[/imath] and [imath]g[/imath] are bijective then [imath]g \circ f[/imath] is bijective. 3) If [imath]g\circ f[/imath] is injective then [imath]f[/imath] is injective. 4) If [imath]g \circ f[/imath] is surjective then [imath]g[/imath] is surjective. 5) If [imath]g \circ f[/imath] is bijective then [imath]f[/imath] is injective and [imath]g[/imath] is surjective. How can I prove the following statements ? I assume that I can say that 2) is bijective If I can prove 1) (I know [imath]g\circ f[/imath] is injective) A little bit help would be awesome thank you... | 477453 | Proof of properties of injective and surjective functions.
I'd like to see if these proofs are correct/have them critiqued. Let [imath]g: A \to B[/imath] and [imath]f: B \to C[/imath] be functions. Then: (a) If [imath]g[/imath] and [imath]f[/imath] are one-to-one, then [imath]f \circ g[/imath] is one-to-one. (b) If [imath]g[/imath] and [imath]f[/imath] are onto, then [imath]f \circ g[/imath] is onto. (c) If [imath]f \circ g[/imath] is one-to-one, then [imath]f[/imath] is one-to-one? (d) If [imath]f \circ g[/imath] is one-to-one, then [imath]g[/imath] is one-to-one? (e) If [imath]f \circ g[/imath] is onto, then [imath]f[/imath] is onto? (f) If [imath]f \circ g[/imath] is onto, then [imath]g[/imath] is onto? (a) Let [imath]a,b \in A[/imath]. If [imath](f \circ g)(a)=(f \circ g)(b)[/imath], then [imath]f(g(a))=f(g(b))[/imath]. Since [imath]f[/imath] is one-to-one, we know that [imath]g(a)=g(b)[/imath]. And, since [imath]g[/imath] is one-to-one is must be that [imath]a=b[/imath]. Hence [imath]f \circ g[/imath] is one-to-one. (b) Since [imath]f[/imath] is surjective we know that for all [imath]c \in C[/imath] there is a [imath]b\in B[/imath] such that [imath]f(b)=c[/imath]. Since [imath]g[/imath] is surjective, there is a [imath]a \in A[/imath] such that [imath]g(a)=b[/imath]. Hence, for all [imath]c \in C[/imath] there is an [imath]a \in A[/imath] such that [imath](f \circ g)(a)=f(g(a))=f(b)=c[/imath]. (c) This is false. Let [imath]A=\{1\}[/imath], [imath]B=\{1,2\}[/imath], and [imath]C=\{1\}[/imath]. Define [imath]g(1)=1[/imath] and [imath]f(1)=f(2)=1[/imath]. Then [imath]f \circ g[/imath] is injective, but [imath]f[/imath] is not injective. (d) This is true. Let [imath]a,b \in A[/imath] where [imath]a \neq b[/imath]. If [imath](f\circ g)(a)\neq(f\circ g)(b)[/imath], then [imath]f(g(a))\neq f(g(b))[/imath]. Hence [imath]g(a) \neq g(b)[/imath]. (e) Since [imath]f \circ g[/imath] is surjective, then for every [imath] c \in C[/imath] there is an [imath]a \in A[/imath] such that [imath]f(g(a))=c[/imath]. Let [imath]b = g(a)[/imath]. Then [imath]b \in B[/imath] and [imath]f(b)=g(f(a))=c[/imath]. Thus [imath]f(b)=c[/imath]. Hence [imath]f[/imath] is surjective. (f) Using the same set up as part (c), we have that [imath]g[/imath] is not surjective since there is nothing in [imath]A[/imath] such that [imath]g[/imath] will go to 1. |
2672151 | Perpendicular vectors of equal length and orthogonal matrices
I have an [imath]n \times n[/imath] orthogonal matrix [imath]A[/imath] and vectors [imath]x,y \in \mathbb{R}^n[/imath] and [imath]\theta \in (0,\pi)[/imath] and this satisfies [imath]Ax = \cos\theta x - \sin\theta y[/imath], and [imath]Ay = \sin\theta x + \cos\theta y[/imath] I must now proof that [imath]x[/imath] and [imath]y[/imath] are perpendicular vectors of equal length. I know this means [imath]|x|^2 = |y|^2[/imath] and [imath]x\cdot y=0[/imath], and I also think I should use the preserved Euclidean inner product. How to continue? | 2671928 | Show that [imath]|x|^2 = |y|^2[/imath] with [imath]A[/imath] orthogonal and [imath]Ax = \cos (θ)x − \sin (θ)y[/imath], [imath]Ay = \sin (θ)x + \cos (θ) y[/imath]
Let [imath]A[/imath] be an orthogonal [imath]n × n[/imath] matrix and suppose that [imath]x, y \in \mathbb R^n[/imath] and [imath]\theta \in (0 , \pi)[/imath] satisfy [imath]Ax = \cos (θ)x − \sin (θ)y,\\ Ay = \sin (θ)x + \cos (θ) y.[/imath] Prove that [imath]x[/imath] and [imath]y[/imath] must be perpendicular vectors of equal length, i.e. show that [imath]|x|^2 = |y|^2[/imath] and [imath]x · y = 0[/imath]. We know that because [imath]A[/imath] is orthogonal then[imath]\langle x,x \rangle =\langle Ax,Ax \rangle, \quad \langle x,y \rangle=\langle Ax,Ay \rangle, \quad \langle y,y \rangle=\langle Ay,Ay \rangle.[/imath] I was already struggling with showing that [imath]|x|^2 = |y|^2[/imath]. So far I just wrote out that [imath]|x|^2= \sum_{i=1}^n (Ax_i)^2= \sum_{i=1}^n \cos^2(\theta)x_i^2-2\cos(\theta)\sin(\theta)x_iy_i+\sin^2(\theta)y_i^2,\\ |y|^2= \sum_{i=1}^n (Ay_i)^2= \sum_{i=1}^n \cos^2(\theta)y_i^2+2\cos(\theta)\sin(\theta)x_iy_i+\sin^2(\theta)x_i^2.[/imath] I don't really see how I could conclude from this that they are equal. |
2673117 | Prove [imath]\left|\begin{smallmatrix} \sin^2x&\cot x&1\\ \sin^2y&\cot y&1\\ \sin^2z&\cot z&1 \end{smallmatrix}\right|=0[/imath] if [imath]x+y+z=\pi[/imath]
If [imath]x, y, z[/imath] are the angles of [imath]\Delta ABC[/imath], then evaluate [imath] \Delta=\begin{vmatrix} \sin^2x&\cot x&1\\ \sin^2y&\cot y&1\\ \sin^2z&\cot z&1 \end{vmatrix}=\quad\color{red}{?} [/imath] My Attempt [imath] \Delta=\frac{1}{\sin x\sin y\sin z}\begin{vmatrix} \sin^3x&\cos x&\sin x\\ \sin^3y&\cos y&\sin y\\ \sin^3z&\cos z&\sin z \end{vmatrix}= [/imath] How do I proceed further and prove that the determinant is zero ? | 2665985 | If [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are the angles of a triangle then find the value of [imath]\Delta[/imath]
I'll state the question from my book below: If [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are the angles of a triangle, then find the determinant value of [imath]\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.[/imath] Here's how I tried solving the problem: [imath]\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}[/imath] [imath]R_2 \to R_2 - R_1[/imath] [imath]R_3 \to R_3 -R_1[/imath] [imath]= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}[/imath] Expanding the determinant along [imath]C_3[/imath] [imath]\begin{align} &= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A) \\ &= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A) \\ &= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C} \\ &= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right] \\ &= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right] \\ &= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C} \end{align}[/imath] I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as [imath]0[/imath]. I don't have any clue about getting the answer. Any help would be appreciated. |
2673237 | Prove that there exists a natural number m such that each eigenvalue of A is an m-th root of unity
Let[imath] A[/imath] and [imath]X[/imath] be invertible complex matrices such that [imath]XAX^{−1} = A^2[/imath] . Prove that there exists a natural number m such that each eigenvalue of A is an m-th root of unity. i was taking the set [imath]\{ \lambda , \lambda^1 , \lambda^{2} , \lambda^{3}, \dots, \lambda^{m} \}[/imath] [imath]\lambda[/imath] is the eigenvalue of [imath] A[/imath] I don't know how to show that there exists a natural number m such that each eigenvalue of A is an m-th root of unity. Pliz help me Thanks in advance | 464064 | A is similar to [imath]A^k[/imath], then each eigenvalue of [imath]A[/imath] is a root of unity
Let [imath]A \in \mathbb{C}(n,n)[/imath] and [imath]k \geq 2[/imath] be an integer such that [imath]A \sim A^k[/imath]. Show that if [imath]A[/imath] is non-singular then each eigenvalue of [imath]A[/imath] is a root of unity. Attempt: Since [imath]A \sim A^k[/imath], [imath]PA = A^kP[/imath] where [imath]P[/imath] is an invertible matrix. Since [imath]A[/imath] is invertible, [imath]0[/imath] cannot be an eigenvalue of [imath]A[/imath]. Suppose [imath]Av = \lambda v \quad v \neq 0[/imath] then [imath]PAv = \lambda Pv[/imath] [imath]\therefore A^k(Pv) = \lambda (Pv) [/imath] but this implies that [imath]Pv[/imath] is an eigenvector of [imath]A[/imath]. But the eigenvalues of [imath]A^k[/imath] are [imath]\lambda^k[/imath] [imath]\therefore \lambda^k=\lambda[/imath] which gives the conclusion required. My questions is: Is the logic correct? If so, Am I missing any details? If not, then how could I approach this? Thanks! |
2672981 | For what value of $p$ would the series $\sum_{n=2}^\infty \frac{1}{n^p\ln(n)}$ converge?
For what value of [imath]$p$[/imath] would the series [imath]$\sum_{n=2}^\infty \frac{1}{n^p\ln(n)}$[/imath] converge? I have tried using the series convergence test, and limit comparison test, which does't help. So i am considering using the divergence test, and finding all values for p for which the limit is not [imath]0[/imath] | 1406936 | For which values of [imath]p[/imath] does the serie [imath]\sum\limits_{n=2}^\infty\frac{1}{n^p\ln(n)}[/imath] converge?
For which values of [imath]p[/imath] does the serie [imath]\sum\limits_{n=2}^\infty\frac{1}{n^p\ln(p)}[/imath] converge? I'm trying to use the ratio test but I can't get a simple term in which use limit easily enough. |
2630285 | Problem about continuous functions and the intermediate value theorem
Let [imath]S^{1} := \lbrace(\cos\alpha, \sin\alpha) \subset \mathbb{R}^{2} | \alpha \in \mathbb{R}\rbrace[/imath] be the circumference of radius [imath]1[/imath] and [imath]f: S^{1} \to \mathbb{R}[/imath] a continuous function. Prove that there exist two points diametrically opposed at which [imath]f[/imath] assumes the same value. My idea for solution: Define [imath]\varphi: S^{1} \to \mathbb{R}[/imath] as[imath]\varphi(\cos\alpha, \sin\alpha) = f(\cos\alpha, \sin\alpha) - f(-\cos\alpha, -\sin\alpha).[/imath] If [imath]f(\cos\alpha, \sin\alpha) = f(-\cos\alpha, -\sin\alpha)[/imath] for all [imath]\alpha \in \mathbb{R}[/imath], the result follows. Otherwise, there exist [imath]\alpha_{1},\alpha_{2}[/imath] such that [imath]f(\cos\alpha_{1}, \sin\alpha_{1}) - f(-\cos\alpha_{1}, -\sin\alpha_{1}) > 0[/imath] and [imath]f(\cos\alpha_{2}, \sin\alpha_{2}) - f(-\cos\alpha_{2}, -\sin\alpha_{2}) < 0.[/imath] Applying the intermediate value theorem, we prove the result. Is this a correct idea? I appreciate your corrections. Thanks! | 2847584 | How to prove [imath]x_0\in S^1[/imath] s.t., [imath]f(x_0)=f(-x_0)[/imath]
Let [imath]f:S^1 \to \mathbb{R}[/imath] continuous function , then there exists [imath]x_0\in S^1[/imath] s.t., [imath]f(x_0)=f(-x_0)[/imath] I tried to calculate by taking [imath]x:=(\cos{t}, \sin{t})[/imath], but [imath]f[/imath] is not specific, I can't. |
1743610 | Spectrum of a positive operator
We know that if [imath]A[/imath] is a self-adjoint unbounded operator on a Hilbert space [imath](H;\left<.,.\right>)[/imath] then [imath]\sigma(A) \subset \mathbb R[/imath]. Now, how it can be shown that if [imath]A[/imath] is more positive i.e. [imath]\left<Au,u\right> \geq 0[/imath], then [imath]\sigma(A) \subset \mathbb [0,+\infty[[/imath] ? Thank you in advance | 1014070 | Positive operator has a positive spectrum?
Let [imath]T : \operatorname{dom}(T) \rightarrow H [/imath] be a positive self-adjoint operator, is it then true that [imath]\sigma(T) \subset [0,\infty)[/imath]? This is something that sounds natural and I guess that it is true. For bounded operator, this can be easily shown, but the only proof I know cannot be adapted to unbounded operator, so I was wondering if anybody here knows how to show this? |
2672596 | Proving periodic points of tent-map are dense in [0,1]
Let [imath]T[/imath] be the tent map defined by [imath]T = \begin{cases} 2x \quad \text{ if } x \le 1/2 \\ 2(1-x) \quad \text{ if } x \ge 1/2\end{cases}[/imath] I am trying to prove that the periodic points of [imath]T[/imath] are dense in the unit interval, [imath]I = [0,1][/imath]. I know that all asymptotically periodic orbits of [imath]T[/imath] are eventually periodic, given that all fixed points are sources. However, I'm struggling in going about solving this given the tools I have. For context, this problem is problem 3.7 in Alligood's text. I was thinking that showing there are infinitely many periodic orbits is sufficient, but I'm unsure. I'm trying to find a more rigorous proof but am at a loss. | 2048525 | Fast way to find period-n points of a tent map?
I need to find a period-3 orbit for a tent map defined by [imath]T(x)=\begin{cases}2x & 0\le x\le0.5 \\[2ex] 2-2x &0.5 < x \le 1 \end{cases}[/imath] So, what is a fast and efficient way of finding it? I only know how to do it by guessing-and-checking. For example, I picked a random point, [imath]x = \frac{2}{7}[/imath] and found that [imath]T\left(\frac{2}{7}\right)=\frac{4}{7}\\[4ex]T\left(\frac{4}{7}\right)=\frac{6}{7}\\[4ex]T\left(\frac{6}{7}\right)=\frac{2}{7}[/imath] which is a period-3 orbit. Is there any better way of finding a period-3 orbit , more specifically, a period-3 point? |
2674353 | Polynomial Doubt
Question : [imath]x^4 + px^3 + qx^2 + px + 1 =0 [/imath] has real roots. Then what is the minimum value of [imath] p^2 +q^2 [/imath] . How I started ? I started by dividing the whole equation by [imath]x^2[/imath] then we get [imath] (x + \frac{1}{x} ) ^2 + p (x + \frac{1}{x} ) + q - 2 = 0 [/imath] Then put [imath](x + \frac{1}{x} ) = t[/imath]. Then discriminant should be greater than equal to zero. But now the problem arises that [imath]t[/imath] does not belong to [imath](-2,2)[/imath] , so taking care of that part leads to solving inequality which I am unable to do . Have I started the right way? One more thing to notice is that the sum of roots of the equation is equal to the sum of reciprocal of the roots . How to proceed further ? | 374505 | Existence of real roots of a quartic polynomial
Question What is the minimum possible value of [imath]a^{2}+b^{2}[/imath] so that the polynomial [imath]x^{4}+ax^{3}+bx^{2}+ax+1=0[/imath] has at least 1 root? Attempt I divided by [imath]x^{2}[/imath] and got [imath]x^{2}+\frac{1}{x^{2}}+2+a\left(x+\frac{1}{x}\right)+b-2=0[/imath] by letting [imath]x+\frac{1}{x}=X[/imath] the equation becomes: [imath]X^{2}+aX+(b-2)=0[/imath] [imath]\therefore X=\frac{-a\pm \sqrt{a^{2}-4b+8}}{2}[/imath] but I am not sure how to continue. If the polynomial has 1 root doesn't that is should be a double root? Are we counting multiplicity or not? |
2672379 | Notation of iterated composition of functions
Let [imath]f[/imath] be a function from [imath]A[/imath] to [imath]B[/imath] such that the image [imath]f(A)\subset A[/imath]. Is there a widely accepted notation for the expression [imath]f\circ\left(f\circ\cdots(f\circ f)\right),[/imath] where [imath]f[/imath] composite with itself [imath]n[/imath] times? I failed to find a natural way to include the information [imath]n[/imath] in the notation. | 795334 | Iterated self-composition of arbitrary function
Does there exist some notation that represents the iterative composition of a single-input, single-output function with itself? As in, say, [imath]f_5(x)=f(f(f(f(f(x)))))[/imath]. In other words, going by the above (incorrect, I'm pretty sure) notation: [imath]f(x)=x+1[/imath], [imath]f_n(m)=m+n[/imath] I'm looking for the correct way to express the notion of "[imath]f_n(x)[/imath]" for any [imath]f[/imath]. |
2674289 | Inequality. Further solutions?
For the community of inequalities. If [imath]z\leq x+y[/imath] ,And [imath]x, y, z \neq -1[/imath], prove : [imath] \frac{z}{1+z} \leq \frac{x}{1+x} + \frac{y}{1+y}. [/imath] I have this proof, but I feel there should be other simpler (direct) proofs My attempt: [imath] z(1-z+z^2-\dots)\leq (x+y)(1-z+z^2-\dots), [/imath] now the idea is that either [imath](1-z+z^2-\dots)\leq(1-x+x^2-\dots)[/imath] and [imath](1-z+z^2-\dots)\leq(1-y+y^2-\dots)[/imath] are true, and thus [imath] z(1-z+z^2-\dots)\leq x(1-x+x^2-\dots)+y(1-y+y^2-\dots), [/imath] which is exactly [imath] \frac{z}{1+z}\leq\frac{x}{1+x} + \frac{y}{1+y} [/imath]. Note: I have added the above condition for [imath]x,y,z[/imath] to define The LHS and RHS of inequality | 1110391 | Show that [imath]\frac c {1+c} \le \frac a {1+a} + \frac b {1+b}[/imath] , for [imath]c \le a+b[/imath] and [imath]a,b,c \ge 0[/imath]
Show that [imath]\frac c {1+c} \le \frac a {1+a} + \frac b {1+b}[/imath] , for [imath]c \le a+b[/imath] and [imath]a,b,c \ge 0[/imath] So need to show [imath]\frac c {1+c} \le \frac {a+b+2ab} {1+a+b+ab}[/imath] We have [imath]\frac c {1+c} \le \frac {a+b} {1+c}\le \frac {a+b+2ab} {1+c}[/imath] So the whole thing reduces to showing [imath]c \ge a+b+ab[/imath] ? Can anyone help me with this, I'm rubbish at inequalities... |
2673308 | Eigenvalues of an idempotent matrix can be [imath]0[/imath] or [imath]1[/imath]
Let [imath]A[/imath] be an idempotent matrix and [imath]c[/imath] be an eigenvalue of [imath]A[/imath]. Then there exists a nonzero vector [imath]v[/imath] such that [imath]Av=cv \Rightarrow A^2v=cAv \Rightarrow Av=c^2v \Rightarrow cv=c^2v \Rightarrow (c-c^2)v=0 \Rightarrow c-c^2=0 \Rightarrow c=0,1[/imath] Is there something wrong with my proof? | 2298922 | If [imath]A[/imath] is idempotent, then the eigenvalues of [imath]A[/imath] are [imath]0[/imath] or [imath]1[/imath]
If [imath]A[/imath] is idempotent (defined by [imath]AA=A[/imath]) then the eigenvalues of [imath]A[/imath] are [imath]0[/imath] or [imath]1[/imath]. Proof: [imath]Ax = \lambda x[/imath] [imath]\Rightarrow Ax = AAx = \lambda Ax = \lambda^{2}x.[/imath] Then in the lecture notes, it says that the above implies that [imath]\lambda^{2} = \lambda[/imath] which implies [imath]\lambda = 0[/imath] or [imath]\lambda = 1[/imath]. I don't quite follow how it is deduced that [imath]\lambda^{2} = \lambda[/imath] from the above calculations. |
2674616 | How can a monoid be a category with just one object?
My book says that: A monoid is a set M equipped with a binary operation [imath].:M\times M\to M[/imath] and a distinguished unit element [imath]u\in M[/imath] such that for all [imath]x,y,z\in M[/imath]: [imath]x\cdot(y\cdot z) = (x\cdot y)\cdot z[/imath] and [imath]u\cdot x = x = x\cdot u[/imath] Equivalently, a monoid is a category with just one object. The arrows of the category are the elements of the monoid. In particular, the identity arrows is the unit element [imath]u[/imath] So, a monoid is a category with just one object. What it means exactly? [imath]M[/imath] has more than one object. Later the book cites [imath]\mathbb{N}[/imath] as being a monoid. Why? | 421215 | Definition of a monoid: clarification needed
I'm only in high school, so excuse my lack of familiarity with most of these terms! A monoid is defined as "an algebraic structure with a single associative binary operation and identity element." A binary operation, to my understanding, is something like addition, subtraction, multiplication, division i.e. it involves 2 members of a set, a single operation, and the resulting third member within that set. And an identity element is a special type of element of a set, with respect to a binary operation on that set, which leaves other elements unchanged when combined with things. Examples are "[imath]0[/imath]" as an additive identity and "[imath]1[/imath]" as a multiplicative identity. How does this definition correspond to a category with a single object? What are some examples of monoids? |
2675277 | Proof verification: For isomorphism [imath]\phi: G\to H[/imath], show that [imath]|\phi (a)|=|a|[/imath] for all [imath]a\in G[/imath].
Could someone please verify my following proof? For isomorphism [imath]\phi: G\to H[/imath] of the groups [imath]G[/imath] and [imath]H[/imath], show that [imath]|\phi (a)|=|a|[/imath] for all [imath]a\in G[/imath]. Proof: Let [imath]a\in G[/imath] with [imath]|a|=n[/imath]. Then [imath]a^{n}=e_{G}[/imath], the identity of [imath]G[/imath]. Let [imath]|\phi(a)|=m[/imath]. Then [imath](\phi (a))^{m}=e_{H}[/imath], the identity of [imath]H[/imath], or [imath]\phi (a^{m})=e_{H}[/imath]. Since [imath]\phi(e_{G})=e_{H}[/imath], it must be that [imath]a^{m}=e_{G}[/imath]. Then it must be that [imath]m=n[/imath]. Therefore, [imath]|\phi (a)|=|a|[/imath] for all [imath]a\in G[/imath]. | 2667653 | Not sure how to go about this group theory proof
Prove that if [imath]f : G \rightarrow H[/imath] is a group isomorphism, then for any element [imath]x \in G[/imath] one has [imath]o(x) = o(f(x))[/imath], where [imath]o(a)[/imath] denotes the order of the element [imath]a[/imath]. I feel that this proof should be online somewhere but haven't been able to find anything so would be helpful if anyone could point me in the right direction. |
2674582 | Complex analysis prove uniform convergence question
[imath] \sum_{n = 1}^\infty \frac{z}{n(1 + n|z|^2)}. [/imath] I need to prove this series converges uniformly on [imath]\mathbb{C}[/imath], so I try to prove this by M-test. I am not sure is this correct, since I am not familiar with complex analysis uniform convergence. | 2673323 | Proving that [imath]f(z)=\sum\limits_{n=1}^{\infty}\frac{z}{n(1+n|z|^{2})}[/imath] converges uniformly on [imath]\mathbb{C}[/imath]
How can one prove that [imath]f(z)=\sum_{n=1}^{\infty}\dfrac{z}{n(1+n|z|^{2})}[/imath] converges uniformly on [imath]\mathbb{C}[/imath] ? My attempt was to let [imath]z=x+iy[/imath], then consider [imath]|\dfrac{z}{n(1+n|z|^{2})}|=\dfrac{\sqrt{x^{2}+y^{2}}}{\sqrt{(n^{2}x^{2}+n)^{2}+(n^{2}y)^{2}}}.[/imath] which looks like a mess. The main idea was to try to find an upper bound, [imath]M_{k}[/imath] so that I can apply the Weierstrass M-test. If I can show that the upper bound, [imath]M_{k}[/imath] is convergent (which is often simple by using root test etc.), I will be able to conclude that [imath]f(z)[/imath] converges uniformly. Is there a simpler or more elegant way I can approach this question and if my approach is correct, can someone kindly guide me to find the required upper bound? |
2675775 | prove [imath]|X-Y|\le|X|+|Y|[/imath]
MY question is how to prove [imath]|X-Y|\le|X|+|Y|[/imath]. My textbook proves another version of triangular inequality, which is [imath]|A|-|B|\le|A-B|[/imath] by substituting X=A-B and Y=B into [imath]|X+Y|\le|X|+|Y|[/imath], but it leaves the above version as an exercise, and I can't solve it. Thank you in advance. :) | 1421512 | Show that [imath]|x-y|\leq |x|+|y|[/imath].
Show that [imath]|x-y|\leq |x|+|y|[/imath]. I know this probably applies the triangle inequality in some way, but I can't figure it out. |
2676398 | Give a example of 2[imath]\times[/imath]2 matrix A such that A has one independent eigenvector while [imath]A^2[/imath] has two independent eigenvectors.
Give a example of 2[imath]\times[/imath]2 matrix A such that A has one independent eigenvector while [imath]A^2[/imath] has two independent eigenvectors. I do found a explaination here. here And I followed it to make X as [imath]\pmatrix{0&1\\1&0}[/imath] and get the A as [imath]\pmatrix{0&0\\1&0}[/imath]. Not really sure whether I am right. Yeah, I checked that post and my problem is I still don't know how to get the A. Because it says x should be nonsingular so I tried [imath]\pmatrix{0&1\\1&0}[/imath] but it seems it's wrong.. | 1015306 | [imath]2\times2[/imath] matrix [imath]A[/imath] such that [imath]A[/imath] has one independent eigenvector while [imath]A^{2}[/imath] has two independent eigenvectors
Give an example of [imath]2\times2[/imath] matrix [imath]A[/imath] such that [imath]A[/imath] has one independent eigenvector while [imath]A^{2}[/imath] has two independent eigenvectors. I would like to know a systematic answer of how to get this. My guess was that any matrix that has 1 independent eigenvector will have the same number of independent eigenvector for the squared matrix. I have tried random selections but obviously couldn't come up with one. What's more useful will be to get the intuition behind this problem. |
2676130 | [imath]\lim_{x\to 0^-}(1+\frac{1}{x})^x[/imath]
I can't remove the indeterminate form [imath]0 \cdot \infty[/imath]. I tried to write the limit as [imath]\lim_{x \to 0^-}e^{x \log \left(1+\frac{1}{x} \right)}[/imath] but it doesn't help. | 1945254 | Finding limit of [imath]\left(1 +\frac1x\right)^x[/imath] without using L'Hôpital's rule.
Is there a way to find this limit without using L'Hôpital's rule. Just by using some basic limit properties. [imath]\lim_{x\to\infty}\left(1+\frac1x\right)^x=e[/imath] |
2676812 | Prove that [imath]\lim \limits_{x \to 0} {\sqrt{x+2}}=2[/imath] (using Epsilon-Delta)
Can someone please guide me through the solving of [imath]\lim \limits_{x \to 0} {\sqrt{x+2}}=2[/imath] using the Epsilon-Delta definition? I understand how to use [imath]\epsilon-\delta[/imath] in general cases and I know, so far, that: Let [imath]\epsilon > 0,[/imath] [imath]\left|\sqrt{x+2}-2\right|<\epsilon\ whenever \left|x-0\right|<\delta[/imath] [imath]-\epsilon<\sqrt{x+2}-2<\epsilon\ whenever -\delta<x<\delta[/imath] Let's solve for [imath]-\epsilon<\sqrt{x+2}-2<\epsilon\\2-\epsilon<\sqrt{x+2}<\epsilon+2\\(2-\epsilon)^2<{x+2}<(\epsilon+2)^2\\\epsilon^2-4\epsilon+2<x<\epsilon^2+4\epsilon+2[/imath] And here I'm stuck. Is this even the right way to do it ? | 560307 | Prove that [imath]\sqrt{x}[/imath] is continuous on its domain [imath][0, \infty).[/imath]
Prove that the function [imath]\sqrt{x}[/imath] is continuous on its domain [imath][0,\infty)[/imath]. Proof. Since [imath]\sqrt{0} = 0, [/imath] we consider the function [imath]\sqrt{x}[/imath] on [imath][a,\infty)[/imath] where [imath]a[/imath] is real number and [imath]s \neq 0.[/imath] Let [imath]\delta=2\sqrt{a}\epsilon.[/imath] Then, [imath]\forall x \in dom,[/imath] and [imath]\left | x-x_0\right | < \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| < \left|\frac{\delta}{2\sqrt{a}}\right|=\epsilon.[/imath] Can I do this? |
2676947 | Prove this function is differentiable at a point but the partial derivatives are not continuous
Let [imath]f:\mathbb{R^2} \rightarrow \mathbb{R}[/imath] be a function defined as \begin{align*} f(x,y) = \begin{cases} (x^2+y^2){\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)} &\quad\text{if } (x,y) \ne(0,0) \\ 0 &\quad\text{otherwise.} \\ \end{cases} \end{align*} Prove that [imath]f[/imath] is differentiable at [imath](0,0)[/imath] but its partial derivatives aren't continuous at [imath](0,0)[/imath]. For the first part, I proved it by definition of differentiability at that point. So, I want that \begin{align*} \lim_{(x,y)\to\ (0,0)} \frac{f(x,y) - f(0,0) - \frac{\partial f}{\partial x}(0,0)x - \frac{\partial f}{\partial y}(0,0)y}{\Vert (x,y) - (0,0)\Vert} = 0 \end{align*} By definition, [imath]f(0,0) = 0[/imath]. Then i calculate [imath]\frac{\partial f}{\partial x}(0,0)[/imath] and [imath]\frac{\partial f}{\partial y}(0,0)[/imath]: \begin{align*} \frac{\partial f}{\partial x}(0,0) = \lim_{h\to\ 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h\to\ 0} \frac{h^2\sin\left(\frac{1}{\mid h \mid}\right)}{h} \leq\lim_{h\to\ 0} \frac{h^2}{h} = 0 \end{align*} \begin{align*} \frac{\partial f}{\partial y}(0,0) = \lim_{h\to\ 0} \frac{f(0,h) - f(0,0)}{h} = \lim_{h\to\ 0} \frac{h^2\sin\left(\frac{1}{\mid h \mid}\right)}{h} \leq\lim_{h\to\ 0} \frac{h^2}{h} = 0 \end{align*} And, \begin{align*} \lim_{(x,y)\to\ (0,0)} f(x,y) = \lim_{(x,y)\to\ (0,0)} (x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) \leq \lim_{(x,y)\to\ (0,0)} (x^2+y^2) = 0 \end{align*} Then the limit is 0 so f is differentiable at [imath](0,0)[/imath]. To prove the partials aren't continuous, I calculate [imath]\frac{\partial f}{\partial x}(x,y)[/imath] using derivative rules around [imath](0,0)[/imath]: \begin{align*} \frac{\partial f}{\partial x}(x,y) = \begin{cases} 2x\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)-x\frac{\cos\left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}} &\quad\text{if } (x,y)\ne(0,0) \\ 0 &\quad\text{otherwise.} \\ \end{cases} \end{align*} Which i have to prove is not continuous but I can't find the right curves to accomplish this. Any suggestions? | 607104 | Proving a scalar function is differentiable at the origin but that its partial derivatives are not continuous at that point.
Let [imath]f: \mathbb R^2 \to \mathbb R[/imath] defined as [imath]f(x,y)=(x^2+y^2)\sin(\dfrac{1}{\sqrt{x^2+y^2}})[/imath] if [imath](x,y) \neq (0,0)[/imath], [imath]f(x,y)=(0,0)[/imath] if [imath](x,y)=(0,0)[/imath] Prove that [imath]f[/imath] is differentiable at [imath](0,0)[/imath] but that the partial derivatives of [imath]f[/imath] are not continuous at the origin. The attempt at a solution: I had some problems at the first part of the exercise: I've calculated [imath]\dfrac{\partial f}{\partial x}(0,0)[/imath] and [imath]\dfrac{\partial f}{\partial y}(0,0)[/imath], both exist and are equal to [imath]1[/imath]. So, to prove [imath]f[/imath] is differentiable at the origin, I have to show that [imath]\lim_{(x,y) \to (0,0)} \dfrac{|f(x,y)-(f(0,0)+<\nabla f(0,0),(x,y)>)|}{\|(x,y)\|}=0[/imath] But [imath]\lim_{(x,y) \to (0,0)} \dfrac{|f(x,y)-(f(0,0)+<\nabla f(0,0),(x,y)>)|}{\|(x,y)\|}[/imath] equals [imath]\lim_{(x,y) \to (0,0)} \dfrac{|(x^2+y^2)\sin(\dfrac{1}{\sqrt{x^2+y^2}})-(x+y)|}{\|(x,y)\|}[/imath]. I don't know how to show this limit equals [imath]0[/imath]. For the second part (I've only analyzed [imath]\dfrac{\partial f}{\partial x}[/imath] since both cases are the same), what I did was try to show that [imath]\lim_{(x,y) \to (0,0)} \dfrac{\partial f}{\partial x}(x,y) \neq \dfrac{\partial f}{\partial x}(0,0)[/imath] I thought of considering the curve [imath]\phi(t)=(t,0)[/imath], so [imath]\lim_{t \to 0} (f\circ \phi)(t)=\lim_{t \to 0} 2t\sin(|t|^{-1})-\dfrac{t\cos(|t|^{-1})}{|t|^{-1}}[/imath]. But [imath]\lim_{t \to 0^{+}} \dfrac{t\cos(|t|^{-1})}{|t|^{-1}}=\lim_{t \to 0^{+}} \cos(\dfrac{1}{t})[/imath] and this limit doesn't exist, so the original limit doesn't exist. This shows the partial derivative with respect to [imath]x[/imath] at the point [imath](0,0)[/imath] is not continuous. I would appreciate if someone could help me to finish the first part and tell me if what I did in the second part of the problem is correct. |
2676790 | [imath]A[/imath] and [imath]B[/imath] are square complex matrices of the same size and [imath]\text{rank}(AB − BA) = 1.[/imath] Show that [imath](AB − BA)^2= 0.[/imath]
[imath]A[/imath] and [imath]B[/imath] are square complex matrices of the same size and [imath]\text{rank}(AB − BA) = 1.[/imath] Show that [imath](AB − BA)^2= 0.[/imath] I am trying to understand the solution given here: Let [imath]C = AB − BA.[/imath] Since [imath]\text{rank} C = 1[/imath], at most one eigenvalue of [imath]C[/imath] is different from [imath]0.[/imath] Also [imath]\text{tr} C = 0[/imath], so all the eigevalues are zero. In the Jordan canonical form there can only be one [imath]2 × 2[/imath] cage and thus [imath]C^2 = 0.[/imath] I do not understand the argument in the bold part. I know that every matrix with complex entries can be put into the Jordan Canonical form but why is there only one [imath]2\times 2[/imath] cage? Please explain. I am still learning this concept. | 861349 | On the nilpotence of the matrix [imath]AB-BA[/imath]
Given [imath]n\times n[/imath] matrices [imath]A,B[/imath] satisfy: [imath]rank(AB-BA)=1[/imath] Prove that [imath](AB-BA)^{2}=0[/imath] Generalize the problem if possible. Any solution not mention Jordan canonical form would be appreciated! |
1995096 | A direct proof that there is a prime between [imath]n[/imath] and [imath]n^2+1[/imath]?
I am trying to prove there is a prime between [imath]n[/imath] and [imath]n^2+1[/imath] without using Bertrand's postulate or Prime number theorem. Do you have any idea? Yuval Filmus's answer for this problem provides a quite useful idea. But since [imath]n^2+1\lt n![/imath] for [imath]n\ge4[/imath], I do not know how to use it for this question. | 1594162 | There is a prime between [imath]n[/imath] and [imath]n^2[/imath], without Bertrand
Consider the following statement: For any integer [imath]n>1[/imath] there is a prime number strictly between [imath]n[/imath] and [imath]n^2[/imath]. This problem was given as an (extra) qualification problem for certain workshops (which I unfortunately couldn't attend). There was a requirement to not use Bertrand's postulate (with which the problem is nearly trivial), and I was told that there does exist a moderately short proof of this statement not using Bertrand. This is my question: How can one prove the above statement without Bertrand postulate or any strong theorems? Although I can only accept one answer, I would love to see any argument you can come up with. I would also want to exclude arguments using a proof of Bertrand's postulate, unless it can be significantly simplified to prove weaker statement. Thank you in advance. |
2654402 | Super-fast growing function exceeding Graham's number
If we define [imath]G_0 = 10^{100}[/imath], and [imath]G_n = 10^{G_{n-1}}[/imath] (hence [imath]G_0[/imath] is a googol, [imath]G_1[/imath] is a googolplex, [imath]G_2[/imath] is a googolplexian), for what first value of n will [imath]G_n[/imath] exceed Graham's number? | 1602702 | How do we compare the size of numbers that are around the size of Graham's number or larger?
When numbers get as large as Graham's number, or somewhere around the point where we can't write them as numerical values, how do we compare them? For example: [imath]G>S^{S^{S^{\dots}}}[/imath] Where [imath]G[/imath] is Graham's number and [imath]S^{S^{S^{\dots}}}[/imath] is [imath]S[/imath] raised to itself [imath]S[/imath] times and [imath]S[/imath] is Skewes number. It appears obvious (I think) that Graham's number is indeed larger, but how does one go about proving that if both numbers are "so large" that they become hard to compare? More generally, how do we compare numbers of this general size? As a much harder problem than the above, imagine a function [imath]G(x,y)[/imath] where [imath]G(64,3)=[/imath] Graham's number. The function [imath]G(x,y)[/imath] is as follows: [imath]G(x,y)=y\uparrow^{(G(x-1,y))}y[/imath] Where [imath]G(0,y)[/imath] is given. I ask to compare [imath]G(60,S)[/imath] and [imath]G(64,3)[/imath] |
2677502 | About a polynomial with a root [imath]\sqrt[3]{2} + \sqrt{5}[/imath].
How to find a polynomial with integer coefficients and a root [imath]\sqrt[3]{2} + \sqrt{5}[/imath]? The degree of this polynomial can't be more than 7. | 1770281 | Prove that [imath]\sqrt[3]{2} +\sqrt{5}[/imath] is algebraic over [imath]\mathbb Q[/imath]
Prove that [imath]\sqrt[3]{2} +\sqrt{5}[/imath] is algebraic over [imath]\mathbb{Q}[/imath] (by finding a nonzero polynomial [imath]p(x)[/imath] with coefficients in [imath]\mathbb{Q}[/imath] which has [imath]\sqrt[3] 2+\sqrt 5[/imath] as a root). I first tried letting [imath]a=\sqrt[3]{2} +\sqrt{5}[/imath] and then square both sides. But I keep on going into a loop by continuing to square it over and over again. Then I tried [imath]a^3=(\sqrt[3]{2} +\sqrt{5})^3[/imath]. Just can't seem to get rid of the radicals. |
525931 | How to generate a basis for a linear mapping in a complex field such that's it's matrix is Upper Triangular?
Let [imath]T:V \to V[/imath] be linear. [imath]V[/imath] is a complex vector space of dimension [imath]k[/imath]. Then there exists a basis so that the matrix generated by [imath]T[/imath] under that basis is upper triangular. The proof is by induction on [imath]k[/imath]. But how to generate such a basis. The first step is that because [imath]v[/imath] is a complex vector space, so there exists a non-zero vector [imath]v[/imath] such that [imath]Tv=\lambda v[/imath] for some [imath]\lambda \in \Bbb F[/imath]. So, take [imath]e_1=v[/imath] because then [imath]T(e_1)=\lambda e_1[/imath]. So, in the first column of the matrix the rows 2 to n will be zero. Now if I extend [imath]e_1[/imath] to any basis of [imath]v[/imath] that basis may not be a basis for which the matrix is upper triangular. Then how to construct it ? | 484640 | How to get the upper triangular form for any linear map?
Let [imath]T:V \to V[/imath] be linear. [imath]V[/imath] is a complex vector space of dimension [imath]k[/imath]. Then there exists a basis so that the matrix generated by [imath]T[/imath] under that basis is upper triangular. The proof is by induction on [imath]k[/imath]. But how to generate such a basis. The first step is that because [imath]v[/imath] is a complex vector space, so there exists a non-zero vector [imath]v[/imath] such that [imath]Tv=\lambda v[/imath] for some [imath]\lambda \in \Bbb F[/imath]. So, take [imath]e_1=v[/imath] because then [imath]T(e_1)=\lambda e_1[/imath]. So, in the first column of the matrix the rows 2 to n will be zero. Now if I extend [imath]e_1[/imath] to any basis of [imath]v[/imath] that basis may not be a basis for which the matrix is upper triangular. Then how to construct it ? |
2677617 | Find the value of [imath]a[/imath] if the value of the limit is [imath]1/\pi[/imath]
Let [imath]a[/imath] and [imath]b[/imath] denote two positive constants. If [imath]\lim_{x\to 0} \Bigg(\frac{\int_0^x \frac{t^2}{\sqrt{a+2t^5}}dt}{bx-e\sin x}\Bigg)= \frac{1}{\pi},[/imath] find the value of [imath]a[/imath]. What I've tried: I've used L'Hospital's rule since it is a [imath]\frac00[/imath] form, to get [imath]\frac{\lim_{x\to 0} \Big(\frac{x^2}{\sqrt{a+2x^5}}\Big)}{{\lim_{x\to 0} ( b-e\cos x)}}[/imath] If [imath]x[/imath] tends to [imath]0[/imath], the equation will become [imath]\frac{\frac{0}{\sqrt{a}}}{b-e}[/imath], making it impossible to become [imath]\frac{1}{\pi}[/imath]. Did I make a mistake somewhere? | 2676167 | How to find integration with Unknown
I am given the following problem which I have problem to know where to even start: The question: [imath]\lim_{x\to 0} \frac{\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt} {bx-esinx}=\frac{1}{\pi}[/imath] The part where I will like to know where to start: [imath]\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt[/imath] I appreciate suggestions on how I may get about solving the integration part before I move on to solve the limit question as a whole. |
2675870 | Existence of a Hessian function for a critical point in a Manifold
I'm trying to prove the following: A critical point of [imath]f \in\mathfrak {F} (M)[/imath] is a point [imath]p \in M [/imath] in which [imath] df_ {p} = 0.[/imath] Then: a) at such a point there is a Hessian function [imath]H: T_ {p} (M) \times T_ {p} (M)\rightarrow\mathbb{R}[/imath] such that [imath]H (X_ { p}, Y_ {p}) = X_ {p} (Yf) = Y_ {p} (Xf)[/imath] for all [imath]X, Y \in \mathfrak {X} (M).[/imath] b) [imath]H[/imath] is symmetric, Bilinear and satisfies [imath]H (\partial_ {i} | _ {p}, \partial_ {j} | _ {p}) = (\partial ^ {2 } / \partial x ^ {i} \partial x ^ {j}) (p)[/imath] relative to a coordinate system. Here [imath]M[/imath] is the differentiable manifold, [imath]\mathfrak {F} (M)[/imath] is the set of all differentiable functions of [imath]M[/imath] in [imath]\mathbb {R}[/imath], [imath]\mathfrak {F} (M)[/imath] corresponds to the set of all vector fields in [imath]M[/imath] and [imath]T_ {p} (M)[/imath] is the set of all vectors tangent to [imath]M[/imath] in [imath]p.[/imath] Assuming part a), the properties of symmetry, bilinearity and equality with the partial derivatives are followed by the definition of vector field and derivation. The part where I'm stuck is a). How do I prove the existence of such a mapping? My idea is that if [imath]p[/imath] is a critical point, it should be possible, by means of charts of the variety [imath]M[/imath] and of [imath]\mathbb {R}[/imath] 'bring the Hessian matrix' from such a point in [imath]\mathbb {R}[/imath] but I can't get this. Any help is appreciated in advance. | 1016764 | Definition of the hessian as a bilinear functional on the tangent space
In Milnor's Morse Theory, the Hessian of a smooth function [imath]f : M \to \mathbb R[/imath] defined on a manifold [imath]M[/imath] at a critical point [imath]p[/imath] is the bilinear functional on [imath]T_p M[/imath] defined as follows: [imath]f_{**}(v, w) = \tilde v_p(\tilde w(f))[/imath] where [imath]\tilde v, \tilde w[/imath] are vector fields that extend [imath]v[/imath] and [imath]w[/imath]. It is written that [imath]\tilde w(f)[/imath] denotes the directional derivative of [imath]f[/imath] in the direction [imath]\tilde w[/imath] and that [imath]\tilde v_p[/imath] is, of course, [imath]v[/imath]. I am assuming that this means [imath]\tilde w(f) = T_p(f)(w)[/imath], but I do not know what it means to write [imath]v(T_p(f)(w))[/imath] it is claimed that [imath]f_{**}[/imath] is symmetric because [imath]\tilde v_p(\tilde w(f)) - \tilde w_p(\tilde v(f)) = [v, w]_p(f) = 0[/imath] since [imath]p[/imath] is critical. If I am not mistaken, the argument used here is that since [imath]p[/imath] is critical, any directional derivative of [imath]f[/imath] at [imath]p[/imath] (e.g. [imath][v, w]_p(f)[/imath]) is zero, but wouldn't that also mean that [imath]\tilde w(f) = \tilde v(f) = 0[/imath]? Hopefully, I am only mixing up notations. I would appreciate any comment. |
2677401 | Sum of combinations of sets
Consider all [imath]1000[/imath]–element subsets of [imath]\{1,2,3,4,\dots 2015\}[/imath]. From each such subset select least element. Find Arithmetic Mean of all these elements. I easily managed the trick but my approach got me a lengthy answer. I selected leaste numbers and multiplied them by their number of subsets possible such as; for [imath]1[/imath], number of subsets possible[imath]=1016[/imath], for [imath]2[/imath], [imath]1015[/imath] and so on. By adding this in series([imath]1\times 1016 + 2\times 1015 + \dots 1016\times 1[/imath]) we would get the answer but it got to be a lengthy one. | 2656657 | IMO/1981 - Combinatorics
IMO/1981 Let [imath]1\le p \le n[/imath] and consider all of the subsets of [imath]p[/imath] elements of the set [imath]\{1, 2, 3,..., n\}[/imath]. Each of these subsets has a minimum element. Let [imath]F(n, p)[/imath] be the arithmetic average of these minimum elements. Prove that [imath]F(n, p) = \frac {n+1}{p+1}[/imath] My attempt: So i initialy tried to think about how many subsets can you have given a minimum value [imath]k_m[/imath], such that [imath]k_m \in \{1, 2,..., n\}[/imath] obviously, but also such that [imath]k \le n - p[/imath], because otherwise that would mean the set can't have a minimum value of [imath]k_m[/imath] and [imath]p[/imath] elements, because that would mean you have less options to construct the set from ( [imath]n - k_m - 1[/imath]) than you are needed ([imath]p-1[/imath]). So i figured that if i fix a [imath]k_m[/imath] i have [imath]{n - k_m - 1} \choose{p - 1}[/imath] different sets in which this fixed element is the minimum, so the arithmetic average asked in the question is equal to: [imath]F(n, p)=\frac{\sum_{k=1}^{n-p} k \binom{n - k - 1}{p - 1}}{\sum_{k=1}^{n-p} \binom{n - k - 1}{p - 1}}[/imath] So i need to prove that: [imath]\frac{\sum_{k=1}^{n-p} k \binom{n - k - 1}{p - 1}}{\sum_{k=1}^{n-p} \binom{n - k - 1}{p - 1}}=\frac{n+1}{p+1}[/imath] But i'm struggling to effectively attack this proof simply because this sum is unlike any other one i've proven before, so i'd like to know: If my reasoning has been correct so far (and if not, why not?) , and also some ideas on how to prove that huge sum. |
2678075 | Abstract Algebra - ideals of the ring [imath]\{a/b : b\text{ odd}\}[/imath]
Let us have a ring [imath]R[/imath] defined as [imath]R=\{a/b:a,b \in \mathbb{Z} \text{ and } b \text{ is odd}\}[/imath]. Show that if an ideal is not the zero set, then it is generated by [imath]2^n\cdot(\text{odd}/\text{odd})[/imath]. | 2669305 | Ideals of the ring of rational numbers with odd denominators
Consider the subring [imath]R\subset\Bbb Q[/imath] , [imath]R=\{\frac ab\ | \ a,b\in\Bbb Z, b\text{ odd}\}[/imath] I am struggling with the following questions: (1) Prove that the ideals of [imath]R[/imath] are [imath]\{0\}[/imath] and [imath]2^nR[/imath] for [imath]n\ge 0[/imath] (2) Prove that [imath]R[/imath] has 2 prime ideals and 1 maximal ideal For (1) I can see that [imath]2^nR[/imath] is an ideal for [imath]n\geq0[/imath] but I'm not sure how to show all ideals other than [imath]\{0\}[/imath] are of this form With question (2), [imath]2^nR\subsetneq 2^{n-1}R\subsetneq...\subsetneq 2R\subsetneq R[/imath] so [imath]2R[/imath] is clearly the only maximal ideal and is therefore prime, but I am not sure where to go from here. Any suggestions? |
2675856 | Find all triangles whose area is 1 and length of sides are all rational numbers.
Problem I am working on a problem to find all triangles whose sides are [imath]a,b,c\in \mathbb{Q}[/imath],and its area [imath]S=1[/imath] My attempt By using Heron’s fomula, [imath] S=\sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}\\=1[/imath] and [imath]x+y> z \mbox{, for} \left\{x,y,z\right\}=\left\{a,b,c\right\}.[/imath] Then,to deal with the complex latter condition, I made the following transformation[imath]\begin{align} &x=-a+b+c\\ &y=a-b+c\\ &z=a+b-c \end{align}[/imath]so that [imath]x+y+z=a+b+c.[/imath] Now,[imath]S=1\\ \Leftrightarrow xyz(x+y+z)=16\\ \mbox{which } x,y,z>0 \mbox{ and } x,y,z\in \mathbb{Q}[/imath] The equation is simple. However, I don’t know how to cope with the condition that [imath]x\, y\,z[/imath] are all positive rational numbers.It seems hard to deform. I have known that [imath](\frac{3}{2},\frac{5}{3},\frac{17}{6})[/imath] is a solution to the problem and there are more. But I don’t know how these numbers are calculated. I am puzzled by these figures. | 2398130 | Does there exist a side-rational triangle of area [imath]1[/imath]?
A side-rational triangle stands for a triangle with each side rational. We know, by cosine theorem and computing the area of the triangle, we can get that each angle of the triangle is of rational [imath]\sin[/imath] and [imath]\cos[/imath]. Then consider ant height of one side, we have [imath]S=h(h\cot \alpha+h\cot \beta)/2[/imath] Then it suffices to show that [imath]t-\frac{1}{t}+T-\frac{1}{T}\notin \mathbb{Q}^2\quad t,T\in \mathbb{Q}[/imath] But i am stuck on it. |
2678284 | Adjoint Functors induce Natural Transformations
Let [imath]L:\mathcal{C} \to \mathcal{D}[/imath] and [imath]R:\mathcal{D} \to \mathcal{C}[/imath] to adjoint functors, therefore for each [imath]X \in \mathcal{C}, Y \in \mathcal{D}[/imath] we have natural bijections [imath]Hom_{\mathcal{D}}(L(X), Y) \cong Hom_{\mathcal{C}}(X, R(Y)) [/imath] My question is how to see that these induce natural transformation [imath]\eta: id_{\mathcal{C}} \to RL[/imath] (resp. [imath]\epsilon: id_{\mathcal{D}} \to LR[/imath])? Here I don't see how to use the fact that [imath]R[/imath] and [imath]L[/imath] are adjoint to see that for arbitrary morphism [imath]f: X \to X'[/imath] in [imath]\mathcal{C}[/imath] following diagram commutates (therefore "natural transformation"): [imath] \require{AMScd} \begin{CD} X @>{\eta_X} >> RL(X) \\ @VVfV @VVRL(f)V \\ X' @>{\eta_{X'}}>> RL(X') \end{CD} [/imath] | 835870 | Proving that the transformation obtained from an adjoint pair is natural
I am reading Homological Algebra by J.J. Rotman and am unable to do this problem. Given an adjoint pair [imath](F,G)[/imath] where [imath]F : \mathcal{C} \to \mathcal{D} [/imath] and [imath]G : \mathcal{D} \to \mathcal{C} [/imath] are two covariant functors, we can obtain a natural transformation [imath] \eta : \mathbb{1_{\mathcal{C}}} \to GF [/imath]. I understand the definition of [imath]\eta[/imath] but am unable to prove that the transformation is natural. Couls someone please help me prove that for [imath]f \in Hom (C,C')[/imath], [imath]GF(f)\eta_C=\eta_{C'}f[/imath] ? Thanks ! |
2678622 | [imath]\sqrt {3} [/imath] is irrational (proof verification)
[imath]\sqrt {3} \in Q[/imath]. Then, [imath]\sqrt{3} = \frac ab[/imath] with the lowest term for [imath]a,b \in Z[/imath]. Then, [imath]3b^2=a^2[/imath], which implies that [imath]a^2[/imath] is divisible by 3. That is, [imath]a[/imath] is also divisible by 3 (by fundamental theorem of arithmetic). I don't understand here [imath]a^2[/imath] divisible by 3 implies [imath]a[/imath] divisible by 3. Could you explain it? | 1427806 | Irrational number [imath]\sqrt2[/imath] proof
I have seen the proof that [imath]\sqrt2[/imath] is irrational However I'm a bit confused. The proof goes something like this. first assume [imath]m[/imath] and [imath]n[/imath] are integers such that [imath] \frac{m}{n} = \sqrt{2} [/imath] [imath] \frac{m^{2}}{n^{2}} = 2 [/imath] [imath]m^{2} = 2 n^{2}[/imath] [imath]\implies m^{2} [/imath] is divisible by [imath]2 \implies m[/imath] is divisible by 2. Let [imath]2q = m[/imath]. Then [imath]m^{2} = 4 q^{2} = 2n^{2}[/imath] [imath] n^{2} = 2 q^{2}[/imath] This means that [imath]n^{2}[/imath] is divisible by [imath]2 \implies n[/imath] is divisible by [imath]2[/imath]...... Hence [imath]\sqrt{2}[/imath] is irrational. [imath]$[/imath][imath]If [/imath] m^{2} [imath] is divisible by [/imath]2[imath] how do we know that [/imath]m[imath] is also divisible by [/imath]2[imath]? If you say it's a prime number then I can argue that I didn't need to do all these steps. I could have just said that its factors don't multiply to [/imath]2[imath] hence [/imath]\sqrt{2}$ is irrational |
2673211 | Show that each vertex [imath]v[/imath] of a tree [imath]T[/imath] appears [imath]\deg(v)-1[/imath] times in Prüfer sequence of [imath]T[/imath].
Show that each vertex [imath]v[/imath] of a tree [imath]T[/imath] appears [imath]\deg(v)-1[/imath] times in Prüfer sequence of [imath]T[/imath]. Could you do this by induction on [imath]n[/imath] where [imath]n[/imath] is the number of vertices in a graph? So far, I have my base case [imath]n=2[/imath] in which case both vertices have degree [imath]1[/imath] and since the Prüfer encoding is empty, the base case holds. How would I go from here? Or am I going about it in completely the wrong way? | 2667780 | Proof using Prüfer encoding
The question is as below: Show that each vertex [imath]v[/imath] of a tree [imath]T[/imath] appears [imath]deg(v)-1[/imath] times in the Prüfer encoding of [imath]T[/imath], where [imath]deg(v)[/imath] is the degree of [imath]v[/imath] in [imath]T[/imath]. For this, I think there are 3 scenarios a leaf should be considered for: (1) when it is a leaf as part of a tree; (2) a leaf that has been discarded; (3) the remaining edge. I feel like this will result in a short proof which is only two (or so) statements long, but at the same time, feel it is too simple. EDIT: I've had a look at this again and narrowed it down to 2 cases: Case 1: The vertex [imath]v[/imath] in question has [imath]deg(v)\ge2[/imath]. [imath]v[/imath] then loses its neighbours as the algorithm is applied repeatedly until [imath]v[/imath] becomes a leaf itseelf. At this point, [imath]v[/imath] is attached to another neighbour, leaving an edge. This means the algorithm can no longer be applied and we have obtained the final Prüfer Encoding for the tree [imath]T[/imath]. Case 2: The vertex [imath]v[/imath] has [imath]deg(v)\ge 2[/imath]. It loses its neighbours as the algorithm is applied repeatedly but is deleted itself at some point in the process. Have I gone about this wrong? Thank you. |
1593036 | Prove that [imath]\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )[/imath]
Let [imath]x_1,x_2,\ldots,x_n > 0[/imath] such that [imath]\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1[/imath]. Prove the following inequality. [imath]\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).[/imath] Attempt I tried using HM-GM and I got [imath]\left ( \dfrac{1}{x_1x_2\cdots x_n}\right)^{\frac{1}{2n}} \geq \dfrac{n}{\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}}} \implies \dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_1}}+\cdots+\dfrac{1}{\sqrt{x_n}} \geq n(x_1x_2 \cdots x_n)^{\frac{1}{2n}}[/imath]. But I get stuck here and don't know if this even helps. | 2677511 | Elementary algebra: inequalities
I have studied the following theorems on inequality: the QAGH (quadratic-arithmetic-geometric-harmonic) mean inequality, the triangle inequality, Cauchy Schwarz inequality, Weierstrass inequality and Chebycheff's sum inequality which states that If [imath]a_1,a_2,\ldots a_n[/imath] are real numbers such that (i) [imath]a_1\le a_2\le\ldots \le a_n[/imath] and [imath]b_1\le b_2 \le \ldots \le b_n[/imath], then: [imath]n(a_1b_1 + a_2b_2 +\ldots + a_nb_n) \ge (a_1 + a_2 + \cdots + a_n)[/imath] (ii) [imath]a_1\ge a_2\ge\ldots\ge a_n[/imath] and [imath]b_1\le b_2 \le \ldots \le b_n[/imath], then: [imath]n(a_1b_1 + a_2b_2 +\ldots + a_nb_n) \le (a_1 + a_2 + \cdots + a_n)[/imath] One question from my assignment for which I need help is: If [imath]0\le x_1\le x_2\le\ldots\le x_n,[/imath] and it is given that [imath]\frac{1}{1+x_1} +\frac{1}{1+x_2}+\ldots+\frac{1}{1+x_n} = 1[/imath] Prove that: [imath] (\sqrt{x_1} + \sqrt{x_2} +\cdots + \sqrt{x_n}) \ge (n - 1) \left(\frac{1}{\sqrt {x_1} } +\frac{1}{\sqrt{x_2} } + \cdots + \frac{1}{\sqrt{x_n} }\right) [/imath] Where n is a natural number, [imath] n\ge 2[/imath] I have tried in many ways to prove this statement using the theorems I know, without any success. Could you please help me? |
2678834 | How would we prove [imath]\frac{n^2 - n}{2}[/imath] is [imath]\mathcal{O}(n^2)[/imath]?
Given a runtime of [imath]\frac{n^2 - n}{2}[/imath], how would we prove that the big-O notation is [imath]\mathcal{O}(n^2)[/imath]. I want to use the [imath]f(n) = \mathcal{O}(g(n))[/imath] formula. | 2677134 | Big-O Notation Proof
Given the definition of Big-O, prove that [imath]f(n) = n^2 - n[/imath] is [imath]O(n^2)[/imath]. When I use the given definition, I get [imath]n^2 - n \leq n^2 - n^2[/imath] which means that [imath]n^2 -n \leq 0[/imath], which is not true. Is there some step I'm missing? |
904323 | Derivation of Lagrange-Charpit Equations
I am working through the derivation of the Lagrange-Charpit equations presented in this Wikipedia article: http://en.wikipedia.org/wiki/Method_of_characteristics#Fully_nonlinear_case I am interested in the "fully" nonlinear case. Where we have: [imath]F(x_1,...,x_n,u,p_1,...,p_n)=0 [/imath] And [imath] p_i=\frac{\partial u}{\partial x_i} [/imath] I am fine with the derivation up until the point where they say that (Also, [imath]\dot{x_i}=dx_i/ds[/imath]): [imath]\sum_i(\dot{x}_idp_i-\dot{p_i}dx_i)=0 [/imath] Follows from taking the exterior derivative of: [imath] du-\sum_ip_idx_i=0 [/imath] From the little I know about exterior derivatives, it seems like doing this would give (for the two dimensional case): [imath]0=\left(\frac{\partial p_2}{\partial x_1}-\frac{\partial p_1}{\partial x_2}\right)dx_1\wedge dx_2 [/imath] Because [imath]ddu=0[/imath] and the anti-symmetry of the wedge product. I don't see how this result leads to the one given. Could someone help me out? I feel like there is something very simple that I am missing. | 2674556 | How to take this exterior derivative of the expression [imath]du - \sum_i p_i dx_i[/imath]?
I am reading the wikipedia page about applying the method of characteristics in the fully nonlinear case. We have the fully nonlinear equation [imath] \tag{1} F(x_1, \cdots, x_n , u, p_1, \cdots, p_n) = 0,[/imath] here [imath]\tag{2} p_i = \frac{\partial u}{\partial x_i}[/imath] is the partial derivative of [imath]u[/imath] with respect to [imath]x_i[/imath]. In the method of characteristics, we wish to reduce the PDE to a family of ODE. Let assume that [imath]u[/imath] is a solution to (1). [imath]s\mapsto (x_1(s), \cdots, x_n(s), u(s), p_1(s), \cdots, p_n(s))[/imath] be a curve so that (1) is satisfied for all [imath]s[/imath]. Then it is claimed that the following holds: \begin{equation} \begin{split} \sum_i (F_{x_i} +F_up_i)\dot x_i + \sum_i F_{p_i}\dot p_i &=0\\ \dot u - \sum_i p_i\dot x_i &=0\\ \sum_i ( \dot x_i dp_i - \dot p_i dx_i) &= 0. \end{split} \end{equation} I can see that the first two equations follow from taking total derivative with respect to [imath]s[/imath] of (1) and the expression [imath]u(s) = u(x_1(s),\cdots, x_n(s))[/imath]. In the wiki page, it is claimed that ... the third follows by taking an exterior derivative of the relation [imath]du - \sum p_i dx_i = 0[/imath]. Unfortunately, I fail to see how the third equation are derived using exterior derivative. Could you give me the steps so I could check my work, please? P.S.: I’ve been struggling a lot lately on Frobenius theorem and systems of total differential equations. If you could explain that, I would deeply appreciate it. |
2679652 | Mean value of a convex function
Let [imath]f[/imath] be a continuous real valued function defined on [imath][a,b][/imath]. Prove that, if [imath]f[/imath] is convex, then: [imath]f\left(\frac{a+b}{2}\right)(b-a) \geq \int_a^b f[/imath] I am looking for a proof of this that doesn’t involve Riemann sums, but I can’t find any. Thanks in advance. | 1827772 | Convex integral inequality
I cannot prove that if [imath]f(x)[/imath] is convex on [imath][a,b][/imath] then [imath]f\Big(\frac{a+b}2\Big) \le \frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 .[/imath] |
2679949 | Prove that if [imath]G[/imath] is a cyclic group of order [imath]n[/imath], then [imath]G[/imath] is isomorphic to [imath]\mathbb{Z}_{n}[/imath].
Could someone please help me correct my following proof? Prove that if [imath]G[/imath] is a cyclic group of order [imath]n[/imath], then [imath]G[/imath] is isomorphic to [imath]\mathbb{Z}_{n}[/imath]. Proof: Let [imath]G=\left \langle a \right \rangle[/imath] with [imath]|G|=n[/imath]. Define the relation [imath]\phi:\mathbb{Z}_{n}\to G[/imath] by [imath]\phi(k)=a^{k}[/imath] with [imath]0\leq k<n[/imath]. Well-defined: For [imath]0\leq k_{1}<n[/imath] and [imath]0\leq k_{2}<n[/imath], if [imath]k_{1}=k_{2}[/imath], then [imath]\phi(k_{1})=a^{k_{1}}=a^{k_{2}}=\phi(k_{2})[/imath]. Operation preserving: Since [imath]\phi(k_{1}+k_{2})=a^{k_{1}+k_{2}}=a^{k_{1}}a^{k_{2}}=\phi(k_{1})\phi(k_{2})[/imath], the operation is preserved. One-to-one: For [imath]m,n\in \mathbb{Z}_{n}[/imath], let [imath]m>n[/imath] and assume [imath]\phi(m)=\phi(n)[/imath]. Then [imath]a^{m}=a^{n}[/imath] or [imath]a^{m-n}=e[/imath]. Then [imath]m-n>0[/imath]. I am not sure how to prove that [imath]m-n<0[/imath]. Is it due to [imath]G[/imath] being abelian? Onto: [imath]\phi[/imath] is onto iff for all [imath]a^{k}\in G[/imath], there exists a [imath]k\in \mathbb{Z}_{n}[/imath] such that [imath]\phi(k)=a^{k}[/imath]. By definition, [imath]\phi[/imath] is onto. | 2194959 | Prove that all cyclic groups of the same order are isomorphic
I want to prove that 2 cyclic groups of the same order are always isomorph. So, let [imath]G,H[/imath] be cyclic groups with [imath]|G| = |H|[/imath]. Then: I suspect the isomorphism would be the one that maps a generator g to a generator h. Thus, consider this function: [imath]f: G \rightarrow H[/imath] defined by [imath]f(g^k) = h^k[/imath]. I already showed that this mapping is surjective and a homomorphism, but I get stuck at the injectivity part. Can someone give a hint (my attempts: I tried to deduce whether the kernel is [imath]e_G[/imath] and I also used the definition of injectivity but I got stuck in both approaches)? |
2679698 | Why is it that [imath]\exists f_1,f_2\in M[f_1, f_2[/imath] have no common factor]?
I am attempting to follow the below solution for part (a) of Chapter 12, Problem M.7 in Artin's Algebra textbook. I am having trouble understanding why [imath]\exists f_1,f_2\in M[f_1, f_2[/imath] have no common factor]. By "common factor," I presume that this really means [imath]\exists f_1,f_2\in M \forall d\in M[/imath][[imath]d|f_1[/imath] & [imath]d|f_2[/imath] [imath]\implies[/imath] [imath]d[/imath] is a unit]. (Correct me if I am wrong.) But how would I go about proving that this is true? | 691708 | Nonprincipal prime ideals contain two relatively prime elements
Let [imath]R[/imath] be a principal ideal domain and let [imath]P[/imath] be a nonprincipal prime ideal of [imath]R[x][/imath]. I'm having trouble seeing why [imath]P[/imath] must contain two elements with no common divisor. Can anyone help me? Thanks. |
2679997 | Proof help: Induction ( [imath] 1/ \sqrt1 + 1/\sqrt2 +....+1/\sqrt{n}[/imath] [imath]< \sqrt{n}[/imath] ) for [imath]n>1[/imath]
Proof: Base case: For [imath]n = 2[/imath], we have that: [imath]\dfrac 12 < 2[/imath] [imath] \implies \sqrt{\dfrac 12}<\sqrt2[/imath] [imath] \implies \dfrac {1}{\sqrt{2}}<\sqrt2[/imath] Assume true for [imath]n = k[/imath] i.e.: [imath] \dfrac 1{\sqrt1} + \dfrac{1}{\sqrt2} +....+\dfrac 1{\sqrt{k}} [/imath] [imath]< \sqrt{k} [/imath] Then, for [imath]n = k+1[/imath]: [imath] \dfrac 1{\sqrt1} + \dfrac 1{\sqrt2} +....+\dfrac 1{\sqrt k} + \dfrac 1{\sqrt{k+1}}[/imath] [imath]< \sqrt{k} + \dfrac 1{\sqrt{k+1}} [/imath] I don't know how to proceed any further and reduce [imath] \sqrt{k} + \dfrac{1}{\sqrt{k+1}} [/imath] to just [imath] \sqrt{k+1}[/imath] Any hints, ideas? | 2149448 | Other Idea to show an inequality [imath]\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n[/imath]
[imath]\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n[/imath] I want to prove this by Induction [imath]n=1 \checkmark\\ n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\ n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}[/imath] so [imath]\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}[/imath]now we prove that [imath]\sqrt k+\dfrac{1}{\sqrt {k+1}} >\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark[/imath] and the second method like below , and I want to know is there more Idia to show this proof ? forexample combinatorics proofs , or using integrals ,or fourier series ,.... Is there a close form for this summation ? any help will be appreciated . |
2678968 | Show that [imath]\bar mf=f\bar m[/imath] for any isometry [imath]f[/imath] of [imath]S^2[/imath]
Let [imath]\bar m[/imath] be the isometry sending a point [imath]P[/imath] to its opposite point [imath]-P[/imath], show that [imath]\bar m[/imath] commutes with any isometry of [imath]S^2[/imath] I know I can use the fact that isometries preserve distances, but do I need to individually apply this to each type of isometry? | 2678951 | Show that antipodal points remain antipodal under any isometry of [imath]S^2[/imath]
"The antipodal map of [imath]S^2[/imath] is the isometry [imath]\bar m[/imath] sending each point [imath]P[/imath] to its opposite point [imath]-P[/imath]" (from textbook) Show that antipodal points remain antipodal under any isometry of [imath]S^2[/imath]. In other words, for each isometry f of [imath]S^2[/imath], if [imath]P[/imath] and [imath]P'[/imath] are antipodal, then so are [imath]f(P)[/imath] and [imath]f(P')[/imath] I am unsure how I can define the [imath]P[/imath] and [imath]P'[/imath] so that they are opposites of each other in the antipodal map, and am I supposed to apply the traditional isometries (translation, reflection and rotation) to the points? I don't understand this problem. |
629122 | Proving that expression is equivalent to the definition of derivative
Let [imath]f[/imath] be differentiable at [imath]x=a[/imath]. Prove that if [imath]x_n \to a^+[/imath] and [imath]y_n \to a^-[/imath] then: [imath]\lim_{n\to \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(a).[/imath] Every option that I think about seems to my very trivial, so I believe that I am doing something wrong. Both numerator and denominator approach zero as [imath]n\to\infty[/imath] as the case of the formal definition of derivative, but it isn't guaranteed that the limits are equal (“[imath]\frac{0}{0}[/imath]”). Any direction? | 2821464 | Sequence [imath]\{a_n\}[/imath] and [imath]\{b_n\}[/imath] satisfy [imath]a_n \to c, b_n \to c [/imath]
Sequence [imath]\{a_n\}[/imath] and [imath]\{b_n\}[/imath] satisfy [imath]a_n \to c, b_n \to c [/imath] , and [imath]a_n\lt c \lt b_n [/imath]. [imath]f'(c) [/imath] exists. Prove [imath]\frac{f(a_n)-f(b_n)}{a_n-b_n}\to f'(c)[/imath] I have a rough idea like this, [imath]\frac{f(a_n)-f(b_n)}{a_n-b_n}=\frac{f(a_n)-f(c)-(f(b_n)-f(c))}{a_n-b_n}=\frac{f(a_n)-f(c)}{a_n-b_n}-\frac{f(b_n)-f(c)}{a_n-b_n}=\frac{f'(c)\cdot(a_n-c)}{a_n-b_n}-\frac{f'(c)\cdot(b_n-c)}{a_n-b_n}=f'(c)[/imath] But obviously not strictly. So how to do this thing strictly? |
2680615 | What is the order of operations for [imath]p \implies q \implies r[/imath]
I've been studying mathematical logic recently and we have briefly covered the order of operations for operators like AND/OR/IMPLIES, etc. However, we have a challenge question regarding how the following statement should be interpreted in terms of order of operations, and I don't believe we have covered this material nor can I find the same question answered online. The statement is [imath]p \implies q \implies r[/imath] The question asks if the above statement is correctly represented by [imath](p \implies q) \implies r[/imath], or [imath]p \implies (q \implies r)[/imath], or neither - i.e. what is the correct order of operations when there are no brackets and the two logic operators are equally weighted. I have used a truth table to determine that the above two statements are not equivalent, but are either the logical equivalent to the first statement, or neither? | 104143 | In what order are chained implications evaluated? (i.e. [imath]a \implies b \implies c[/imath])
Implication does not appear to be associative: a b c | (a -> b) -> c | a -> (b -> c) F T F | F | T F F F | F | T Is [imath]a \implies b \implies c[/imath] evaluated as [imath]a \implies (b \implies c)[/imath] or [imath](a \implies b) \implies c[/imath]? |
2680495 | Ceiling Function Proof from First Principles?
My question is whether it is possible to prove that a ceiling function exists ie. If [imath]a\in\mathbb R[/imath], then there exists [imath]n\in\mathbb Z[/imath] satisfying [imath]a\le n< a+1[/imath]. The only solutions to this that I have found are those which involve floor/ceiling functions within the proof itself. Where would one start in solving a proof like this? Possibly proof by contradiction? | 2679291 | Prove that if [imath]x \in R,[/imath] then there exists [imath]n \in Z[/imath] satisfying [imath]x \leq n < x+1[/imath]
So this question in my book looks like it's essentially asking me to prove the ceiling function exists. This question is slightly different to other things I found in related question because we're asked to prove the existence of the value in the middle of the inequality. My initial thoughts are to define an [imath]n \in N[/imath] such that [imath]n > x[/imath] and [imath]-n < x[/imath] Following from this one can say [imath]-n < x \leq n < x + 1[/imath]. Now here I just lose any sort of direction and instead think of trying another method defining a set and using the properties of [imath]inf[/imath] and [imath]sup[/imath] to move from there. The tricky thing I encounter as a roadblock is once again the proving of the existence of the [imath]n[/imath] which is different to many related questions. Any help or solutions are appreciated, ready and rearing to reply :) |
2681331 | Proof from Stein and Shakarchi's Complex Analysis that if [imath]\{f_n\}[/imath] converges uniformly compacta to [imath]f[/imath] then so does [imath]\{f_n'\}[/imath] to [imath]f'[/imath]
I am looking at the proof of the following theorem from Stein and Shakarchi's Complex Analysis. However, in the proof, I don't understand why we can assume without loss of generality that the sequence of functions converges uniformly on all of [imath]\Omega[/imath]. [imath]\Omega[/imath] here is an open set, but the assumption only holds for compact sets, so how is there no loss in generality? If [imath]\{f_n\}_{n=1}^\infty[/imath] is a sequence of holomorphic functions that converges uniformly to a function [imath]f[/imath] in every compact subset of an open connected set [imath]\Omega[/imath], then the sequence of derivatives [imath]\{f_n'\}_{n=1}^\infty[/imath] converges uniformly to [imath]f'[/imath] on every compact subset of [imath]\Omega[/imath]. The proof proceeds as follows. Without loss of generality assume that the sequence of functions in the theorem converges uniformly on all of [imath]\Omega[/imath]. Given [imath]\delta>0,[/imath] let [imath]\Omega_\delta[/imath] denote the subset of [imath]\Omega[/imath] defined by [imath]\Omega_\delta = \{z \in \Omega: \bar{D_\delta}(z)\subset \Omega\}.[/imath] In other words, [imath]\Omega_\delta[/imath] consists of all points in [imath]\Omega[/imath] which are at distance [imath]>\delta[/imath] from its boundary. To prove the theorem, it suffices to show that [imath]\{f_n'\}[/imath] converges uniformly to [imath]f'[/imath] on [imath]\Omega_\delta[/imath] for each [imath]\delta[/imath]. This is achieved by proving the following inequality: [imath]\sup_{z\in \Omega_\delta} |F'(z)|\le \frac{1}{\delta} \sup_{\zeta \in \Omega} |F(\zeta)|[/imath] whenever [imath]F[/imath] is holomorphic in [imath]\Omega,[/imath] since it can then be applied to [imath]F=f_n -f[/imath] to prove the desired fact. | 469848 | holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.
In Stein's complex analysis, there is one theorem 5.3 of Chapter 2 which says that if holomorphic functions [imath]f_n[/imath] converges uniformly to [imath]f[/imath] on any compact subset of an open set [imath]\Omega[/imath], then so does [imath]f'_n[/imath]. In his proof, he said "without loss of generality we can assume [imath]f_n[/imath] converges uniformly to [imath]f[/imath] on all of [imath]\Omega[/imath]". So I think what he meant was that from the conditions of the theorem, we can get [imath]f_n[/imath] converges uniformly to [imath]f[/imath] on all of [imath]\Omega[/imath]. However, I don't know how to prove this assumption. Any help would be welcome. |
871020 | Zorn's lemma and maximal ideals
Let's consider two statements: Zorn's lemma and theorem about existence of maximal ideals in commutative ring with [imath]1[/imath]. It's easy to prove that Zorn's lemma implies existence of maximal ideals. I wonder if the converse proposition is true, i. e., does existence of maximal ideals implies Zorn's lemma? My approach was to construct a ring structure on an ordered set which satisfies Zorn's lemma condition such that existence of maximal ideal implies existence of maximal element in the set, but I didn't succeed. | 2737192 | Zorn's Lemma and Theorem of Ring Theory.
[imath]\bf{Theorem.}[/imath] Let [imath]R[/imath] be a commutative ring with identity. If [imath]I[/imath] is a proper ideal of [imath]R[/imath], then [imath]I[/imath] is contained in a maximal ideal. and [imath]\bf{Zorn's\;Lemma.}[/imath] Suppose a partially ordered set [imath]P[/imath] has the property that every chain in [imath]P[/imath] has an upper bound in [imath]P[/imath]. Then the set [imath]P[/imath] contains at least one maximal element. I have never seen a demonstration of this theorem without the use of the Zorn's Lemma. I know that Zorn's Lemma is a much more general than rings and I don't see a reason for the theorem to imply the Zorn's Lemma, but it makes me wonder if there can be any kind of equivalence between these two things. |
2680718 | Connected subsets of [imath]\mathbb{R^3}[/imath]
Question: Let [imath]V[/imath] be span of [imath](1,1,1)[/imath] and [imath](0,1,1)∈\mathbb{R^3}[/imath], let [imath]u_1=(0,0,1)[/imath], [imath]u_2=(1,1,0)[/imath], [imath]u_3=(1,0,1)[/imath] then which of the following are correct? [imath]1.(\mathbb R^3\backslash V)\cup \{(0,0,0)\}[/imath] is not connected. [imath]2.(\mathbb R^3\backslash V)\cup \{ tu_1+(1-t)u_2:0\le t\le 1\}[/imath] is connected. [imath]3.(\mathbb R^3\backslash V)\cup \{ tu_1+(1-t)u_3:0\le t\le 1[/imath]} is connected. [imath]4.(\mathbb R^3\backslash V)\cup \{(t,2t,2t):t\in \mathbb R[/imath]} is connected. My attempt: I know what is definition of connected sets, but usually I had applied the definition to subsets of [imath]\mathbb{R}[/imath] but yet i didn't applied it subsets of bigger space [imath]\mathbb{R^3}[/imath]. But to start the problem, first I had to find [imath]V[/imath]. Now, [imath]V= span\{(1,1,1),(0,1,1)\}[/imath] [imath]= \{x(1,1,1)+y(0,1,1): x, y ∈\mathbb{R^3}\}[/imath] [imath]=\{(x,x+y,x+y): x,y ∈\mathbb{R}\}[/imath] Now, what will be, [imath](\mathbb R^3\backslash V)[/imath]? it's complement of [imath]V[/imath] in bigger space [imath]\mathbb {R^3}[/imath], beyond my imagination!! Got stuck! From hours, Please help me.. | 2044516 | If [imath]\mathbb R^3\setminus V[/imath] connected where [imath]V[/imath] is the subspace generated by [imath]\{(1,1,1),(0,1,1)\} [/imath]
Let [imath]V[/imath] be a subspace of [imath]\mathbb R^3[/imath] generated by the vectors [imath]\{(1,1,1),(0,1,1)\}.[/imath] There are [imath]3[/imath] vectors [imath]u_1=(0,0,1),u_2=(1,1,0),u_3=(1,0,1)[/imath].Then which of the following sets are connected? [imath]1.\left(\mathbb R^3\setminus V\right)\cup\{(0,0,0)\}[/imath] [imath]2.\left(\mathbb R^3\setminus V\right)\cup \{tu_1+(1-t)u_3:t\in [0,1]\}[/imath] [imath]3.\left(\mathbb R^3\setminus V\right)\cup\{tu_1+(1-t)u_2:t\in [0,1]\}.[/imath] [imath]4.\left(\mathbb R^3\setminus V\right)\cup \{(t,2t,2t):t\in \mathbb R\}[/imath] Now [imath]V[/imath] is generated by two linearly independent vectors so is like [imath]\mathbb R^2.[/imath] So [imath]\left(\mathbb R^3\setminus V\right)[/imath] will be disconnected but giving the [imath](0,0,0)[/imath] with them will make it connected. No. [imath]4[/imath], I think will be connected also because [imath]\{(t,2t,2t),t\in \mathbb R\}[/imath] is in [imath]V[/imath] and contains the point [imath](0,0,0)[/imath] Now, any neighbourhood of this point will have points from both [imath]V[/imath] and [imath]\left(\mathbb R^3\setminus V\right)[/imath] and so the union is connected.I'm not sure about no. [imath]2[/imath] and [imath]3.[/imath] |
2681195 | Complex series that diverges at three points of [imath]|z|=1[/imath]
Does there exist a complex series that the radius of convergence is [imath]R=1[/imath] but the series diverges at exactly three points of a set [imath]\{z \in \Bbb{C} : \space |z|=1 \}?[/imath] I found this exercise and I wonder how should it be solved. I know that in real analysis there exist such series that the radius of convergence is [imath]1[/imath] in the interval [imath]X[/imath] but the same series diverges at the endpoints of [imath]X[/imath]. But it has only two points of divergence. Can anyone show me the way? | 122881 | Power series that diverges in specified number of points on the circle of convergence
I want to construct a complex power series with the radius of convergence [imath]R=1[/imath] that diverges in: 1) [imath]k[/imath] points on the circle [imath]\{|z|=1\}[/imath]. 2) countable number of points on the circle I have already crafted series that converges on the whole circle and diverges on the whole circle. Can anybody help with the problem? Update: I have googled a series [imath]\sum{\frac{z^{kn}}{kn}}[/imath] that is an answer for 1). Though I still need help for the countable case. |
2681709 | [imath]\sum a_n[/imath] converges conditionally and [imath]\sum b_n[/imath] converges absolutely, then will [imath]\sum a_nb_n[/imath] converge absolutely?
Suppose [imath]\sum a_n[/imath] converges conditionally and [imath]\sum b_n[/imath] converges absolutely, then will [imath]\sum a_nb_n[/imath] converge absolutely? I know that [imath]\sum|a_n|[/imath] does not converge while [imath]\sum a_n[/imath] does and that [imath]\sum|b_n|[/imath] does converge, and also that [imath]\lim_{n \to \infty} |a_nb_n|=0[/imath] but I'm not sure how to proceed from here. | 1110101 | To show that [imath]\sum a_nb_n[/imath] is absolutely convergent
Assume that [imath]\sum a_n[/imath] is convergent and [imath]\sum b_n[/imath] is absolutely convergent.To show that [imath]\sum a_nb_n[/imath] is absolutely convergent My try::Consider the sequence of partial sums of [imath]\sum |a_nb_n|[/imath] [imath]S_n=|a_1b_1|+|a_2b_2|+......+|a_nb_n|[/imath] Now [imath](a_1b_1+a_2b_2+......+a_nb_n)\leq ( (a_1^2+a_2^2+..+a_n^2))^\frac{1}{2}((b_1^2+b_2^2+..+b_n^2)^\frac{1}{2})[/imath] How to proceed from here? |
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