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579198 | Convergence of sequence defined with other sequence
[imath]a_n[/imath] is a sequence with [imath]\lim\limits_{n \rightarrow \infty}{\frac{a_{n+1}}{a_n}}=L[/imath], [imath]a_n>0[/imath]. So the task is to show that [imath]c_n:=(a_n)^\frac{1}{n}[/imath] converges and [imath]\lim\limits_{n \rightarrow \infty}{c_n}=L[/imath]. I've been working on this for hours now and don't have any useful result. Please help :) | 287932 | Convergence of Ratio Test implies Convergence of the Root Test
In Elias Stein and Rami Shakarchi's Complex Analysis textbook, we have the following exercise: Show that if [imath]\{a_n\}_{n=0}^\infty[/imath] is a sequence of complex numbers such that [imath]\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L,[/imath] then [imath]\lim_{n\to\infty}|a_n|^{1/n}=L.[/imath] I've been trying to prove this with no luck. The only thing I've thought of doing is [imath]\lim_{n\to\infty}\left(\frac{|a_{n+1}|^n}{|a_n|^n}\right)^{1/n},[/imath]but this hasn't lead me anywhere except dead ends. Will someone provide a hint for me about how to proceed? Thanks! Minor update: I don't know if it's helpful yet, but I know we can write the limit as [imath]\lim_{n\to\infty}\left(\frac{|a_{n+1}a_n\cdots a_0|}{|a_n\cdots a_0|}\cdot\frac{1}{|a_n|}\right).[/imath]This reminds me a lot of the geometric mean, which even has the exponents I'm trying to get... |
13582 | Show that a continuous function has a fixed point
Question: Let [imath]a, b \in \mathbb{R}[/imath] with [imath]a < b[/imath] and let [imath]f: [a,b] \rightarrow [a,b][/imath] continuous. Show: [imath]f[/imath] has a fixed point, that is, there is an [imath]x \in [a,b][/imath] with [imath]f(x)=x[/imath]. I suppose this has to do with the basic definition of continuity. The definition I am using is that [imath]f[/imath] is continuous at [imath]a[/imath] if [imath]\displaystyle \lim_{x \to a} f(x)[/imath] exists and if [imath]\displaystyle \lim_{x \to a} f(x) = f(a)[/imath]. I must not be understanding it, since I am not sure how to begin showing this... Should I be trying to show that [imath]x[/imath] is both greater than or equal to and less than or equal to [imath]\displaystyle \lim_{x \to a} f(x)[/imath] ? | 512786 | Continuous function from the unit interval into the unit interval has a fixed point
Show that if [imath]f[/imath] is continuous on [imath][0,1][/imath] and [imath]0\leq f(x) \leq 1[/imath] for all [imath] x \in [0,1][/imath], then there exists one point [imath]c \in [0,1][/imath] at which [imath]f(c)=c[/imath]. (Hint: Apply the Intermediate Value Theorem to the function [imath]g(x)=x−f(x)[/imath].) I have some idea how to do this, but the area where I am getting stumped is...the value of [imath]f(x)[/imath]? How can I find the value of [imath]f(x)[/imath]? Do I even need to find [imath]f(x)[/imath]? Edit: okay I got something... I only need f(0)>0 and f(1)<1 to show g(0)<0 So then what happens for f(0)=0 or f(1)=1 ? |
579379 | If [imath]f[/imath] is an entire function with a non dense image then it is constant
Is it true that if [imath]f[/imath] is entire with a non dense image then it is constant? Can anybody help with the proof? Or give a counter example. Thank you | 337154 | how to show image of a non constant entire function is dense in [imath]\mathbb{C}[/imath]?
how to show image of a non constant entire function is dense in [imath]\mathbb{C}[/imath]? is there any smallest proof? I have seen this as a theorem in some books but I want some elementary proof. |
26591 | How would you solve the diophantine [imath]x^4+y^4=2z^2[/imath]
I would like to know any way of solving the diophantine equation [imath]x^4+y^4=2z^2[/imath]. Or ideas that seem worth trying out. By solving I mean fining all solutions and proving there are no more. Keith Conrad showed how to reduce this equation to a different one which was solved by Fermat in his notes about Fermat descent, other than I have no ideas how to solve it. I tried to do descent on it directly but that seems completely impossible so I am interested in other techniques. Thanks very much. | 886372 | Solution to the Diophantine equation [imath]x^4+y^4=2z^2[/imath]
Does there exists a nontrivial positive integer solution with [imath]x\ne y,[/imath] of [imath]x^4+y^4=2z^2.[/imath] |
579677 | Show that [imath]H[/imath] is a normal subgroup of [imath]G[/imath] is it has index [imath]2[/imath].
I'm stuck on this question for hours. I think I need a little help and a different point of view to complete it. Let [imath]H[/imath] be a subgroup of a group [imath]G[/imath] and suppose that [imath]H[/imath] has index two in [imath]G[/imath], i.e., that there are precisely two elements in [imath]G/H[/imath]. Show that [imath]H[/imath] is a normal subgroup of [imath]G[/imath]. Thanks in advance. | 84632 | Subgroup of index 2 is Normal
Please excuse the selfishness of the following question: Let [imath]G[/imath] be a group and [imath]H \le G[/imath] such that [imath]|G:H|=2[/imath]. Show that [imath]H[/imath] is normal. Proof: Because [imath]|G:H|=2[/imath], [imath]G = H \cup aH[/imath] for some [imath]a \in G \setminus H[/imath]. Let [imath]x\in G[/imath]. Then [imath]x \in H[/imath] or [imath]x \in aH[/imath]. Suppose [imath]x \in H[/imath]. Then [imath]xhx^{-1} \in H[/imath] because [imath]H[/imath] is a group. Suppose [imath]x \in aH[/imath]. Then [imath]ahH(ah)^{-1} = ahHh^{-1}a^{-1} = a(hHh^{-1})a^{-1}[/imath] ***1 ***1: Can I say now that [imath]x \in H[/imath], based on the that [imath]hHh^{-1} \in H[/imath] and that [imath]aa^{-1} = e[/imath] ? Thank you for the time you spent reading my doubts. |
579953 | Evaluation of a function
Given [imath]f(t)[/imath] satisfies [imath]\dfrac{d^2f(t)}{dt^2}-(2-3i)f(t)=0,\,f(0)=1,\,\lim_{t \rightarrow \infty}f(t)=0[/imath]. What is [imath]f(\pi)[/imath] ? | 579396 | Solve a differential equation and evaluate the solution at a particular value of independent variable
If [imath]\frac{dy(x)}{dx}=(2-3i)y(x)[/imath] where [imath]i=\sqrt{-1}[/imath], what is the value of [imath]y(\pi)[/imath]? |
507930 | How prove this inequality: [imath]\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|[/imath]?
Let [imath]x_{1},x_{2},\cdots,x_{n}[/imath] be real numbers. Show that [imath]\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|.[/imath] I think this problem may be solved using nice methods, but I can't find them yet; I know this may be of use: [imath]|a|+|b|\ge |a+b|.[/imath] But I can't make it work. Thank you everyone. | 73945 | Stuck trying to prove an inequality
I have been trying to prove (the left half of) the following inequality: [imath] \underbrace{\sum_i \sum_j |x_i| \le \sum_i \sum_j |x_i + x_j|}_\textrm{?} \le 2 \sum_i \sum_j |x_i|[/imath] (All [imath]x_i[/imath]s are arbitrary reals and sums are over [imath]1, 2, \dots, n[/imath]) The right half follows by a simple application of the triangle inequality, but the left half isn't as simple. I tried induction, and I could reduce the above problem to proving the following statement: [imath] \sum_i |x_i + x_{n+1}| + \sum_{i=1}^{n+1} |x_i + x_{n+1}| \ge \sum_{i=1}^{n+1} |x_i| + n|x_{n+1}|[/imath] or more simply (replacing [imath]x_{n+1}[/imath] by [imath]y[/imath]), [imath] 2\sum_i |x_i + y| \ge \sum_i |x_i| + (n-1)|y|[/imath] However, I don't know if the above "reduction" is any simpler than the original problem itself! Can I continue the above line of thought, or is there an easier way to solve this problem? Any help is appreciated! :) |
580037 | Examples of canonical projections that are not epimorphisms and canonical injections that are not
Although in [imath]\mathsf{Set}[/imath], canonical projections from a product are surjective and canonical injections to a coproduct are in fact injections, there seems to be nothing forcing this to be the case elsewhere, and indeed Wikipedia indicates without proof that they needn't be epic/monic. Unfortunately, I'm a rank beginner and have not yet knowingly read about a category less rich in monomorphisms and epimorphisms than [imath]\mathsf{Set}[/imath] (though Aluffi has already introduced a couple examples that are richer in them), so I don't know where to try to find a counterexample. Any hints? To clarify: I am seeking canonical injections to coproducts that are not monomorphisms and/or canonical projections from products that are not epimorphisms. | 238995 | Are all projection maps in a categorical product epic?
Are all [imath]\Pi A_\alpha \stackrel{\pi_i}\longrightarrow A_\alpha[/imath] projection maps epic, given that [imath]\Pi A_\alpha[/imath] be the product of [imath]A_\alpha[/imath]s? Of course, assuming that the product exists. |
580359 | Show that [imath]||x|-|y||≤|x-y|[/imath]
Can anyone help me show that: [imath]||x|-|y||≤|x-y|[/imath] I am new to proofs and I am not sure how I can show something as trivial as this! | 91181 | Proof for triangle inequality for vectors
Generally,the length of the sum of two vectors is not equal to the sum of their lengths. To see this consider the vectors [imath]u[/imath] and [imath]v[/imath] as shown below. By considering [imath]u[/imath] and [imath]v[/imath] as two sides of a triangle, we can see that the lengths of the third side is [imath]\| u + v \|[/imath] and we have [imath]\| u + v \| \leq \|u\| + \|v\|[/imath]. Under what circumstance equality occurs and how can one prove that? |
579284 | Prove that if [imath]R[/imath] is a principal ideal ring and [imath]S[/imath] is a multiplicatively closed subset of [imath]R[/imath] then [imath]S^{-1}R[/imath] is also a principal ideal ring.
Prove that if [imath]R[/imath] is a principal ideal ring and [imath]S[/imath] is a multiplicatively closed subset of [imath]R[/imath] then [imath]S^{-1}R[/imath] is also a principal ideal ring. Thanks for any insight. | 536624 | Is the localization of a PID a PID?
Let [imath]A[/imath] be a commutative ring and [imath]S[/imath] a multiplicative closed subset of [imath]A[/imath]. If [imath]A[/imath] is a PID, show that [imath]S^{-1}A[/imath] is a PID. I've taken an ideal [imath]I[/imath] of [imath]S^{-1}A[/imath] and I've tried to see that is generated by one element; the ideal [imath]I[/imath] has the form [imath]S^{-1}J[/imath] with [imath]J[/imath] an ideal of [imath]A[/imath]. [imath]J[/imath] is generated by one element but I can't see why [imath]I[/imath] has to be generated by one element, maybe I'm wrong. |
580003 | Module over a ring satisfying ACC
Let [imath]R[/imath] be a ring that satisfies ACC on the set of its left ideals and [imath]M[/imath] be a finite generated [imath]R[/imath]-module. Prove that every submodule of [imath]M[/imath] is finitely generated. I know that if [imath]M[/imath] satisfies ACC condition then every submodules of it are finite generated. I only want to show that [imath]M[/imath] satisfies ACC condition. Help me some hints. Thanks a lot. | 492832 | Suppose [imath]M[/imath] is a finitely generated module over [imath]R[/imath] where [imath]R[/imath] is a Noetherian ring. Prove that [imath]M[/imath] is a noetherian module.
Suppose [imath]M[/imath] is a finitely generated module over [imath]R[/imath] where [imath]R[/imath] is a Noetherian ring. Prove that [imath]M[/imath] is a noetherian module. Proof: Since [imath]M[/imath] is finitely generated module, there exists a free module [imath]N[/imath] such that [imath]M \cong N/\ker(f)[/imath] where [imath]f:N \rightarrow M[/imath]. Since [imath]N[/imath] is free, it can be expressed as direct sum of isomorphic copy of underlying ring, i.e. [imath]N \cong\bigoplus R[/imath]. Since [imath]R[/imath] is noetherian ring, [imath]\bigoplus R[/imath] is also noetherian. Since [imath]\ker(f)[/imath] is a submodule of [imath]\bigoplus R[/imath], [imath]\ker(f)[/imath] is also noetherian, which implies that [imath]M[/imath] is noetherian. Is my proof correct? Is it true that direct sum of noetherian ring is noetherian? Remark: Since [imath]N[/imath] and [imath]\ker(f)[/imath] are both noetherian, then [imath]N/\ker(f)[/imath] is also noetherian, and hence [imath]M[/imath] is also noetherian. Since [imath]M[/imath] is finitely generated, the free module [imath]N[/imath] is of finite rank, and hence is isomorphic to finitely many copies of [imath]R[/imath]. |
580607 | Generating pair for PSL(2,q)
Could anyone help me with an example of a pair of generating elements for the group [imath]\mathrm{PSL}(2,q)[/imath]? Here [imath]\mathrm{PSL}(2,q)[/imath] denotes the group of [imath]2\times 2[/imath] matrices with determinant [imath]1[/imath] and whose entries are elements of a finite field of order [imath]q[/imath] (alternatively, the quotient group of [imath]\mathrm{SL}(2,q)[/imath] by its center). Many thanks! | 460653 | Generators and relations for [imath]\text{PSL}(2, \mathbb{F}_q)[/imath]
Is there a nice presentation for the group [imath]\text{PSL}(2,\mathbb{F}_q)[/imath], for every prime [imath]q[/imath]? (I don't have any particular definition in mind for "nice", other than, say, a small number of generators and relations.) |
580788 | Prove by induction that [imath]a-b|a^n-b^n[/imath]
Given [imath]a,b,n \in \mathbb N[/imath], prove that [imath]a-b|a^n-b^n[/imath]. I think about induction. The assertion is obviously true for [imath]n=1[/imath]. If I assume that assertive is true for a given [imath]k \in \mathbb N[/imath], i.e.: [imath]a-b|a^k-b^k[/imath], I should be able to find that [imath]a-b|a^{k+1}-b^{k+1}[/imath], but I can't do it. Any help is welcome. Thanks! | 188657 | Why is [imath]a^n - b^n[/imath] divisible by [imath]a-b[/imath]?
I did some mathematical induction problems on divisibility [imath]9^n[/imath] [imath]-[/imath] [imath]2^n[/imath] is divisible by 7. [imath]4^n[/imath] [imath]-[/imath] [imath]1[/imath] is divisible by 3. [imath]9^n[/imath] [imath]-[/imath] [imath]4^n[/imath] is divisible by 5. Can these be generalized as [imath]a^n[/imath] [imath]-[/imath] $b^n[imath] = (a-b)N$, where N is an integer? But why is $a^n$ $-$ $b^n[/imath] = (a-b)N$ ? I also see that [imath]6^n[/imath] [imath]- 5n + 4[/imath] is divisible by [imath]5[/imath] which is [imath]6-5+4[/imath] and [imath]7^n[/imath][imath]+3n + 8[/imath] is divisible by [imath]9[/imath] which is [imath]7+3+8=18=9\cdot2[/imath]. Are they just a coincidence or is there a theory behind? Is it about modular arithmetic? |
577873 | Is [imath]xy = 1[/imath] connected ?
The graph of [imath]xy = 1[/imath] is connected in [imath]\mathbb{C}^2[/imath]. The above statement is true. Why? Please show reason. In [imath]\mathbb{R}^2[/imath] [imath]xy = 1[/imath] is not connected as it has two disjoint components in [imath]1[/imath]-st and [imath]3[/imath]-rd quadrant. I have no idea about this function in [imath]\mathbb{C}^2[/imath]. I do not know multivariate complex analysis. Thank you for your help. | 180615 | The graph of xy = 1 is connected or not
The graph of [imath]xy = 1[/imath] in [imath]\Bbb C^2[/imath] is connected. True or false? I know that it is not connected in [imath]\Bbb R^2[/imath], but what is the case of [imath]\Bbb C^2[/imath]? |
580799 | Calculus Related Rates Question
A baseball diamond is a square with side length [imath]90[/imath] ft. A batter hits the ball and runs toward first base with a speed of [imath]f(t)[/imath] ft/s after [imath]t[/imath] seconds. At what rate is the batter's distance to second base decreasing when the batter is halfway to first base? This is not a duplicate, because the speed is varying in this question compared to the one you guys linked it to. | 520676 | Speed of object towards a point not in the object's trajectory?
Trying to study for my mid-term, but I'm having slight difficulties understanding what I'm supposed to do in this one problem: A batter starts running towards first base at a constant speed of 6 m/s. The distance between each adjacent plate is 27.5 m. After running for 20 m, how fast is he approaching second base? At the same moment, how fast is he running away from third base? (see image below) This is what I have so far: Let [imath]d[/imath] be the distance the batter has run thus far The distance between the batter and first base is 7.5 m The distance between the batter and second base is [imath]\sqrt {27.5^2 + (27.5-d)^2}\ [/imath], or approx. 28.5044 m when [imath]d = 20[/imath] The distance between the batter and third base is [imath]\sqrt {27.5^2 + d^2}\ [/imath], or approx. 34.0037 m when [imath]d = 20[/imath] No need to hand feed me the answer, I'd just like a bit of insight on how to solve the problem. |
581775 | Complex numbers inequality
Let [imath]z[/imath] , [imath]w[/imath] [imath] \ \in \mathbb{C}[/imath] with [imath]|z|[/imath] , [imath]|w| < 1[/imath]. Then prove that [imath]|\frac{z-w}{1-z\bar{w}}| < 1[/imath]. What I have noticed : [imath]|z-w| = |w||\frac{z\bar{w}}{|w|} -1| [/imath] but I don't know how to proceed. | 506058 | Show that [imath]\left|\frac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1[/imath] when [imath]|\alpha|,|\beta| < 1[/imath]
This is the question I'm stumbling with: When [imath]|\alpha| < 1[/imath] and [imath]|\beta| < 1[/imath], show that: [imath]\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1[/imath] The chapter that contains this question contains (among others) the triangle inequalities: [imath]\left||z_1| - |z_2|\right| \le |z_1 + z_2| \le |z_1| + |z_2| [/imath] I've tried to use the triangle inequalities to increase the dividend and/or decrease the divisor: [imath]\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < \cfrac{|\alpha| +|\beta|}{\left|1-|\bar{\alpha}\beta|\right|}[/imath] But it's not clear if or why that would be smaller than one. I've also tried to multiply the equation by the conjugated divisor [imath]\cfrac{1-\alpha\bar{\beta}}{1-\alpha\bar{\beta}}[/imath], which gives a real divisor, but the equation does not appear solvable. Any hint would be much appreciated. |
581733 | The Solutions [imath](a,b,c,d)[/imath] of [imath]a^0+a^1+a^2+\cdots+a^{b-1}+a^b=c^d[/imath]
Let [imath]a\ge2, b\ge2,c\ge2,d\ge2\in\mathbb N[/imath]. Also, let [imath]N(a,b,c,d)[/imath] be the number of the solutions [imath](a,b,c,d)[/imath] of [imath]a^0+a^1+a^2+\cdots+a^{b-1}+a^b=c^d.[/imath] Then, here is my question. Question : Are the followings true? [imath](1)[/imath] [imath]N(a,b,c,2)=2[/imath]. [imath](2)[/imath] [imath]N(a,b,c,3)=1[/imath]. [imath](3)[/imath] There exist [imath]d[/imath] such that [imath]N(a,b,c,d)=0[/imath]. [imath](4)[/imath] For every [imath]d\ge2[/imath], [imath]N(a,b,c,d)[/imath] is finite. Example : [imath]{3}^{0}+{3}^1+{3}^2+3^3+3^4={11}^2,[/imath] [imath]{7}^{0}+{7}^1+{7}^2+7^3={20}^2,[/imath] [imath]{18}^{0}+{18}^1+{18}^2=7^3.[/imath] Motivation : I have not been able to find the other solutions even by using computer (I hope I don't miss anything). I don't have any good idea for answering these questions. I would like to know any relevant references. Can anyone help? | 152530 | Finding solutions to equation of the form [imath]1+x+x^{2} + \cdots + x^{m} = y^{n}[/imath]
Exercise [imath]12[/imath] in Section [imath]1.6[/imath] of Nathanson's : Methods in Number Theory book has the following question. When is the sum of a geometric progression equal to a power? Equivalently, what are the solutions of the exponential diophantine equation [imath]1+x+x^{2}+ \cdots +x^{m} = y^{n} \qquad \cdots \ (1)[/imath] in integers [imath]x,m,n,y[/imath] greater than [imath]2[/imath]? Check that \begin{align*} 1 + 3 + 3^{2} + 3^{3} + 3^{4} & = 11^{2}, \\\ 1 + 7 + 7^{2} + 7^{3} &= 20^{2}, \\\ 1 + 18 +18^{2} &= 7^{3}. \end{align*} These are the only known solutions of [imath](1)[/imath]. The Wikipedia link doesn't reveal much about the above question. My question here would be to ask the following: Are there any other known solutions to the above equation. Can we conjecture that this equation can have only finitely many solutions? Added: Alright. I had posted this question on Mathoverflow some time after I had posed here. This user by name Gjergji Zaimi had actually given me a link which tells more about this particular question. Here is the link: https://mathoverflow.net/questions/58697/ |
581630 | Question about cyclic (Renew Question)
At my Finding the "square root" of a permutation(Finding the "square root" of a permutation I explain it wrong , now I try to explain it better (I edit the old Q to). ========================================================= [imath]r[/imath] is odd, and [imath]ord(\alpha)=r[/imath]. ([imath]\alpha,\beta[/imath] are cycles ). Now, [imath]\alpha=(a_1\cdots a_n)[/imath]. I need to find [imath]\beta[/imath] That will make [imath]\alpha = \beta^2[/imath] How can I show it? What I tried: [imath]ord(\alpha)=r[/imath] so [imath]a^{r+1}=\alpha[/imath]. Let say that [imath]\beta=\alpha^{\frac{r+1}{2}}[/imath], [imath]\frac{r+1}{2}[/imath] is an integer, how do I prove that [imath]\alpha^\frac{r+1}{2}[/imath] is cyclic? Because then [imath]\alpha=\beta^2[/imath].... This is where I stuck... Thank you and sorry for the another Q, but now thing are more clear... | 581258 | Finding the "square root" of a permutation
Suppose [imath]r[/imath] is odd, and [imath]{\rm ord}(\alpha)=r[/imath]. ([imath]\alpha,\beta[/imath] are cycles.) Now, [imath]\alpha=(a_1\cdots a_n)[/imath]. I need to find [imath]\beta[/imath] that will make [imath]\alpha = \beta^2[/imath] How can I show it? What I tried? [imath]{\rm ord}(\alpha)=r[/imath] so [imath]a^{r+1}=\alpha[/imath]. Let say that [imath]\beta=\alpha^{\frac{r+1}{2}}[/imath], [imath]\frac{r+1}{2}[/imath] is an integer, how do I prove that [imath]\alpha^\frac{r+1}{2}[/imath] is a cycle? Because then [imath]\alpha=\beta^2[/imath]... Thank you! |
581810 | What the symbol [imath]\subseteq[/imath] represents generally?
My book says that [imath]\subset[/imath] is used to represente any subset, proper or improper, needing in this case to show the anti symmetric property of sets. ([imath]A = B \iff A \subset B \, \, \wedge \,\, B \subset A)[/imath] And [imath]\subseteq[/imath] is used specifically to represent improper subsets. (In other words, [imath]A \subseteq B \iff A = B) [/imath] But i saw so many articles using [imath]\subseteq[/imath] and not [imath]\subset[/imath] then i am kinda suspicious about this definition. If this definition is correct, why [imath]\subseteq[/imath] it's so used? Just because it helps to prove the equality betwen the sets? Thanks for the help. | 50253 | [imath]\subset[/imath] vs [imath]\subseteq[/imath] when *not* referring to strict inclusion
Inspired by the confusion in the comments on this question: I always thought that the standard was to read [imath]\subset[/imath] as "is a strict subset of", and [imath]\subseteq[/imath] could mean proper or improper inclusion. Was I wrong? |
582073 | Prove that [imath]\sum_{k=0}^n\frac{1}{k!}\geq \left(1+\frac{1}{n}\right)^n[/imath]
It basically says it all in the title. I tried solving the inequality using the bernoulli inequality somehow [imath]\dfrac{\displaystyle\sum_{k=0}^n\frac{1}{k!}}{(1+\frac{1}{n})^n}\geq 1,[/imath] but the factorial gets me everytime. Also I've tried induction without success: [imath]\displaystyle\sum_{k=0}^{n+1}\frac{1}{k!}\stackrel{IV}\geq \left(1+\frac{1}{n}\right)^n+\frac{1}{(n+1)!}[/imath] | 404916 | A binomial inequality with factorial fractions: [imath]\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}[/imath]
Prove that [imath]\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}[/imath] for [imath]n>1 , n \in \mathbb{N}[/imath]. |
582017 | How do i prove an *infinite* product of connected spaces is connected?
Let [imath](X_i,\tau_i)_{i\in I}[/imath] be a family of connected spaces. I couls prove this when [imath]I[/imath] is finite, but how do i prove this when [imath]I[/imath] is infinite? | 257981 | Product of connected spaces - Proof
I'm working on the "iff"-relation given by: [imath]X=\prod_{i\in I}X_i[/imath] is connected iff each [imath]X_i[/imath] non-empty is connected for all [imath]i\in I[/imath]. I could prove the "[imath]\Rightarrow[/imath]"-direction very easyly. I also proved that a finite product of connected spaces is connected. Now i want to prove the following: Choose [imath]z=(z_i)\in\prod_{i\in I}X_i[/imath]. For every finite subset [imath]J\subset I[/imath] is the set [imath]X_J:=\left\{x\in X:x_i=z_i\ \forall I-J\right\}[/imath] connected. I have given the follwoing proof: This set is homeomorphic with a finite product [imath]\prod_{j\in J}X_j[/imath] given by the map defined by: [imath]x=(x_j)_{j\in J}[/imath] mapped on [imath]y=(y_i)_{i\in I}[/imath] such that [imath]y_j=x_j[/imath] if [imath]j\in J[/imath] and [imath]y_j=z_j[/imath] if [imath]j\notin J[/imath]. This mapping is continuous and injective (and also the inverse is continous since it is the projection map). But then we know that [imath]X_J[/imath] is connected since every finite product is connected (if the components are connected). Is this proof correct? The only thing i have to prove now is that [imath]Y=\cup_{J\subset I,J\ finite}X_J[/imath] is dense in X. How do I do that? Can someone help? Thank you |
577917 | Find four groups of order 20 not isomorphic to each other.
Find four groups of order 20 not isomorphic to each other and prove why they aren't isomorphic. So far I thought of [imath]\mathbb Z_{20}[/imath], [imath]\mathbb Z_2 \oplus\mathbb Z_{10}[/imath], and [imath]D_{10}[/imath] (dihedral group), but I can't find another one. Would [imath]U(50)[/imath] work? I know it has order 20 and is cyclic but I'm not exactly sure how to move from there. Can someone to point me on the right direction? | 585379 | Non-Isomorphic groups of order 20
How many and which distinct groups of order 20 that are not isomorphic? Why are they not isomorphic? I have the [imath]Z_{20}, Z_2+Z_{10}, D_{10}[/imath]. There are more than 4. I don't know the limit. |
11085 | Height of a tetrahedron
How do I calculate the height of a regular tetrahedron having side length [imath]1[/imath] ? Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table? | 2922521 | Finding the height of a regular tetrahedron
The length of an edge of a regular tetrahedron is [imath]8cm[/imath]. Find the total area of surfaces and the volume I've been able to find the area of all surfaces, which is [imath](4\times27.713) cm^2[/imath] but I'm not able to find the volume. There's no height given. How can I derive the height? |
580288 | Dimension of direct sum of vector spaces
Let [imath]V[/imath] and [imath]W[/imath] be finite dimensional vector spaces on a field [imath]F[/imath]. Show that [imath]\dim(V\oplus W) = \dim V +\dim W[/imath]. My idea: let [imath]\dim V=n[/imath] and [imath]\dim W=n[/imath]. So [imath]\mathcal{A}=[/imath] {[imath]{v_1 , v_2 ,... , v_n} [/imath]} and [imath]\mathcal{B}=[/imath]{[imath]{w_1, w_2 ,... ,w_m}[/imath]} this means [imath]\forall v\in V[/imath], [imath]v=x_1 v_1 + ... +x_n v_n [/imath] such that [imath]x_1 , ... , x_n \in F[/imath] . Similarly, [imath]\forall w\in W[/imath] , [imath]w= y_1 w_1 + ... + y_m w_m[/imath] such that [imath]y_1 , ... , y_m \in F[/imath] . I know as [imath]V[/imath] , [imath]W[/imath] have finite basis [imath]V\oplus W[/imath] has finite basis, I mean, [imath]\forall (v,w)\in V\oplus W[/imath] , [imath](v,w)=a_1 (v_1 ,w_1 ) + ... + a_p (v_p , w_p )[/imath] such that [imath]a_1 , ... , a_p \in F[/imath] . Let [imath]p[/imath] is minimum between [imath]{m,n}[/imath] .Beside, [imath]\forall (v,w)\in V\oplus W[/imath] , [imath](v,w)=(v,0)+(0,w)[/imath] . So we can say, [imath](v,w)= x_1 (v_1 ,0) + y_1 (0,w) +...+ x_p ( v_p , 0) + y_p (0 , w_p)[/imath]. I want to know how I have to continue to get correct proof. Also, I want to write in mathematical way. I will appreciate any help. | 395130 | Dimensions of vector subspaces in a direct sum are additive
[imath]V = U_1\oplus U_2~\oplus~...~ \oplus~ U_n~(\dim V < ∞)[/imath] [imath]\implies \dim V = \dim U_1 + \dim U_2 + ... + \dim U_n.[/imath] [Using the result if [imath]B_i[/imath] is a basis of [imath]U_i[/imath] then [imath]\cup_{i=1}^n B_i[/imath] is a basis of [imath]V[/imath]] Then it suffices to show [imath]U_i\cap U_j-\{0\}=\emptyset[/imath] for [imath]i\ne j.[/imath] If not, let [imath]v\in U_i\cap U_j-\{0\}.[/imath] Then \begin{align*} v=&0\,(\in U_1)+0\,(\in U_2)\,+\ldots+0\,(\in U_{i-1})+v\,(\in U_{i})+0\,(\in U_{i+1})+\ldots\\ & +\,0\,(\in U_j)+\ldots+0\,(\in U_{n})\\ =&0\,(\in U_1)+0\,(\in U_2)+\ldots+0\,(\in U_i)+\ldots+0\,(\in U_{j-1})+\,v(\in U_{j})\\ & +\,0\,(\in U_{j+1})+\ldots+0\,(\in U_{n}). \end{align*} Hence [imath]v[/imath] fails to have a unique linear sum of elements of [imath]U_i's.[/imath] Hence etc ... Am I right? |
582862 | Why does [imath]( \operatorname e^x)' = \operatorname e^x?[/imath]
It's known the the derivative of exponential function [imath]a^x[/imath] is [imath]xa^{x-1}[/imath]. If I play [imath]e[/imath] as [imath]a[/imath], we'll get [imath](a^x = \operatorname e^x)' = x \operatorname e^{x-1}[/imath]. Why does [imath](\operatorname e^x)' = \operatorname e^x[/imath]? | 403884 | Why isn't the derivative of [imath]e^x[/imath] equal to [imath]xe^{(x-1)}[/imath]?
When we take a derivative of a function where the power rule applies, e.g. [imath]x^3[/imath], we multiply the function by the exponent and subtract the current exponent by one, receiving [imath]3x^2[/imath]. Using this method, why is it that the derivative for [imath]e^x[/imath] equal to itself as opposed to [imath]xe^{x-1}[/imath]? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help. |
583281 | Number rings and (round) parentheses versus (square) brackets
Is there a reason for the difference in the use of parentheses versus brackets as used in algebraic extensions. For example, when the field rational numbers [imath]{\mathbb{Q}}[/imath] extended with [imath]i = \sqrt{-1}[/imath] is denoted [imath]{\mathbb{Q}}(i)[/imath], whereas the ring of integers [imath]{\mathbb{Z}}[/imath] extended with [imath]i[/imath] is denoted [imath]{\mathbb{Z}}[i][/imath]? In particular, does this difference in notation in itself carry any meaning beyond the applications to rings versus fields, for example? Could, for example, '[imath]{\mathbb{Q}}[i][/imath]' and '[imath]{\mathbb{Z}}(i)[/imath]' (or similar) have useful, but different, meanings? Brackets are also used in the context of rings of polynomials (such as [imath]{\mathbb{Z}}[X][/imath], [imath]{\mathbb{Q}}[X][/imath] and [imath]{\mathbb{R}}[X][/imath], etc.). Is this related? Thanks. | 472497 | Field Extension Notation
I've seen similar questions asked here, but I've not been able to find a comprehensive answer. I know that for a ring [imath]R[/imath], [imath]R[X][/imath] denotes the ring of polynomials over [imath]R[/imath] and [imath]R(X)[/imath] denotes the field of fractions of [imath]R[X][/imath]. But if [imath]\alpha \in S[/imath], where [imath]S \supseteq R[/imath] are rings, what is the distinction between [imath]R[\alpha][/imath] and [imath]R(\alpha)[/imath]? It is my understanding that if [imath]R[/imath] is a field then [imath]R[\alpha] \cong R(\alpha)[/imath], but that this is not generally true for any ring. Is this correct? Adding to my confusion is the fact that [imath]\mathbb Z[i][/imath] and [imath]\mathbb Z(i)[/imath] are used interchangeably, despite [imath]\mathbb Z[/imath] not being a field. However, it's clear to me that they are equivalent ( i.e. [imath]\mathbb Z[i] = \mathbb Z(i) = \{ a + bi | a,b \in \mathbb Z \}[/imath]); is this because [imath]i[/imath] is algebraic over [imath]\mathbb Z[/imath]? Or for some other reason? Thanks for your help! |
403924 | [imath]x^p-c[/imath] has no root in a field [imath]F[/imath] if and only if [imath]x^p-c[/imath] is irreducible?
Hungerford's book of algebra has exercise [imath]6[/imath] chapter [imath]3[/imath] section [imath]6[/imath] [Probably impossible with the tools at hand.]: Let [imath]p \in \mathbb{Z}[/imath] be a prime; let [imath]F[/imath] be a field and let [imath]c \in F[/imath]. Then [imath]x^p - c[/imath] is irreducible in [imath]F[x][/imath] if and only if [imath]x^p - c[/imath] has no root in [imath]F[/imath]. [Hint: consider two cases: char[imath](F) = p[/imath] and char[imath](F)[/imath] different of [imath]p[/imath].] I have attempted this a lot. Anyone has an answer? | 1006141 | [imath]f(x)=x^p-a[/imath] is either ireducible or has a root?
Let [imath]p[/imath] be a prime number. Prove that for any field [imath]k[/imath] and any [imath]a\in k[/imath], the polynomial [imath]f(x)=x^p-a[/imath] is either irreducible or has a root. I think if [imath]\operatorname{Char}k=0[/imath] then [imath]f[/imath] is an irreducible polynomial and if [imath]\operatorname{Char}k=p[/imath] for [imath]p[/imath] a prime number then [imath]f=(x-a)^p[/imath] so [imath]f[/imath] has a root. This problem is in Galois Theory, by Miles Reid. |
584198 | (Q,+) and (C*,+) has no finite index subgroup
How to prove ([imath]\mathbb Q[/imath],+) and ([imath]\mathbb C^*[/imath],+) has no finite index subgroup? | 182311 | Subgroup of [imath]\mathbb{Q}[/imath] with finite index
Consider the group [imath]\mathbb{Q}[/imath] under addition of rational numbers. If [imath]H[/imath] is a subgroup of [imath]\mathbb{Q}[/imath] with finite index, then [imath]H = \mathbb{Q}[/imath]. I just saw this on our exam earlier and was stumped on how to show this. Any ideas? |
584111 | Integrating this complicated integral for statistics
I want to show that : [imath] \int_{-\infty}^{\infty} e^\frac{-u^2}{2} du = \sqrt{2\pi} [/imath] Is there an elementary way using the tools of Calculus II to do this type of integration? I have not studied numerical analysis yet. | 429632 | Gaussian Integral
Consider the following Gaussian Integral [imath]I = \int_{-\infty}^{\infty} e^{-x^2} \ dx[/imath] The usual trick to calculate this is to consider [imath]I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \ dx \right) \left(\int_{-\infty}^{\infty} e^{-y^{2}} \ dy \right)[/imath] and convert to polar coordinates. We get [imath]\sqrt{\pi}[/imath] as the answer. Is it possible to get the same answer by considering [imath]I^{3}, I^{4}, \dots, I^{n}[/imath]? |
490946 | Proving that all sets of size n have [imath]2^n[/imath] subsets
I've tried to construct a proof for this using recursion. My knowledge of recursion and set theory in general is quite limited so I'd appreciate some feedback! The claim in symbolic logic: [imath]\forall n \in \mathbb N, \exists u \in U, S(u) \wedge |u| = n \Rightarrow \mathcal |P(u)| = 2^n[/imath] where: [imath]U:[/imath] the set of everything [imath]S(x):[/imath] x is a set Initial Value Assume [imath]n = 0[/imath] Then [imath]\exists u \in U, |u| = 0[/imath] Then [imath] u = \varnothing [/imath] Then [imath] |\mathcal P(u)| = | \{ \varnothing \} | = 1 = 2^0 [/imath] Prove: [imath]\forall n \in \mathbb N, \exists u, x \in U, [S(u) \wedge (x \notin u) \wedge (|u| = n) \wedge \mathcal |P(u)| = 2^n] \Rightarrow [(|u \cup \{x\}| = n + 1) \wedge (|\mathcal P(u \cup \{x\})| = 2^{n + 1})][/imath] Assume [imath]|u| = n[/imath] Then [imath] |\mathcal P(u)| = 2^n[/imath] Then [imath] |u \cup \{x\} | = n + 1[/imath] because [imath] |\{x\}| = 1[/imath] Then [imath] \mathcal |P(u \cup \{x\})| = |\mathcal P(u)| * |\mathcal P(\{x\})| = 2^n * 2 = 2^{n+1} [/imath] Is this convincing enough or do I need to add more? EDIT: Fixed notation | 62172 | The number of subsets of a set of cardinality [imath]n[/imath]
Please help with this question. Show that for a finite set [imath]A[/imath] of cardinality [imath]n[/imath], the cardinality of [imath]$P(A)$[/imath] is [imath]2^n[/imath], where [imath]P(A)[/imath] is the power set of [imath]A[/imath]. Thank you in advance for any help that is given. |
584537 | Does there exist an infinite [imath]S \subset \Bbb R^3[/imath] such that any three vectors in [imath]S[/imath] are linearly independent?
Does there exist an infinite subset [imath]S \subset \Bbb R^3[/imath] such that any three vectors in [imath]S[/imath] are linearly independent? | 257479 | There exist an infinite subset [imath]S\subseteq\mathbb{R}^3[/imath] such that any three vectors in [imath]S[/imath] are linearly independent.
Could anyone just give me hint for this one? There exist an infinite subset [imath]S\subseteq\mathbb{R}^3[/imath] such that any three vectors in [imath]S[/imath] are linearly independent. True or false? |
584882 | [imath]\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} +\cdots+{1 \over {\sqrt n }}\right)[/imath]
[imath]\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }}+\cdots+{1 \over {\sqrt n }}\right)[/imath] ( Without use of integrals ). I was able to squeeze it from the bottom to [imath]{\lim }=1[/imath], but that's not good enough. I'd be glad for help. | 238753 | Evaluation of the limit [imath]\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)[/imath]
Evaluate the limit : [imath]\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)[/imath] I can use the sandwich principle, certain convergence criteria, Cesaro mean theorem, limit arithmetic.. things around this area. Any help would be greatly appreciated, thanks! Sorry for not elaborating more at the beginning, rookie first-post mistake I suppose. :) |
584863 | Binary string with even and odd number of 1s
How could it be shown that the number of binary string of length k with an even number of 1s is the same as those with an odd number of 1s. Eg. for [imath]k = 3[/imath] : Binary string length 3 with even amount of 1s : [imath]000 ; 011 ; 110 ; 101[/imath] Binary string length 3 with odd amount of 1s : [imath]111 ; 100 ; 010 ;001[/imath] Both of these set have the same cardinality. I'd like to see a generalization for any [imath]k\in\mathbb{N}[/imath]. If think we need to show there exist a bijection between the set of binary string of length [imath]k[/imath] with even number of 1s and and the one with odd number of 1s in order to prove they have the same cardinality, but I'm not sure how to define such a function. Thanks | 176114 | Count the number of n-bit strings with an even number of zeros.
I am currently self-studying introductory combinatorics by reading Introduction to combinatorial mathematics. I am currently in the first chapter, and I have a question regarding one of the examples. The question was asking to count the number of n-bit strings with an even number of zeros. The answer is of course [imath]2^{n-1}[/imath]. The author gave 2 solutions. I however didn't completely understand what I think is the straightforward one. The solution I got was that he took out 1 bit, leaving [imath](n-1)[/imath] bits, if the number of zeros is even in the [imath](n-1)[/imath]-bit number, then he will just append a 1, if not then he will append a zero. So in the end we just needed to count the number of [imath](n-1)[/imath]-bit strings. The other solution (the straightforward one) that I didn't understand examined the symmetry that half of the [imath]2^n[/imath] must have an even number of zeros, and the other half will have an odd number of zeros. I just don't get why this property must hold. I can understand that half of the [imath]2^n[/imath] numbers will have even parity, but I can't see how it holds for the parity of the number of zero or one bits. If anyone can show me how that property holds, I'd be very grateful. I'd also be interested to see different explanations and proofs if possible. Thank you. |
584728 | Proving the sum of two functions is [imath]\Theta[/imath] of the max of those functions?
Suppose that the functions [imath]\ f_1, f_2, g_1, g_2: \mathbb{N}\to\mathbb{R}[/imath] (set of real numbers greater or equal to 0) are such that [imath]\ f_1 \in \Theta \ (g_1)[/imath] and [imath]\ f_2 \in \Theta \ (g_2)[/imath]. Prove that [imath] \ f_1 + f_2 \in \Theta (\max\{\ g_1, g_2\})[/imath] I have no idea how to go about this proof. Could someone please explain? Thank you! | 582889 | Suppose [imath]f_1 \in \Theta(g_1) \land f_2 \in \Theta(g_2)[/imath]. Prove [imath]f_1 + f_2 \in \Theta(\max\{g_1, g_2\})[/imath].
I need to prove that [imath]f_1 \in \Theta(g_1) \land f_2 \in \Theta(g_2) \implies f_1 + f_2 \in \Theta(\max\{g_1, g_2\})[/imath] This question is relevant, but I have a slightly different case, so I don't know how to translate it into this one, because I need to turn the [imath]f[/imath]'s into [imath]g[/imath]'s. Am I allowed to say [imath]\max\{f_1, f_2\} \in \Theta(\max\{g_1, g_2\})[/imath]? Or even [imath]\Theta(\max\{f_1, f_2\}) = \Theta(\max\{g_1, g_2\})[/imath]? |
585092 | How to prove [imath](\frac{n+1}{n})^n[/imath]
Prove that [imath](\frac{n+1}{n})^n<n\quad [/imath] for any n=3,4,5... by using induction. For n=3 is true. and lets assume [imath](\frac{n+1}{n})^n<n[/imath] is true. we must show that [imath](\frac{n+2}{n+1})^{n+1}<n+1[/imath] is true. How can I continue? | 582459 | Math Induction Proof: [imath](1+\frac1n)^n < n[/imath]
So I have to prove: For each natural number greater than or equal to 3, [imath](1+\frac1n)^n<n[/imath] My work: Basis step: [imath]n=3[/imath] [imath]\left(1+\frac13\right)^3<3[/imath] [imath]\left(\frac43\right)^3<3[/imath] [imath]\left(\frac{64}{27}\right)<3[/imath] which is true. Now the inductive step, assume [imath]P(k)=\left(1+\frac1k\right)^k<k[/imath] to be true and prove [imath]P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}<k+1[/imath]. This is where I am stuck because usually you add or multiply by [imath]k+1[/imath] or some similar term. |
585387 | Intuitive explanation for [imath]E[X]=\int_{0}^\infty P(X\ge x) dx[/imath]
I have come across two proofs for this on wikipedia, but I cannot wrap my head around the intuition behind this. Why should this hold? I am looking for a possible geometric interpretation of the area under the curve. | 536442 | Intuitive explanation for [imath]\mathbb{E}X= \int_0^\infty 1-F(x) \, dx[/imath]
I can see by manipulating the expression why [imath]\mathbb{E}X[/imath] works out to be [imath]\int_0^\infty 1-F(x)\,dx[/imath], where [imath]F[/imath] is the distribution function of [imath]X[/imath], but what is an intuitive explanation for why that is true? If at each point we sum the probability [imath]\mathbb{P}(X>x)[/imath], why should we end up with the expectation? Thanks |
584924 | characteristic function of stable symmetric distribution
I'm stuck with one problem of my homework: "Show that [imath]\phi\left(t\right) = {\rm e}^{−\left\vert\,t\,\right\vert^{\,\alpha}}\,, t \in {\mathbb R}[/imath], is a characteristic function only for [imath]0 \leq \alpha \leq 2[/imath]". I'm trying with Polya's criterion but I cannot use it since the second derivative depends on t and it shouldn't [imath]\left(~\mbox{it should be bigger than zero on}\ {\mathbb R}_{+}~\right)[/imath]. So, i tried with Bochner’s theorem, and again I cannot understand why it should be positive definite only for 0 ≤ α ≤ 2. Could someone give me an idea on how to solve this problem? Infinitely thanks | 76369 | For symmetric stable distributions, why is [imath]\alpha \le 2[/imath]?
I'm preparing a lecture on stable distributions, and I'm trying to find a simple explanation of the following fact. Suppose we are trying to come up with stable distributions. From the definition, it's clear that a distribution is stable iff its characteristic function [imath]\phi[/imath] satisfies [imath]\phi(t)^n = e^{i t b_n} \phi(a_n t)[/imath]. The normal distribution, with chf [imath]\phi(t) = e^{-t^2/2}[/imath] clearly satisfies this with [imath]b_n = 0[/imath], [imath]a_n = \sqrt{n}[/imath]. This suggests that we look for distributions with chfs of the form [imath]\phi(t) = e^{-c |t|^\alpha}[/imath]. For [imath]0 \le \alpha \le 2[/imath], this is indeed a chf, and there is a nice proof in Durrett's book, constructing it as a weak limit using Lévy's continuity theorem. But: For [imath]\alpha > 2[/imath], is there a simple reason why [imath]\phi(t) = e^{-c |t|^\alpha}[/imath] cannot be a chf? Breiman's Probability proves a general formula for the chf of a stable distribution, using a representation formula for infinitely divisible distributions, but it's more work than I want to do for this. |
366714 | Sequence of continuous functions which converges to a continuous limit
Any help with this: construct a sequence of continuous functions defined on [imath] [0,1] [/imath] which converges pointwise but not uniformly to a continuous limit ? Thank you. | 82190 | Uniform convergence for sequences of functions
Please help me out with this problem: Let [imath]f_n(x) : [0,1] \to \mathbb R[/imath] be a sequence of continuous functions convergent at every [imath]x \in [0,1][/imath] to a continuous function [imath]f: [0,1] \to \mathbb R[/imath] . Does [imath]f_n[/imath] converge uniformly to [imath]f[/imath]? My first solution: I read this example in a different question posted earlier and I think it works: Take a sequence [imath]f_n:[0,1] \to \mathbb R[/imath] such that [imath]f_n[/imath] increases linearly from [imath]0[/imath] to [imath]1[/imath] on the interval [imath]\left[ 0,\frac{1}{n} \right][/imath], decreases linearly from [imath]1[/imath] to [imath]0[/imath] on the intreval [imath]\left[ \frac{1}{n}, \frac{2}{n} \right][/imath], and is [imath]0[/imath] on [imath]\left[ \frac{2}{n},1 \right][/imath]. Then [imath]f_{n} \to 0[/imath] nonuniformly. Is this answer true? Also, if anyone can give some examples of such function with detailed proof, I will very appreciative? |
530594 | [imath]C^1[/imath] condition implies locally Lipschitz
Let [imath]A[/imath] be open in [imath]\mathbb{R}^m[/imath]; let [imath]g:A\rightarrow\mathbb{R}^n[/imath]. If [imath]S\subseteq A[/imath], we say that [imath]S[/imath] satisfies the Lipschitz condition on [imath]S[/imath] if the function [imath]\lambda(x,y)=|g(x)-g(y)|/|x-y|[/imath] is bounded for [imath]x\neq y\in S[/imath]. We say that [imath]g[/imath] is locally Lipschitz if each point of [imath]A[/imath] has a neighborhood on which [imath]g[/imath] satisfies the Lipschitz condition. Show that if [imath]g[/imath] is of class [imath]C^1[/imath], then [imath]g[/imath] is locally Lipschitz. If [imath]g[/imath] is of class [imath]C^1[/imath], that means the component derivatives of [imath]g[/imath] are continuous. The [imath]i[/imath]th-component derivative ([imath]i=1,2,\ldots,m[/imath]) at a point [imath]r\in A[/imath] is [imath]D_ig(r)=\lim_{t\rightarrow 0}\dfrac{g(r+te_i)-g(r)}{t}[/imath]. So this function is continuous over all points [imath]r\in A[/imath]. How can I deduce local Lipschitz from here? | 200637 | Locally continuously differentiable implies locally Lipschitz
I am interesting in the following result: Let [imath]X[/imath] be a normed space and [imath]f : X \to \mathbb{R}[/imath]. If [imath]f[/imath] is continuously differentiable in a neighborhood [imath]V[/imath] of a point [imath]x_0 \in X[/imath], then [imath]f[/imath] is locally Lipschitz at [imath]x_0[/imath]. Using mean value theorem, for all [imath]x,y[/imath] in an open ball [imath]B \subset V[/imath], there exists [imath]z \in [x,y][/imath] such that [imath]f(x)-f(y)= \langle f'(z), x-y \rangle[/imath] so [imath]|f(x)-f(y)| \leq \max\limits_{z \in B} ||f'(z)||. ||x-y||[/imath]. In finite dimension, we can suppose that the closure of [imath]B[/imath] is in [imath]V[/imath] so that [imath]\max\limits_{z \in B} ||f'(z)|| < + \infty[/imath] because [imath]z \mapsto f'(z)[/imath] is continuous and the closure of [imath]B[/imath] is compact. But what happens in infinite dimension? Is the result still true? |
585919 | Series [imath]\sum a_n[/imath] and [imath]\sum a_n^3[/imath]
Is possible to find series that: [imath]\sum_{n=1}^\infty a_n[/imath] is converges but [imath]\sum_{n=1}^\infty a_n^3[/imath] is diverges I thought about something with [imath](-1)^n[/imath] and Leibniz criterion but don't have idea. | 96666 | If [imath]\sum_{1}^{\infty}(a_n)^3[/imath] diverges, does [imath]\sum_{1}^{\infty}(a_n)[/imath]?
Per the title, if [imath]\sum_{1}^{\infty}(a_n)^3[/imath] diverges, does this imply that [imath]\sum_{1}^{\infty}(a_n)[/imath] diverges? I'd appreciate hints (!) for dealing with this excercise. EDIT Per the contrapositive, it is not given that [imath]a_n[/imath] converges absolutely, or that it is nonnegative for all [imath]n[/imath]. Thank you! |
586074 | Maximal value of an infinite set.
Consider a continuous function [imath]f[/imath] over an interval [imath][a,b][/imath]. Let [imath]S[/imath] be the set of all values that [imath]f(x)[/imath] takes over [imath]I[/imath]. Intuitively speaking, I believe this set has a maximal and minimal value. Is this the case? How can I prove it? | 436676 | Prove that a continuous function on a closed interval attains a maximum
As the title indicates, I'd like to prove the following: If [imath]f:\mathbb R\to\mathbb R[/imath] is a continuous function on [imath][a,b][/imath], then [imath]f[/imath] attains its maximum. Now, I do have a working proof: [imath][a,b][/imath] is a connected, compact space, which means that because [imath]f[/imath] is continuous, [imath]f([a,b])[/imath] is compact and connected as well. Therefore, [imath]f([a,b])[/imath] is a closed interval, which means it has both a minimum and, as desired, a maximum. What I would like, however, is a proof that doesn't require such general or sophisticated framework. In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness. Any comments, hints, or solutions are welcome and apreciated. |
351719 | Prove: [imath] \lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1 [/imath]
x is a real number between 0 and 1 [imath] \lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1 [/imath] SO far i know you need to do this to prove it: let b>1, prove that [imath]\root n \of b \rightarrow 1[/imath] as [imath]n \rightarrow\infty[/imath] I was given this hint: let [imath]a_n = \root n \of b - 1[/imath] , prove that [imath]a_n \rightarrow 0[/imath] Thanks! | 125588 | Limit of [imath]2^{1/n}[/imath] as [imath]n\to\infty[/imath] is 1
How do I prove that: [imath]\lim \limits_{n\to \infty}2^{1/n}=1[/imath] Thank you very much. |
586540 | Why [imath]\int_0^\infty e^{-t}\log tdt =-\gamma[/imath]?
What is [imath]\int_0^\infty e^{-t}\log tdt?[/imath] Plugging into Wolfram Alpha yields that it's [imath]-\gamma[/imath], where [imath]\gamma[/imath] is Euler's constant. How can we get it? Maybe by some contour integration? | 112304 | Showing that [imath]\gamma = -\int_0^{\infty} e^{-t} \log t \,dt[/imath], where [imath]\gamma[/imath] is the Euler-Mascheroni constant.
I'm trying to show that [imath]\lim_{n \to \infty} \left[\sum_{k=1}^{n} \frac{1}{k} - \log n\right] = -\int_0^{\infty} e^{-t} \log t \,dt.[/imath] In other words, I'm trying to show that the above definitions of the Euler-Mascheroni constant [imath]\gamma[/imath] are equivalent. In another post here (which I can't seem to find now) someone noted that [imath]\int_0^{\infty} e^{-t} \log t \,dt = \left.\frac{d}{dx} \int_0^{\infty} t^x e^{-t} \,dt \right|_{x=0} = \Gamma'(1) = \psi(1),[/imath] where [imath]\psi[/imath] is the digamma function. This may be a good place to start on the right-hand side. For the left-hand side I was tempted to represent the terms with integrals. It is not hard to show that [imath]\sum_{k=1}^{n} \frac{1}{k} = \int_0^1 \frac{1-x^n}{1-x} \,dx,[/imath] but I'm not sure this gets us anywhere. Any help would be greatly appreciated. |
586085 | Prove that [imath]\operatorname{dist}(x,M)=\frac {|f(x)|}{\| f \|}[/imath]
Let [imath]f\in E'[/imath] (dual space of [imath]E[/imath]), [imath]f\neq 0[/imath], [imath]M=\{f=0\}[/imath]. Prove that [imath]\operatorname{dist}(x,M)=\dfrac {|f(x)|}{\|f\|}[/imath]. My idea: [imath]\operatorname{dist}(x,M)=\inf\limits_{v\in M}\|x-v\|[/imath], [imath]|f(x-v)|=|f(x)|\leq \|f\|\cdot\|x-v\|\implies \|x-v\|\geq\dfrac {|f(x)|}{\|f\|}[/imath]. But I can't prove the opposite inequality. Can anyone help me? Thank you. | 253953 | Distance between point and linear Space
Suppose [imath]E[/imath] is a normed vector space. Let [imath]f[/imath] be a continuous linear functional on [imath]E[/imath] and denote by [imath]M[/imath] the Kernel of [imath]f[/imath]. Let [imath]x\in E[/imath]. How to show that [imath]\operatorname{dist}(x,M)=\displaystyle\inf_{y\in M}\|y-x\|=\frac{|f(x)|}{\|f\|}\, ? [/imath] |
549111 | What is [imath]\lim\limits_{n\to\infty}(\sqrt2-\sqrt[3]2)\cdots(\sqrt2-\sqrt[n]2)[/imath]? How to approach?
[imath]\lim\limits_{n\to\infty}(\sqrt2-\sqrt[3]2)(\sqrt2-\sqrt[4]2)(\sqrt2-\sqrt[5]2)\cdots(\sqrt2-\sqrt[n]2)[/imath] Could you tell me how to approach this kind of question? How do I find the limit of this sequence? I know that for very large [imath]n[/imath] the each bracket is more than [imath]1[/imath], so my guess is its going to infinity, how do I prove such a thing? | 452173 | Finding the limit of roots products [imath](\sqrt{2}-\sqrt[3]{2})(\sqrt{2}-\sqrt[4]{2})(\sqrt{2}-\sqrt[5]{2})\cdot \cdot \cdot (\sqrt{2}-\sqrt[n]{2})[/imath]
I need to find: [imath] \lim_{n \to \infty } (\sqrt{2}-\sqrt[3]{2})(\sqrt{2}-\sqrt[4]{2})(\sqrt{2}-\sqrt[5]{2})\cdot \cdot \cdot (\sqrt{2}-\sqrt[n]{2}) [/imath] So far, I think that [imath]0<\sqrt{2}-\sqrt[n]{2}<1[/imath], and it seems to me that the limit will approach zero but I can't figure how to show it mathematically. |
462065 | [imath]\int_{I^n}(\min_{1\leq i\leq n} x_i)^a dx[/imath]
Let [imath]I=[0,1][/imath]. Find [imath]\int_{I^n} \left(\min_{1\leq i\leq n} x_i \right)^a dx,[/imath] for [imath]a\in \mathbb{R}[/imath]. Please help to solve this. Thank you. | 181678 | Evaluating [imath]\int_{I^n} \left( \min_{1\le i \le n}x_i \right)^{\alpha}\,\, dx[/imath]
Let [imath]\alpha \in \mathbb R[/imath] and let's call [imath]I:=[0,1][/imath]. Evaluate [imath] \int_{I^n} \left( \min_{1\le i \le n}x_i \right)^{\alpha}\,\, dx. [/imath] Well, the case [imath]n=1[/imath] is easy and the integral equals [imath]\frac{1}{\alpha+1}[/imath], for every [imath]\mathbb R \ni \alpha \ne - 1[/imath]. I've done also the case [imath]n=2[/imath] and, if I'm not wrong, it's [imath]\displaystyle \frac{2}{(\alpha+1)(\alpha+2)}[/imath]. My big problem is that I cannot understand how to deal with the general case. Any ideas? Thanks in advance. |
587442 | cauchy product of two series diverges
For [imath]n \in \mathbb{N}[/imath] be [imath]a_n=b_n=\frac {(-1)^n}{\sqrt{n+1}}[/imath] and [imath]c_n=\sum\limits_{k=0}^{n}a_kb_{n-k} [/imath] Show that the series [imath]\sum\limits_{n=0}^{\infty}a_n[/imath], [imath]\sum\limits_{n=0}^{\infty}b_n[/imath] converge, but their cauchy produkt [imath]\sum\limits_{n=0}^{\infty}c_n[/imath] diverges. I've showed so far, that [imath]\frac {(-1)^n}{\sqrt{n+1}}[/imath] converges, but how can i show that their cauchy product [imath]\sum\limits_{n=0}^{\infty}c_n[/imath] diverges? | 519762 | Introduction to Analysis: Multiplication Theorem for Series
I've been stuck on this problem over the weekend so I decided to ask for some direction. The problem reads: "The multiplication theorem for series requires that the two series be absolutely convergent; if this condition is not met, their product may be divergent. Show that the series [imath]\sum_0^∞ \frac{(-1)^i}{\sqrt{1 + i}}[/imath] gives an example: it is conditionally convergent, but its product with itself is divergent. (Estimate the size of the odd terms [imath]c_{2n+1}[/imath] in the product.)" Someone earlier suggest I show the Cauchy product and show that it diverges. If I understand correctly, the two series must converge absolutely. If I can show that two do not converge, then I have shown what the problem ask for. So I tried and had this in mind. I reckon the prove is not correct because I do not fully understand how to prove Cauchy product, but it's an idea. Consider [imath]a_n[/imath] = [imath](-1)^n[/imath], [imath]b_n = \frac{1}{\sqrt{n+1}}[/imath]. The sum of their product, [imath]C_n[/imath] = [imath]\sum_0^{\infty} c_i[/imath] = [imath]\sum_0^{\infty}\frac{(-1)^i}{\sqrt{1 + i}}[/imath] conditional converges. Thus [imath]\forall[/imath][imath]\epsilon[/imath] > 0, [imath]\exists[/imath]N s.t. n, m [imath]\geq[/imath] N [imath]\rightarrow[/imath] |[imath]\sum_n^m c_i[/imath]| < [imath]\epsilon[/imath]. We assume without loss of generality that the series of [imath]a_n[/imath] and the series of [imath]b_n[/imath] does not converge absolutely. |[imath]\sum_0^{\infty} (-1)^i[/imath]| [imath]\leq[/imath] [imath]\sum_0^{\infty} |(-1)^i|[/imath] [imath]\leq[/imath] [imath]\sum_0^{\infty} 1^i[/imath] As n approaches infinity, the [imath]\sum_0^{\infty} 1^i[/imath] = [imath]1 + 1 + ... + 1 = \infty[/imath], thus diverging. |[imath]\sum_0^{\infty} \frac{1}{\sqrt{i+1}}[/imath]| = [imath]\sum_0^{\infty} |\frac{1}{\sqrt{i+1}}|[/imath] = [imath]\sum_0^{\infty} \frac{1}{\sqrt{i+1}}[/imath]. As n approaches infinity, the [imath]\sum_0^{\infty} \frac{1}{\sqrt{i+1}}[/imath] = [imath]\infty[/imath], thus diverging. So I understand the idea, I just don't know how to go about finishing up the proof using cauchy. Thank you for taking the time to read this and thanks in advance for commenting. |
587734 | Let n be a positive integer. Prove that:
Let n be a positive integer. Prove that: [imath]\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor[/imath] | 6087 | For [imath]n \in \mathbb{N}[/imath] [imath]\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}[/imath]
This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with. For [imath]n \in \mathbb{N}[/imath], prove that this identity is true: [imath]\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}[/imath] |
561228 | Prove a certain condition is satisfied by all non-abelian groups
How can I prove following problem in abstract algebra? Let [imath]G[/imath] is a finite non-abelian group. show that there exist elements [imath]a,g,h\in G[/imath] such that [imath]g\neq h, h=aga^{-1}[/imath] and [imath]gh=hg[/imath]. Please help me. Thanks in advance. | 401522 | Prove that [imath]\exists a,g,h\in G\colon h=aga^{-1}, g\neq h ,gh=hg[/imath] in a finite non-abelian group [imath]G[/imath].
Let [imath]G[/imath] be a finite and non-abelian group. How do I prove the following statement? [imath]\exists a,g,h\in G \colon\quad h=aga^{-1},\ g\neq h ,\ gh=hg.[/imath] Thanks in advance. |
587767 | How do you calculate this sum [imath]\sum_{n=1}^\infty nx^n[/imath]?
I can not find the function from which I have to start to calculate this power series. [imath]\sum_{n=1}^\infty nx^n[/imath] Any tips?. Thanks. | 333192 | Solve [imath]\sum nx^n[/imath]
I am trying to find a closed form solution for [imath]\sum_{n\ge0} nx^n\text{, where }\lvert x \rvert<1[/imath]. This solution makes sense to me: [imath]\sum_{n\ge0} x^n=(1-x)^{-1} \\ \frac{d}{d x} \sum_{n\ge0} x^n = \frac{d}{d x} (1-x)^{-1} \\ \sum_{n\ge0} nx^{n-1} = (1-x)^{-2} \\ x \sum_{n\ge0} n x^{n-1} = x(1-x)^{-2} \\ \sum_{n\ge0} nx^n=\frac x{(1-x)^2}[/imath] However, a book I am reading used the following method: [imath]\sum_{n\ge0}nx^n=\sum_{n\ge0}x\frac d{dx}x^n= x\frac d{dx}\sum\limits_{n\ge0}x^n=x\frac d{dx}\frac1{1-x}=\frac x{(1-x)^2}[/imath] This seems closely related to the solution I described above, but I am having difficulty understanding it. Can someone explain the method being used here? |
587888 | How find if a number is the sum of 2 perfect squares?
I'm looking for a way to find if a number is the sum of 2 perfect squares. given x find if there exists non negative integers a, b, m, n such that: [imath]a^m + b^n = x[/imath] | 587880 | Express as sum of powers
How to find if a number can be expressed as sum of two perfect powers. That is, given [imath]x[/imath], I have to find if there exists non negative integers [imath]a, b, m, n[/imath] such that [imath]a^m + b^n = x[/imath]. |
587900 | Taking a const out of a series
[imath]\sum\limits_{n = 1}^\infty {2{a_n}} [/imath] is a series. is it right to say the following? [imath]\mathop {\lim }\limits_{n \to \infty } {S_{n = }}\sum\limits_{n = 1}^\infty {2{a_n}} = 2\sum\limits_{n = 1}^\infty {{a_n}} [/imath] | 418150 | Elementary proof, convergence of a linear combination of convergent series
Could you tell me how to prove that if two series [imath] \sum_{n=0} ^{\infty}x_n, \sum_{n=0} ^{\infty} y_n[/imath] are convergent, then [imath]\sum_{n=0} ^{\infty}(\alpha \cdot x_n + \beta \cdot y_n)[/imath] is also convergent and its sum equals [imath]\alpha \cdot \sum_{n=0} ^{\infty} x_n + \beta \cdot \sum_{n=0} ^{\infty} y_n[/imath]? |
587992 | Proving that the ring [imath]R[/imath] of [imath]2\times 2[/imath] matrices over [imath]\Bbb{Q}[/imath] contains only two ideals: [imath](0)[/imath] and [imath]R[/imath].
There's a question in Herstein: Prove that the ring [imath]R[/imath] of [imath]2\times 2[/imath] matrices defined over [imath]\Bbb{Q}[/imath] contains only two ideals: [imath](0)[/imath] and [imath]R[/imath]. This seems to say that if I take any non-zero element, say [imath]a=\begin{pmatrix} 1&1\\1&1\end{pmatrix}[/imath], then [imath]aR=R[/imath]. This implies that for any [imath]\begin{pmatrix} p&q\\r&s\end{pmatrix}\in R[/imath], there exists [imath]\begin{pmatrix} f&g\\k&l\end{pmatrix}\in R[/imath] such that [imath]\begin{pmatrix} f&g\\k&l\end{pmatrix}\begin{pmatrix} 1&1\\1&1\end{pmatrix}=\begin{pmatrix} p&q\\r&s\end{pmatrix}[/imath]. It is clear that such a matrix [imath]\begin{pmatrix} f&g\\k&l\end{pmatrix}[/imath] need not exist in [imath]R[/imath], as the system of equations [imath]f+g=p[/imath] and [imath]f+g=q[/imath] does not have a solution unless [imath]p=q[/imath]. Is th question wrong then? Thanks in advance! | 22629 | Why is the ring of matrices over a field simple?
Denote by [imath]M_{n \times n}(k)[/imath] the ring of [imath]n[/imath] by [imath]n[/imath] matrices with coefficients in the field [imath]k[/imath]. Then why does this ring not contain any two-sided ideal? Thanks for any clarification, and this is an exercise from the notes of Commutative Algebra by Pete L Clark, of which I thought as simple but I cannot figure it out now. |
588310 | [imath]x^{p-1} + ... + x^2 + x + 1[/imath] is irreducible using Eisenstein's criterion?
Eisenstein's criterion says that for a prime [imath]p[/imath] if the following conditions are satisfied for a primitive polynomial, [imath]f(x)[/imath], then that polynomial is irreducible in [imath]\mathbb{Z}[x][/imath] [imath]p \mid a_0, a_0, ..., a_{n-1}[/imath] [imath]p \not\mid a_n[/imath] [imath]p^2 \not\mid a_0^2[/imath] Now I am almost certain I read somewhere that this lemma can be used to prove that the polynomial [imath]x^{p-1} + x^{p-2} + ... + x^2 + x + 1[/imath] is irreducible. All coefficients are [imath]1[/imath] though and we have have an awkward amount of terms, [imath]p - 1[/imath] terms as opposed to [imath]p[/imath] terms - so how can Eisenstein's criterion be applied here? | 87609 | Eisenstein Criterion with a twist
As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ([imath]a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x][/imath] is irreducible in [imath]\mathbb{Q}[/imath]) how do we prove that if [imath]p[/imath] is a prime, [imath]x^{p-1}+x^{p-2}+\dots+x+1[/imath] is irreducible over [imath]\mathbb{Q}[/imath]? |
588418 | Proof by Induction for a [imath]f_3 + f_6 + · · · + f_{3n} = \frac{1}{2} (f_{3n+2} - 1)[/imath]
The Fibonacci numbers are defined as follows: [imath]f_0 = 0,\ f_1 = 1[/imath], and for [imath]n ≥ 2,\ f_n = f_{n−1} +f_{n−2}[/imath]. Prove that for every positive integer [imath]n[/imath], [imath]f_3 + f_6 + \ldots + f3_n = \frac{1}{2} (f_{3n+2} - 1) [/imath] Here are my steps: Basis Step: For [imath]n ≥ 1[/imath] [imath]f_{3\cdot1} = \frac{1}{2} (f_{3\cdot 1+2} - 1) [/imath] LHS [imath]= f_3 = 2[/imath] RHS [imath]= \frac{1}{2} (f_5 - 1) = \frac{1}{2} (5 – 1) = 2[/imath] Induction Step: Assume that [imath]f_3 + f_6 + \ldots + f_{3n} = \frac{1}{2} (f_{3n+2} - 1)[/imath] for some [imath]n ≥ 1[/imath]. Show that [imath]f_3 + f_6 + \ldots + f_{3n} + f_{3(n+1)} = \frac{1}{2} (f_{3(n+1)+2} – 1)[/imath]. [imath]\frac{1}{2} (f_{3n+2} – 1) = \frac{1}{2} (f_{3(n+1)+2} – 1) + f_{3(n+1)}\ \text{(Basis)}\\ \text{LHS}=\frac{1}{2}(f_{3n+5}-1)+f_{3n+3}[/imath] I don't know how to proceed further. I understand what I have to do, but I don't know where to start reducing it using the fib definition. | 587326 | Fibonacci sequence proof
Prove the following: [imath]f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ [/imath] For [imath]n \ge 2[/imath] Well I got the basis out of the way, so now I need to use induction: So that [imath]P(k) \rightarrow P(k+1)[/imath] for some integer [imath]k \ge 2[/imath] So, here are my first steps: [imath] \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = \\ & = \frac 12(f_{3k+2}-1) + f_{3k+1} + f_{3k} + f_{3k+1} \\ & = \frac 12(f_{3k+2}-1) + f_{3k+1} + \frac 12(f_{3k+2}-1) + f_{3k+1} \\ & = f_{3k+2}-1 + 2 \cdot f_{3k+1} \end{align} [/imath] And the fun stops around here. I don't see how to get to the conclusion: [imath]\frac 12(f_{3k+5}-1) \\ [/imath]. Any help from this point would be great. |
588648 | What is easiest way to prove that [imath]\sqrt 8/2[/imath] is equal to [imath]\sqrt 2[/imath].
What is the most easy way I can prove [imath]\sqrt 8/2[/imath] is equal to [imath]\sqrt 2[/imath] I have done [imath](\sqrt 8/2)^2[/imath] but at the end it gives me [imath]\pm \sqrt 2[/imath] and not positive [imath]\sqrt 2[/imath] So how? | 235998 | Why is [imath]\sqrt{8}/2[/imath] equal to [imath]\sqrt{2}[/imath]?
I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why [imath]\sqrt{8}/2[/imath] equal to [imath]\sqrt{2}[/imath]. Thanks in advance |
35462 | What is the difference between "family" and "set"?
What is the difference between "family" and "set"? The definition of "family" on mathworld (http://mathworld.wolfram.com/Family.html) is a collection of objects of the form [imath]\{a_i\}_{i \in I}[/imath], where [imath]I[/imath] is an index set. But, I think a set can also be represented in this form. So, what is the difference between the concept family and the concept set? Is there any example of a collection of objects that is a family, but not a set, or reversely? Many thanks! | 2338870 | What is difference between set and family?
I am curious of that words; 'family' and 'set. Some use the set of the blah-blah-blah~. However, the other some use the family of the blah-blah-blah~. For example, we express [imath]X=\{v_i\}_{i=1}^N[/imath] as [imath]X[/imath] is a set of [imath]N[/imath] vectors [imath]v_1, v_2, \ldots, v_N[/imath]. [imath]X[/imath] is a family of [imath]N[/imath] vectors [imath]v_1, v_2, \ldots, v_N[/imath]. How can I understand the diffence between them? |
589892 | How to prove geometrically the limit [imath]\lim_{x \to 0}\frac{1-\cos{x}}{x}[/imath] using squeeze theorem
Here, I ask how to geometrically prove the limit \begin{equation}\lim_{x \to 0}\frac{1-\cos{x}}{x}\end{equation} using squeeze theorem. I see so many proofs for the other important limit \begin{equation}\lim_{x \to 0} \frac{\sin{x}}{x}\end{equation} But it seems that no one discussed the first one yet. | 36299 | Finding the limit of [imath](1-\cos(x))/x[/imath] as [imath]x\to 0[/imath] with squeeze theorem
How do I find: [imath] \lim_{x\to0}\frac{1-\cos(x)}{x} [/imath] Using the squeeze theorem. Particularly, how would I arrive at its bounding functions? If possible, please try not to use derivatives. |
589936 | Find a limit [imath]\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)[/imath]
I am to find the limit of [imath]\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)[/imath] so I used: [imath]\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)[/imath] but I just can't solve it to the end... Please show me all steps, or at least most of them, so I'll know how to solve it. Thank you. This question was posted on: Find [imath]\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)[/imath], and got 3 answers, but I still don't know how should I solve it, because when I try to solve it (with help of those 3 answers) I get : [imath]0−12/0[/imath] every time and that goes to minus infinity... | 588335 | Find [imath]\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)[/imath]
I am to find the limit of [imath]\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)[/imath] so I used: [imath]\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)[/imath] but I just can't solve it to the end. |
589498 | Conjecture about some rings and roots of unity.
Let [imath]\Bbb R_{\geqslant 0}[X_n][/imath] be a polynomial semiring. More precisely [imath]\Bbb R_{\geqslant 0}[X_n][/imath] are the polynomials of [imath]X_n[/imath] with positive real coefficients with [imath](X_n)^n = 1[/imath]. Let [imath]F(n)[/imath] be the ring [imath]\Bbb R_{\geqslant 0}[X_n]/(1+X_n+(X_n)^2+...+(X_n)^{n-1})[/imath] For example [imath]F(3)[/imath] is equal to [imath]\Bbb R_{\geqslant 0}[X_3]/(1+X_3+(X_3)^2)[/imath]. We can give the isomorphism explicit by giving [imath] X_3 = ( 1^{1/3} )[/imath] [imath]1^{1/3} = \exp\left(\frac{2 \pi i}{3}\right)[/imath] Thus we write : [imath]F(3) => X_3 = (1^{1/3})[/imath] Likewise we get [imath]F(4) => X_4 = ( 1^{1/4} , 1^{2/4} )[/imath] The question becomes what the explicit isomorphisms for [imath]X_n[/imath] look like. [imath]F(n) => X_n = ( "?" , "?" , "?" , ... )[/imath] ([imath]"?"[/imath] are roots of unity but which ones ? ) The simplest conjecture is [imath]F(n) => X_{2n} = ( 1^{1/2n} , 1^{2/2n} , 1^{3/2n} , 1^{4/2n} , ... , 1^{n/2n} )[/imath] [imath] F(2n+1)[/imath] => [imath]X_{2n+1} = ( 1^{\frac{1}{2n+1}} , 1^{\frac{2}{2n+1}} , 1^{\frac{3}{2n+1}} , 1^{\frac{4}{2n+1}} , ... , 1^{\frac{n}{2n+1}} ) [/imath] But is this conjecture true? EDIT To avoid confusion [imath]X_4 = ( 1^{1/4} , 1^{2/4} )[/imath] DOES NOT MEAN [imath]X_4 = i[/imath] !! [imath]X_4 = ( 1^{1/4} , 1^{2/4} )[/imath] is like a diagonal 2x2 matrix with entries on the diagonal being [imath]( 1^{1/4} , 1^{2/4} )[/imath]. Likewise ( * , * , *) is like a a diagonal matrix of size 3x3. etc etc. I have been asked NOT TO WORK WITH matrices, but this led to the confusions such as [imath]X_4 = i[/imath] Hence against advice of others I mention it here yet again ! This edit was necessary to avoid people to think that the question made no sense, or the conjecture is trivially false. Its not that simple. | 587458 | Conjecture about some group semiring representations ( and roots of unity ).
Let [imath]\Bbb R_+=[0,\infty)[/imath] be a semiring. [imath]\Bbb R_+[C_n][/imath] is the group semiring formed by the semiring [imath]\Bbb R_+[/imath] and the cylic group [imath]C_n[/imath]. Let [imath]\Bbb R_+[X_n][/imath] be the polynomial semiring. (polynomials of [imath]X_n[/imath] with positive real coef. ) [imath]\Bbb R_+[C_n][/imath] is isomorphic with [imath]\Bbb R_+[X_n]/(1+X_n+(X_n)^2+...+(X_n)^{n-1})[/imath] For example [imath]\Bbb R_+[C_3][/imath] is isomorphic to [imath]\Bbb R_+[X_3]/(1+X_3+(X_3)^2)[/imath]. We can give the isomorphism explicit by giving [imath] X_3 = ( 1^{1/3} )[/imath] [imath]1^{1/3} = \exp\left(\frac{2 \pi i}{3}\right)[/imath] Thus we write : [imath]\Bbb R_+[C_3] => X_3 = (1^{1/3})[/imath] Likewise we get [imath]\Bbb R_+[C_4] => X_4 = ( 1^{1/4} , 1^{2/4} )[/imath] Or in matrix form : [imath]X_4 = \begin{pmatrix} 1^{1/4} & 0 \\ 0 & 1^{2/4} \end{pmatrix}[/imath] Let [imath](*, * , ... , * )[/imath] denote the so-called diagonal matrix : [imath]\begin{pmatrix} * & 0 & \cdots & 0 \\ 0 & * & \cdots& 0\\ \vdots & \vdots & \ddots & 0\\ 0 & 0 & 0 & * \end{pmatrix}[/imath] (by which I mean a square matrix filled with zero's apart from [imath] * [/imath] on the diagonal.) The question becomes what the explicit isomorphisms for [imath]X_n[/imath] look like. [imath]\Bbb R_+[C_n] => X_n = ( "?" , "?" , "?" , ... )[/imath] ([imath]"?"[/imath] are roots of unity but which ones ? ) The simplest conjecture is [imath]M[C_{2n}] => X_{2n} = ( 1^{1/2n} , 1^{2/2n} , 1^{3/2n} , 1^{4/2n} , ... , 1^{n/2n} )[/imath] [imath] \Bbb R_+[C_{2n+1}][/imath] => [imath]X_{2n+1} = ( 1^{\frac{1}{2n+1}} , 1^{\frac{2}{2n+1}} , 1^{\frac{3}{2n+1}} , 1^{\frac{4}{2n+1}} , ... , 1^{\frac{n}{2n+1}} ) [/imath] But is this conjecture true? |
121128 | When does the set enter set theory?
I wonder about the foundations of set theory and my question can be stated in some related forms: If we base Zermelo–Fraenkel set theory on first order logic, does that mean first order logic is not allowed to contain the notion of sets? The axioms of Zermelo–Fraenkel set theory seem to already expect the notion of a set to be defined. Is there are pre-definition of what we are dealing with? And where? In set theory, if a function is defined as a set using tuples, why or how does first order logic and the axioms of Zermelo–Fraenkel set theory contain parameter dependend properties [imath]\psi(u_1,u_2,q,...)[/imath], which basically are functions? | 1334678 | Does mathematics become circular at the bottom? What is at the bottom of mathematics?
I am trying to understand what mathematics is really built up of. I thought mathematical logic was the foundation of everything. But from reading a book in mathematical logic, they use "="(equals-sign), functions and relations. Now is the "=" taken as undefined? I have seen it been defined in terms of the identity relation. But in order to talk about functions and relations you need set theory. However, set theory seems to be a part of mathematical logic. Does this mean that (naive) set theory comes before sentential and predicate logic? Is (naive)set-theory at the absolute bottom, where we can define relations and functions and the eqality relation. And then comes sentential logic, and then predicate logic? I am a little confused because when I took an introductory course, we had a little logic before set-theory. But now I see in another book on introduction to proofs that set-theory is in a chapter before logic. So what is at the bottom/start of mathematics, logic or set theory?, or is it circular at the bottom? Can this be how it is at the bottom? naive set-theory [imath]\rightarrow[/imath] sentential logic [imath]\rightarrow [/imath] predicate logic [imath]\rightarrow[/imath] axiomatic set-theory(ZFC) [imath]\rightarrow[/imath] mathematics (But the problem with this explanation is that it seems that some naive-set theory proofs use logic...) (The arrows are of course not "logical" arrows.) simple explanation of the problem: a book on logic uses at the start: functions, relations, sets, ordered pairs, "=" a book on set theory uses at the start: logical deductions like this: "[imath]B \subseteq A[/imath]", means every element in B is in A, so if [imath]C \subseteq B, B \subseteq A[/imath], a proof can be "since every element in C is in B, and every element in B is in A, every element of C is in A: [imath]C \subseteq A[/imath]". But this is first order logic? ([imath](c \rightarrow b \wedge b \rightarrow a)\rightarrow (c\rightarrow a)[/imath]). Hence, both started from each other? |
590774 | Improper integral of [imath]\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\ dx[/imath]
For [imath]a>b>0[/imath], calculate [imath]\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\ dx[/imath] My try : By Taylor series, [imath]\int\ \frac{e^{-ax} - e^{-bx}}{x} \ dx= \sum_{n=1}^\infty \frac{[\ (-a)^n -(-b)^n\ ]}{n}\frac{x^n}{n!} +C [/imath] Note that from ratio test, this series converges absolutely for [imath]x\in [0,\infty)[/imath]. So give me a hint. Thanks Seocond Try : Recall [imath] \Gamma(t) = \int_0^\infty s^{t-1}e^{-s}\ ds\ (t>0)[/imath] So [imath]\Gamma(t) = a^t \int_0^\infty \frac{e^{-as}}{s^{1-t}}\ ds[/imath] So integral we want to calculate is [imath] \lim_{t\rightarrow 0}\ [a^{-t} - b^{-t}]\ \Gamma(t) [/imath] Right ? See I found the following article proving of Integral [imath]\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}dx = \ln\left(\frac{a}{b}\right)[/imath] | 552384 | Proving of Integral [imath]\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}dx = \ln\left(\frac{a}{b}\right)[/imath]
Prove that [imath] \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right) [/imath] My Attempt: Define the function [imath]I(a,b)[/imath] as [imath] I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx [/imath] Differentiate both side with respect to [imath]a[/imath] to get [imath] \begin{align} \frac{dI(a,b)}{da} &= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &= \int_{0}^{\infty}e^{-ax}\,dx\\ &= -\frac{1}{a}(0-1)\\ &= \frac{1}{a} \end{align} [/imath] How can I complete the proof from here? |
590862 | [imath]G[/imath] a finite [imath]p-[/imath]group, [imath]H < G[/imath] (i.e, [imath]H[/imath] a subgroup of [imath]G[/imath], but [imath]H \neq G[/imath]), prove that [imath]H \neq N_G(H)[/imath]
I have one problem, that I think it's pretty hard to solve. The problem reads: Let [imath]G[/imath] be a finite [imath]p-[/imath]group, and [imath]H < G[/imath] (i.e [imath]H \le G[/imath], and [imath]H \neq G[/imath]). Prove that [imath]H \neq N_G(H)[/imath]. Here are my thoughts on this problem: I know that if [imath]G[/imath] is a finite [imath]p-[/imath]group, then its center is non-trivial, i.e [imath]\mathcal{Z}(G) \neq \{e\}[/imath]. Then, I start off by Proof by Contradiction (note that we already have [imath]H \lhd N_G(H)[/imath]), i.e, assume that [imath]H = N_G(H)[/imath], then for every [imath]g \notin H[/imath], there exists [imath]h_g\in H[/imath], such that [imath]g^{-1}h_gg \notin H[/imath]. And from here, I come up with the fact that, in order for [imath]H[/imath] to satisfy [imath]H = N_G(H)[/imath], we must have [imath]\mathcal{Z}(G) \lhd H[/imath]. However, the two facts above don't seem to help much. :( Can you guys give me a little push on this problem? Thanks so much, And have a good day, | 562180 | Normal subgroup where G has order prime
If [imath]o(G) = p^n,[/imath] p a prime number, and H is a subgroup of G, show that there exists an [imath]x\in G[/imath], but [imath]x\notin H[/imath] such that [imath]x^{-1}Hx = H[/imath] How can I prove this? |
591014 | prove [imath]f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)[/imath]
I need help with this proof: [imath]f: X\rightarrow Y[/imath] [imath]C,D\subseteq Y[/imath] [imath]f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)[/imath] Thanks. | 228711 | What are the strategies I can use to prove [imath]f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)[/imath]?
[imath]f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)[/imath] I think I have to show that the LHS is a subset of the RHS and the RHS is a subset of the LHS, but I don't know how to do this exactly. |
591036 | Combination Confusion - equal
I'm not understanding why the below are equal, can someone please explain? Thanks! [imath]\frac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose 4}x^n[/imath] | 570238 | How is [imath]\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n[/imath]
Can someone please explain to me how is [imath]\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n[/imath] Thanks! |
591400 | Show that there is no simple group G, |G|=300
I have absolutely no idea how to go about this. I have started to find how many sylow p subgroups there are. No simple group of order [imath]300[/imath] I've looked at this but I don't know why they've even started the answer in the way they have. | 13111 | No simple group of order [imath]300[/imath]
So I've been trying to prove that there's no simple group of order [imath]300[/imath]. This is what I did and I was wondering if it was enough. [imath]|G|=2^2 \cdot 3 \cdot 5^2[/imath]. Suppose [imath]G[/imath] is simple. Then there would be [imath]6[/imath] Sylow [imath]5[/imath]-subgroups, one of which will have an index of [imath]6[/imath]. But then [imath]|G|=300[/imath] does not divide [imath]6![/imath] which leads to a contradiction. So, the number of Sylow [imath]5[/imath]-subgroups is [imath]1[/imath] and [imath]\exists[/imath] a proper normal Sylow [imath]5[/imath]-subgroup in [imath]G[/imath]. Hence [imath]G[/imath] is not simple. |
589710 | Prove that point share section on equal parts
There is incircle [imath]\Gamma[/imath] of triangle [imath]ABC[/imath] tangent to [imath]AB,BC,CA[/imath] respectively at [imath]K,L,M[/imath]. Point [imath]D[/imath] is the midpoint of section [imath]MK[/imath]. [imath]DL[/imath] is diameter of another circle which intersects with [imath]\Gamma[/imath] at [imath]L,P[/imath] and with [imath]MK[/imath] at [imath]D,R[/imath]. Point of intersection line [imath]PR[/imath] with [imath]AD[/imath] is [imath]S[/imath] show that [imath]DS=AS[/imath] . So I have a huge problem with the adoption of the concept, what I noticed so far is [imath]\angle PLD = \angle DRS[/imath] , [imath]\angle RDS = 90 = \angle LPD [/imath] hence [imath]\triangle LPD \sim \triangle DRS[/imath] | 580912 | prove that line bisect section
There is incircle [imath]\Gamma[/imath] of triangle [imath]ABC[/imath] tangent to [imath]AB,BC,CA[/imath] respectively at [imath]K,L,M[/imath]. Point [imath]D[/imath] is the centre of section [imath]MK[/imath]. [imath]|DL|[/imath] is diameter of another circle which intersects with [imath]\Gamma[/imath] at [imath]L,P[/imath] and with [imath]MK[/imath] at [imath]D,R[/imath]. Show that line [imath]PR[/imath] divide [imath]AD[/imath] on equal parts. Actually I have no idea how to prove it, the only thing that that occured to me is to show that AD is a diameter of circle, have anyone idea how to link the line [imath]PR[/imath] with section [imath]AD[/imath] ? |
466950 | Characteristic polynomial divides minimal polynomial if and only if all eigenspaces are one-dimensional
Prove that characteristic polynomial of a complex matrix [imath]A[/imath] divides its minimal polynomial if and only if all eigenspaces of [imath]A[/imath] are one-dimensional. As far as I can see I the only possible case is when minimal polynomial equals characteristic one. All distinct eigenvalues with multiplicity 1 grant us that the eigenspaces would be one-dimensional, I thought that this is the key to the solution, however we have the theorem stating that on the other side, eigenspace dimension could be less then or equal to its eigenvalue algebraic multiplicity. Everything now mixed up, will be thankful for any help. | 81467 | When are minimal and characteristic polynomials the same?
Assume that we are working over a complex space [imath]W[/imath] of dimension [imath]n[/imath]. When would an operator on this space have the same characteristic and minimal polynomial? I think the easy case is when the operator has [imath]n[/imath] distinct eigenvalues, but what about if it is diagonalizable? Is that sufficient, or can there be cases (with repeated eigvals) when char poly doesn't equal min poly? What are the general conditions when the equality holds? Is it possible to define them without use of determinant? (I am working by Axler and he doesn't like it.) Thanks. |
592324 | For what numbers [imath]n[/imath] every group of order [imath]n[/imath] is abelian?
What are the numbers [imath]n[/imath] such that every group of order [imath]n[/imath] is abelian? For every prime [imath]p[/imath], every group of order [imath]p[/imath] or [imath]p^2[/imath] is abelian. If there is a prime [imath]p[/imath] such that [imath]p^3\mid n[/imath], then I can take a direct product of a nonabelian group of order [imath]p^3[/imath] with a cyclic group of order [imath]\frac{n}{p}[/imath] and get a nonabelian group. So, we are left with values of [imath]n[/imath] divisible by more than one prime, but not divisible by [imath]p^3[/imath] for any prime [imath]p[/imath]. For some orders of this type (i.e. [imath]35[/imath]) all are abelian, but not for all ofcourse. | 156487 | For which [imath]n[/imath], [imath]G[/imath] is abelian?
My question is: For Which natural numbers [imath]n[/imath], a finite group [imath]G[/imath] of order [imath]n[/imath] is an abelian group? Obviouslyو for [imath]n≤4[/imath] and when [imath]n[/imath] is a prime number, we have [imath]G[/imath] is abelian. Can we consider any other restrictions or conditions for [imath]n[/imath] to have the above statement or the group itself should have certain structure as well? Thanks. |
592821 | Find all rational solutions to [imath]x^3 - y^2 = 2[/imath].
Find all rational solutions to [imath]x^3 - y^2 = 2[/imath]. The only integers solutions are [imath](3,\pm5)[/imath]: http://mathforum.org/library/drmath/view/51569.html | 91437 | How to find all rational points on the elliptic curves like [imath]y^2=x^3-2[/imath]
Reading the book by Diophantus, one may be led to consider the curves like: [imath]y^2=x^3+1[/imath], [imath]y^2=x^3-1[/imath], [imath]y^2=x^3-2[/imath], the first two of which are easy (after calculating some eight curves to be solved under some certain conditions, one can directly derive the ranks) to be solved, while the last , although simple enough to be solved by some elementary consideration of factorization of algebraic integers, is at present beyond my ability, as my knowledge about the topic is so far limited to some reading of the book Rational Points On Elliptic Curves, by Silverman and Tate, where he did not investigate the case where the polynomial has no visible rational points. By the theorem of Mordell, one can determine its structure of rational points, if the rank is at hand. So, according to my imagination, if some hints about how to compute ranks of elliptic curves of this kind were offered, it would certainly be appreciated. Thanks in advance. |
591045 | Proving that [imath]L=\{w\in \Sigma^*: |w|_a= 2^n +273[/imath], [imath]n\in \mathbb{N} \}[/imath] is irregular.
I am trying to prove that [imath]L=\{w\in \Sigma^*: |w|_a= 2^n +273[/imath], [imath]n\in \mathbb{N} \}[/imath] is irregular, whereas: [imath]\Sigma=\{a,b\}[/imath]. I tried to use the pumping lemma with no success. I have also tried to prove that there are infinite equivalency classes (thus, [imath]L[/imath] doesn't have a finite automata which accepts it) but I got stuck, any help? | 589884 | Determine whether [imath]L=\{w:|w|_a=2^n+273\text{ for }n\in \mathbb{N}\}[/imath] is regular.
Given the alphabet [imath]\Sigma=\{a, b\}[/imath] and for the next Language [imath]L=\{w:|w|_a=2^n+273\text{ for }n\in \mathbb{N}\}[/imath] determine whether the language is regular. Firstly, I think this language is regular. And i was thinking about two options of solving this one, But i'm not not sure if i use them correctly. Lets say i wanna write a regular expression for this kind of language: [imath]b^*ab^*ab^*\dots ab^*[/imath] And i need to repeat it [imath]2^n+273[/imath] times. And altough [imath]n\in\mathbb{N}[/imath], The expression above gives me all the the words that are in language. Or am i looking at this question the wrong way? |
593334 | Is it true that [imath]|\mathbb{R}|=2^\omega=\omega_1[/imath]?
Is it true that [imath]|\mathbb{R}|=2^\omega=\omega_1[/imath]? Note that [imath]\omega_1[/imath] is the successor of [imath]\omega[/imath] and [imath]2^\omega[/imath] is |all functions from [imath]\omega[/imath] to 2|. | 154985 | Do [imath]\omega^\omega=2^{\aleph_0}=\aleph_1[/imath]?
As we know, [imath]2^{\aleph_0}[/imath] is a cardinal number, so it is a limit ordinal number. However, it must not be [imath]2^\omega[/imath], since [imath]2^\omega=\sup\{2^\alpha|\alpha<\omega\}=\omega=\aleph_0<2^{\aleph_0}[/imath], and even not be [imath]\sum_{i = n<\omega}^{0}\omega^i\cdot a_i[/imath] where [imath]\forall i \le n[a_i \in \omega][/imath]. Since [imath]\|\sum_{i = n<\omega}^{0}\omega^i\cdot a_i\| \le \aleph_0[/imath] for all of them. Besides, [imath]\sup\{\sum_{i = n<\omega}^{0}\omega^i\cdot a_i|\forall i \le n(a_i \in \omega)\}=\omega^\omega[/imath], and [imath]\|\omega^\omega\|=2^{\aleph_0}[/imath] since every element in there can be wrote as [imath]\sum_{i = n<\omega}^{0}\omega^i\cdot a_i[/imath] where [imath]\forall i \le n[a_i \in \omega][/imath] and actually [imath]\aleph_{0}^{\aleph_0}=2^{\aleph_0}[/imath] many. Therefore [imath]\omega^\omega[/imath] is the least ordinal number such that has cardinality [imath]2^{\aleph_0}[/imath], and all ordinal numbers below it has at most cardinality [imath]\aleph_0[/imath]. Hence [imath]\omega^\omega=2^{\aleph_0}=\aleph_1[/imath]? |
438933 | Finite group is abelian if the representatives of its conjugacy classes commute
Let [imath]G[/imath] be a finite group and let [imath]g_1 , g_2 ,...,g_r[/imath] be the representatives of its conjugacy classes. If [imath]g_i g_k=g_k g_i[/imath] for every [imath]i,k \in[/imath] {[imath]1,2,...,r[/imath]}, then prove that [imath]G[/imath] is abelian. My original trial was to prove this in arbitrary subgroup of [imath]S_n[/imath] and using Cayley's theorem we can prove this easily, and the reason is we can take advantage of the fact that if [imath]\tau , \sigma \in S_n[/imath], [imath]\sigma = (a_{11} ... a_{1 n_{1}})(a_{21} ... a_{2 n_{2}}) ... (a_{r1} ... a_{r n_{r}})[/imath] then [imath]\tau \sigma \tau ^{-1} = (\tau(a_{11}) ... \tau(a_{1 n_{1}}))(\tau(a_{21}) ... \tau(a_{2 n_{2}})) ... (\tau(a_{r1}) ... \tau(a_{r n_{r}}))[/imath] But this didn't give me any valuable results, so any hints which can be useful ? I found a proof of this exercise here, but I want to construct my own proof. Any hints ? | 704890 | Suppose G is finite, Show G is abelian
Suppose [imath]G[/imath] is finite, let [imath]A_1,A_2,...,A_n[/imath] be the conjugacy classes in [imath]G[/imath] and they are distinct. how do I show that if [imath]a_ib_j = b_ja_i[/imath] where [imath]a_i[/imath] is in [imath]A_i[/imath] and [imath]b_j[/imath] is in [imath]A_j[/imath], then [imath]G[/imath] is abelian Thanks |
593072 | If s=∅ , T≠∅ then SxT = ∅ . Why?
I recently studied about Cartesian products and I thought that I understood its concept, Until I ran into this expression: If [imath]S=\emptyset[/imath], [imath]\ne\emptyset[/imath], then [imath]S\times T = \emptyset[/imath] . Is an empty set the same as zero? in the sense that in nullifies a non empty set? Thanks! | 51401 | How do the sets [imath]\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset [/imath] look like?
If we have a function [imath]f:A \rightarrow B[/imath], then one way to give meaning, I think, to this function, in terms of set theory, is to say, that [imath]f[/imath] is actually a binary relation [imath]f=(A,B,G_f)[/imath], where [imath]G_f \subseteq A \times B[/imath] is the graph of the function. Now my question is: what is [imath]f[/imath] if [imath]\bullet \ A=\emptyset, \ B\neq\emptyset[/imath],? [imath] \bullet \ B=\emptyset, \ A\neq\emptyset[/imath] ? [imath] \bullet \ B=\emptyset, \ A=\emptyset[/imath] ? (Another way to formulate this, I think, would be: How do the sets [imath]\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset [/imath] look like? Are they all [imath]\emptyset[/imath] ?) |
593658 | Linear transformations and eigenvalues
Let [imath]T: \mathbb C^n \rightarrow \mathbb C^n[/imath] be linear. Let [imath]\beta[/imath] and [imath]\gamma[/imath] be any two ordered bases. Prove that the eigenvalues of [imath][T]_\beta[/imath] and [imath][T]_\gamma[/imath] are the same. Can anyone provide tips/hints in the right direction? I'm struggling as I try to understand this intuitively....thank you | 8339 | Similar matrices have the same eigenvalues with the same geometric multiplicity
Suppose [imath]A[/imath] and [imath]B[/imath] are similar matrices. Show that [imath]A[/imath] and [imath]B[/imath] have the same eigenvalues with the same geometric multiplicities. Similar matrices: Suppose [imath]A[/imath] and [imath]B[/imath] are [imath]n\times n[/imath] matrices over [imath]\mathbb R[/imath] or [imath]\mathbb C[/imath]. We say [imath]A[/imath] and [imath]B[/imath] are similar, or that [imath]A[/imath] is similar to [imath]B[/imath], if there exists a matrix [imath]P[/imath] such that [imath]B = P^{-1}AP[/imath]. |
587846 | If an [imath]H\leq G[/imath] has an irreducible representation of dimension [imath]d[/imath], then show [imath]G[/imath] has an irreducible representation of atleast dimension [imath]d[/imath].
Let [imath]H[/imath] be a subgroup of a group [imath]G[/imath], and let [imath]\rho:H\rightarrow GL(V )[/imath] be an irreducible representation of dimension [imath]d[/imath]. Prove that there is an irreducible representation of [imath]G[/imath] whose dimension is at least [imath]d[/imath]. This is a homework problem I would appreciate it if someone could give me a hint on how to begin. I am assuming I have to apply properties of induced representation, but I am not sure where to begin. | 588260 | If an [imath]H\le G[/imath] has an irreducible representation of dimension [imath]d[/imath], then show [imath]G[/imath] has an irreducible representation of at least dimension [imath]d[/imath].
Let [imath]H[/imath] be a subgroup of a group [imath]G[/imath], and let [imath]\rho :H\to GL(V)[/imath] be an irreducible representation of dimension [imath]d[/imath]. Prove that there is an irreducible representation of [imath]G[/imath] whose dimension is at least [imath]d[/imath]. This is a homework problem I would appreciate it if someone could give me a hint on how to begin. I am assuming I have to apply properties of induced representation, but I am not sure where to begin. It should be noted we have covered both Frobineus reciprocity and Mackey's irreducibility criterion. I tried taking the induced representation of [imath]\rho[/imath], since that is all I can do. But the problem is that it may not be irreducible. I want to use Mackey, but I don't feel like I have enough information to apply it. Again any help would be great. |
593796 | Find all groups of order 121 up to isomorphism
Find all groups of order [imath]121[/imath] up to isomorphism. if i'm right, all such groups are abelian groups, since this is [imath]p^2[/imath] for [imath]p[/imath] = [imath]11[/imath] and [imath]11[/imath] is prime. there's only one, isn't there? isn't it just [imath]\mathbb{Z}_{121} [/imath] which is isomorphic to [imath]\mathbb{Z}_{11} \times \mathbb{Z}_{11}[/imath] ? | 317356 | There exists only two groups of order [imath]p^2[/imath] up to isomorphism.
I just proved that any finite group of order [imath]p^2[/imath] for [imath]p[/imath] a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is [imath]G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z[/imath]. This is abelian and has order [imath]p^2[/imath]. I think the other is [imath]\Bbb Z/p^2 \Bbb Z[/imath]. Now, it should follow from the fact that there is only one cyclic group of order [imath]n[/imath] up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that [imath]G[/imath] as before is not cyclic. But this is easy to see, since we cannot generate any [imath](x,y)[/imath] with [imath]x\neq y[/imath] by repeated addition of some [imath](z,z)[/imath]. Now, it suffices to show that any other group of order [imath]p^2[/imath] is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that [imath]G=\langle (1,0) ,(0,1)\rangle[/imath]. How can I move on? |
594194 | Evaluate [imath]\sum_{n=1}^{\infty}\arctan (\frac{2}{n^2})[/imath]
Question is to Evaluate [imath]\sum_{n=1}^{\infty}\arctan (\frac{2}{n^2})[/imath] What I have done so far is i tried taking sequence of patial sums. [imath]S_1=\arctan (\frac{2}{1})[/imath] [imath]S_2=\arctan (\frac{2}{1})+\arctan (\frac{2}{2^2})=\arctan 2+\arctan (\frac{1}{2})[/imath] [imath]\tan S_2=\tan(\arctan 2+\arctan (\frac{1}{2}))=\frac{***}{1-2.\frac{1}{2}}=\frac{***}{0}\Rightarrow S_2 =\frac{\pi}{2}[/imath] Now there is a problem for [imath]S_3[/imath] [imath]S_3=\arctan (\frac{2}{1})+\arctan (\frac{2}{2^2})+\arctan (\frac{2}{2^3})=\frac{\pi}{2}++\arctan (\frac{2}{2^3})[/imath] I can not apply [imath]\tan[/imath] function on both sides as this would again give me : [imath]\tan S_3=\tan(\frac{\pi}{2}+\arctan (\frac{2}{2^3}))=\frac{\infty+**}{1-\infty**}[/imath] which does not makes much sense. I would be thankful if some one can help me with this. Thank you. P.S : I would be thankful if some one can suggest me another way to solve this than that of the originally posted solution in quoted duplicate... | 243802 | Showing that [imath] \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}[/imath]
I would like to show that: [imath] \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}[/imath] We have: [imath] \sum_{n=1}^N \arctan \left( \frac{2}{n^2} \right) =\sum_{n=1}^N \arctan (n+1)-\arctan(n-1) =-\frac{\pi}{4}+\arctan N+\arctan(N+1)\rightarrow \frac{3\pi}{4}[/imath] Do you agree with my proof? |
594089 | Is [imath]\sum_{n=0}^{\infty}\frac{(\log\log2)^n}{n!}>\frac{3}{5}?[/imath]
I tried to find bounds for the sum using regular techniques, but couldn't get any result nor contradiction. | 513252 | Is [imath]\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}>\frac{3}{5}[/imath]
Question is to check if : [imath]\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}>\frac{3}{5}[/imath] the problem is that i am sure that this series[imath]\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}[/imath] is convergent (if it is not then i would be very happy as i will then be done.) I would be thankful if someone can help me out with solving this. |
594228 | Proof that [imath]n^3-n[/imath] is a multiple of [imath]3[/imath].
I'm struggling with this problem of proof by induction: For any natural number [imath]n[/imath], prove that [imath]n^3-n[/imath] is a multiple of [imath]3[/imath]. I assumed that [imath]k^3-k=3r[/imath] I want to show that [imath](k+1)^3-(K+1)=3r[/imath] The final statement is [imath]K^3 +3K^2+2K[/imath] Am I missing something ? | 533396 | Prove by mathematical induction that [imath]n^3 - n[/imath] is divisible by [imath]3[/imath] for all natural number [imath]n[/imath]
I'm working on a task where I'm a bit unsure if the answer I've got is correct. Here is the task: Show by induction that the following assertion is true for all natural numbers [imath]n[/imath] [imath]n^3 - n[/imath] is divisible by [imath]3[/imath] Here is my answer: For [imath]n = 1[/imath], [imath]n^3 - n = 1 - 1[/imath] which is divisible by [imath]3[/imath] Assume the statement is true for some number [imath]n[/imath], that is, [imath]n^3 - n[/imath] is divisible by [imath]3[/imath]. Now, [imath](n + 1)^3 - (n + 1) = n^3 + 3n^2 + 3n + 1 - n - 1 = (n^3 - n) + 3(n^2 + n) = (n^3 - n) + 3n(n+1)[/imath] which is [imath]n^3 - n[/imath] plus a multiple of [imath]3[/imath]. Since we assumed that [imath]n^3 - n[/imath] was a multiple of [imath]3[/imath], it follows that [imath](n + 1)^3 - (n + 1)[/imath] is also a multiple of [imath]3[/imath]. So, since the statement "[imath]n^3 - n[/imath] is divisible by [imath]3[/imath]" is true for [imath]n = 1[/imath], and its truth for [imath]n[/imath] implies its truth for [imath]n + 1[/imath], the statement is true for all whole number [imath]n[/imath]. I would appreciate if someone could go through the task and the answer and see if I've done this correctly. Thanks a lot! |
594537 | normal subgroup of a group [imath]G[/imath]
Suppose we have a prime number [imath]p[/imath] as well as a positive integer [imath]n[/imath]. Consider a group [imath]G[/imath] with [imath]p^n[/imath] elements. I want to prove that [imath]G[/imath] has a normal subgroup with [imath]p^{n-1}[/imath] elements. I found this in the book I am reading. When I think about it, it makes sense, but I was a little iffy on how to prove it. | 549635 | A [imath]p[/imath]-group of order [imath]p^n[/imath] has a normal subgroup of order [imath]p^k[/imath] for each [imath]0\le k \le n[/imath]
This is problem 3 from Hungerford's section about the Sylow theorems. I have already read hints saying to use induction and that [imath]p[/imath]-groups always have non-trivial centres, but I'm still confused. This is what I have so far: Suppose [imath]|G| = p^n[/imath]. For [imath]k = 0[/imath], [imath]\{e\}[/imath] is a normal subgroup of order [imath]p^0[/imath]. Suppose [imath]N_1, ..., N_k[/imath] are normal subgroups of [imath]G[/imath] with orders as described (ie. [imath]|N_i| = p^i[/imath]) and [imath]k < n[/imath]. [imath]G/N_k[/imath] is a [imath]p[/imath]-group, so it has a non-trivial centre [imath]C(G/N_k)[/imath]. [imath]\pi^{-1}(C(G/N_k))[/imath] is a subgroup of [imath]G[/imath], where [imath]\pi : G \to G/N_k[/imath] is the quotient map. It is normal in [imath]G[/imath] because [imath]\pi[/imath] is a homomorphism and [imath]C(G/N_k)[/imath] is a normal subgroup of [imath]G/N_k[/imath]. I know it contains [imath]N_k[/imath] and some stuff not in [imath]N_k[/imath], so it has to have order at least [imath]p^{k+1}[/imath], but I do not know how to argue it must be equal, or indeed if it even is equal. |
594601 | [imath]L^2[/imath] norm convergence with continuity assumption
Let [imath]f,f_1,f_2,\ldots\colon\mathbb{R}\rightarrow\mathbb{R}[/imath] be continuous functions in [imath]L^2(\mathbb{R})[/imath]. Suppose that [imath]\|f_n-f\|_2\rightarrow 0[/imath] as [imath]n\rightarrow 0[/imath]. Is it true that [imath]f_n(x)\rightarrow f(x)[/imath] for almost every [imath]x[/imath]? Without the continuity assumption, I remember this is not true. But what about with continuity? | 504172 | Does [imath]L^p[/imath]-convergence imply pointwise convergence for [imath]C_0^\infty[/imath] functions?
It is stated in my professor's notes that, given a sequence [imath]\{f_j\}[/imath] of [imath]C_0^\infty(\Omega)[/imath] functions (infinitely differentiable with compact support), and a function [imath]g\in C_0^\infty(\Omega)[/imath], all defined in an open set [imath]\Omega\in\mathbb{R}^n[/imath]: [imath]\|f_j-g\|_{L_p(\Omega)}\to0\implies|f_j(x)-g(x)|\to0\,\forall x\in\Omega[/imath] I was not able to prove it, however. Searching on the Internet, I found many counterexamples to the statement that [imath]L^p[/imath]-convergence implies pointwise convergence, but they do not deal with [imath]C_0^\infty(\Omega)[/imath] functions, so it may also be true. Can anyone point to a proof or a counterexample? |
594649 | If [imath]G[/imath] is a group and [imath]f: G \to G[/imath] is defined by [imath]f(x) = x^{-1}[/imath] is it a homomorphism?
I don't even know where to start here. Also, if it is a group homomorphism, I have to determine whether they are injective or surjective. Any help would be great, thanks! | 511956 | When is mapping [imath]g[/imath] to [imath]g^{-1}[/imath] a group homomorphism?
When is mapping [imath]g[/imath] to [imath]g^{-1}[/imath] a group homomorphism? It just means that the map maps the identity to the identity and inverses to inverses. So does that mean it's only a homomorphism if the inverse is itself? |
594328 | How to prove or disprove statements about sets
Prove or disprove: The set: [imath]\{ \emptyset\}^{\Bbb N}[/imath] has only one element. The set [imath]\emptyset^{\Bbb N}[/imath] is empty. I'm pretty sure both are true, for 1. all the natural numbers go to the empty set and that is the only element. Where as 2. all the natural numbers go to nothing so that's why it's empty. (Same as [imath] {\Bbb N}^{\emptyset} [/imath] nothing goes to something is nothing (I think)). The problem is I have no idea how to show this so advice would be appreciated. Edit: My defintion of a function is: [imath]f:B\to A[/imath] if [imath](a_1,b),(a_2,b)\in f \ \Rightarrow \ a_1=a_2[/imath] So for 2. I can simply say that: [imath](1,\emptyset),(2,\emptyset)\in f \Rightarrow a_1\neq a_2[/imath] ? | 596859 | Elementary set theory - are these sets empty?
we are asked to answer if the following statements are true or false, and why: 1) The set [imath]{\{\emptyset\}^{\mathbb N}}[/imath] has exactly [imath]1[/imath] element. 2) The set [imath]{{\emptyset}^{\mathbb N}}[/imath] is empty. 3) The set [imath]{\mathbb N}^{\emptyset}[/imath] is empty 4) [imath](\{1,2,3\}^{\mathbb N})[/imath]-[imath]\{1,2\}^{\mathbb N}[/imath]=[imath]\{3\}^{\mathbb N}[/imath] What I tried doing: I tried using the fact that [imath]|A^B|=|A|^{|B|}[/imath], so for the first question we get [imath]1^{\aleph0}[/imath] = 1 For the second question, same thing, we get [imath]0[/imath], but for the third question, we have a problem...I think that set is empty, but when I look at [imath]\aleph0 ^0[/imath] it should be 1... |
595008 | field extension [imath]F(x)=F(x^2)[/imath]
Let [imath]x[/imath] be algebraic over [imath]F[/imath] such that the field extension [imath]F(x):F[/imath] satisfies [imath][F(x):F][/imath] odd. Then prove [imath][F(x):F]=[F(x^2):F][/imath] hence [imath]F(x)=F(x^2)[/imath]. How to prove? I only obtained the proof for the case [imath]x^{2m+1}=1[/imath]. How to prove for a general polynomial [imath]p(x)[/imath] of odd degree such that [imath]p(x)=0[/imath]? | 451014 | Prove that if [imath][F(\alpha):F][/imath] is odd then [imath]F(\alpha)=F(\alpha^2)[/imath]
Prove that if [imath][F(\alpha):F][/imath] is odd then [imath]F(\alpha)=F(\alpha^2)[/imath]. My justification for this question is as follows; Suppose [imath]F(\alpha^2)\subsetneq F(\alpha)[/imath], we have [imath]F \subsetneq F(\alpha^2) \subsetneq F(\alpha)[/imath]. As [imath][F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F][/imath] and [imath][F(\alpha):F(\alpha^2)]=2[/imath] we would be end up with the case that [imath][F(\alpha):F][/imath] is a multiple of 2 i.e., even which contradicts given condition of [imath][F(\alpha):F][/imath] being odd. So, [imath]F(\alpha^2) =F(\alpha)[/imath]. After doing this, i have seen for a solution on some webpage where he/she used criterion of minimal polynomial. Please let me know Is my justification clear or there are any gaps in between? |
594969 | For [imath]p,q[/imath] prime and [imath]p \neq q[/imath] show that [imath]C_p \times C_q \cong C_{pq}[/imath]
Here are my thoughts so far: [imath]C_p = \langle x \mid x^p =e\rangle[/imath] [imath]C_q = \langle y \mid y^q =e\rangle[/imath] [imath]C_p \times C_q[/imath] has elements of the form [imath](x^a,y^b)[/imath] There are [imath]p[/imath] possible values for [imath]x^a[/imath] and [imath]q[/imath] possible values for [imath]y^b[/imath]. So there are [imath]pq[/imath] possible elements in [imath]C_p \times C_q[/imath]. [imath]C_{pq} = \langle z\mid z^{pq} = e\rangle[/imath] and there are [imath]pq[/imath] elements in this group. For an isomorphism from [imath]C_p \times C_q[/imath] to [imath]C_{pq}[/imath] we send [imath](e,e)[/imath] to [imath]e[/imath] to ensure that the identities are mapped to each other. However, I'm not sure how to define an isomorphism for the other elements. | 5969 | Product of two cyclic groups is cyclic iff their orders are co-prime
Say you have two groups [imath]G = \langle g \rangle[/imath] with order [imath]n[/imath] and [imath]H = \langle h \rangle[/imath] with order [imath]m[/imath]. Then the product [imath]G \times H[/imath] is a cyclic group if and only if [imath]\gcd(n,m)=1[/imath]. I can't seem to figure out how to start proving this. I have tried with some examples, where I pick [imath](g,h)[/imath] as a candidate generator of [imath]G \times H[/imath]. I see that what we want is for the cycles of [imath]g[/imath] and [imath]h[/imath], as we take powers of [imath](g,h)[/imath], to interleave such that we do not get [imath](1,1)[/imath] until the [imath](mn)[/imath]-th power. However, I am having a hard time formalizing this and relating it to the greatest common divisor. Any hints are much appreciated! |
595149 | Prove that either [imath](2^{10500} + 15)[/imath] or [imath](2^{10500} + 16)[/imath] is not a perfect square.
Prove that either [imath](2^{10500} + 15)[/imath] or [imath](2^{10500} + 16)[/imath] is not a perfect square. how should I solve this problem? what is the idea for solving this kind of problems? Thank you so much | 590354 | How to prove that either [imath]2^{500} + 15[/imath] or [imath]2^{500} + 16[/imath] isn't a perfect square?
How would I prove that either [imath]2^{500} + 15[/imath] or [imath]2^{500} + 16[/imath] isn't a perfect square? |
595162 | Find the value of [imath]\cos(2\pi /5)[/imath] using radicals
This is homework so if there is another example that can illustrate the technique I would happily accept that as guidance. The only thing I have been able to find is a question asking about [imath]\cos(2\pi/7)[/imath], which I think is a much harder problem. I dont have the faintest idea how to solve this and the textbook (Hungerford) doesn't have any examples at all. Ive tried looking for resources online but havent found any that I was able to understand. So can anyone show me how to solve these types of problems? Thanks a bunch. | 7695 | How to prove [imath]\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}[/imath]?
I would like to find the apothem of a regular pentagon. It follows from [imath]\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.[/imath] But how can this be proved (geometrically or trigonometrically)? |
170611 | Why is the product of all units of a finite field equal to [imath]-1[/imath]?
Suppose [imath]F=\{0,a_1,\dots,a_{q-1}\}[/imath] is a finite field with [imath]q=p^n[/imath] elements. I'm curious, why is the product of all elements of [imath]F^\ast[/imath] equal to [imath]-1[/imath]? I know that [imath]F^\ast[/imath] is cyclic, say generated by [imath]a[/imath]. Then the product in question can be written as [imath] a_1\cdots a_{q-1}=1\cdot a\cdot a^2\cdots a^{q-2}=a^{(q-2)(q-1)/2}. [/imath] However, I'm having trouble jumping from that product to [imath]-1[/imath]. What is the key observation to make here? Thanks! | 2358872 | Product of non-zero elements of a field is [imath]-1[/imath]
This looks very much like something which is a duplicate, but I couldn't find the "desired duplicate". Sorry if this has been asked before. Let [imath]F[/imath] be a field of [imath]q[/imath] elements and [imath]F\setminus \{0\} = \{a_1,\ldots,a_{q-1}\}[/imath]. Show that [imath]\prod a_i = -1[/imath]. If [imath]F[/imath] has characteristic [imath]2[/imath] then this is trivial. So let's assume it has characteristic [imath]p>2[/imath]. I tried to use the fact that [imath]F\setminus \{0\}[/imath] is a cyclic group with no avail. Any hint? |
595338 | Need a proof to show all the units are satisfied [imath]\mathbf{Z}\sqrt{2}[/imath] is the all the integer solution in Pell equation
We know the integer solutions of Pell's equation [imath]a^2-2b^2=\pm1[/imath] correspond to the units of [imath]\textbf{Z}[\sqrt{2}][/imath]. How can we prove this? | 590775 | Need a proofreading why all the units are satisfied [imath]a^2-2b^2 =\pm1[/imath] for [imath]\mathbf{Z}[\sqrt{2}][/imath]
All the units are satisfied Pell's equation [imath]a^2-2b^2=\pm1[/imath] for [imath]\mathbf{Z}[\sqrt{2}][/imath], [imath]a,b\in\mathbf{Z}[/imath]. Here is my proof: Let [imath]a+b\sqrt{2}[/imath] be a unit [imath]\in\mathbf{Z}[\sqrt{2}][/imath]. This implies [imath](a+b\sqrt{2})(c+d\sqrt{2})=1, c+d\sqrt{2}\in\mathbf{Z}\sqrt{2}[/imath] [imath]\implies \mathrm{norm}((a+b\sqrt{2})(c+d\sqrt{2}))=1[/imath] [imath]\implies \mathrm{norm}(a+b\sqrt{2})\cdot \mathrm{norm}(c+d\sqrt{2})=1[/imath] [imath]\implies (a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=1[/imath] [imath]\implies (a^2-2b^2)(c^2-2d^2)=1[/imath] [imath]\implies \left[ a^2-2b^2=1\text{ and } {c}^2-2{d}^2=1\right]\text{ or } \left[ a^2-2{b}^2=-1\text{ and }{c}^2-2{d}^2=-1\right][/imath] [imath]\implies {a}^2-2{b}^2=\pm1[/imath] |
595202 | Positive definite symmetric matrix satisfying a certain property concerning a group homomorphism
We have a finite group [imath]G[/imath] and a homomorphism [imath]\rho: G \rightarrow \mathbb{GL}_n(\mathbb{R})[/imath] where [imath]n[/imath] is a positive integer. I need to show that there's an [imath]n\times n[/imath] positive definite symmetric matrix that satisfies [imath]\rho(g)^tA\rho(g)=A[/imath] for all [imath]g \in G[/imath], where [imath]t[/imath] means the transpose. So far, all I could do was write definitions...not sure where to start. So for all non-zero vectors [imath]v[/imath] with [imath]n[/imath] real entries, [imath]v^tAv>0[/imath], and I know [imath]A=A^t[/imath] Also, I know [imath]A[/imath] is positive-definite if and only if there exists an invertible [imath]n\times n[/imath] matrix [imath]P[/imath] such that [imath]A=P^tP[/imath]. But I'm not really sure where to go with that. I was wondering maybe I can start with [imath]\rho(g)^tA\rho(g)=A[/imath] and then show that [imath]A[/imath] is positive-definite, but I don't really see that going anywhere. | 587932 | Existence of a G-invariant matrix
Let [imath]\phi: G \to GL(\mathbb{R}^n)[/imath] be a homomorphism, [imath]G[/imath] finite. Prove that there is a positive-definite matrix [imath]M[/imath] such that [imath]\phi(g)^tM \phi(g) =M[/imath] [imath]\forall g \in G [/imath]. This looks really interesting. Any idea is appreciated. |
595402 | A surjective ring homomorphism [imath]\phi : C([0,1]) \rightarrow \mathbb{R}[/imath] is evaluation at a point
Let [imath]\phi : C([0,1]) \rightarrow \mathbb{R}[/imath] be a surjective ring homomorphism. How would I prove that [imath]\phi[/imath] is the evaluation map [imath]\phi(f) = f(x)[/imath] for some [imath]x \in [0,1][/imath]? I'm not even sure this conclusion is true. All I've done is started a proof by contradiction, assuming there is no such x, played around with quantifiers to determine that I need to show that [imath]\forall x\in[0,1], \exists f\in C([0,1])[/imath] such that [imath]\phi(f) \neq f(x)[/imath] leads to a contradiction. How do I go about finishing this? | 159801 | Characterizing all ring homomorphisms [imath]C[0,1]\to\mathbb{R}[/imath].
This is something I've been trying to work out this evening. Let [imath]R[/imath] be the ring of continuous real-valued functions on [imath][0,1][/imath] with pointwise addition and multiplication. For [imath]t\in [0,1][/imath], the map [imath]\phi_t\colon f\to f(t)[/imath] is a ring homomorphism of [imath]R[/imath] to [imath]\mathbb{R}[/imath]. I'm trying to show that every ring homomorphism of [imath]R\to\mathbb{R}[/imath] has this form. Suppose otherwise, that there is some [imath]\phi\neq\phi_t[/imath], and thus there is some [imath]f_t\in R[/imath] such that [imath]\phi(f_t)\neq \phi_t(f_t)=f_t(t)[/imath]. Define [imath]g_t=f_t-\phi(f_t)1\in R[/imath]. Here [imath]\phi(f_t)1[/imath] is the constant function sending [imath][0,1][/imath] to [imath]\phi(f_t)[/imath]. Then [imath]g_t(t)\neq 0[/imath]. My first small question is why does [imath]\phi(g_t)=0[/imath]? It seems only that [imath]\phi(g_t)=\phi(f_t)-\phi(\phi(f_t)1)[/imath]. I would like to conclude that there are only finitely many [imath]t_i[/imath] such that [imath]g(x)=\sum g_{t_i}^2(x)\neq 0[/imath] for all [imath]x[/imath]. Then [imath]g^{-1}=1/g(x)\in R[/imath], but [imath]\phi(g)=0[/imath], contradicting the fact that homomorphisms map units to units. How can we be sure there are only finitely many [imath]g_{t_i}[/imath] such that the sum of their squares is never [imath]0[/imath]? Thanks. |
493791 | Finite ring has only zero divisors and units
If a nonzero element [imath]a[/imath] of a ring [imath]\mathbb{D}[/imath] does not have a multiplicative inverse, then [imath]a[/imath] must be a zero divisor. If [imath]\mathbb{D}[/imath] has finitely many elements, then the statement is true. However, I don't see why, as for the infinite case, 7 isn't a zero divisor only because every element [imath]a > 0[/imath] when multiplied by a [imath]b > 0[/imath] gives you [imath]ab > 0[/imath] which has nothing to do with infinity or finite amounts of elements. | 60969 | Every nonzero element in a finite ring is either a unit or a zero divisor
Let [imath]R[/imath] be a finite ring with unity. Prove that every nonzero element of [imath]R[/imath] is either a unit or a zero-divisor. |
484828 | Normal subgroups of dihedral groups
In relation to my previous question, I am curious about what exactly are the normal subgroups of a dihedral group [imath]D_n[/imath] of order [imath]2n[/imath]. It is easy to see that cyclic subgroups of [imath]D_n[/imath] is normal. But I suspect that case analysis is needed to decide whether dihedral subgroups of [imath]D_n[/imath] is normal. A little bit of Internet search suggests the use of semidirect product [imath](\mathbb Z/n\mathbb Z) \rtimes (\mathbb Z/2\mathbb Z) \cong D_n[/imath], but I do not know the condition for subgroups of a semidirect product to be normal. I would be grateful if you could suggest a way to enumerate the normal subgroups of [imath]D_n[/imath] that does not resort to too much of case analysis. | 1469029 | Normal subgroup in Dihedral group
How do you show that {[imath] 1,a,a^{2}, ..., a^{n-1} [/imath]} is a normal subgroup of the nth dihedral group? I thought maybe it is the only subgroup of n elements , so then it's a normal subgroup. But I don't know how to prove it. |
596356 | Prove that [imath]R\left[x\right]/I\left[x\right]\cong\left(R/I\right)\left[x\right][/imath]
Let [imath]I[/imath] be an ideal of a ring [imath]R[/imath], define [imath]I[x][/imath] to be the set of all polynomials whose coefficients are in [imath]I[/imath]. Prove that [imath]R\left[x\right]/I\left[x\right]\cong\left(R/I\right)\left[x\right][/imath] Help me some hints | 594934 | [imath](R/I)[x]=R[x]/I[/imath]
If [imath]R[/imath] is a ring and [imath]I[/imath] an ideal of this ring, is that true that [imath](R/I)[x]=R[x]/I[/imath]? Note that can [imath]I[/imath] be identified as a subset of [imath]R[x][/imath] and we need that in order to [imath]R[x]/I[/imath] makes sense. I think it's true because every polynomial [imath](a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n[/imath] of [imath](R/I)[x][/imath] can be written as a polynomial of [imath]R[x]/I[/imath] and vice versa since we have this equality: [imath](a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n=(a_0+a_1x+\ldots a_nx^n) +I[/imath] Is my reasoning right? Thanks |
596320 | Units in finite polynomial rings
Are the units of the quotient ring [imath]\mathbb{F}_2[x]/\langle x^k+1 \rangle[/imath] known in general, where [imath]\mathbb{F}_2[/imath] is the finite field with two elements? I'm specifically interested in the case where [imath]k[/imath] is divisible by two, such as say [imath]k=8[/imath], or [imath]k=12[/imath]. If so, can they be constructed easily? I'd be grateful if you could provide a few examples. | 595603 | Units in finite polynomial rings
Are the units of the quotient ring [imath]\mathbb{F}_2[x]/\langle x^k+1 \rangle[/imath] known in general, where [imath]\mathbb{F}_2[/imath] is the finite field with two elements? I'm specifically interested in the case where [imath]k[/imath] is divisible by two, such as say [imath]k=8[/imath], or [imath]k=12[/imath]. If so, can they be constructed easily? I'd be grateful if you could provide a few examples. |
596472 | Proof by induction and divisibility [imath]21 | (4^{n+1} + 5^{2n-1}) [/imath]
Prove by induction: [imath]21 | (4^{n+1} + 5^{2n-1}) [/imath] Skipping through the basis and onto the induction: [imath]4\cdot 4^{n+1}+5^2 \cdot 5^{2n-1}=21a [/imath] for some integer [imath]a[/imath] The following steps were: [imath]4\cdot (21a-5^{2n-1})+ 125 \cdot5^{2n-1} \\= 84a-4 \cdot5^{2n-1} + 125 \cdot5^{2n-1} [/imath] But I can't factor out a 21 from here... Any input? | 227728 | Induction: Prove that [imath]4^{n+1}+5^{2 n - 1}[/imath] is divisible by 21 for all [imath]n \geq 1[/imath].
Induction: Prove that [imath]4^{n+1}+5^{2 n - 1}[/imath] is divisible by 21 for all [imath]n \geq 1[/imath]. |
597184 | If [imath]\gcd(a,b)=1[/imath] then, [imath]\gcd(a^2,b^2)=1[/imath]
Prove or disprove 'If [imath]\gcd(a,b)=1[/imath] then, [imath]\gcd(a^2,b^2)=1[/imath], with [imath]a,b\not= 0[/imath]' I need to prove this statement. I think it is true and also the converse is true. I took some examples such as [imath]\gcd(2,5)=1[/imath], and [imath]\gcd(4,25)=1[/imath]... But how can I prove this? I know that [imath]1=ax+by[/imath] for some [imath]4x,y\in \mathbb{Z}[/imath].. But not sure How to go from there.. Any clue? | 536623 | Show that if [imath]a[/imath] and [imath]b[/imath] are positive integers with [imath](a,b)=1[/imath] then [imath](a^n, b^n) = 1[/imath] for all positive integers n
Show that if [imath]a[/imath] and [imath]b[/imath] are positive integers with [imath](a,b)=1[/imath] then [imath](a^n, b^n) = 1[/imath] for all positive integers n Hi everyone, for the proof to the above question, Can I assume that since [imath](a, b) = 1[/imath], then in the prime-power factorization of a and b, they have no prime factor in common, when they are taken to the [imath]nth[/imath] power, they will still have no prime factors in common, and so [imath](a^n, b^n) = 1[/imath] for all positive integers n. I think I'm jumping to conclusions here again, if so, leave some tips on how to do the proof properly, thanks :) And also, I do not know how to approach the reverse problem where if [imath](a^n, b^n) = 1 then (a, b) = 1[/imath], any guidance will be much appreciated! |
524454 | How to prove $\gcd(a^m,b^m) = \gcd(a,b)^m$ using Bézout's Lemma
The problem is to prove the following If [imath]\gcd(a,b) = c[/imath], then [imath]\gcd(a^m, b^m) = c^m[/imath] I know that this can be solved easily by proving that [imath]c\mid a \implies c^m \mid a^m[/imath] and [imath]c\mid b \implies c^m \mid b^m[/imath]. So the greatest common divisor of [imath]a^m[/imath] and [imath]b^m[/imath] is [imath]c^m[/imath]. But the problem is that I want to use Bézout's Lemma (Identity) to prove this. I first introduce the following notation: [imath]a = ca' \quad \quad b = cb', \quad \quad \text{where $(a',b') = 1$.}[/imath] Now from Bézout's Lemma, becaus [imath]a'[/imath] and [imath]b'[/imath] are coprime integers then we have integer solutions for the following equation: [imath]a'x + b'y = 1[/imath] It's obvious that also the following equation has an integer solution: [imath]a'^mx_1 + b'^my_1 = 1[/imath] Although I'm not able to prove this using Bézout's Lemma. Now multiply both sides by [imath]c^m[/imath] we get: [imath](a'c)^mx + (b'c)^my = c^m[/imath] [imath]a^mx + b^my = c^m[/imath] But Bézout's Lemma states that for given [imath]c^m[/imath] is the greatest common multiplier of [imath]a^m[/imath] and [imath]b^m[/imath] or it is a multiplier of it. But how to prove that it's not a multiplier of the greatest common multiper, and in fact is the greatest common multiplier. Is it possible to solve this problem with Bézout's Lemma? | 696444 | For all [imath]n \geq 1[/imath], and positive integers [imath]a,b[/imath] show: If [imath]\gcd (a,b)=1[/imath], then [imath]\gcd(a^n,b^n)=1[/imath]
For all [imath]n \geq 1[/imath], and positive integers [imath]a,b[/imath] show: If [imath]\gcd (a,b)=1[/imath], then [imath]\gcd(a^n,b^n)=1[/imath] So, I wrote the [imath]gcd (a,b)=1[/imath] as a linear combination: [imath]ax+by=1[/imath] And, I wrote the [imath]gcd(a^n,b^n)=1[/imath] as a linear combination: [imath]a^n (u)+b^n (v)=1[/imath] can I write the second linear combinations with [imath]x,y[/imath] and then raise the first equation to the nth power or not? |
594198 | Aut[imath](K_4)[/imath] regarding
I want to show that Aut[imath]K_4\cong S_3[/imath], where [imath]K_4[/imath] is the Klein's 4-group. I have shown that there exist 6 automorphsms from [imath]K_4[/imath] to [imath]K_4[/imath]. But how to show that Aut[imath]K_4[/imath] is nonabelian? | 98291 | [imath]\operatorname{Aut}(V)[/imath] is isomorphic to [imath]S_3[/imath]
I'm currently working my way through Harvard's online abstract algebra lectures (if you're interested, you can find them here). The lectures come complete with notes and homework problems. Of course, since I don't actually go to Harvard, I can't hand in the homework assignments to find out if I'm doing them correctly. So I've decided to try to crowd source the grading of my solutions. I'll post individual questions as I finish them and wait for comments. I would like to get critiques of not just my reasoning but the style of my write ups as well. Also, any alternative approaches to the problem would be welcome. I've looked for forums dedicated to these kinds of OCW courses but have not been able to find any. This surprises me. It seems like, with the advent of these free online educational resources, an online meeting place for those who take advantage of them would be a natural offshoot. So this might be a bit of an experiment. Unless I'm the 873rd person to post something like this here. If that's the case, sorry for being so long winded. Anyway, on with the question. This one is assigned in the 4th lecture. Let [imath]V[/imath] denote the Klein 4-group. Show that [imath]\operatorname{Aut}(V)[/imath] is isomorphic to [imath]S_3[/imath]. (I've moved my solution below to keep this question from showing up in the unanswered list.) Thanks... |
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