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585001
|
Show that if [imath]n[/imath] is not divisible by [imath]2[/imath] or by [imath]3[/imath], then [imath]n^2-1[/imath] is divisible by [imath]24[/imath].
Show that if [imath]n[/imath] is not divisible by [imath]2[/imath] or by [imath]3[/imath], then [imath]n^2-1[/imath] is divisible by [imath]24[/imath]. I thought I would do the following ... As [imath]n[/imath] is not divisible by [imath]2[/imath] and [imath]3[/imath] then [imath]n=2k+1\;\text{or}\\n=3k+1\;\text{or}\\n=3k+2\;\;\;\;[/imath]for some [imath]n\in\mathbb{N}[/imath]. And then make induction over [imath]k[/imath] in each case.[imath]24\mid (2k+1)^2-1\;\text{or}\\24\mid (3k+1)^2-1\;\text{or}\\24\mid (3k+2)^2-1\;\;\;\;[/imath]This question is in a text of Euclidean Division I'm reviewing, and I wonder if there is a simpler, faster, or direct this way.
|
2795551
|
If [imath]a[/imath] is an odd positive integer [imath]>[/imath] [imath]2[/imath] such that 3 [imath]\not[/imath] | [imath]a[/imath], then [imath]12[/imath] [imath]|[/imath] [imath](a^2 -[/imath]1 ))
So I have the following proposal: [imath]a[/imath] [imath]\in[/imath] [imath]\mathbb{Z}_{+}[/imath] : [imath]a[/imath] = [imath]2k +1[/imath] [imath]>[/imath] [imath]2[/imath] : 3 [imath]\not[/imath] | [imath]a[/imath] and [imath]>2[/imath] then [imath]12[/imath] [imath]|[/imath] [imath](a^2 -[/imath]1 ). My initial idea was to show: [imath]a^2 -1\equiv 0 \mod 12[/imath] Logically it follows: [imath](2k+1)^2 -1\equiv 0 \mod 12[/imath] [imath]4k^2 + 4k\equiv 0 \mod 12[/imath] However I do not know where to go from here. This proof also doesn't incorporate the condition 3 [imath]\not |[/imath] a . Any tips? Sorry for formatting, this is my first time using MathJax.
|
1991740
|
Prove using the definition of the limit that [imath]\lim_{x\to 0}\frac{1}{x}[/imath] fails to exist.
Prove, using the [imath]\epsilon-\delta[/imath] definition of the limit, that [imath]\displaystyle\lim_{x\to 0}\frac{1}{x}[/imath] fails to exist. It is obvious that [imath]\displaystyle\frac{1}{x}[/imath] approaches [imath]+\infty[/imath] from the right of [imath]0[/imath] and approaches [imath]-\infty[/imath] from the left of [imath]0[/imath]. But how do I show that the limit does not exist using the [imath]\epsilon-\delta[/imath] definition?
|
583363
|
Showing that [imath]\lim\limits_{x \rightarrow 0} \frac{1}{x}[/imath] does not exist.
Forgive me if this has been asked before, but I searched and could not find an answer. I am trying to show that [imath]\lim\limits_{x \rightarrow 0} \frac{1}{x}[/imath] does not exist. If the limit did exist, and was equal to [imath]l[/imath], then for every [imath]\varepsilon > 0[/imath] there would exist some [imath]\delta > 0[/imath] such that for all [imath]x[/imath], [imath]0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| < \varepsilon[/imath] If the limit doesn't exist, then for every number [imath]l[/imath], there exists some [imath]\varepsilon > 0[/imath] such that for all [imath]\delta > 0[/imath], [imath]0 < |x| < \delta \implies \left| \frac{1}{x} - l \right| \geq \varepsilon[/imath] So, in my efforts to find some unattainable [imath]\varepsilon[/imath], I have found that \begin{align} \varepsilon &\leq \left| \frac{1}{x} - l \right| \\\ &= \left| \frac{1}{x} - \frac{lx}{x} \right| \\\ &= \left| \frac{1-lx}{x} \right| \\\ &= \frac{|1-lx|}{|x|} \\\ &\implies \varepsilon |x| \leq |1 - lx| = |1 + (-lx)| \leq |1| + |-lx| \\\ &\implies 1 \geq \varepsilon |x| - |l| |x| = (\varepsilon - |l|) |x| \\\ &\implies |x| \leq \frac{1}{\varepsilon - |l|} \end{align} So I have found that if [imath]|x| < \frac{1}{\varepsilon - |l|}[/imath], then [imath]\left| \frac{1}{x} - l \right| \geq \varepsilon[/imath], for any [imath]\varepsilon[/imath]. But, I need this to be true for any choice of [imath]\delta[/imath], for some specific [imath]\varepsilon[/imath]. Intuitively, I can think "as [imath]\varepsilon[/imath] gets closer to [imath]|l|[/imath], then the quotient becomes larger, meaning that [imath]|x| < \frac{1}{\varepsilon - |l|}[/imath] is satisfied for more choices of [imath]\delta[/imath]", but I don't know how to formalize this. Any ideas for a next step? Am I completely off base with my reasoning?
|
1994890
|
Is [imath]A = \{ B \subset E: B \text{ countable or } B^c \text{ countable}\}[/imath] a sigma algebra?
Is [imath]A = \{ B \subset E: B \text{ countable or } B^c \text{ countable}\}[/imath] a sigma algebra? I would want someone to verify if my proof below is valid. a countable union of elements in [imath]A[/imath] is still in [imath]A[/imath] I divided it into three cases, let [imath](H_j)_{j \in \mathbb{N}} \in A[/imath] All [imath]H_j[/imath] countable, a countable union of countable sets is countable, and it lies in [imath]A[/imath]. all [imath]H_{j}^{c}[/imath] countable then [imath](\cup_{j \in \mathbb{N}} H_j )^c = \cap_{j \in \mathbb{N}} H_{j}^{c} [/imath] countable and it lies in [imath]A[/imath]. Either [imath]H_j[/imath] or [imath]H_{j}^{c}[/imath] is countable, then [imath]|(\cup_{j \in \mathbb{N}} H_j)^c| = | \cap_{j \in \mathbb{N}} H_{j}^{c}| = min\{|H_1|, \dots , |H_n|\} \leq |\mathbb{N}|[/imath] countable, and it lies in [imath]A[/imath]
|
663442
|
Show that [imath]\left\{A\subset\Omega: A~\text{ is countable or }A^C\text{ is countable}\right\}[/imath] is a [imath]\sigma[/imath]-algebra
Let [imath]\Omega[/imath] a set and [imath] C=\left\{A\subset\Omega: A~\text{ is countable or }A^C\text{ is countable}\right\}. [/imath] Show that [imath]C[/imath] is a [imath]\sigma[/imath]-algebra and that [imath] \sigma(\left\{\left\{ x\right\}: x\in\Omega\right\})=C. [/imath] Hello! First I have to show that [imath]C[/imath] is a [imath]\sigma[/imath]-algebra. Here are my results; I have some problems to finish the proof. (1) If [imath]\Omega[/imath] is countable, then [imath]\Omega\in C[/imath]. If [imath]\Omega[/imath] is not countable, then [imath]\Omega^C=\emptyset[/imath] is. So in any case [imath]\Omega\in C[/imath]. (2) Consider [imath]A\in C[/imath]. If [imath]A[/imath] is countable, then [imath]A^C[/imath] is in [imath]C[/imath], because [imath](A^C)^C=A[/imath] is countable then. If [imath]A[/imath] is not countable, it has to be by definition [imath]A^C[/imath] countable (otherwise [imath]A[/imath] could not be in [imath]C[/imath]), so [imath]A^C[/imath] in [imath]C[/imath]. (3) Consider [imath]A_1,A_2,\ldots\in C[/imath]. First case: All [imath]A_i, i=1,2,\ldots[/imath] are countable. Then [imath]\bigcup_{i\geq 1}A_i[/imath] is countable as countable union of countable sets. So [imath]\bigcup_{i\geq 1}A_i\in C[/imath]. Second case: All [imath]A_i, i=1,2,\ldots[/imath] are not countable. Then all [imath]A_i^C, i=1,2,\ldots[/imath] are countable and for any [imath]i\in\left\{1,2,.\ldots\right\}[/imath] it is [imath] \left(\bigcup_{i\geq 1}A_i\right)^C=\bigcap_{i\geq 1}A_i^C\subset A_i^C [/imath] which means that [imath]\bigcap_{i\geq 1}A_i^C[/imath] is countable. So it follows that [imath]\bigcup_{i\geq 1}A_i[/imath] is in [imath]C[/imath], because its complement is countable. Now what is with the third case, that some [imath]A_i[/imath] are countable and others not? Do not know how to handle it. Overmore, I have problems to show [imath]\supset[/imath] of the identity that is to show. The other implikation [imath]\subset[/imath] follows, because [imath]C[/imath] is a [imath]\sigma[/imath]-algebra (which I have not completely shown) and [imath]\left\{\left\{ x\right\}: x\in\Omega\right\}\subset C[/imath]. So would you please help me to finish point (3) (case 3) and the implication [imath]\supset[/imath]?
|
1994955
|
Prove sum of 2 little 'o is equal to the little 'o from the sum
I need to prove if for [imath]f(x) = o(h(x))[/imath] and [imath]g(x) = o(q(x))[/imath] then [imath]f(x) + g(x) = o(|h(x)| + |q(x)|)[/imath] I think that this is false but I am struggling to find a counter example.
|
1258661
|
Why sum of two little "o" notation is equal little "o" notation from sum?
Why sum of two little "o" notation is equal little "o" notation from sum? [imath]o( f(n) ) + o( g(n) ) = o( f(n) + g(n) ) ?[/imath] For example: [imath]f(n) = n^3[/imath] [imath]g(n) = 1/n[/imath] so [imath]o(f(n)) = n^2[/imath] [imath]o(g(n)) = 1/n^2[/imath] and [imath]o( f(n) ) + o( g(n) ) = n^2 + 1/n^2[/imath] [imath]o( f(n) + g(n) ) = n^2[/imath] Of course, I could write it like [imath]o( f(n) ) + o( g(n) ) = n^2 + o( g(n) )[/imath] [imath]o( f(n) + g(n) ) = n^2 + o( g(n) )[/imath] My question is why? I don't understand it, because in first we always get two parameters.
|
1994926
|
Proving the countability of a set of polynomials
I want to show that the set of all polynomials with rational coefficients is countable. I know that to prove this I need to find a mapping [imath]f[/imath] from the natural numbers to this set that is a bijection. But I'm struggling conceptually of how to think of this. In fact, my intuition is that this isn't even true since can't we plug in an uncountable set of numbers into our polynomial? It's because of this thought that I can't even think of what a bijection would be like. If anyone could help untangle these conceptual issues I'm having that would be good -- I don't necessarily need a proof of the claim, though that's fine too. Edit: not really a duplicate.
|
1574436
|
Show that the set of polynomials with rational coefficients is countable.
Problem: Show that the set of polynomials with rational coefficients is countable. Idea: We know that the set of rational numbers is denumerable. This implies that the set of rational numbers is countable. We also know that the degree that each polynomial can be is a natural number (I think, and I'm not sure how to word this.) Therefore, I think we can reason through this somehow, by showing that the plane [imath]Q X N[/imath] Is denumerable. I'm just not sure how to do this. Any ideas. Note: this is for my introduction to proofs class study quide. So, I would prefer not to go to over the top.
|
1995246
|
Prove that [imath]\gcd(\det(A),26)=1[/imath] if a [imath]2x2[/imath] matrix [imath]A[/imath] with entries in [imath]Z_{26}[/imath] is invertible
Prove that [imath]gcd(det(A),26)=1[/imath] if a [imath]2x2[/imath] matrix [imath]A[/imath] with entries in [imath]Z_{26}[/imath] is invertible. We know that [imath]A[/imath] is invertible. Let the inverse of [imath]A[/imath] be [imath]A^{-1}=(det(A))^{-1}A^{*}[/imath], where [imath]A^{*}[/imath] is the adjoint matrix of [imath]A[/imath]. However, this is where I am stuck, I cannot see how the formula for [imath]A^{-1}[/imath] is useful. Any help is greatly appreciated
|
1046314
|
Prove a residue matrix [imath]A[/imath] (with coefficients in [imath]\mathbb Z_n)[/imath] has an inverse if and only if [imath]\gcd(\det A,n) = 1[/imath]
Prove a residue matrix [imath]A[/imath] (with coefficients in [imath]\mathbb Z_n)[/imath] has an inverse if and only if [imath]\gcd(\det A,n) = 1[/imath]. I've always done matrix arithmetic in a field [imath]\mathbb F[/imath] and that is what every Linear Algebra book has taught me to do. However, I've started looking into Cryptography and here the term residue matrix come up. Can someone prove the above result or tell me where to look it up ? Where can I find books about Cryptography-related Linear Algebra ? I've studied several books on Linear Algebra, but they are all equal in the sense that the ring of interest is a field. Normally a matrix [imath]A[/imath] (with coefficients in a field [imath]\mathbb F[/imath]) has an inverse if and only if [imath]\det A \neq 0[/imath], and in this case [imath]\det A[/imath] is neccesarily invertible, since we are in a field. So the above result is a more general result than this.
|
1995447
|
If [imath]19|x^2[/imath] then [imath]19|x[/imath]
[imath]x\notin 19\mathbb{Z}[/imath] If [imath]19|x^2[/imath] then [imath]19|x[/imath]. Proof by contraposition: If [imath]19\not|x[/imath], then [imath]19\not|x^2[/imath] If [imath]19\not|x[/imath] then [imath]x[/imath] must take on one of the following forms: [imath]x=19k+1, x=19+2, x=19k+3,......,x=19k+18[/imath] for some [imath]k[/imath] 1) [imath]x=19k+1[/imath] [imath]x^2=(19k+1)^2=361k^2+38k+1=19(19k^2+2k)+1[/imath] 2)[imath]x=19k+2[/imath] [imath]x^2=(19k+2)^2=361k^2+76k+1=19(19k^2+4k)+1[/imath] . . . . continuing this route 5)[imath]x=19k+5[/imath] [imath]x^2=(19k+5)^2=361k^2+192k+25=19(19k^2+10k+1)+6[/imath] again continuing 18)[imath]x=19k+18[/imath] [imath]x^2=(19k+18)^2=361k^2+684k+324=19(19k^2+36k+17)+1[/imath] Note how for each case, [imath]x^2[/imath] is not divisible by [imath]19[/imath] as we are always left with a remainder [imath]\therefore\;[/imath] The original statement is false.
|
1162373
|
If [imath]p[/imath] is a prime and [imath]p \mid ab[/imath], then [imath]p \mid a[/imath] or [imath]p \mid b[/imath].
The proof is already given in the textbook but I tried other way around. Proof by contradiction: Let's assume that [imath]p[/imath] doesn't divide [imath]a[/imath] and [imath]p[/imath] doesn't divide [imath]b[/imath], but [imath]p[/imath] divides [imath]ab[/imath]. So [imath]\gcd(p,a)=1[/imath] and [imath]\gcd(p,b)=1[/imath]. Given that we can construct linear combinations [imath]sp+ta=1[/imath] together with [imath]up+wb=1[/imath]. Multiplying the left and the right sides of the equations we get [imath]spup+spwb+taup+tawb=1[/imath] or [imath]p(sup+swb+tau)+ab(tw)=1 \implies \gcd(p,ab)=1[/imath]. This is a contradiction. Is it rigorous and sound proof?
|
1983169
|
How to calculate [imath]\lim_{n \rightarrow \infty} e^{-n} (\frac{1}{n}+1)^{n^2}[/imath]?
How to calculate [imath]\lim_{n \rightarrow \infty} e^{-n} (\frac{1}{n}+1)^{n^2}[/imath]? The result should be [imath]\frac{1}{\sqrt{e}}[/imath] But I have no idea how to get it. I've been only thinking of rules to multiply the [imath]e^{-n}[/imath] through somehow, but I don't seem to find anything. The different quantities also seem to work in different directions making interpretation difficult, since: [imath]e^{-n}\rightarrow 0[/imath] [imath](\frac{1}{n}+1) \rightarrow 1[/imath] [imath]n^2 \rightarrow \infty[/imath] So these interactions make it difficult to say anything about this. Perhaps there's a way to develop inequalities that show which terms will dominate when approaching infinity?
|
422892
|
Find the limit [imath]\displaystyle\lim_{n\rightarrow\infty}{(1+1/n)^{n^2}e^{-n}}[/imath]?
Find the limit [imath]\displaystyle\lim_{n\rightarrow\infty}{(1+1/n)^{n^2}e^{-n}}[/imath]? I found the limit as [imath]e^{-1/2}[/imath] using l'Hospital rule. I guess I made a mistake. Because the limit seems to be 1. Also, can we find the limit without L'Hospital rule?
|
575587
|
Noetherian topological subspaces
I'm trying to prove that any subset of a noetherian topological space is noetherian in its induced topology. MY ATTEMPT OF SOLUTION Let [imath]X[/imath] be a topological space and [imath]Y[/imath] a subspace of [imath]X[/imath]. If [imath]F=\{U_i\cap Y\}[/imath] is a family of open subsets of [imath]Y[/imath], where the [imath]U_i[/imath] are open subsets of [imath]X[/imath], can I say that [imath]M\cap Y[/imath] is a maximal element of [imath]F[/imath], where [imath]M[/imath] is a maximal element of the [imath]U_i[/imath]'s? If don't, I don't know how to solve this question. I really need help. Thanks in advance.
|
314374
|
Subspace of Noetherian space still Noetherian
I see this everywhere, but cannot find a proof of it. Is it just this easy; it feels wrong. Suppose [imath]X[/imath] is noet and [imath]Y\subset X[/imath] with subspace topology. If [imath]Y[/imath] has a descending chain of closed sets [imath]Y_1 \supset Y_2 \supset Y_3 \supset \ldots[/imath], where each [imath]Y_i[/imath] is formed by intersection some closed set in [imath]X[/imath] with [imath]Y[/imath]. That is [imath](X_1\cap Y)\supset (X_2\cap Y) \supset \ldots[/imath] but I can't seem to finish this because it is not as if I can just remove the intersections with [imath]Y[/imath] right?
|
1994584
|
Stability inquiry of [imath]x(t) = 0.1e^{-t} + 0.1e^t - 0.1\sin(2t) + 0.1e^3 (\sin(2(t-3)) + e^{-t+3}-e^{t-3})\times H(t-3)[/imath]
Determine if [imath]x(t)[/imath] is stable, where [imath]x(t)[/imath] is a solution to the differential equation: [imath]x'''+x''+4x'+4x = e^t(H(t) - H(t-3))[/imath] and x(t) is defined as: [imath]x(t) = 0.1e^{-t} + 0.1e^t - 0.1\sin(2t) + 0.1e^3 (\sin(2(t-3)) + e^{-t+3}-e^{t-3})\times H(t-3)[/imath] How do I approach this? Do I use eigenvalues? If so, I get 2i, - 2i and 1, is that correct and what do these values imply? Any help is greatly appreciated!
|
1994487
|
Determining stability of the differential equation
Determine wether the differential equation is stable or not for the following conditions: [imath]x'''(0) = x''(0) = x'(0) = x(0) =0[/imath] [imath]\frac{d^3x(t)}{dt^3}+\frac{d^2x(t)}{dt^2}+4\frac{dx(t)}{dt}+4x(t) = e^t(H(t)-H(t-3))[/imath] Now, I got that the solution to this differential equation reduces x(t) by use of Laplace Transforms, however, how do I determine whether or not it is stable? [imath]x(t) = 0.1e^{-t} + 0.1e^t - 0.1\sin(2t) + 0.1e^3 (\sin(2(t-3)) + e^{-t+3}-e^{t-3})*H(t-3)[/imath] Now I first thought I would look at it graphically and I used Matlab software to get a plot as shown below: And from the plot above it clearly appears to diverge (though this is not a proof). Anyone have an idea as to an elegant proof to show the solution is unstable? Anyone help would be greatly appreciated!
|
1981539
|
How to solve this limit problem? [imath]\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )[/imath]
I am stuck with the following question from my homework: [imath]\lim_{n \to \infty} \left ( \sqrt{n + \sqrt{n}} - \sqrt{n - \sqrt{n}} \right )[/imath] Using wolfram alpha gives me 1 for the solution. However, I would like to know how you come up with this result. I hope someone can explain this to me.
|
394517
|
How to evaluate [imath]\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)[/imath]?
How can I evaluate [imath]\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)[/imath]?
|
1995927
|
[imath]n^2+(n+1)^2+n^2\cdot(n+1)^2[/imath] is a perfect square
Prove that [imath]n^2+(n+1)^2+n^2\cdot(n+1)^2[/imath] is a perfect square, for all [imath]n\in\mathbb N[/imath]. I have not tried anything but numerical calculations.
|
1909974
|
If [imath]a[/imath] and [imath]b[/imath] are consecutive integers, prove that [imath]a^2 + b^2 + a^2b^2[/imath] is a perfect square.
Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form [imath]k^2l^2[/imath] but nothing's clicking. I tried Bézout but the same expression can be written as [imath]a^2 + (a^2 + 1)(a+1)^2[/imath] which would imply that there is no real root. Would really appreciate some help, thanks.
|
1995582
|
How do I find the recurrence equation solution for [imath]T(n) = T(n-1) + n + 2?[/imath]
Okay so I am supposed to find the recurrence equation of [imath]T(n) = T(n-1)+n+2[/imath], where [imath]T(1) = 1[/imath]. I know the answer should come out to be [imath]\dfrac12(n(n+5)-4)[/imath] but I don't understand how to get that answer.
|
1995278
|
Mathematical Induction for Recurrence Relation
I have solved the following recurrence relationship: [imath]T(1) = 1[/imath] [imath]T(n) = T(n-1) + n + 2[/imath] so [imath]T(n) = \frac{1}{2}n^2+\frac{5}{2}n -2[/imath] I am now trying to perform mathematical induction to prove this. [imath]Basis:[/imath] [imath]T(1)=1=3-2=\frac{1}{2} + \frac{5}{2} - 2 [/imath] [imath]Induction:[/imath] [imath]T(k+1) = T(k) + k+1 + 2[/imath] [imath]= \frac{1}{2}k^2 + \frac{5}{2}k -2 + k+1 +2[/imath] [imath]= \frac{1}{2}k^2 + \frac{7}{2}k +1[/imath] What can I do next?
|
1982348
|
[imath]\mathop{\lim}\limits_{x\to+\infty} \dfrac{\ln x}{x^\varepsilon}, \varepsilon>0[/imath] without L'Hospital's rule
I got stuck when proving [imath]\ln x = o\left( {{x^\varepsilon }} \right)[/imath] ([imath]\varepsilon>0[/imath]). Any ideas without L'Hospital's rule? [imath]\mathop{\lim}\limits_{x\to+\infty} \dfrac{\ln x}{x^\varepsilon}, \varepsilon>0[/imath] Thanks in advance.
|
717924
|
Determining a tricky limit
What is the limit of [imath]\lim_{n \to \infty} \frac{\log n}{n^\delta}[/imath] where [imath]\delta > 0[/imath]?
|
817362
|
if [imath]f(x) \sim g(x)[/imath] is [imath] \sum f(k) \sim \sum g(k)[/imath]
if [imath]f(x) \sim g(x)[/imath] as [imath]x \to \infty[/imath] then is [imath]\sum_{k=1}^N f(k) \sim \sum_{k=1}^N g(k)[/imath] as [imath]N \to \infty[/imath]? Intuitively, i should think so because as [imath]k[/imath] gets larger [imath]f[/imath] and [imath]g[/imath] get closer so it should 'even out'?
|
1845165
|
Does [imath]a_n \sim b_n[/imath] imply [imath]\sum_n a_n \sim \sum_n b_n[/imath] for [imath]a_n, b_n>0[/imath]?
I am almost embarrassed to asked to this question, but after considering it for a while I realize I need some help. In the following [imath]a_n, b_n >0[/imath]. So, by limit comparison test if [imath]a_n = O(b(n))[/imath] or, for a stricter case that I am interested in, [imath]\frac{a_n}{b_n} \to_n 1[/imath] both [imath]S_1 = \sum_n a_n[/imath] and [imath]S_2 = \sum_n b_n[/imath] either diverge or converge or, in other words, [imath]S_1 = O(S_2)[/imath] or [imath]\frac{S_1}{S_2} \to_n O(1)[/imath]. How about a stricter condition: does this convergence imply that [imath]\frac{S_1}{S_2} \to_n 1[/imath] or [imath]S_1 \sim S_2[/imath]?. For example, [imath]S_1 \sim n \log n + O(1)[/imath] and [imath]S_2 \sim n \log n + n + \frac{1}{n}[/imath]. I think the answer is not in general, I tried to compare [imath]\sum_k \log (1+\frac{k}{n})[/imath] and [imath]\sum_k \frac{k}{n}[/imath] (they are [imath]\sim[/imath] if Taylor series is used). And I got the first constant to be [imath]\log \frac{4}{e}[/imath], and the second, of course, [imath]\frac{1}{2}[/imath]. Any suggestions? EDIT: After reading the solution, I went through my calculation again and what I get is [imath] \sum_{k=1}^{n} \log (1+\frac{k}{n}) = \log (2n)! - \log n! - n \log n [/imath] Either using Stirling's approximation or [imath]\sum_{k=1}^{n} \log k \sim \int_{1}^{n} \log x dx[/imath] I get the front term equal to [imath]\log \frac{4}{e}[/imath], which is different to [imath]\frac{1}{2}[/imath] from [imath]\sum_{k=1}^{n} \frac{k}{n}[/imath]. Where is the mistake?
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1996063
|
Starting the epsilon-delta definition of continuity from delta
The classical way the [imath]\epsilon[/imath]-[imath]\delta[/imath] definition for continuity (limits) proceeds is by starting with a given [imath]\epsilon>0[/imath], and our ability to then find a number [imath]\delta>0[/imath], such that when [imath]|x-x_{0}|<\delta[/imath], we get [imath]|y-y_{0}|<\epsilon[/imath]. My question is: why cannot we do exactly the same definition but starting from a given [imath]\delta[/imath] (instead of starting with a given [imath]\epsilon[/imath])? That is, why not say: given [imath]\delta>0[/imath], we can find a number [imath]\epsilon>0[/imath], such that when [imath]|x-x_{0}|<\delta[/imath] we get [imath]|y-y_{0}|<\epsilon[/imath] ? Note: this question is different from other questions posted regarding the direction of implication of the definition (e.g. this post)
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1198831
|
Slightly changing the formal definition of continuity of [imath]f: \mathbb{R} \to \mathbb{R}[/imath]?
I'm curious for some perspectives on why it would be wrong to change the definition of continuity of [imath]f: \Bbb R \to \Bbb R[/imath] in the following way: Original definition. [imath]f : \Bbb R \to \Bbb R[/imath] is said to be continuous at [imath]x \in \Bbb R[/imath] if [imath]\forall \epsilon > 0[/imath] [imath]\exists \delta > 0[/imath] such that [imath]|x - a| < \delta \implies |f(x) - f(a)| < \epsilon[/imath]. Altered definition. [imath]f : \Bbb R \to \Bbb R[/imath] is said to be continuous at [imath]x \in \Bbb R[/imath] if [imath]\forall \delta > 0[/imath] [imath]\exists \epsilon > 0[/imath] such that [imath]|x - a| < \delta \implies |f(x) - f(a)| < \epsilon[/imath]. The altered definition is more in line with what I think when I think about continuity intuitively: nearby points are sent to nearby points. It only makes sense to me to be able to choose "nearness" in the domain (i.e., [imath]\forall \delta > 0[/imath]) and show there is nearness in the codomain (i.e., [imath]\exists \epsilon > 0[/imath]) to prove intuitively that "nearby points are sent to nearby points". Similarly, if [imath]X, Y[/imath] are topological spaces, we say [imath]f: X \to Y[/imath] is continuous if the preimages of open sets are open. What would be wrong about changing the definition to say that a map is continuous if the images of open sets are open (i.e., [imath]f[/imath] is continuous if it is an open map)? This is more inline with the intuitive idea of "nearby points being sent to nearby points" -- you pick nearness in the domain (i.e., an arbitrary open set) an show nearness in the codomain (i.e., the image is open). Does anyone have any useful remarks?
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1996964
|
Entire function and injectivity
So if [imath]f(z)=az+b[/imath], is an entire injective map from [imath]\Bbb C[/imath] to [imath]\Bbb C[/imath], for both a and b are complex numbers and a is not equal to 0. I've proved that if f is an injective entire function, it cannot have an essential singularity at infinity, but then how to show that [imath]f(z)[/imath] has to be a polynomial? And why [imath]f(z)[/imath] has to satisfy that [imath]f(z)=az+b[/imath]?
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964534
|
Proving that all entire & injective functions take the form [imath]f = ax + b[/imath]?
I'm a little confused at both the overall logic in this proof. Are we simply using [imath]g(z)[/imath] to make conclusions about [imath]f(z)[/imath], because [imath]g(z)[/imath] is the reciprocal of [imath]f[/imath]? Is the proof assuming that [imath]f[/imath] is injective and entire (all the while knowing that it has some sort of singularity at [imath]z = 0[/imath]), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that [imath]z = 0[/imath] is a pole singularity, it reaches the conclusion that [imath]f[/imath] must be of the form [imath]f(z) = az + b[/imath]? Also, more specific questions about the different cases: Removable singularity case: Why is [imath]f[/imath] bounded on the closed circle if [imath]f[/imath] is continuous? Am I missing something simple? Essential singularity case: Why exactly is [imath]f(\{|z| > r \} \cap f(\{|z|<r\}) \neq \emptyset[/imath]? [imath]f(\{|z| > r \})[/imath] is dense, but how does [imath]f(\{|z|<r\}[/imath] being open guarantee that their union is non-empty?
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1997007
|
Show that [imath]E[x]=\int_0^\infty(1-F_X(x))dx[/imath]
Show that [imath]E[x]=\int_0^\infty(1-F_X(x))dx[/imath] The solution is as follows: [imath]\begin{align} \int_0^\infty P(X>x)dx &=\int_0^\infty \int_x^\infty f_X(y)dydx\\ &=\int_0^\infty \int_0^y dxf_X(y)dy\\ &=\int_0^\infty yf_X(y)dy\\ &=E[X] \end{align}[/imath] I don't understand step 3 in the solution, specifically while the limit of integration changed from [imath]\infty[/imath] to [imath]y[/imath].
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1690740
|
Prove that [imath]E(X) = \int_{0}^{\infty} P(X>x)\,dx = \int_{0}^{\infty} (1-F_X(x))\,dx[/imath].
Let [imath]X[/imath] be a continuous non-negative random variable (i.e. [imath]R_x[/imath] has only non-negative values). Prove that [imath]E(X) = \int_{0}^{\infty} P(X>x)\,dx = \int_{0}^{\infty} (1-F_X(x))\,dx[/imath] where [imath]F_X(x)[/imath] is the CDF for [imath]X[/imath]. Using this result, find [imath]E(X)[/imath] for an exponential ([imath]\lambda[/imath]) random variable. I know that by definition, [imath]F_X(x) = P(X \leq x)[/imath] and so [imath]1 - F_X(x) = P(X>x)[/imath] The solution is: [imath]\int_{0}^{\infty} \int_{x}^{\infty} f(y)\,dy dx = \int_{0}^{\infty} \int_{0}^{y} f(y)\,dy dx = \int_{0}^{\infty} yf(y) dy.[/imath] I'm really confused as to where the double integral came from. I'm also rusty on multivariate calc, so I'm confused about the swapping of [imath]x[/imath] and [imath]\infty[/imath] to [imath]0[/imath] and [imath]y[/imath]. Any help would be greatly appreciated!
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1997244
|
Cardinality of [imath] X\times \mathbb{N} [/imath] when [imath]X[/imath] is infinite
Suppose [imath]X[/imath] is infinite. I think we can show that [imath]|X\times\mathbb{N}| = |X|[/imath]. How do we construct an injective function [imath]f:X\times\mathbb{N} \to X[/imath]?
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1896440
|
Cardinality of cartesian product of an infinite set with N
Here the problem 1.4.26 from Sohrab, Basic Real Analysis Duplicate This are not duplicate of this question: Cardinality of Cartesian Product of Uncountable Set with Countable Set the answer gives no proof How to represent the Cartesian product of an infinite set? despite the title is not really connected Cardinality of cartesian product only finite set Problem Show that, if [imath]A[/imath] is an infinite set, then [imath]|A\times\mathbb{N}|=|A|[/imath]. Hint: Let [imath]\mathcal{F}[/imath] denote the set of all bijective maps [imath]f:S\times\mathbb{N}\to S[/imath], where [imath]S\subset A[/imath]. Since [imath]|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|[/imath], we have [imath]\mathcal{F}\neq \varnothing[/imath] (Why?) Show that Zorn's Lemma can be applied in [imath]\mathcal{F}[/imath] to produce a maximal bijection [imath]h:B\times\mathbb{N}\to B[/imath], with [imath]B\subset A[/imath], and that we must have [imath]B=A[/imath], by examining the cases where [imath]S\setminus B[/imath] is finite or infinite. Proof Let [imath]\mathcal{F}[/imath] denote the set of all bijective maps [imath]f:S\times\mathbb{N}\to S[/imath], where [imath]S\subset A[/imath]. Since [imath]|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|[/imath] (see exercise 1.4.10(i)), we have [imath]\mathcal{F}\neq \varnothing[/imath] (From Exercise 1.4.22(d) we know that for a infinite set [imath]A[/imath], we have [imath]|A|\geq\aleph_0[/imath], and [imath]|\mathbb{N}|=\aleph_0[/imath], thus there is an injection from [imath]g:\mathbb{N}\to A[/imath], and, by defining [imath]S:=g(\mathbb{N})\subset A[/imath], we get a bijective map (composition of two bijective maps) [imath]S\times\mathbb{N}\to\mathbb{N}\times\mathbb{N}\to\mathbb{N}\in\mathcal{F}[/imath]). If we consider the inclusion as partial order, then every chain [imath]S_i\subset A[/imath], [imath]i\in I[/imath], has [imath]A[/imath] as upper bound and applying Zorn's Lemma, we get the existence of a maximal [imath]\bar{S}[/imath] and hence a bijection [imath]h:\bar{S}\times\mathbb{N}\to\bar{S}[/imath], with [imath]\bar{S}\subset A[/imath]. Now we prove that [imath]\bar{S}=A[/imath]. [imath]R:=A\setminus \bar{S}[/imath]. [imath]\bar{S}[/imath] is infinite, since the example of element of [imath]\mathcal{F}[/imath], that we gave above is bijective to [imath]\mathbb{N}[/imath]. We have to do this. The second is trying to follow the hint and is incomplete: Consider that [imath]A\neq\bar{S}[/imath], then there is [imath]r\in R:=A\setminus\bar{S}[/imath]. We chose an element [imath]s\in\bar{S}[/imath] and define the bijection [imath]h':\bar{S}\cup\{r\}\times\mathbb{N}\to \bar{S}\cup\{r\}[/imath] as follow [imath]h'(a,n)=\begin{cases}h(s,2n-1)&\text{for }a=s\\ r&\text{for }a=r,n=1\\ h(s,2n-2)&\text{for }a=r,n\neq 1\\ h(a,n)&\text{otherwise}\end{cases} [/imath] Since [imath]\bar{S}\subsetneq \bar{S}\cup\{r\}[/imath] we get a contradiction to the maximality of [imath]\bar{S}[/imath]. Now either [imath]R[/imath] is finite or infinite. 1) [imath]R[/imath] finite, let say [imath]R=\{r_1,\cdots,r_n\}[/imath]. Then chose [imath]n[/imath] elements [imath]s_1,\cdots,s_n\in\bar{S}[/imath] and we define the bijection [imath]h':A\times\mathbb{N}\to A[/imath] as follow [imath]h'(a,n)=\begin{cases}h(a,2n-1)&\text{for }a\in\{s_1,\cdots,s_n\}\\ a&\text{for }n=1,a\in\{r_1,\cdots,r_n\}\\ h(s_i,2n-2)&\text{for }a=r_i\\ h(a,n)&\text{otherwise}\end{cases}[/imath] In case [imath]n>0[/imath] we get a contradiction with [imath]\bar{S}[/imath] being the maximal element. Thus we have only to exclude the case 2) [imath]R[/imath] infinite. HOW TO DO THAT? Question Above I tried two proofs. The first part is the same. The two second parts are given by the two bullets. The second proof tries to follow the hint. My questions are: - Is my proof correct? - In which direction should go the proof from the hint?
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1997329
|
How to get the limit of this function?
How am I suppose to handle this? [imath]\lim_{x\rightarrow \infty}\left(1+\frac{\pi}{x}\right)^{2x}[/imath]
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244540
|
Proof that [imath]\lim_{m\to\infty}(1+\frac{r}{m})^{mt}=e^{rt}[/imath]
Can someone show me a straightforward proof that [imath]\lim_{m\to\infty}(1+\frac{r}{m})^{mt}=e^{rt}[/imath] Thanks!
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1997819
|
A normal subgroup [imath]H[/imath] of [imath]G[/imath], [imath][G:H] = n[/imath]
Let [imath]H\triangleleft G[/imath], and H is a subgroup of G such that [imath][G:H] = k[/imath], where [imath][G:H][/imath] states the number of left cosets of [imath]H[/imath] in [imath]G[/imath]. 1) How may I show that for all [imath]a\in G, a^k\in H[/imath]? 2) If we do NOT assume the normality of [imath]H[/imath], can I have any counterexample showing that 1) is not true?
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660832
|
Normal Subgroup Counterexample
Im having trouble with the second part of this question, Let [imath]H[/imath] be a normal subgroup of [imath]G[/imath] with [imath]|G:H| = n[/imath], i) Prove [imath]g^n \in H[/imath] [imath]\forall g \in G[/imath] (which i have done) ii) Give an example to show that this all false when [imath]H[/imath] is not normal in [imath]G[/imath].(which I am having trouble with showing) Any suggestions?
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1998106
|
limit of [imath]\sqrt{n+k}-\sqrt{n}[/imath] without Taylor series
Let [imath]a_n = \sqrt{n+k}-\sqrt{n}[/imath] be a sequence where [imath]k\in \Bbb R^+[/imath]. Find the limit of [imath]a_n[/imath]. I want to evaluate this limit without doing a series expansion of the terms (it seems like overkill). Intuitively, it should be [imath]0[/imath] because as [imath]n[/imath] becomes large the [imath]k[/imath] becomes insignificant. But how can I show it? I can't just factor out [imath]\sqrt{n}[/imath] because [imath]\sqrt{n}(\sqrt{1+\frac{k}{n}}-1)[/imath] goes to [imath]\infty\cdot 0[/imath].
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1457129
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How do I calculate [imath]\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}[/imath]?
I've seen a handful of exercises like this: [imath]\lim_{x\to+\infty}(\sqrt{x+a}-\sqrt{x})[/imath] I've never worked with limits to infinity when there is some arbitrary number [imath]a[/imath]. I am not given any details about it. Apparently the answer is [imath]0[/imath]. How was that conclusion reached? My guess is that since [imath]x = +\infty[/imath], the result of [imath]x + a[/imath] will still be [imath]+\infty[/imath] so we would have [imath]\sqrt{x}-\sqrt{x} = 0[/imath]. But that doesn't convince me. For starters, we don't know what [imath]a[/imath] is: it could be [imath]-\infty[/imath] or something, so [imath]\infty - \infty[/imath] would be indeterminate...
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1997522
|
If [imath]\lim_{x \to c} f(x)[/imath] does not exist and [imath]\lim_{x \to c} g(x)[/imath] does not exist, then [imath]\lim_{x \to c} (f(x).g(x))[/imath] does not exist.
I need to prove the theorem If [imath]\lim_{x \to c} f(x)[/imath] does not exist and [imath]\lim_{x \to c} g(x)[/imath] does not exist, then [imath]\lim_{x \to c} (f(x).g(x))[/imath] does not exist. but I stuck. First, I said that this statement is equivalent to the statement If [imath]\lim_{x \to c} (f(x)g(x))[/imath] exists, then [imath]\lim_{x \to c} f(x)[/imath] exits or [imath]\lim_{x \to c} g(x)[/imath] exists. but after assuming [imath]\lim_{x \to c} (f(x)g(x))[/imath] exists, I could not reach anywhere. Any help would be appreciated. Edit: From the answer, I saw that it is a wrong statement, but still without giving a counterexample, how can we prove it that it is wrong ?
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513822
|
Can the limit of a product exist if neither of its factors exist?
Show an example where neither [imath]\lim\limits_{x\to c} f(x)[/imath] or [imath]\lim\limits_{x\to c} g(x)[/imath] exists but [imath]\lim\limits_{x\to c} f(x)g(x)[/imath] exists. Sorry if this seems elementary, I have just started my degree... Thanks in advance.
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1487783
|
Sum of multiples of [imath]2[/imath] coprime numbers cover positive integers
I have a conjecture that any two positive coprime integers [imath]a[/imath] and [imath]b[/imath] can be added to form any integer greater than or equal to [imath](a-1)(b-1)[/imath]. This is true for [imath]3[/imath] and [imath]5[/imath], which can form any integer greater than or equal to [imath]8[/imath], where [imath]8=3+5, 9=3+3+3, 10=5+5; k=3n+8[/imath] or [imath]3n+9[/imath] or [imath]3n+10[/imath]. It is true for [imath]4[/imath] and [imath]7[/imath], which can form any integer greater than or equal to [imath]18[/imath], where [imath]18=4+7+7, 19=4+4+4+7, 20=4+4+4+4+4, 21=7+7+7[/imath], etc. How can it be proven that this will work for any two coprime numbers?
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535649
|
the set [imath]\{xm + ny : x,y \in Z^+\}[/imath] contains all but a finite number of the positive integers.
If [imath]m[/imath] and [imath]n[/imath] are relatively prime then the set [imath]\{xm + ny : x,y \in Z^+\}[/imath] contains all but a finite number of the positive integers. What I tried : We know that there exist integers [imath]x,y[/imath] such that [imath]mx+ny=1[/imath]. But both can't be negative/positive. So, we need to separate the positive and negative part of each coefficient. Let [imath]a_1 =x^+[/imath] and [imath]b_1=x^-[/imath] and [imath]a_2 =y^+[/imath] and [imath]b_2=y^-[/imath]. Then [imath]ma_1 + na_2 = 1 + mb_1 + nb_2[/imath]. So, I have got two consecutive integers. Stuck after that.
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180283
|
Why is the probability that a continuous random variable takes a specific value zero?
My understanding is that a random variable is actually a function [imath]X: \Omega \to T[/imath], where [imath]\Omega[/imath] is the sample space of some random experiment and [imath]T[/imath] is the set from which the possible values of the random variable are taken. Regarding the set of values that the random variable can actually take, it is the image of the function [imath]X[/imath]. If the image is finite, then [imath]X[/imath] must be a discrete random variable. However, if it is an infinite set, then [imath]X[/imath] may or may not be a continuous random variable. Whether it is depends on whether the image is countable or not. If it is countable, then [imath]X[/imath] is a discrete random variable; whereas if it is not, then [imath]X[/imath] is continuous. Assuming that my understanding is correct, why does the fact that the image is uncountable imply that [imath]Pr(X = x) = 0[/imath]. I would have thought that the fact that the image is infinite, regardless of whether it is countable or not, would already imply that [imath]Pr(X = x) = 0[/imath] since if it is infinite, then the domain [imath]\Omega[/imath] must also be infinite, and therefore [imath]Pr(X = x) = \frac{\text{# favorable outcomes}}{\text{# possible outcomes}} = \frac{\text{# outcomes of the experiment where X = x}}{|\Omega|} = \frac{\text{# outcomes of the experiment where X = x}}{\infty} = 0[/imath] What is wrong with my argument? Why does the probability that a continuous random variable takes on a specific value actually equal zero?
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2298610
|
If [imath]X[/imath] is a continuous random variable, then [imath]P(a \le X \le b) = P(a < X \le b) = P(a \le X < b) = P(a < X < b)[/imath]. Why?
When introducing the concept of continuous random variables, my textbook states the following: If [imath]X[/imath] is a continuous random variable, then [imath]P(a \le X \le b) = P(a < X \le b) = P(a \le X < b) = P(a < X < b)[/imath]. However, it does not give a justification for why this is true. I would greatly appreciate it if people could please take the time to explain why this is true.
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1999647
|
True or false. Order of [imath]\phi (a)[/imath] is equal to the order of [imath]a[/imath]
If it's true write a proof. If it's false, give a counter example. If [imath]\phi : G_1 \rightarrow G_2[/imath] is a homomorphism and [imath]a\in G[/imath] then the order of [imath]\phi(a)[/imath] then is equal to the order of [imath]a[/imath]. My attempt: This is false. Consider [imath]\phi:Z_{15} \rightarrow Z_6[/imath] difined by [imath]\phi([a]_{15})[/imath]=[imath][a]_6[/imath]. This is homomorphism since [imath]\phi([a]_{15}+[b]_{15})= \phi([a+b]_{15})=[a+b]_6=[a]_6+[b_6]= \phi[a]_{15}+\phi[b]_{15}.[/imath] Let [imath]a\in Z_{15}=3[/imath]. The order of [imath]3[/imath] is [imath]5[/imath] since [imath]3+3+3+3+3=0mod15[/imath]. The order of [imath]\phi(3)[/imath] is [imath]2[/imath] since [imath]3+3=0mod6[/imath]. EDIT: My counter example is not well-defined. What if i change it to [imath]\phi([a]_{15}) = [3a]_6[/imath] and follow the same steps?
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76882
|
Is it true that [imath]|g| = |\phi(g)|[/imath] for all homomorphisms [imath]\phi: G \to G[/imath] and [imath]g \in G[/imath]?
True or false. (Prove or give a counterexample.) Let [imath]G[/imath] be a group. Then [imath]|g| = |\phi(g)|[/imath], for all homomorphisms [imath]\phi: G \to G[/imath] and all [imath]g \in G[/imath]. Solution. False. [imath]\phi: \mathbb Z_{10} \to \mathbb Z_{12}[/imath] defined by [imath]\phi(x)=0[/imath] for all [imath]x \in \mathbb Z_{10}[/imath] is a counterexample. This function is a homomorphism because [imath]\phi(x+y) = 0 = 0+0 = \phi(x) + \phi(y)[/imath] for all [imath]x, y \in \mathbb Z[/imath] (this function is discussed in problem 3 in assignment 7 as the function sending [imath][1][/imath] to [imath][0][/imath]). The order of [imath]g=1[/imath] is infinity. The order of [imath]\phi(1)=0[/imath] is one. The problem is that he said [imath]G \to G[/imath] is equivalent to [imath]\mathbb Z_{10} \to \mathbb Z_{12}[/imath], which is think is not right. Explain please? EDIT: Please look at this test and give me your honest opinion, http://zimmer.csufresno.edu/~ovega/teaching/151/Exam2Solutions.pdf
|
2000092
|
How to prove [imath]\sqrt{18}[/imath] is irrational without using proof by contradiction?
I want to know how to prove [imath]\sqrt{18}[/imath] is irrational using method other than proof by contradiction. I have always been taught to prove irrationality using proof by contradiction. So when I was asked this in my exam I was really surprised. Can anyone think of other methods to prove this? Please help. Thank you.
|
2001385
|
using proof by contradiction.
I am wondering whether there is another method to show that [imath]\sqrt{118}[/imath] is irrational. I have always been taught to use proof by contradiction for showing irrationality. Can anyone think of other methods? Please help.
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1997729
|
Diffeomorphisms act transitively on a connected, smooth manifold
Prove that if [imath]M[/imath] is a connected, smooth manifold, then [imath]\text{Diff}(M)[/imath] acts transitively on [imath]M[/imath]. What I'm trying to do is to take an arbitrary [imath]p\in M[/imath] and show that the orbit of [imath]p[/imath] [imath]\mathcal{O}_p:=\{q\in M \mid \phi(q)=p \text{ for some }\phi\in\text{Diff}(M)\}[/imath] is both open and closed. Since [imath]M[/imath] is connected, [imath]\mathcal{O}_p=M[/imath] for any [imath]p[/imath], which proves the proposition. But I'm having difficulty to prove [imath]\mathcal{O}_p[/imath] is open in the first place. My initial idea was to take a chart [imath](U,\psi)[/imath] about [imath]p[/imath] and, for any [imath]q\in U[/imath], take some [imath]g\in\text{Diff}(\mathbb{R}^n)[/imath] such that [imath]g(\psi(q))=\psi(p)[/imath] (that's possible because [imath]\text{Diff}(\mathbb{R}^n)[/imath] is transitive). Then [imath]\phi:=\psi^{-1}\circ g\circ \psi \in \text{Diff}(U)[/imath] has the property that [imath]\phi(q)=p[/imath]. But that is not enough, since we need [imath]\phi\in\text{Diff}(M)[/imath]. I've tried to prove that there is an extension of [imath]\phi\in\text{Diff}(U)[/imath] to some [imath]\bar{\phi}\in\text{Diff}(M)[/imath], but I couldn't. Besides, I have no idea how to prove [imath]\mathcal{O}_p[/imath] is closed. Any ideas? Thanks!
|
1196147
|
Proving that given any two points in a connected manifold, there exists a diffeomorphism taking one to the other
Suppose [imath]M[/imath] be a connected manifold and [imath]x, y \in M[/imath] are two points. Then I'm trying to show that there is a diffeomeorphism [imath]f[/imath] of [imath]M[/imath] that takes [imath]x[/imath] to [imath]y[/imath]. Since the set of points for which there is a diffeomorphism of [imath]M[/imath] taking [imath]x[/imath] to that point is clopen and [imath]M[/imath] is connected, I think it should be enough to consider the case when both [imath]x[/imath] and [imath]y[/imath] lie in the same chart. But how do I proceed from here? I think I should somehow take a vector field and consider flows, but I'm not really sure. Can someone please provide a solution? Thanks in advance.
|
2001247
|
Show that [imath]\mathbb Z_{12} \times \mathbb Z_5[/imath] is isomorphic to [imath]\mathbb Z_{60}[/imath]
There is a theorem that states that [imath]\mathbb Z_{m} \times \mathbb Z_n[/imath] is isomorphic to [imath]\mathbb Z_{mn}[/imath] if the [imath]gcd(m,n) = 1[/imath] because otherwise [imath]\mathbb Z_{m} \times \mathbb Z_n[/imath] would not be cyclic. Would deriving that proof and saying that [imath]\mathbb Z_{12} \times \mathbb Z_5[/imath] is isomorphic to [imath]\mathbb Z_{60}[/imath] by stating that [imath]gdc(12,5) = 1[/imath] be a sufficient proof?
|
375348
|
Prove that [imath]\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}[/imath] implies [imath]\gcd(m,n)=1[/imath].
Prove that [imath]\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}[/imath] implies [imath]\gcd(m,n)=1[/imath]. This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated. Thanks!
|
2001243
|
Expected number of turns to get 6 tails
Suppose I have [imath]6[/imath] fair coins, all facing heads ups. For each turn I will flip each of the [imath]6[/imath] coins. If a coin lands tails up, I set it aside. The next turn I take the remaining coins facing heads up and repeat the process. Now I want to know what is the expected number of turns [imath]T[/imath] until I have exactly [imath]6[/imath] tails. I know [imath]P(T=1)=\frac{1}{2^6}[/imath] But how do I find [imath]P(T=x)[/imath] with [imath]x\gt 1[/imath]? For [imath]T=2[/imath] it seems I should count the different ways this might happen i.e. [imath]1[/imath] tail on turn [imath]1[/imath], [imath]5[/imath] tails on turn [imath]2[/imath], etc. There are [imath]5[/imath] different ways, so is the probability [imath]5/64[/imath]? That does not seem right to me. Any hint is appreciated.
|
2000829
|
Expected days to finish a box of cookies
Adam has a box containing 10 cookies. Each day he eats each cookie with a probability of [imath]\frac12[/imath]. Calculate the expected number of days it takes Adam to complete the cookies. As a start we can set [imath]X[/imath] as the expected days it takes for Adam to finish eating the cookies. However I'm unable to progress further.
|
2001466
|
How to prove reflexivity, symmetry and transitivity for the following relation?
I would like to know how to prove reflexivity, symmetry and transitivity for [imath]\sim[/imath] according to the following definition: Suppose [imath]\sim[/imath] is defined on the set of the integers as follows : [imath]a\sim b[/imath] iff [imath]ab ≤ a|b|[/imath] Please help me. Thanks!
|
2001500
|
How to prove the symmetrical relation for the following statement?
I would like to know how to prove the symmetrical relation for [imath]\sim[/imath] according to the following definition: Suppose [imath]\sim[/imath] is defined on the set of the integers as follows : [imath]a\sim b[/imath] iff [imath]ab ≤ a|b|[/imath] Please explain to me. Thank you.
|
2001274
|
An isomorphic map from [imath]\mathbb R^{*} \rightarrow \mathbb R^{*}[/imath]
Say the map [imath]\phi[/imath] is an isomorphism from [imath]\mathbb R^{*} \rightarrow \mathbb R^{*}[/imath], where [imath]\mathbb R^{*}[/imath] is the group of all real numbers not including [imath]0[/imath] under the multiplication operation. The map [imath]\phi[/imath] is a bijection such that [imath]\phi (ab) = \phi(a) \phi(b)[/imath]. How would I prove that if [imath]r > 0[/imath] then [imath]\phi (r) > 0[/imath] and if [imath]r < 0[/imath] then [imath]\phi (r) < 0[/imath]? I can think qualitatively why this is true. For instance, the identity element of [imath]\mathbb R^{*}[/imath] is [imath]1[/imath] and will remain [imath]1[/imath] for any bijection of [imath]\mathbb R^{*} \rightarrow \mathbb R^{*}[/imath] otherwise the mapping will not be a group, thus, not an isomorphism. In order for [imath]1[/imath] to exist in the group after the mapping, you cannot multiply the elements of [imath]\mathbb R^{*}[/imath] by a constant. However, I don't know how to prove this mathematically. Does anyone know a way?
|
2001215
|
Is this an isomorphism possible?
I am working on the following homework problem: Let [imath]\phi[/imath] be an isomorphism from [imath]\mathbb{R}^*[/imath] to [imath]\mathbb{R}^*[/imath] (nonzero reals under multiplication). Show that if [imath]r>0[/imath], then [imath]\phi(r) > 0[/imath]. I was trying to prove this by contradiction, but I haven't been able to find any problems. I know that we must have [imath]\phi(1) = 1[/imath], and [imath]\phi(-1) = -1[/imath] no matter what, but despite playing around with those values, as well as [imath]\pm r[/imath] and [imath]\pm \frac{1}{r}[/imath], I haven't gotten any contradictions, and I'm starting to think this might be possible. What if we defined [imath]\phi(x)[/imath] as: [imath]x\ \text{if}\ x=\pm 1 \\ -x \text{, otherwise}[/imath] Would this not be an isomorphism? Any hints for this question would be much appreciated.
|
101214
|
Structure of ideals in the product of two rings
[imath]R[/imath] and [imath]S[/imath] are two rings. Let [imath]J[/imath] be an ideal in [imath]R\times S[/imath]. Then there are [imath]I_{1}[/imath], ideal of [imath]R[/imath], and [imath]I_{2}[/imath], ideal of [imath]S[/imath] such that [imath]J=I_{1}\times I_{2}[/imath]. For me is obvious why [imath]\left\{ r\in R\mid \left(r,s\right)\in J\text{ for some } s\in S\right\}[/imath] is an ideal of [imath]R[/imath] (and the same for [imath]S[/imath]) so I can prove [imath]J[/imath] is a subset of the product of two ideals and also that the product of these two ideals is also an ideal. But I can't see how to prove equality without assuming existence of unity or commutativity. Trying basically to show that if [imath]\left(r,s\right)\in J[/imath] then also [imath]\left(r,0\right),\left(0,s\right)\in J[/imath]. Thank you for your help!
|
2027894
|
Ideals of the real numbers
What would be the ideals of the real numbers? I have to figure out the ideals of a ring R, which is a product of integers and reals like so: [imath]R = \mathbb{Z}\times \mathbb{R} \times \mathbb{Z}[/imath] Are the ideals simply 0 and [imath]\mathbb{R}[/imath]?
|
2001036
|
Kripke models - prove that formula is not tautology in intuitionistical logic
My teacher proved that following formula is not tautology in intuitionistical logic: [imath]\neg(p\wedge q)\Leftrightarrow (\neg p\vee\neg q)[/imath] Is it sufficient to show (and we try to do it) that [imath]\neg(p\wedge q)\Rightarrow (\neg p\vee\neg q)[/imath] is not tautology. My teacher wrote red and black formulas. My formulas are green - there are formulas that to my eye my teacher forgot to write. I know that teacher give arbitrar Kripke model. Tell me please: 1. If my green "supplement" is correct ? 2. From what my teacher conclude red formulas ? When it comes to black things I know that it is about kripke model.
|
1999070
|
Prove that formula is not tautology.
Show Kripke's model proving that the following formula isn't tautology in intuitionistic logic. [imath] \neg ( p \wedge q ) \implies (\neg p \vee \neg q) [/imath] Please help/hint me ;)
|
2001518
|
Embedding any finite group into [imath]A_n[/imath]
I was reading a proof that any finite group can be embedded into the alternating group [imath]A_{n}[/imath], and it mentioned that in order to show this, it is sufficient to show that [imath]Sym(n)[/imath] can be embedded into [imath]A_{2n}[/imath]. Could somebody please explain to me why this is?
|
1056283
|
Embedding [imath]S_n[/imath] in [imath]A_{2n}[/imath]
I want to embed the symmetric group [imath]S_n[/imath] into the bigger alternating group [imath]A_{2n}[/imath]. How could I find such an injective homomorphism?
|
2001996
|
How do I derive a reduction formula for [imath]\int \frac{x^n}{\sqrt{x^2 + a^2}}dx[/imath]?
How do I evaluate a reduction formula for the indefinite integral of [imath]\int \frac{x^n}{\sqrt{x^2 + a^2}}dx[/imath] , for any positive integer n? I have tried integration by parts, with [imath]u=x^2[/imath] and [imath]\frac{dv}{dx}[/imath] = [imath]\frac{1}{a\sqrt{\frac{x^2}{a^2}+1}}[/imath] but still unable to see any recurring pattern. Can someone please guide me/show me a sample solution?
|
1997753
|
Reduction Integration of [imath]I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}[/imath]
How do I derive this using reduction formula? [imath]I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}[/imath] I got [imath]\sqrt{x^2+a^2}(x^n) - \frac{1}{n-1}(x^{n-1})\sqrt{x^2+a^2} + \frac{1}{n-1}(I_n) [/imath] . I'm not too sure if this is correct. I would appreciate any form of help or correction to my answer. Thank you.
|
1970634
|
Evaluating [imath]\frac1{n^2+1^2}+\frac2{n^2+2^2}+\dots+\frac n{n^2+n^2}[/imath] using Riemann sum
Not sure how to get started on this question. Do I have to form the summation? If yes, how do I go about doing it? Any help will be really appreciated!
|
1446408
|
limit [imath]\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}[/imath]
How do I evaluate this? [imath]\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}[/imath] I got concerned for that, I've tried make it integral for Riemann but it still undone.
|
2002077
|
Pythagoras always divisible by 3
Lets look at the pythagoras [imath]a^2+b^2=c^2[/imath] for the integers [imath]a,b,c[/imath]. Proof that one of components [imath]a,b,c[/imath] is always divisible by 3. How do i prove that?
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1995091
|
Pythagorean triples [imath](a,b,c)[/imath] and divisibility of [imath]a[/imath] or [imath]b[/imath] by [imath]3[/imath].
Call a triple of integers [imath](a, b, c)[/imath] a Pythagorean triple if [imath]a^2 + b^2 = c^2[/imath] , i.e., if [imath]a, b, c \in \mathbb{N^*}[/imath] are the (measures of) sides of a right triangle. Examples of Pythagorean triples are (3, 4, 5), (5, 12, 13), (8, 15, 17) and (3312, 16766, 17090). Show that if [imath](a, b, c)[/imath] is a Pythagorean triple if at least one of a and b is divisible by 3. I think to begin with [imath]a^2 + b^2 = c^2 \rightarrow a^2 - c^2 = b^2[/imath] to show this. Then I get stuck to prove this?
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2002090
|
Ideals of [imath]\mathbb Z\times\mathbb F_4[/imath]
Let [imath]R_1=\mathbb{Z}[/imath] and [imath]R_2=\mathbb{F}_4[/imath]. 1) Find all ideals in [imath]R_1 \times R_2[/imath]. Since [imath]\mathbb{Z}[/imath] has only principal ideals as [imath](0), (1), (2)...[/imath] The ideals of [imath]\mathbb{F}_4[/imath] is where I get confused, I think they are also principal ideals which are [imath](0), (1), (2), (3)[/imath]. Let [imath]I_1 \rhd R_1[/imath] and [imath]I_2 \rhd R_2[/imath]. Then, the ideals of [imath]R_1 \times R_2[/imath] are [imath]I_1 \times I_2=\{((i_1),(i_2)): i_1 \in \mathbb{Z}, i_2 \in \{0,1,2,3\}\}[/imath] 2) Which of them are principal ideal? I tend to think that they are all principal ideals. 3) Which of them are prime ideals, which are maximal ideals? I have no clue about this. Any suggestion would be appreciated! Thank you!
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2001454
|
Finding the Ideals of a Direct Product of Rings
Let [imath]R_1 = \mathbb{Z}[/imath], and [imath]R_2 = \mathbb{F}_4[/imath]. Find all ideals in [imath]R_1\times R_2[/imath]. Which of them are principal ideals? Which are prime? Which are maximal? I know the ideals in [imath]\mathbb{Z}[/imath] and [imath]\mathbb{F}_4[/imath], but I am having trouble finding the ideals in the cartesian product of both of these rings. If someone could help me out it would be appreciated.
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2001735
|
Using Addition Theorem for binomial identities
I know the addition theorem for binomial coefficients: [imath]\binom{x^*+y^*}{n}=\sum_{k=0}^n \binom{x^*}{n-k} \binom{y^*}{k}[/imath] for [imath]x^*,y^*\in \mathbb{R}[/imath] and [imath]n \in \mathbb{N}[/imath]. Using this theorem i can prove identities such [imath]\binom{2n}{n}=\sum_{k=0}^n \binom{n}{k}^2[/imath] easily by applying [imath]x^*=y^*=n[/imath]. But what about the identity: [imath]\binom{x+y+n-1}{n}=\sum_{k=0}^n \binom{x+n-k-1}{n-k} \binom{y+k-1}{k}[/imath] Of course we see that it results from the Theorem but is it a problem that [imath]k[/imath] is "part of [imath]x^*[/imath] and [imath]y^*[/imath] here?
|
1867972
|
Proving that [imath]{x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} [/imath]
How can I prove that [imath]{x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} [/imath] I tried the following: We use the falling factorial power: [imath]y^{\underline k}=\underbrace{y(y-1)(y-2)\ldots(y-k+1)}_{k\text{ factors}},[/imath] so that [imath]\binom{y}k=\frac{y^{\underline k}}{k!} .[/imath] Then [imath]{x +y+n- 1 \choose n} = \frac{(x +y+n- 1)!}{n! ((x +y+n- 1) - n)!} = \frac{1}{n!}. (x +y+n \color{#f00}{-1})^{\underline n} [/imath] And [imath] {x+n-k-1 \choose n-k}{y+k-1 \choose k}[/imath] [imath]\frac{1}{(n-k)!}.(x+n-k-1)^{\underline{n-k}}.\frac{1}{k!}.(y+k-1)^{\underline{k}}[/imath] [imath]\frac{1}{k!.(n-k)!}.(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}[/imath] [imath]\sum_{k=0}^n{n \choose k}(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}[/imath] According to the Binomial-coefficients: [imath] ((x+n-k-1) + (y+k-1))^{\underline{n}}[/imath] [imath] (x+y+n\color{#f00}{- 2})^{\underline{n}}[/imath] What is wrong ? und How can I continue? :/
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1212482
|
Sum of Power of Two Fibonacci reciprocals
Evaluate [imath]\sum_{k=0}^{\infty} \frac{1}{F_{2^n}} \;.[/imath] I'm thinking of using a relation from a term to another.
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157479
|
Evaluate the sum: [imath]\sum\limits_{n=0}^{\infty} \frac1{F_{(2^n)}}[/imath]
Evaluate the sum: [imath]\sum_{n=0}^{\infty} \frac{1}{F_{(2^n)}}[/imath] where [imath]F_{m}[/imath] is the [imath]m[/imath]-th term of the Fibonacci sequence. I need some support here. Thanks.
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2002507
|
Eigenvalue of stochastic matrix
Why is the eigenvalue of a stochastic matrix always [imath]1[/imath]? I have found lots of articles simply saying it is obvious that the eigenvalue is [imath]1[/imath] but can't get my head around the proofs.
|
40320
|
Proof that the largest eigenvalue of a stochastic matrix is [imath]1[/imath]
The largest eigenvalue of a stochastic matrix (i.e. a matrix whose entries are positive and whose rows add up to [imath]1[/imath]) is [imath]1[/imath]. Wikipedia marks this as a special case of the Perron-Frobenius theorem, but I wonder if there is a simpler (more direct) way to demonstrate this result.
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2002561
|
Determine [imath]\mathrm{Aut}(\Bbb Z_{2}\times\Bbb Z_{2})[/imath]
The title is the whole problem, but I don't know what is that mean and what it want me to do? To find out the properties of [imath]\Bbb Z_{2}\times \Bbb Z_{2}[/imath]? Can anyone explain this to me?
|
541026
|
Let V denote the Klein 4-group. Show that [imath]\text{Aut} (V)[/imath] is isomorphic to [imath]S_3[/imath]
After a week in my Abstract Algebra class, the professor proposed this as a problem. I'm not entirely sure where to begin. [imath] V = \{ e, \tau, \tau_1, \tau_2 \}[/imath], so I'm not sure exactly what is meant by [imath]\text{Aut} (V)[/imath]. Is it simply saying that the only way to have an automorphism on [imath]V[/imath] is to rearrange the order of the elements of [imath]V[/imath] and therefore [imath]\text{Aut} (V) \cong S_3[/imath]?
|
2000249
|
show matrix completion is SDP
I am working on showing a matrix completion problem is a SDP. Specifically, we want to show \begin{array} \underset{minimize}_{ X\in \mathbb{R}^{m \times n}} & \sum_{(i,j) \in \Omega} ( X_{ij} - Z_{ij} )^2 + \lambda \| X \|_{tr} \end{array} is a SDP. I know we can formulate [imath]\|X\|_{tr}[/imath] using its dual hence the above problem could be written as: \begin{array} \underset{minimize}_{X\in\mathbb{R}^{m \times n},W_1\in\mathbb{S^{m}},W_2\in\mathbb{S^n}} &\sum_{(i,j) \in \Omega} ( X_{ij} - Z_{ij} )^2 + \lambda(\text{tr}(W_1)+\text{tr}(W_2))\\ \text{subject to} & \begin{pmatrix} W_1 & \frac{1}{2} X\\ \frac{1}{2}X^{T} W_2\\ \end{pmatrix}\succeq0 \end{array} But I do not see how to show this is a SDP, either in usual form or canonical form. Thanks.
|
1993751
|
How to convert the matrix completion problem to the standard SDP form?
Given the matrix completion problem defined below. \begin{equation} \begin{array}{ll} \text{minimize }{X \in \mathbb{R}^{m \times n}} & \sum_{(i,j) \in \Omega} ( X_{ij} - Z_{ij} )^2 + \lambda \text{trace}(\sqrt[]{X^TX}) , \end{array} \end{equation} where, [imath]Z[/imath] is the observed matrix, [imath]X[/imath] is a low rank matrix that we want to find, [imath]\Omega[/imath] stores the row and column indices of the observed set. with tuning parameter [imath]\lambda > 0[/imath]. However it is a SDP problem and could be formulated as, \begin{equation*} \begin{array}{ll} \text{minimize }{x \in \mathbb{R}^p} & c^T x \\ \text{subject to } & x_1 A_1 + \cdots + x_p A_p \preceq B, \end{array} \end{equation*} for some fixed [imath]c, B, A_i, \; i=1,\ldots,p[/imath]. As the matrix completion problem utilizes the trace norm, F.Y.I, there are two ways of formulating the trace norm through optimization, \begin{equation} \begin{array}{ll} \max_{Y \in \mathbb{R}^{m \times n}} & \text{trace}(X^T Y) \\ \text{subject to} & \left[ \begin{array}{cc} I_m & Y \\ Y^T & I_n \end{array} \right] \succeq 0, \end{array} \label{eq:aa:primal} \end{equation} and its alternative dual, \begin{equation} \begin{array}{ll} \min_{\substack{W_{1} \in \mathbb{S}^{m}, \\ W_{2} \in \mathbb{S}^{n}}} & \text{trace}(W_{1}) + \text{trace}(W_{2}) \\ \text{subject to} & \left[ \begin{array}{cc} W_{1} & (1/2) X \\ (1/2) X^T & W_{2} \end{array} \right] \succeq 0, \end{array} \label{eq:aa:dual} \end{equation} What I have achieved so far, \begin{align*} \min_{X \in \mathbb{R}^{m \times n}} \|P_\Omega(X-Z)\|_F^2 + \lambda \| X \|_{\text{tr}} \\ = \text{trace}(\left(P_\Omega(X-Z)\right)^TP_\Omega(X-Z)) + \lambda \| X \|_{\text{tr}}\\ = \|\left(P_\Omega(X-Z)\right)^TP_\Omega(X-Z)\|_{tr} + \lambda \| X \|_{\text{tr}} \end{align*} [imath]\| X \|_{\text{tr}} = \text{trace}(\sqrt[]{X^TX})[/imath]
|
2002576
|
To find [imath]|Aut(K_{4})| = |Aut(\mathbb{Z}_{2} \times \mathbb{Z}_{2})|[/imath]
I'm in the process of trying to find [imath]|Aut(K_{4})|[/imath], where [imath]K_{4}[/imath] is the Klein-4 group, isomorphic to [imath]\mathbb{Z}_{2} \times \mathbb{Z}_{2}[/imath]. [imath]K_{4} = \{ id, (12)(34), (13)(24), (14)(23)\}[/imath] is a subgroup of [imath]S_{4}[/imath], the group of permutations of [imath]4[/imath] objects. Right now, I have already established that [imath]K_{4} \simeq \mathbb{Z}_{2} \times \mathbb{Z}_{2}[/imath] and have proven that if [imath]G,H[/imath] groups such that [imath]G \simeq H[/imath], then [imath]Aut(G) \simeq Aut(H)[/imath]. So, I am at the point where I am ready to check which of the permuations in [imath]K_{4}[/imath] are homorphisms, and I am unsure of how to do that. For example, for the permuation [imath](12)(34) = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{pmatrix}[/imath], I want to show that the map [imath]f: \begin{pmatrix}1 & 2 & 3 & 4 \end{pmatrix} \mapsto \begin{pmatrix} 2 & 1 & 4 & 3 \end{pmatrix} [/imath] is a homomorphism, but I am not sure what the appropriate group operations to use are. Until now, I have only ever calculated [imath]Aut[/imath] of [imath]\mathbb{Z}_{n}[/imath], where the operation I would use was addition modulo [imath]m[/imath], but I've never dealt with a case where the underlying group was a permutation group and not an additive group. Could somebody please demonstrate for me how to show that [imath]f: \begin{pmatrix}1 & 2 & 3 & 4\end{pmatrix} \mapsto \begin{pmatrix} 2 & 1 & 4 & 3 \end{pmatrix} [/imath] meets the criteria for being homomorphic? If I see one worked out example, I can probably figure out how to do the rest on my own, but as it is right now, I am at a loss. Thank you.
|
103390
|
Follow-up to question: Aut(G) for G = Klein 4-group is isomorphic to [imath]S_3[/imath]
This is most likely a lack of understanding of wording on my part. I was considerind the Klein 4-group as the set of four permutations: the identity permutation, and three other permutations of four elements, where each of those is made up of two transposes, (i.e., 1 [imath]\rightarrow[/imath] 2, 2 [imath]\rightarrow[/imath] 1 and 3 [imath]\rightarrow[/imath] 4, 4 [imath]\rightarrow[/imath] 3) taken over the three possible such combinations of four elements. Here, then, is my question. I am assuming (?) that Aut(G) in this case is the set of permutations of the four elements of the Klein 4-group - or the three non-identity ones for the purpose of showing isomorphic to [imath]S_3[/imath]. If this is the case, what does it mean to have a permutation of these three permutations that I mentioned above? As always, thanks for your help.
|
2002287
|
How to show that this subgroup is normal?
[imath]H = \left\{ id, (12)(34), (13)(24), (14)(23)\right\}[/imath] is a subgroup of [imath]A_4[/imath]. A normal subgroup is a subgroup where the right and left cosets are the same. To show that [imath]H[/imath] is a normal subgroup would I have to compute every single coset and show that the left and right cosets are the same? Is there a clever way to do this problem?
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1124697
|
Check whether subgroup is normal in [imath] S_4[/imath]
Let H={e,(12)(34)} and K ={e, (12)(34), (13)(24), (14)(23)} be subgroups of [imath] S_4[/imath] where e is identity element. Then which of following is true H and K are normal subgroups of [imath]S_4[/imath] H is normal in K and K is normal in[imath] A_4[/imath] H is normal in [imath]A_4[/imath], but not normal in [imath] S_4[/imath] K is normal in [imath]S_4[/imath] but H is not How should this question be approached, checking left and right cosets of these subgroup is very time consuming. Checking gh[imath] g^{-1}[/imath] [imath]\in [/imath] H for every g in [imath] S_4[/imath] is also time consuming.
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1920650
|
Proof that it always exists a prime between [imath]\sqrt n[/imath] and [imath]n[/imath]
I'm looking for a proof that ensures that for every natural number [imath]n > 1[/imath] there is always a prime [imath]p[/imath] such that [imath]\sqrt{n} \leq p \leq n[/imath] holds. I know you can deduce my previous claim as a corollary of some famous results like "Bertrand's postulate", but I was wondering if it does exist any easy self-contained proof.
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669017
|
At least one prime between [imath]\sqrt{n}[/imath] and [imath]n[/imath]?
Prove that for all [imath]n>2[/imath] there exists at least one prime [imath]p[/imath] such that [imath]\sqrt{n}<p<n[/imath] using elementary methods. My try: If not, [imath]\sum_{p<\sqrt{n}} (\lfloor n/p \rfloor-\lfloor \sqrt{n}/p \rfloor) \geq n-\sqrt{n}[/imath] then...?
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2002674
|
How to show that any two equivalence classes of non-zero elements in [imath]\mathbb{Z}_{p}[/imath] have the same number of elements
My question is related to this question I asked last year. I am just getting around to picking up the problem again, and normally, I would just ask a follow-up question to the answer that was given, but the person who wrote it is away from MSE due to medical reasons, and I am in urgent need of an answer. My original question was: Let [imath]p[/imath] be a prime number and [imath]\gcd(p,n)=1[/imath]. Define an equivalence relation on [imath]\mathbb{Z}_{p}[/imath] as follows: [imath]x \sim y[/imath] iff [imath]n^{r}x = n^{t}y[/imath] for some [imath]r,t \geq 0[/imath]. Let [imath]m[/imath] be the number of equivalence classes of this equivalence relation. Prove that [imath]m-1[/imath] is a divisor of [imath]p-1[/imath]. I followed the answerer's advice and showed that [imath]x \sim 0[/imath] iff [imath]x = 0[/imath], and now am working on the next step,which is to show that any two equivalence classes of non-zero elements of [imath]\mathbb{Z}_{p}[/imath] have the same number of elements, call it [imath]k[/imath]. However, I am not sure how to do this part. Could someone please let me know how I am supposed to do this? Thank you.
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1510659
|
Prove that 1 less than the number of equivalence classes divides [imath]p-1[/imath] where [imath]p[/imath] is prime
I am faced with the following problem: Let [imath]p[/imath] be a prime number and [imath]\gcd(p,n)=1[/imath]. Define an equivalence relation on [imath]\mathbb{Z}_{p}[/imath] as follows: [imath]x \sim y[/imath] iff [imath]n^{r}x = n^{t}y[/imath] for some [imath]r,t \geq 0[/imath]. Let [imath]m[/imath] be the number of equivalence classes of this equivalence relation. Prove that [imath]m-1[/imath] is a divisor of [imath]p-1[/imath]. I must admit that I am at a loss as even where to begin with this one: the only thing I know for certain is that [imath]m-1[/imath] is certainly not a divisor of [imath]p[/imath]! (Unless of course, [imath]m-1[/imath] is equal to either [imath]1[/imath] or [imath]p[/imath]). Well, that and that since there are [imath]m[/imath] equivalence classes, one of them must be [imath][m-1][/imath]. I even tried working backwards, saying to myself okay, if [imath]m-1|p-1[/imath], then [imath]\exists c[/imath] such that [imath]p-1 = c(m-1)\, \to \, p = cm-c+1[/imath]. But that also doesn't seem to get me anywhere. It seems as though I have been staring at this forever with nothing to show for it. Thus, any assistance anyone out there could give me would be greatly appreciated.
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2002912
|
How to show that if [imath]H \leq G[/imath] and [imath]H[/imath] is the only subgroup of [imath]G[/imath] of order n than [imath] H\trianglelefteq G[/imath]
Say [imath]G[/imath] is a finite group and [imath]H[/imath] is a subgroup of [imath]G[/imath] such that [imath]H[/imath] is the only subgroup of [imath]G[/imath] of the order [imath]n[/imath]. How would I prove that [imath]H[/imath] is a normal subgroup of [imath]G[/imath]? I was thinking of using the fact that [imath]|gHg^{-1}| = |H|[/imath], but I did not get anywhere.
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2002382
|
Prove that [imath]H[/imath] is normal.
Let [imath]G[/imath] be a finite group and let [imath]H[/imath] be a subgroup of [imath]G[/imath] with the property that no other subgroup of [imath]G[/imath] has order [imath]|H|[/imath]. Prove that [imath]H[/imath] is normal. My attempt: To show that [imath]H[/imath] is normal, I need to show that for any [imath]g,h \in G[/imath], [imath]ghg^{-1}\in H[/imath]. Let the order of [imath]G=n[/imath] and let the order of [imath]H=k[/imath]. If [imath]H[/imath] is a subgroup of [imath]G[/imath] then [imath]k[/imath] divides [imath]n[/imath] and the identity element [imath]e[/imath] is both [imath]G[/imath] and [imath]H[/imath]. Also for any [imath]a,b \in H[/imath], [imath]ab^{-1}\in H[/imath]. Since [imath]k[/imath] divides [imath]n[/imath], [imath]n=qk[/imath] for some integer [imath]q[/imath]. [imath]a^{n}=a^{qk}=e[/imath]... (I don't think this proof is getting me anywhere. I think I need a hint about what does "[imath]H[/imath] being a subgroup of [imath]G[/imath] such that no other subgroup of [imath]G[/imath] has order [imath]|H|[/imath]" imply...)
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1102646
|
closure of a convex set in a normed linear space is convex ?
Is it true that if [imath]A[/imath] is a convex set in a normed linear space [imath]V[/imath] , then the closure of [imath]A[/imath] is also convex ? (I know that the interior is convex )
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202378
|
Is closure of convex subset of [imath]X[/imath] is again a convex subset of [imath]X[/imath]?
Today I was going through my text book exercise and got hold of the following question. Let [imath]X[/imath] be a normed linear space with norm [imath]\lVert\cdot\rVert[/imath] and [imath]A[/imath] is a non empty convex subset of [imath]X[/imath] then prove that [imath]\operatorname{Closure}(A)[/imath] is also a convex subset of [imath]X[/imath]. I gave it a try but could not succeed.
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2002654
|
How to prove an analytic function is analytic
I know the fact that not every [imath]C^{\infty}[/imath] function is analytic, for which there is the famous example: [imath]f(x):= \begin{cases} e^{-1/x} & x>0 \\ 0 & x\leq 0 \end{cases} [/imath] In that case, [imath]f[/imath] is [imath]C^{\infty}[/imath] but its Taylor series is identically zero, which is clearly different from [imath]f[/imath] itself. But how can I prove a function is actually analytic? Take [imath]\sin(x)[/imath] or [imath]\cos(x)[/imath], for example. We can easily calculate each Taylor series [imath]T_{\sin}(x):=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}[/imath] and [imath]T_{\cos}(x):=\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{(2k)!}[/imath] and check the convergence of both. But how do we prove that [imath]T_{\sin}(x)=\sin(x)[/imath] and [imath]T_{\cos}(x)=\cos(x)[/imath] for all [imath]x\in\mathbb{R}[/imath]? What about other examples ([imath]\tan(x)[/imath], [imath]e^x[/imath] etc)? Do we really have to treat each case separately? Is there any theorem that makes this task easier? Thanks!
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590455
|
How to check the real analyticity of a function?
I recently learnt Taylor series in my class. I would like to know how is to possible to distinguish whether a function is real-analytic or not. First thing to check is if it is smooth. But how can I know whether the taylor series converges to the function? For example: [imath]f(x)=\frac{1}{1-x}, x\in(0,1)[/imath] has [imath]n^{th}[/imath] degree taylor polynomial [imath]\sum_{k=0}^n x^k[/imath]. In this case, I understand that [imath]f[/imath] is analytic in its domain since the geometric series [imath]\sum_{k=0}^\infty x^k[/imath] for [imath]x\in(0,1)[/imath] converges to [imath]\frac{1}{1-x}[/imath]. In general, what is the trick? For example, how to know if [imath]\sin(x),\cos(x)[/imath] are analytic?
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2003039
|
Does a compact embedded manifold achieve all unit normals?
Fix some compact [imath]n-1[/imath] dimensional manifold [imath]M[/imath] without boundary embedded in [imath]\mathbb{R}^n[/imath]. For any unit vector [imath]v\in \mathbb{R}^n[/imath], does there exist a point [imath]p\in M[/imath] at which the unit normal vector is [imath]v[/imath]? I think the result holds if [imath]n=2[/imath], because the only compact [imath]1[/imath]-dimensional manifolds embedded in [imath]\mathbb{R}^2[/imath] are diffeomorphic to the circle, so should achieve all unit normals. Intuitively, it seems reasonable that this should hold for [imath]n=3[/imath] as well. This feels like it's just basic geometry and there should be a proof for it around somewhere, but I haven't found it yet. I think an equivalent question goes something like: if the level set [imath]S=f^{-1}(y)[/imath] of a differentiable function [imath]f:\mathbb{R}^n\to \mathbb{R}[/imath] is non-empty and 'finite' in some sense, does [imath]\frac{\nabla f}{\left|\left|\nabla f\right|\right|}[/imath] achieve all values of the unit sphere [imath]\mathbb{S}^{n-1}[/imath] on [imath]S[/imath]?
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389196
|
Prove that Gauss map on M is surjective
Let [imath]M[/imath] be a closed, orientable, and bounded surface in [imath]\mathbb{R}^3[/imath]. (a) Prove that the Gauss map on [imath]M[/imath] is surjective. (b) Let [imath]K_+(p) = \max \{0, K(p)\}[/imath]. Show that [imath] \int K_+dA \ge 4\pi. [/imath] in the area of [imath]M[/imath]. Do not use the Gauss-Bonnet Theorem to prove this.
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2002751
|
Prove that [imath]I=\{r \in R: f(r) \in J\}[/imath] is a prime ideal in [imath]R[/imath].
Let [imath]f:R\to S[/imath] be a surjective homomorphism of commutative rings. If [imath]J[/imath] is a prime ideal in [imath]S[/imath], and [imath]I=\{r \in R: f(r) \in J\}[/imath], prove that [imath]I[/imath] is a prime ideal in [imath]R[/imath]. I am perfectly fine with proving that [imath]I[/imath] is prime, however, I am struggling to prove that [imath]I[/imath] is an ideal in [imath]R[/imath] in the first place. Clearly [imath]I \neq \emptyset[/imath], since if [imath]j \in J\subseteq S[/imath], by surjectivity of [imath]f[/imath], there exists an [imath]r \in R[/imath] s.t. [imath]f(r)=j[/imath], i.e. [imath]r \in I[/imath]. Now let [imath]a,b \in I[/imath], then [imath]f(a), f(b) \in J[/imath]. Then [imath]f(ab)=f(a)f(b) \in J[/imath], i.e. [imath]ab \in I[/imath]. Also notice that [imath]f(a-b)=f(a)-f(b) \in J[/imath] (since [imath]J[/imath] is an ideal, and hence a subring, of [imath]S[/imath]), and so [imath]a-b \in I[/imath]. We thus have that [imath]I[/imath] is a subring of [imath]R[/imath]. In order to prove that [imath]I[/imath] is an ideal in [imath]R[/imath], we require the following If [imath]a \in I[/imath] and [imath]r \in R[/imath], then [imath]ar \in I[/imath]. Since [imath]I[/imath] is commutative (since [imath]R[/imath] is commutative), this will be enough to ensure that [imath]ra=ar \in I[/imath] is also satisfied. This is the part I am struggling with: Let [imath]a \in I[/imath], then [imath]f(a) \in J[/imath].Now consider any [imath]r \in R[/imath], then clearly [imath]f(r) \in S[/imath] and so [imath]f(ar)=f(a)f(r) \in S[/imath]. The problem, however, is I require [imath]f(ar) \in J[/imath] to have that [imath]ar \in I[/imath]. How can I do this?
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409999
|
Prove that the preimage of a prime ideal is also prime.
Let [imath]f: R \rightarrow S[/imath] be a ring homomorphism, with [imath]R[/imath] and [imath]S[/imath] commutative. If [imath]P[/imath] is a prime ideal of [imath]S[/imath], show that the preimage [imath]f^{-1}(P)[/imath] is a prime ideal of [imath]R[/imath]. Define [imath]g: S \rightarrow S/P[/imath] with kernel [imath]s[/imath]. Let [imath]h = g \circ f: R \rightarrow S/P[/imath]. Since [imath]h[/imath] is a ring homomorphism, the kernel is an ideal of [imath]R[/imath]. Also, from the first isomorphism theorem, we know that [imath]R/\ker(h) \cong S/P[/imath]. Since [imath]P[/imath] is a prime ideal of [imath]S[/imath], we know that [imath]S/P[/imath] is an integral domain. Since [imath]R/\ker(h)[/imath] is isomorphic to [imath]S/P[/imath], it must also be an integral domain, which implies that the kernel of [imath]h[/imath] (which is the preimage of [imath]P[/imath]) is a prime ideal of [imath]R[/imath]. Do you think my answer is correct? The reason why I was a bit skeptical of my answer is because I did not use the fact that [imath]R[/imath] and [imath]S[/imath] are commutative. So I'm wondering if I missed something [imath]\dots[/imath] Thank you in advance.
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2003838
|
If [imath]b [/imath] divides [imath]ad [/imath] with[imath] \mathrm {gcd}(a,b)=1[/imath], then [imath]b [/imath] divides [imath]d [/imath]
Let [imath]a\in \mathbb {Z} [/imath] and [imath]a,b \in \mathbb {N} [/imath]. Suppose that [imath]b [/imath] divides [imath]ad [/imath] where [imath]\mathrm {gcd} (a,b)=1[/imath]. What can you conclude from the above assumptions? I have a feeling that we must have [imath]b [/imath] divides [imath]d [/imath], but I am unsure and have no intuition guiding me as to a formal way to express what I believe is the conclusion.
|
1984311
|
Suppose [imath]a,b,m,n[/imath] are natural numbers, prove that if [imath]a({m \over n})=b[/imath] and [imath](a,n)=1[/imath], then [imath]{m \over n} \in \mathbb{N}[/imath]
Suppose that [imath]a,b,m,n[/imath] are natural numbers, prove the following statement: If [imath]a({m \over n})=b[/imath] and [imath](a,n)=1[/imath], then [imath]{m \over n} \in \mathbb{N}[/imath]
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2004152
|
Let [imath]\sum_{n=0}^{\infty} a_n[/imath] be a convergent series such that [imath]a_n \downarrow 0[/imath]. Prove that, [imath] \lim_{n\to \infty} na_n = 0 [/imath]
This is my proof so far, Let [imath]\varepsilon > 0[/imath]. By CCC, there is [imath]N > 0[/imath], such that, if [imath]n,m>N[/imath], then [imath]|\Sigma_{i=m+1}^n a_i| < \varepsilon[/imath]. We can drop the absolute values as [imath]a_i \geq 0[/imath], and, [imath]a_{m+1}\geq a_{m+2}\geq ... \geq a_n[/imath]. now I'm stuck.
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1983651
|
Let [imath](x_n)\downarrow 0[/imath] and [imath]\sum x_n\to s[/imath]. Then [imath](n\cdot x_n)\to 0[/imath]
Let [imath](x_n)[/imath] a decreasing sequence and [imath]\sum x_n\to s[/imath]. Then [imath](n\cdot x_n)\to 0[/imath] Check my proof please, Im not completely sure about it correctness. If [imath]\sum_{k=h}^\infty x_k= s[/imath] then we can rewrite the sum for starting index [imath]1[/imath] with the change [imath]k-h=j[/imath], then [imath]\left(\sum_{j=1}^n x_j\right)-n\cdot x_n=\sum_{j=1}^n (x_j-x_n)[/imath] Then taking limits [imath]\color{red}{\lim_{n\to\infty}\left[\left(\sum_{j=1}^n x_j\right)-n\cdot x_n\right]}=\lim_{n\to\infty}\sum_{j=1}^n (x_j-x_n)=\sum_{j=1}^\infty (x_j-0)=\color{red}{\lim_{n\to\infty}\sum_{j=1}^n x_j}=s[/imath] where I used the fact that [imath](x_n)\to 0[/imath]. Thus equating the colored expressions this implies that [imath]\lim_{n\to\infty} nx_n=0[/imath]. The proof, to my eyes, seems correct but I dont needed in any moment to use the fact that [imath](x_n)[/imath] is a monotonic sequence so it is possible that I make a mistake somewhere or that the proof is incorrect. My second attempt Because [imath]\sum x_k[/imath] converges and is positive (cause [imath](x_n)\downarrow 0[/imath]) we can write [imath]\sum_{k=n+1}^{2n+m}x_k=\left|\sum_{k=n+1}^{2n+m}x_k\right|<\epsilon/2,\quad \forall n,m\ge N[/imath] Then, cause [imath](x_n)[/imath] is decreasing [imath](n+m)x_{2n+m}\le\sum_{k=n+1}^{2n+m}x_k<\epsilon/2\\\implies (2n+m)x_{2n+m}\le2(n+m)x_{2n+m}<\epsilon,\quad\forall n,m\ge N[/imath] Because [imath]m[/imath] is arbitrary setting [imath]M=2N>N[/imath] we can finally write [imath]nx_n<\epsilon,\quad\forall n\ge M[/imath] It is this proof correct? Thank you.
|
1757818
|
If G contains a normal subgroup [imath]N \cong \mathbb{Z}_2[/imath] and [imath]G/N \cong \mathbb{Z}[/imath], then [imath]G\cong \mathbb{Z}\times \mathbb{Z_2}[/imath].
If G contains a normal subgroup [imath]N \cong \mathbb{Z}_2[/imath] and [imath]G/N \cong \mathbb{Z}[/imath], then [imath]G\cong \mathbb{Z}\times \mathbb{Z_2}[/imath]. I'm trying to create an isomorphism [imath]\phi : G \rightarrow \mathbb{Z}\times\mathbb{Z_2}[/imath] as [imath]g \mapsto (gN, ??)[/imath]. What can I use for my second coordinate?
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1778257
|
If [imath]G[/imath] contains a normal subgroup [imath]H \cong \mathbb{Z_2}[/imath] such that [imath]G/H[/imath] is infinite cyclic, then [imath]G \cong \mathbb{Z} \times \mathbb{Z_2}[/imath]
Let [imath]G[/imath] be a group containing a normal subgroup [imath]H \cong \mathbb{Z_2}[/imath] such that [imath]G/H[/imath] is infinite cyclic. Then [imath]G \cong \mathbb{Z} \times \mathbb{Z_2}[/imath]. Since [imath]G/H[/imath] is infinite cyclic it's isomorphic to [imath]\mathbb{Z}[/imath]. And since [imath]H \cong \mathbb{Z_2}[/imath], we have [imath]G/H \times H \cong \mathbb{Z} \times \mathbb{Z_2}[/imath]. So the question now is can we make [imath]G/H \times H \cong G[/imath] (without any recourse to the Isomorphism theorems please)? Any hints appreciated.
|
2003384
|
Binomial equality proof
[imath] \sum_{k=0}^{n} \binom{2n}{2k} = 2^{2n-1} [/imath] Now, I tried: [imath] (a+b)^n= \sum_{k=0}^{n}a^{n-k} b^{k} \binom{n}{k}[/imath] as I used: [imath] (1+1)^{2n-1}=(a+b)^{n}[/imath] But then I'm stuck because I'm not sure what to do with the binom itself. Or I tried using the binom identities, but I'm not too sure about that too.
|
2001785
|
What is wrong with the proof?
I try to prove the identity: [imath]\sum_{k=0}^n \binom{2n+1}{2k}=2^{2n}[/imath] but have troubles with the last step: [imath]\sum_{k=0}^n \binom{2n+1}{k}=\sum_{k=0}^n \left( \binom{2n}{2k} + \binom{2n}{2k-1}\right) =\sum_{j=-1}^{2n}\binom{2n}{j}=2^{2n}[/imath] But should not [imath]j[/imath] start from [imath]0[/imath]? Is something wrong?
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2003300
|
Limits and exponential: [imath]\lim\limits_{x \to\infty} (x-x^2 \ln(1+1/x))[/imath]
Evalulate [imath]\lim\limits_{x \to ∞} (x-x^2 \ln(1+1/x))[/imath] I know the answer is 1/2 and we have to make use of the exponential function but i can't seem to simplify the expression to get the answer, please help
|
1998703
|
How do I determine [imath]\lim_{x\to\infty} \left[x - x^{2} \log\left(1 + 1/x\right)\right][/imath]?
How do I determine this limit ?: [imath] \lim_{x \to \infty}\left[x - x^{2}\log\left(1 + {1 \over x}\right)\right] [/imath] I have tried to decompose the [imath]\log[/imath] function but I don't know how to proceed from there.
|
2003507
|
Show that [imath]X_n\to0[/imath] in probability does not imply [imath]\frac{S_n}{n}\to 0[/imath] in probability, even if the [imath]X_n[/imath] are independent.
Show that [imath]X_n\to0[/imath] in probability does not imply [imath]\dfrac{S_n}{n}\to 0[/imath] in probability, even if the [imath]X_n[/imath] are independent. How to approach this problem. Any hints will be appreciated. I am not able to even start the problem.
|
368467
|
Is Cesaro convergence still weaker in measure?
I've encountered a question I couldn't answer, and I would appreciate any help: Is it true that [imath]f_n \xrightarrow{m}0\,\,\implies \,\,\frac{1}{n} \sum_{k=1}^{n}f_k \xrightarrow{m}0[/imath] where [imath](X,{\mathcal{B}},m)[/imath] is a probability space, and [imath]\xrightarrow{m}[/imath] means: [imath]\forall \epsilon>0,\, m\left(|f_n|\geq \epsilon\right) \rightarrow 0.[/imath] This is similar to Cesaro convergence, which is weaker than regular convergence, but it is not the case here. I've tried to prove the proposition, or to give a counter example, but didn't manage to succeed in either. Any insights will be appreciated!
|
2003407
|
A question about a limit involving nested radicals
[imath]\lim\limits_{n\rightarrow \infty}1＋\sqrt{2+\sqrt[3]{3＋…\sqrt[n]{n}}}[/imath] Any hint will be appreciated
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1633066
|
How do I calculate this limit: [imath]\lim\limits_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}}[/imath]?
I have seen this question on the internet and was interested to know the answer. Here it is : Calculate [imath]\lim\limits_{n\to\infty}(1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}})[/imath]? Edit : I really tried doing it but wasn't able to get somewhere. I know how to do questions like [imath] y = (1+\sqrt{1+\sqrt{1+\dotsb+\sqrt 1}}) [/imath] and then we write [imath] (y-1)^2 = y [/imath] and solve. But for this I have no method. So I would like even a sort of a hint to try to get me started, no need for answer.
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2004463
|
Help with Limit: [imath]\lim_{n \to \infty} \frac{(2 n)! (n)^n}{n! (2 n)^{2 n}}[/imath]
Honestly, I am just plain stuck, I have been hitting my head against it for [imath]2[/imath] days straight. I know the solution should be [imath]2^{-2}[/imath] but... Help would be appreciated. If anyone has a Wolfram Alpha PRO account I am sure that thing churns out the solution as it is something quite basic I just don't see it and its driving me crazy. Edit: Stacks seemed to correct my formating from this sloppy one "Limit[(2 n)! n^n (E^n/(n! (2 n)^(2 n))), n -> Infinity]" [imath]\displaystyle\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}[/imath] to this one [imath]\displaystyle\lim_{n \to \infty} \frac{(2 n)! (n)^n}{n! (2 n)^{2 n}}[/imath] somewhere dropping the [imath]e^n[/imath] term. I should have learned to propperly use latex in stackexchange first.
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2004739
|
Help with Limit: [imath]\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}[/imath]
Help with Limit: [imath]\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}[/imath] EDIT: Posted a similar question just yesterday but bad formating from my part dropped a term, and lead me to strange places. Honestly, I am just plain stuck, I have been hitting my head against it for [imath]2[/imath] days straight. I know the solution should be [imath]2^{-1/2}[/imath] but... Help would be appreciated. If anyone has a Wolfram Alpha PRO account I am sure that thing churns out the solution as it is something quite basic I just don't see it and its driving me crazy.
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2004414
|
Please help I have this and I didn't get it well
How to solve [imath]\lim_{x \rightarrow 2} \frac{1}{x-2}[/imath]? I tried to solve it by dividing the both sides on [imath]x[/imath] but it doesn't work and I tried to get the dam [imath]-2[/imath] out but also didnt work so how to solve it I know that there is no limit for this but how to Remove the indeterminacy
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921633
|
What does [imath]\lim_{x\to4} \frac{1}{x-4}[/imath] equal?
[imath]\lim_{x\to4} \frac{1}{x-4}[/imath] Would it be correct to say that the limit is undefined because the denominator would be [imath]0[/imath]?
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2005599
|
10001 and in general [imath]10^k + 1[/imath], how can we identify without brute force that such numbers are not prime?
I know that [imath]137 \times 73 = 10001 [/imath]. I am looking for a properly reasoned approach. I believe that there is some general result also which says that all numbers of the form [imath]10^k + 1[/imath] (for [imath]k>2[/imath]) are NOT prime. Why? I mean can the general expression for the factors of such numbers be found?
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1995847
|
All numbers of form [imath]10^{k} + 1[/imath] are composite for [imath]k{\gt}2[/imath] proof
I have observed that all numbers of the form [imath]10^{k} + 1[/imath] are composite for [imath]k{\gt}2[/imath], and am trying to prove the same. Progress For odd [imath]k[/imath] , it is fairly obvious, as we get 11 as a divisor. I am having difficulty with [imath]k[/imath] even. Have no clue how to approach. Would prefer a non-inductive approach, if possible Developments Thanks to crostul, we can say it is composite for [imath]k[/imath] having an odd divisor. Hence, it only remains to prove it for [imath]k = 2^n[/imath]
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1998172
|
If [imath]x \leq y \leq z[/imath], and both sides of the equation are defined, then [imath]\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.[/imath]
Show that if [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath] are consecutive terms of an arithmetic sequence, with [imath]x \leq y \leq z[/imath], and both sides of the equation are defined, then [imath]\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.[/imath] I have no idea how to even start this problem, I'm stuck. Solutions are greatly appreciated.
|
1050711
|
If [imath]x[/imath], [imath]y[/imath], [imath]z[/imath] are in arithmetic progression, show that [imath]\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y. [/imath]
Show that if [imath]x, y,[/imath] and [imath]z[/imath] are consecutive terms of an arithmetic sequence, and [imath]\tan y[/imath] is defined, then [imath]\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y. [/imath] I'm not sure what trig identities I would use and how to use them. Could I get some help? Thanks.
|
2005467
|
How to tackle this question of bounded variation?
Let [imath]f : [0,1] \longrightarrow \mathbb R[/imath] be a function defined by [imath]f(x) = x^\alpha \sin \frac 1 {x^\beta}[/imath] , whenver [imath]x \in (0,1][/imath] and [imath]f(0) = 0[/imath] where [imath]\alpha > \beta[/imath]. Then for what values [imath]\alpha[/imath] and [imath]\beta[/imath], [imath]f[/imath] is a function of bounded variation on [imath][0,1][/imath]? I think the theorem which states that "If [imath]f : [a,b] \longrightarrow \mathbb R[/imath] be a continuous function on [imath][a,b][/imath] and if [imath]f'[/imath] exists and is bounded on [imath](a,b)[/imath] then [imath]f[/imath] is a function of bounded variation on [imath][a,b][/imath]" can be applied to solve this kind of problem. But we also know that there exist functions which are of BV but do not satisfy the above theorem. So, my question is what is the actual way to proceed? Please help me. Thank you in advance.
|
1999502
|
Find the values of [imath]\alpha[/imath] and [imath]\beta[/imath] for which the function [imath]f[/imath] is of bounded variation on [imath][0,1][/imath]?
The question is : Let [imath]f : [0,1] \longrightarrow \mathbb {R}[/imath] be defined by [imath]f(x) = x^{\alpha} \sin {\frac {1} {x^{\beta}}}[/imath] , whenever [imath]x \in (0,1][/imath] and [imath]=0[/imath] , whenever [imath]x = 0[/imath]. Then find the values of [imath]\alpha[/imath] and [imath]\beta[/imath] for which [imath]f[/imath] becomes a function of bounded variation on [imath][0,1][/imath]. Please give me a right way to proceed.Thank you in advance. I have just obtained the result which is [imath]\alpha > 2[/imath] and [imath]\beta < \alpha - 2[/imath].Is it correct at all?Please verify it.
|
1998338
|
X ~ N(0,1) What is the CDF of random vector [imath] \begin{pmatrix} X\\ X\\ \end{pmatrix}[/imath]
Is this the degenerate case? Would the CDF just be that of the standard normal? Thanks for the help.
|
1996810
|
What is the distribution function of [imath]\left( \begin{smallmatrix} X \\ (-1)^nX \\ \end{smallmatrix}\right)[/imath] where [imath]X\sim N(0,1)[/imath]
Find the distribution function of this multivariate normal variable where [imath]X \sim N(0, 1)[/imath] [imath] \begin{pmatrix} X \\ (-1)^nX \\ \end{pmatrix} [/imath] I need this for a proof that marginal convergence in distribution does not imply joint convergence in distribution, but I am really rusty. Any help is much appreciated. Edit: Explaining notation. This is a sequence of two-dimensional random variable vectors, where [imath]n[/imath] takes values in the natural numbers. The distribution function should differ for odd and even values of [imath]n[/imath].
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2005014
|
If [imath]g \circ f[/imath] is surjective, what can you say about the surjectivity of [imath]f[/imath] and [imath]g[/imath]?
Given [imath]f : A \to B[/imath] and [imath]g: B \to C[/imath], if [imath]g \circ f[/imath] is surjective, what can you say about the surjectivity of [imath]f[/imath] and [imath]g[/imath]? My Attempted Proof Suppose [imath]g \circ f[/imath] is surjetive and assume either [imath]f[/imath] or [imath]g[/imath] not surjective. Then [imath]g(f(A)) \subset C[/imath], and there exists [imath]c \in C[/imath], such that [imath]c \neq g(f(a))[/imath] for any [imath]a \in A[/imath], contradicting our assumption. Thus [imath]g \circ f[/imath] is surjective if and only if both [imath]f[/imath] and [imath]g[/imath] are surjective. [imath]\square[/imath] Is my proof correct? If so how rigorous is it?
|
986980
|
What can you say about [imath]f[/imath] and [imath]g[/imath] in the case that [imath]fg[/imath] is 1) Injective, 2) Surjective - Cohn - Classic Algebra Page 15
Question: Are my proofs below valid? In both cases we are using: [imath]f:A\to B, g: B\to C[/imath] Notation of your type converted: [imath](g\circ f)(x)=g(f(x))=xfg[/imath] If [imath]fg[/imath] is injective what can be said about [imath]f[/imath] and [imath]g[/imath]? If [imath]fg[/imath] is injective then [imath]f[/imath] is injective: Proof: [imath](x_1fg = x_2fg \implies x_1=x_2)[/imath] [imath]x_1fg=x_2fg[/imath] is clearly true when [imath]x_1f=x_2f[/imath] Which requires [imath]x_1=x_2[/imath], hence [imath]f[/imath] is injective. If [imath]f[/imath] were not injective, then [imath]x_1fg\ne x_2fg[/imath] [imath]\blacksquare[/imath] If [imath]fg[/imath] is surjective what can be said about [imath]f[/imath] and [imath]g[/imath]? If [imath]fg[/imath] is surjective then [imath]g[/imath] is surjective: Proof: [imath](\forall c\in C)(\exists a\in A) |( a(fg))=c[/imath] This means for [imath]g[/imath] every element must have had been mapped to. How can this be proved Mathematically? well [imath](af)g=c[/imath] is surjective, hence [imath](af)=(b\in B)[/imath] and [imath]bg=c[/imath] [imath](\forall c\in g)(\exists b\in B)|(bg=c)[/imath] Hence [imath]g[/imath] is surjective. [imath]\blacksquare[/imath]
|
2004357
|
Are there continuous functions for which the epsilon-delta property doesn't hold?
The standard definition of continuity is as follows: A function is continuous if [imath]\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon [/imath] This may sound silly, but is the converse true? In other words, is there an example of a function which is continuous, but for which this property doesn't hold? I thought I remembered reading something about the exponential function being an example of a continuous function for which the epsilon-delta implication is false, since it gets arbitrarily steep as x approaches infinity, but I could be wrong. Thanks!
|
1060655
|
What does "if and only if" mean in definitions?
Consider the following definition: A sequence [imath]\{p_n\}[/imath] is Cauchy if we have that for every [imath]n, m \ge N[/imath]: [imath]|p_n - p_m| < \epsilon[/imath] Although if and only if is not used, we know that if a sequence is Cauchy, the Cauchy criterion holds. Consider now the following definition: An action of a group [imath]G[/imath] on a set [imath]X[/imath] is transitive if and only if for each [imath]x, y \in X[/imath] there is a [imath]g \in G[/imath] such that [imath]gx = y[/imath]. What motivates the use of "if and only if" here? As far as definitions go, my head has automatically thought "if and only if." The latter example is taken from Ratcliffe's "Foundations of Hyperbolic Manifolds", and he switches a lot between "if" and "if and only if" in his definitions.
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1673842
|
Simplify and find [imath]\lim_{n\to \infty}\frac{(2n-1)!}{(2n+1)!}[/imath]
So I was calculating [imath]\lim_{n\to \infty}\frac{(2n-1)!}{(2n+1)!}[/imath] and couldn't solve it, so I saw the answer sheet and it said that the limit was [imath]0[/imath], I checked the process and they simplified the expression to [imath]\lim_{n\to\infty} \frac{(2n-1)!}{(2n + 1)(2n)(2n − 1)!}[/imath] and then [imath]\frac{1}{(2n + 1)(2n)}[/imath] and said it was [imath]0[/imath] How did they achieved this?
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1510409
|
Find limit with factorial
I am trying to calculate the limit of [imath]\lim_{n \to \infty} \frac{(2n-1)!}{(2n+1)!}[/imath] and I am stuck. I tried n division but that didn't work. Any help is appreciated
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1207250
|
Limit of function using L'Hôpital's rule: [imath]\lim_{x \to \infty}{x-x^2\ln(1+\frac{1}{x})}[/imath]
I have to calculate [imath]\lim_{x \to \infty}{x-x^2\ln(1+\frac{1}{x})}[/imath]. I rewrote it as [imath]\lim_{x \to \infty}{\frac{x-x^3\ln^2(1+\frac{1}{x})}{1 + x\ln(1+\frac{1} {x})}}[/imath] and tried to apply L'Hôpital's rule but it didn't work. How to end this?
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1702282
|
using L'Hospital solve [imath]\lim_{x \to \infty} x - x^{2}\ln(1 + \frac{1}{x})[/imath]
I can't get this to [imath] = \frac{0}{0}[/imath] form so I can use l'Hospital rule [imath]\lim_{x \to \infty} x - x^{2}\ln\left(1 + \frac{1}{x}\right)[/imath] tips? [EDIT] [imath]\lim_{x \to 0} \frac{1}{x} - \frac{\ln(1 + x)}{x^{2}}[/imath] second term [imath]\lim_{x \to 0} \frac{\ln(1 + x)}{x^{2}} = \lim_{x \to 0} \frac{\frac{1}{1 + x}}{2x} = 2\lim_{x \to 0} \frac{x}{x+1} = 0[/imath] first term [imath] \lim_{x \to 0} \frac{1}{x} = \infty[/imath] is this okey?
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2006305
|
Limit with subsequences
I have the following limit [imath]\lim \limits_{n \to \infty}\left({1 \over 2^1}+{3 \over 2^2} + {5 \over 2^3} + \dots + {2n-1 \over 2^n}\right)[/imath] My idea is to separate it into geometric progressions and rewrite them using the equation for the sum of first terms of a geometric progression. I manage to get one geometric progression but the other elements of the sum don't follow a geometric pattern. Do you have any hints as to how I might separate the terms? Or, alternatively, if there is a simpler way to do this?
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903504
|
Sum of an unorthodox infinite series
[imath] \frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots [/imath] This is a pretty unorthodox problem, and I'm not quite sure how to simplify it. Could I get a solution? Thanks.
|
2002617
|
Show if there is an analytic function [imath]f \in \mathbb{C}[/imath] such that [imath]f\left(\frac{1}{n}\right)=\frac{n}{1+n}[/imath]
That's it. I don't really know what to do. I've tried to used analytic continuation with the function [imath]g(z)=\frac{1}{1+z}[/imath] but of course, such a function is not analytic in [imath]\mathbb{C}[/imath] so i can't use the theorem. And i can't find the contradiction
|
311099
|
Can a holomorphic function satisfy f(1/n)=1/(n+1)?
Does there exist a function [imath]f[/imath] which is holomorphic on [imath]B_0(2)[/imath] (open disc of radius 2 in the complex plane) such that [imath]f(1/n)=1/(n+1) \forall n \in \mathbb{N}[/imath]? At the moment I'm thinking not but a proof is seeming elusive. Any hints would be appreciated.
|
2006569
|
Using a test (any but the comparison test) to solve ∑(k=1 to ∞) 1/(2^k-1)
[imath]\sum_{k=1}^{∞}\frac{1}{(2^k-1)}[/imath] I need to show that the series is convergent. If there is a method that can be used to solve this (not the comparison test), what test would that be, and how would I go about solving it using that test?
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1978310
|
Find [imath]\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}[/imath]
Calculate [imath]\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}.[/imath] I used Wolfram|Alpha to compute it and got it to be approximately equal to [imath]0.6[/imath]. How to compute it? Can someone give me a hint or a suggestion to do it?
|
1999019
|
Prove that [imath]\limsup_{n \to \infty} x_n \cdot y_n = x \cdot\limsup_{n \to \infty} y_n[/imath]
Let [imath](x_n)_n[/imath] and [imath](y_n)_n[/imath] be sequences in [imath]\mathbb{R}[/imath], such that [imath](x_n)_n[/imath] converges to [imath]x \in \mathbb{R}^+[/imath]. Suppose that [imath](y_n)_n[/imath] is bounded. Prove that [imath]\limsup_{n \to +\infty} x_n \cdot y_n = x \cdot \limsup_{n \to +\infty} y_n[/imath] I know that [imath]x=\limsup_{n \to +\infty} x_n[/imath], so it basically comes down to proving [imath]\limsup_{n \to +\infty} x_n \cdot y_n = \limsup_{n \to +\infty} x_n \cdot \limsup_{n \to +\infty} y_n[/imath], but that's where I'm stuck...
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1244661
|
If [imath]\lim_{n\to \infty}a_n = a\in \mathbb{R}[/imath] . Prove that [imath]\limsup_{n\to \infty}a_n x_n=a\limsup_{n\to \infty}x_n[/imath] .
Note: [imath]x_n[/imath] is a sequence which is not necessarily convergent. The following was my attempt. Since [imath]\lim_{n\to \infty}a_n=a[/imath] then [imath]\limsup_{n\to \infty}a_n=a[/imath] . Also [imath]\sup(a_nx_n)=\sup(a_n)\sup(x_n)[/imath]. Therefore, [imath]\limsup_{n\to \infty}a_nx_n=\limsup_{n\to \infty}a_n\limsup_{n\to \infty}x_n=a\limsup_{n\to \infty}x_n[/imath]. I am not sure if this is correct. So could someone please show me how it is done. Thanks
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2006142
|
Multivariable Limit - Does it exist?
I know that the limit below [imath]\lim_{(x,y)\to (1,0)}[/imath] [imath]\frac{xy-y}{(x-1)^2+y^2}[/imath] does not exist. However, I haven't found a path that is not equal to 0. I have tried: [imath]y=mx[/imath], [imath]x=cy^2[/imath],[imath]y=0[/imath], and [imath]x=0[/imath]. Any suggestions?
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1800138
|
How can I prove this limit doesn't exist?
Right now, I'm doing a question: [imath]\lim_{(x,y)\to(1,0)}\frac{xy-y}{(x-1)^2 +y^2}[/imath] I know the limit doesn't exist, but I can't figure out how to prove it. I tried putting [imath]x=1[/imath], and getting [imath]0/y^2[/imath], and put [imath]y=0[/imath], got [imath]0/(x^2-2x+1)[/imath], but I don't think that does it. (edit: this is not a duplicate; I'm having a hard time getting good explanations, hence why I asked)
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803221
|
$\operatorname{ord}(u)=r,\,\operatorname{ord}(v)=s$ $\,\Rightarrow\,\operatorname{ord}(uv)=rs\,$ if $\,r,s\,$ (co)primes
Let u and v be elements of a commutative group, and suppose that their orders are r and s, respectively. Show that if r and s are distinct primes then the order of uv is rs. I will use the follwing theorem to proof this. Theorem: Let x be an element of order m in a finite group G. Then [imath]x^s=1[/imath] in G if and only if s i a multiple of m. My attempt We can here see that [imath](uv)^{rs}=[commutative]=u^{rs}v^{rs}=1^s1^r=1[/imath] By the theorem, this tells us that this is true only if only rs is a multiple of the order of [imath]uv[/imath]. Let the order be [imath]ord(uv)=k[/imath]. We know that every positive integer can be written as [imath]a=bm+l, 0<=l<b[/imath] and since [imath]r[/imath] and [imath]s[/imath] is primes (positive integers), [imath]rs[/imath] can be written as [imath]rs=km+l, 0<=l<k[/imath] This gives ut that [imath]1=(uv)^{rs}=(uv)^{km+l}=1^muv^l=(uv)^l[/imath] If l>0 then the equation [imath](uv)^l=1[/imath] contradicts that the positive integer k is the least positive integer for which [imath](uv)^k=1[/imath]. Therefore l=0 and [imath]rs=km[/imath] . We can here see that [imath]rs[/imath] is a multiple of k (and this proof also the theorem). [imath]rs=km => k | rs => k|r ORk|s[/imath] But we know that [imath]r[/imath] and [imath]s[/imath] are distinct primes, so k must be equal to one of the primes. [imath]k=r => (uv)^k=(uv)^r=1v^r=v^r=1[/imath] And since we know that [imath]ord(v)=s[/imath] the theorem tells us that [imath]r[/imath] must be a multiple of [imath]s[/imath]. But since both of them is primes, we get that [imath]r=s[/imath] but this is a contradiction (r and s are distincts as the exercise tells us). Now, we try with [imath]k=s[/imath]. Same as above, we get that [imath]s=r[/imath] which is a contradiction. Neither k=r nor k=s is right since by the theorem we get that [imath]k=r=s[/imath] and we have that [imath]r[/imath] and [imath]s[/imath] are distincts. This tells us that k cannot be equal to r nor s, since this gives us that [imath]r=s[/imath]. And since we know (and as i did proof)that [imath]rs=km[/imath], this tells us, as i wrote before, that [imath]k|rs[/imath] Since k is neither equal to [imath]r[/imath] or [imath]s[/imath], this tells us that k must be equal to [imath]rs[/imath]. Therefore [imath]ord(uv)=k=rs[/imath]. Edit: i know that this can be prooved by l.c.m(x,y) but we have not used it so far. That's why i did not proof it with l.c.m(x,y)
|
2089339
|
a and b commutes of order p q with p and q co-prime show that ab is order pq
Let [imath](G,*)[/imath] be an abelian group with two elements a and b of order p,q co-prime. What is the order of a*b ? At most pq as [imath](a*b)^{pq}=a^{pq}*b^{pq}=1_G[/imath] I guess it is exactly pq but I do not see why ?
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2008092
|
How to evaluate [imath]\int {{x^2-1}\over{(x^4+3x^2+1)\arctan(\frac{x^2+1}x )}}[/imath]
How do I go about doing this: [imath]\displaystyle \int {{x^2-1}\over{(x^4+3x^2+1)\arctan(\frac{x^2+1}x )}}[/imath] I have tried integration by parts but it seems to be making the problem more complicated. Also, I have thought of substitution but I couldn't find any suitable substitution.
|
2007485
|
How do you evaluate the integral [imath]\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx[/imath]?
i'm required to evaluate this integral. I've tried factorizing but it doesn't lead me to anywhere. [imath]\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx[/imath] I've also tried letting [imath]u = \frac{x^2+1}{x}[/imath], [imath]du/dx[/imath] gets me [imath]1-\frac{1}{x^2}[/imath] but it doesn't seem to be working either. Hope to receive some advise/ solutions on how to start tackling the question
|
2007225
|
Is it possible for [imath]x^3-3[/imath] to be a perfect square?
Is it possible for [imath]x^3-3[/imath] to be a perfect square (or the negative of a perfect square), where [imath]x[/imath] is an integer? I tried using modular arithmetic to find a contradiction, but couldn't find any. How can we find a contradiction?
|
1520668
|
how to solve the following mordell equation:[imath] y^2 = x^3 - 3[/imath]
i just started solving mordell's equations and get a little bit stuck. For example: [imath] y^2 = x^3 - 3[/imath]. I know that [imath]x[/imath] must be odd, for if [imath]x[/imath] is even [imath]y^2 \equiv 5 \pmod{8}[/imath]. So [imath]x \equiv \{1,3\} \pmod{4}[/imath]. Now note the following if we add 4 at both sides: [imath]y^2 + 4 = (x+1)(x^2 -x +1)[/imath]. The right hand side is the same as [imath](x-1)^2 + x[/imath] I now know [imath](x^2 -x +1)[/imath] is divisible by a prime divisor [imath]3 \pmod{4}[/imath] if [imath]x \equiv 3 \pmod {4}[/imath] and divisible by [imath]1 \pmod{4}[/imath] if [imath]x \equiv 1 \pmod{4}[/imath]. So we get [imath]y^2 + 4 \equiv 0\pmod{p}[/imath], which is solvable if [imath]\left(\frac{-2}{p}\right)[/imath] = 1. (I don't think this step is valid). this is only true if the following occurs: [imath]p \equiv 1 \pmod{4}[/imath] and [imath]p \equiv +- 1\pmod{8}[/imath]. or [imath]p \equiv 3 \pmod{4}[/imath] and [imath]p \equiv +- 3\pmod{8}[/imath]. so [imath]p \equiv \{1,3\}\pmod{8}[/imath] but [imath]y^2 + 4[/imath] is either [imath]4 \pmod{8}[/imath] or [imath]5 \pmod {8}[/imath], so there are no solutions? Am i solving this the correct way or is there a shorter path? Kees
|
2007946
|
Reasons for limits in triple integral of sphere
I know that the volume of a sphere is given by the triple integral [imath] \int_0^{2\pi} \int_0^{\pi} \int_0^r \rho ^2 \sin \phi d\rho d\phi d\theta [/imath]. I am having trouble on how the limits for the angles are derived. Why don't they both go up to [imath]2\pi[/imath]. Please help.
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1499802
|
Triple integration, Spherical coordinates
How do we get limit such as [imath]0\le\theta\le\pi[/imath], [imath]0\le\phi\le2\pi[/imath] in spherical coordinate system where [imath]x=r \sin\theta\cos\phi, y=r \sin\theta\sin\phi, z=r \cos\theta[/imath] Why is the [imath]\theta[/imath]-limit [imath][0,\pi][/imath] and not [imath]2 \pi[/imath]? I think that the angle of the foot of perpendicular from the point taken in spherical coordinate can lie within [imath][0,2 \pi][/imath]
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1124812
|
Given that a,b,c are distinct positive real numbers, prove that (a + b +c)( 1/a + 1/b + 1/c)>9
Given that [imath]a,b,c[/imath] are distinct positive real numbers, prove that [imath](a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9[/imath] This is how I tried doing it: Let [imath]p= a + b + c,[/imath] and [imath]q=\frac1{a}+ \frac1{b} + \frac1{c}[/imath]. Using AM>GM for [imath]p, q[/imath], I get: [imath]\frac{p+q}{2} > {(pq)}^{1/2}[/imath] [imath]\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2[/imath] And for any [imath]x\in \mathbb{R}, \space \space x+\frac1{x}≥2.[/imath] Thus, [imath](a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)<9, [/imath] which is the opposite of what had to be proven. What did I do wrong?
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904056
|
If [imath]a,b,c[/imath] are positive, then [imath](a+b+c)(1/a+1/b+1/c)\ge 9[/imath]
The question asks to prove that if "[imath]x_1,x_2,x_3[/imath] are positive numbers show that: [imath](x_1+x_2+x_3) \left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3} \right)\ge 9[/imath] I've tried to use the fact that the arithmetic mean is greater than the harmonic mean since it looks like it's kind of in that form but it doesn't seem to work. I tried expanding it too and simplifying a little to get: [imath] x_1^2(x_2+x_3)+x_2^2(x_1+x_3)+x_3^2(x_1+x_2)\ge 6 x_1x_2x_3[/imath] but I can't seem to see anyway to further get it into a form that easily proves the required inequality. Any hints as to a way to do this?
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2008442
|
[imath]\sum_\limits{n=0}^{\infty} a_n[/imath] converges [imath]\implies \sum_\limits{n=0}^{\infty} a_n^2[/imath] converges
Let [imath](a_n)[/imath] be a sequence of positive terms and suppose that [imath]\sum_\limits{n=0}^{\infty} a_n[/imath] converges. Show that [imath]\sum_\limits{n=0}^{\infty} a_n^2[/imath] converges. This is in the section on the Comparison Test so that must be what I'm supposed to use. But I don't see how. [imath](a_n)^2[/imath] might be smaller or larger than [imath]a_n[/imath] depending on [imath]a_n[/imath]. And I can't use the Comparison Test with some other series because there's no info here about how fast [imath]\sum a_n[/imath] converges. Hmm. Any hints?
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2747074
|
Prove if the sum is convergent, then the square is convergent
I've seen multiple ways on how to solve it online (and most likely the majority are wrong), but I don't know how to solve this fully so that I could get the full grade during my exam. The question is as follows: Prove that if [imath]\sum a_n[/imath] is convergent with [imath]a_n > 0[/imath] for all [imath]n[/imath], then [imath]\sum a^2_n[/imath] is also convergent. I understand that the starting point to prove this is that [imath]\lim a_n =0[/imath], but after that I really don't know what I'm supposed to say.
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2008773
|
Is there a function so that [imath]f(f(x))=1/x[/imath]?
I tried to look for a function that fulfills this rule; [imath]f(f(x))=\frac{1}{x}[/imath] I tried a lot, and I didn't find anything. Can anyone help me find such a function? Or prove that there isn't? It doesn't have to be a real function.
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1387367
|
Is there a continuous function [imath]f(x)[/imath] such that the inverse function is [imath]1/f(x)[/imath]?
A student came to me with this question and I cracked my head for one hour but I couldn't unambiguously prove that it exists or it doesn't exist. Continuous and invertible function such that [imath]f^{-1}(x) =1/f(x)[/imath] on its domain of definition.
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1997029
|
Prove :(P → Q) ∨ (Q → P) using natural deduction
Allowed inference rules: ∨-I, ∨-E, ∧-I, ∧-E, →-I, →-E, ¬-I, ¬-E I tried to prove a contradiction by assuming [imath]¬ ((P → Q) ∨ (Q → P))[/imath] but got stuck, or am I doing it in the wrong way? Edit: My proof attempt [imath]1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum[/imath] [imath]2.\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ Assum[/imath] [imath]3.\qquad(P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; ∨-I(2)[/imath] [imath]4.\qquad(¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\; ∧-I(1,3)[/imath] [imath]5.\qquad ¬ (P → Q)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\; ¬-I(2,4)[/imath] [imath]6.\qquad Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\quad\; Assum[/imath] [imath]7.\qquad (P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\; ∨-I(6)[/imath] [imath]8.\qquad (¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\;\; ∧-I(1,7)[/imath] [imath]9.\qquad ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; ¬-I(6,8)[/imath] [imath]10.\qquad ¬ (P → Q) ∧ ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\;\qquad\qquad\quad\;\;\, ∧-I(5,9)[/imath] Stuck at this line, can't spot any contradiction to line 1
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2011829
|
How can you prove [imath](P\rightarrow Q)\lor(Q \rightarrow P)[/imath] using natural deduction?
I am a Physics major interested in logic, but I cannot seem to find how to put the rules together for a proof. I am reading a book that uses a tree-style!
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2009041
|
Suppose [imath]0,0,1,0,0,1,0,0,1,...[/imath] is a sequence. What is the general term of this sequence?
Let [imath]0,0,1,0,0,1,0,0,1,...[/imath] be a sequence. What is the general term of this sequence?
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51762
|
What Is The General Term Of These Sequences?
[imath]a_{n,3}[/imath] [imath]1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,\cdots \cdots [/imath] [imath]a_{n,5}[/imath] [imath]1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,\cdots \cdots [/imath] [imath]a_{n,7}[/imath] [imath]1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,\cdots \cdots [/imath] [imath]\vdots [/imath] [imath]a_{n,2k+1}[/imath] [imath]1,\cdots \cdots\cdots \cdots,1,\cdots \cdots\cdots \cdots,\overbrace{1,\cdots \cdots\cdots \cdots}^{2k+1},1,\cdots \cdots\cdots \cdots,1,\cdots \cdots[/imath] I found [imath]a_{n,k}= \left \lfloor \left | \cos\frac{\pi (n-1)}{2k+1} \right | \right \rfloor[/imath] but this term include abs and floorfucntion. I mean i want the General Term without abs and floorfucntion.(plus sum) Any hints will be appreciated, thank you.
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2010502
|
Show that if [imath]p=2^a -1 [/imath] is prime then a is prime
Show that if [imath]p=2^a -1[/imath] is prime then [imath]a[/imath] is prime. I found this problem very hard to solve and I only managed to show that [imath]a[/imath] must be odd; if [imath]a[/imath] was even we could rewrite [imath]p[/imath] as [imath]p=2^{2k}-1=(2^k-1)(2^k+1)[/imath] with [imath]k[/imath] different than [imath]2[/imath]. Other than this I couldn't do anything else.
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263640
|
How can I algebraically prove that [imath]2^n - 1[/imath] is not always prime?
This question is from Elementary Number Theory by W. Edwin Clark. Is [imath]2^n - 1[/imath] always prime, or not? Prove. Is this a start? [imath]x^n - 1 = ( x - 1)(1 + x + x^2 \cdots x^{n - 1})[/imath]. So, [imath]2^n - 1 = \sum \limits _{i = 0}^{n - 1} 2^i.[/imath] Will I reach a solution through the above, or is there any other way? I know that the property doesn't hold true for [imath]n = 1,4,6[/imath] et al but I want an algebraic proof.
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2010751
|
Show that there exists a ring homomorphism [imath]f: Z/(m)\rightarrow Z/(n)[/imath] sending 1 to 1 if and only if [imath]n|m[/imath].
Let [imath]f[/imath] such that [imath]f(a+(m))=a+(n)[/imath] I do not know how to prove that is well defined and that [imath]n|m[/imath] ,Could anyone help me, please?
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1519187
|
Ring homomorphisms from [imath]\mathbb{Z}_n[/imath] to [imath]\mathbb{Z}_m[/imath]
I'm trying to find the ring homomorphisms from [imath]\mathbb{Z}_n[/imath] to [imath]\mathbb{Z}_m[/imath]. What I think is the answer is two cases: 1) If [imath]m \gt n [/imath] there is no ring homomorphism. 2) Otherwise, the homomorphism sends each element [imath]x[/imath] in [imath]\mathbb{Z}_n[/imath] to [imath](x \bmod m)[/imath] in [imath]\mathbb{Z}_m[/imath]. Is this correct? EDIT: Part of the ring homomorphism definition is that 1 is sent to 1
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2011069
|
Localisation : an example where [imath]r's-rs'\neq 0[/imath] but [imath](r,s)\sim(r',s')[/imath]
Let [imath]R[/imath] a commutative ring and [imath]S\subset R[/imath] a subset closed multiplicatively with [imath]1\in S[/imath]. We say that [imath](r,s)\sim (r',s')[/imath] if there is a [imath]t\in s[/imath] s.t. [imath]t(r's-rs')=0[/imath]. The thing is I don't understand the presence of this [imath]t[/imath], and I don't find any example where it's important. I tried for example for [imath]R=\mathbb Z/6\mathbb Z[/imath] and [imath]S=\{1, a\}[/imath] where [imath]a\in \{1,...,5\}[/imath], then with [imath]S=\{0,1,2,3\}[/imath], but still, it doesn't work. What I thought was something like: [imath]\frac{2}{1}=\frac{6}{3}[/imath] but since [imath]\frac{6}{3}=\frac{0}{1}[/imath], we would have that [imath]\frac{2}{1}=\frac{0}{1}[/imath], but is there something around that ? By the way, it's not precise in my course that [imath]0\notin S[/imath] (that's why I included it), but may be it can't be in [imath]S[/imath], no ?
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1992449
|
[imath]f(x)=x/1[/imath] is not injective?
In chapter 3 of Atiyah and Macdonalds Intro to Comm Algebra there is the statement "We also have a ring homomorphism [imath]f:A\rightarrow S^{-1}A[/imath] defined by [imath]f(x)=x/1[/imath]. This is not in general injective." (Note: [imath]A[/imath] denotes an integral domain and [imath]S[/imath] a multiplicatively closed subset of [imath]A[/imath].) How could this possibly not be injective? If [imath]f(x)=f(y)[/imath] then [imath]x/1=y/1[/imath] so [imath]1x=1y[/imath] hence [imath]x=y[/imath]. What am I missing here?
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2010957
|
Let A be right ideal of R. define:
I think I must first prove that I(A) is a subring and then prove that A is in the ideal I(A) and I(A) is the largest sub-ring containing A. but I have problems to show that I(A) is a subring because let x and y in I(A), so [imath]xa\in A[/imath] and [imath]ya \in A[/imath] but I do not know how to prove that [imath]xy\in I(A)[/imath] and the rest.
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329611
|
Idealizer of one-sided ideal
Let [imath]A[/imath] be a ring and let [imath]J[/imath] be a right-sided ideal of [imath]A[/imath]. We call the set [imath]I_{A}(J)=\lbrace a \in A \mid aJ\subset J\rbrace[/imath] the idealizer of [imath]J[/imath]. Show that [imath]I_{A}(J)[/imath] is the largest subring of that [imath]A[/imath] containing [imath]J[/imath] as an ideal.
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2010969
|
Let [imath](G, \circ )[/imath] be a group where [imath]a\circ a =e, \ \forall a\in G [/imath]. Show that [imath](G, \circ )[/imath] is abelian.
I have a question I'm trying to answer. The question is Let [imath](G, \circ )[/imath] be a group where [imath]a\circ a =e, \ \forall a\in G [/imath]. Show that [imath](G, \circ )[/imath] is abelian. [Here [imath]e[/imath] is the identity element.] How would you do this?
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2010788
|
Let X be a set with an associative binary operation ◦ with identity e. Suppose every element of X satisfies x ◦ x = e. Prove that ◦ is commutative.
I'm having problems trying to figure this out. It seems very straight forward, but I don't know what I should do to prove it. Let [imath]X[/imath] be a set with an associative binary operation [imath]\circ[/imath] with identity [imath]e[/imath]. Suppose every element of [imath]X[/imath] satisfies [imath]x \circ x = e[/imath]. Prove that [imath]\circ[/imath] is commutative.
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2011128
|
Prove that exactly [imath]\phi(d)[/imath] elements have order [imath]d[/imath]
Let [imath]p[/imath] be a prime number. Prove that among [imath]1,2,\ldots,p-1,[/imath] exactly [imath]\phi(d)[/imath] elements have order [imath]d[/imath], given that [imath]d \mid (p-1)[/imath]. (Note: Here [imath]\phi(d)[/imath] denotes Euler's Totient Function.) I didn't see a way of proving this statement. How can we count solutions [imath]k[/imath] to [imath]k^d \equiv 1 \pmod{p}[/imath]?
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1857315
|
When [imath]d \mid n[/imath], the number of elements of order [imath]d[/imath] in a cyclic subgroup of order [imath]n[/imath] is [imath]\phi (d)[/imath]
If [imath]d[/imath] is a positive divisor of [imath]n[/imath], the number of elements of order [imath]d[/imath] in a cyclic subgroup of order [imath]n[/imath] is [imath]\phi (d)[/imath]=the number of positive natural numbers less than [imath]d[/imath] which are coprime to [imath]d[/imath]. The question I have concerns a part of the proof: If [imath]d | n[/imath] then there exists exactly one subgroup of order [imath]d[/imath] -- call it [imath]\langle a \rangle[/imath]. Then every element of order [imath]d[/imath] also generates the subgroup [imath]\langle a \rangle[/imath] and an element [imath]a^k[/imath] generates [imath]\langle a \rangle[/imath] iff [imath]gcd(k,d) = 1[/imath] implies that the number of such elements is precisely [imath]\phi (d)[/imath]. How does every element of order [imath]d[/imath] also generate the subgroup [imath]\langle a \rangle[/imath], wouldn't it be only one [imath]a[/imath] since [imath]|\langle a \rangle| = |a|[/imath]? And how does this fact imply [imath]\phi(d)[/imath] is the correct number?
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939904
|
A question regarding the definition of localization of ring
Let [imath]S[/imath] be a multiplicative system. Then the localization of [imath]R[/imath] is defined by an equivalence relation on [imath]R \times S[/imath]. The relation is [imath](a,b) \sim (c,d)[/imath] if there is an [imath]s \in S[/imath] such [imath]s(ad-bc)=0[/imath]. Regarding this, I can't show that transitivity works. Could anyone show me how to prove that [imath](a,b) \sim (c,d)[/imath] and [imath](c,d) \sim (e,f)[/imath] then [imath](a,b) \sim (e,f)[/imath]?
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327294
|
Localization of a ring which is not a domain
Let [imath]A[/imath] be a ring (commutative with [imath]1[/imath]), let [imath]S[/imath] be a multiplicatively closed subset of [imath]A[/imath], i.e [imath]S[/imath] is contained in [imath]A[/imath] , [imath]1\in S[/imath] and [imath]a,b\in S[/imath] implies [imath]ab\in S[/imath], for every [imath]a,b\in A[/imath]. Consider the following relation defined in [imath]A\times S[/imath]: [imath](a,s)\equiv(b,t) \iff \exists u\in S:u(at-bs)=0[/imath] I proved this relation to be reflexive and symmetric, but i have difficulties in proving its transitivity. Can someone suggest me a trick?
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625684
|
A commutative ring whose all proper ideals are prime is a field.
Let [imath]R[/imath] be a commutative ring with [imath]1[/imath]. Suppose that all ideals [imath]I \neq R[/imath] are prime. Prove that [imath]R[/imath] is a field. Help me some hints. Thanks a lot.
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2845868
|
Let [imath]R[/imath] be a commutative ring with [imath]1\neq 0[/imath]. Prove that if every proper ideal of [imath]R[/imath] is prime, then [imath]R[/imath] is a field.
I'm stuck on a detail in the proof of the following: Let [imath]R[/imath] be a commutative ring with [imath]1\neq 0[/imath]. Prove that if every proper ideal of [imath]R[/imath] is prime, then [imath]R[/imath] is a field. Since [imath]0 \neq 1[/imath] we have that [imath]\{0 \}[/imath] is a proper ideal of [imath]R[/imath]. Further, [imath]R[/imath] is an integral domain since if [imath]a\cdot b \in \{0\}[/imath], then either [imath]a=0[/imath] or [imath]b=0[/imath] as [imath]\{0\}[/imath] is prime. Now I would like to prove that [imath]\{0\}[/imath] is the only proper ideal of [imath]R[/imath] which would imply that [imath]\{0\}[/imath] is maximal and [imath]F/\{0\} \cong F[/imath] is a field. My idea is that if [imath]A,B \subseteq R[/imath] are two proper ideals then it must be that [imath]A = B[/imath], and therefore any proper ideal of [imath]R[/imath] is [imath]\{0\}[/imath]. Notice that [imath]AB \subseteq B[/imath] so that [imath]AB[/imath] is also a proper ideal of [imath]R[/imath]. Since [imath]AB \subseteq AB[/imath], and [imath]AB[/imath] is prime, we have that either [imath]A \subseteq AB \subseteq A[/imath] or [imath]B \subseteq AB \subseteq B[/imath], which implies that either that [imath]B = AB \subseteq A[/imath] or [imath]A = AB \subseteq B[/imath]. Say [imath]B = AB \subseteq A[/imath]. I'm now stuck trying to show that [imath]A \subseteq B[/imath]. Is there a better way to do this? I have also tried picking [imath]r \in R\setminus\{0\}[/imath], and showing that [imath]r^{-1} \in R[/imath] without success.
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2011308
|
How to find limit of function: [imath]\lim_{x\to 0}\left(x{{\left\lfloor{ \frac{1}{x}} \right\rfloor}}\right)[/imath]
Find [imath]\displaystyle \lim_{x\to 0}\left(x{{\left\lfloor{ \frac{1}{x}} \right\rfloor}}\right)[/imath]. When [imath]x\to0^{+}[/imath] I have [imath]\left(x{\left\lfloor{ \frac{1}{x}} \right\rfloor}\right)\to 0\cdot \infty[/imath]. When [imath]\to 0^{-}[/imath] I have [imath]\left(x{\left\lfloor{ \frac{1}{x}} \right\rfloor}\right)\to 0\cdot \infty[/imath]. I have no idea,
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879530
|
Evaluate [imath]\lim_{x \to 0} (x\lfloor\frac{1}{x}\rfloor)[/imath]
Evaluate [imath]\lim_{x \to 0} (x\lfloor\frac{1}{x}\rfloor)[/imath] I'm trying to solve it by using the squeeze theorem but I'm stuck. I'm looking for a function [imath]g(x)[/imath] such that [imath]g(x) \leq x\lfloor\frac{1}{x}\rfloor \leq x(\frac{1}{x}) = 1[/imath] Can anybody give me a hint?
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1997461
|
An entire function whose imaginary part is nonnegative must have a constant imaginary part
Let [imath]f(z)[/imath] be entire such that [imath] \text{Im} (f(z)) \geq 0.[/imath] Show that [imath]\text{Im}(f(z))[/imath] is constant. So far what I have done is this: Consider [imath]|e^{if(z)}| = |e^{v(x,y)}| \geq 1.[/imath] Then this implies [imath]|e^{-v(x,y)}| \leq 1.[/imath] I have shown that [imath]e^{-v(x,y)}[/imath] is bounded. Since f is entire, does it follow that its real and imaginary parts are also entire?
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1995399
|
Let [imath]f[/imath] be entire and suppose that [imath]\text{Im} f\ge0[/imath]. Show that [imath]\text{Im} f[/imath] is constant.
How do you show that if [imath]f[/imath] is entire and if [imath]\text{Im} f\ge0[/imath], then [imath]\text{Im} f[/imath] is constant? Is this even true in the first place?
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2010989
|
Prob. 4, Sec. 1.2 in Kreyszig's Functional Analysis: Example of a sequence converging to [imath]0[/imath] which is not in any [imath]\ell^p[/imath]
Here's Prob. 4, Sec. 1.2 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig: Find a sequence which converges to [imath]0[/imath], but is not in any space [imath]\ell^p[/imath], where [imath]1 \leq p < +\infty[/imath]. My effort: For any [imath]n \in \mathbb{N}[/imath], we have the following chain of inequalities. [imath] 0 < \sqrt[p]{2n - \sqrt[n]{n}} \leq 2n - \sqrt[n]{n} < 2n;[/imath] so [imath] 0 < \frac{1}{2n} < \frac{1}{2n - \sqrt[n]{n}} \leq \frac{1}{\sqrt[p]{2n - \sqrt[n]{n}} }.[/imath] Now as the series [imath]\sum \frac{1}{2n}[/imath] diverges to [imath]+\infty[/imath], so does each of the series [imath]\sum \frac{1}{2n - \sqrt[n]{n}} [/imath] and [imath]\sum \frac{1}{\sqrt[p]{2n - \sqrt[n]{n}} }[/imath]. Now we take the sequence [imath]\left( \xi_n \right)_{n \in \mathbb{N}}[/imath], where [imath]\xi_n \colon= \frac{1}{\sqrt[p]{2n - \sqrt[n]{n}} }[/imath] for each [imath]n \in \mathbb{N}[/imath]. Then we note that [imath]\lim_{n \to \infty} \xi_n = 0,[/imath] but this sequence is not in [imath]\ell^p[/imath], for any fixed [imath]p[/imath] such that [imath]1 \leq p < +\infty[/imath]. Am I right? If so, is this example simple enough? Or, can we come up with one which is even simpler? If not, then where am I going wrong?
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1040000
|
find a sequence converging to zero but not the element of lp space for every 1<_p<infinity
I am studying functional analysis and I have a problem about finding a sequence converging to zero such that this sequence is not in [imath]l_p[/imath] for every [imath]p[/imath]. By [imath]l_p[/imath] I mean [imath]l_p=\{(x_k)=(x_1,x_12,...):Σ|x_k|^p<\infty\}[/imath] where [imath]1<p<\infty[/imath]. First I thought very simple sequence [imath]i/k[/imath] which converges to zero but then I realized when [imath]p>2[/imath] it is element of [imath]l_p[/imath] space. I thought couple more examples but they did not work either. Can somebody help me out here? Thanks.
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1999395
|
Let [imath]f : [0, +\infty) \rightarrow \mathbb{R}[/imath] be uniformly continuous. Show that [imath]f[/imath] has sublinear growth
Let [imath]f : [0, +\infty) \rightarrow \mathbb{R}[/imath] be uniformly continuous. Show that [imath]f[/imath] has sublinear growth I need to show that there exists [imath]A,B > 0[/imath] such that |[imath]f(x)| ≤ Ax + B[/imath], for all [imath]x ≥ 0[/imath] I was thinking of using the definition of uniform continuity for [imath]\epsilon = 1[/imath] to show that all the values of [imath]x \in [0, +\infty)[/imath] are obtained in tiny steps without changing the value of [imath]f[/imath] much.
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1961828
|
If f uniformly continous then [imath]|f(x)|\leq C\cdot (1+|x|)[/imath] for constant C
Let f: [imath]\mathbb R\to\mathbb R[/imath] be an uniformly continuous function. Show that there is a constant [imath]C[/imath] with [imath]|f(x)|\leq C\cdot (1+|x|)[/imath]. I can´t seem to find the right idea to solve this problem. Thanks for the help in advance.
|
1997117
|
Trying to prove (A ∩ B) - C = (A - C) ∩ (B - C)
Trying to prove [imath](A ∩ B) - C = (A - C) ∩ (B - C)[/imath] Im stuck here on this problem: [imath](A ∩ B) - C = \{x\mid x \in (A ∩ B) ∧ x \not\in C\}[/imath] [imath](A ∩ B) - C = \{x\mid(x ∈ A ∧ x ∈ B) ∧ x ∉ C\}[/imath] [imath](A ∩ B) - C = \{x\mid(x ∈ A ∧ x ∉ C) \lor (x ∈ B ∧ x ∉ C\}[/imath] I'm stuck right about here because how do I change the or ([imath]\wedge[/imath]) sign to the intersection sign ([imath]\cap[/imath]). Edit: step 3, changed the and to an or
|
1997184
|
Trying to prove (A ∩ B) − C = (A − C) ∩ (B − C)
Trying to prove [imath](A ∩ B) − C=(A − C) ∩ (B − C)[/imath] Im stuck here on this problem: [imath](A∩B)−C=\{x | x ∈ (A ∩ B)∧ x ∉ C\}[/imath] [imath](A∩B)−C=\{x|(x ∈ A ∧ x ∈ B) ∧ x ∉ C\}[/imath] [imath](A∩B)−C=\{x|(x ∈ A ∧ x ∉ C) \lor (x ∈ B ∧ x ∉ C)\}[/imath] I'm stuck right about here because how do I change the or (v) sign to the intersection sign (∩). Please don't link me to the other question like this, I had a typo in it.
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2008464
|
Find a limit of this sequence: [imath]\lim_{x\to\infty }\frac{q^{x}}{x!}[/imath]
[imath]\lim_{x\to\infty }\frac{q^{x}}{x!}[/imath] I need to find a limit of this sequence and prove it using the definition. I know that we can have several variants of the limit of [imath]q^{x}[/imath]: if [imath]1>q>-1[/imath] the limit is [imath]0[/imath], for [imath]q=1[/imath] the limit is [imath]1[/imath], for [imath]q>1[/imath] the limit is [imath]\infty [/imath], and for [imath]q\leq -1[/imath] it doesn't exist. The limit of factorial [imath]x![/imath] should be [imath]\infty [/imath], but I'm not sure and I don't know how to prove it. Can you help me please with the whole fraction. I don't know how to find a limit if we have something like [imath]\frac{0}{\infty }[/imath] or [imath]\frac{1}{\infty }[/imath] or [imath]\frac{\infty}{\infty }[/imath].
|
1401760
|
Find [imath]\lim_{n\to\infty}\frac{a^n}{n!}[/imath]
First I tried to use integration: [imath]y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}[/imath] [imath]\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}[/imath] But I could not express it as a riemann integral. Now I am thinking about sandwich theorem. [imath]\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}[/imath] Since [imath]\frac{a}{t+1}>\frac{a}{t+2}>\frac{a}{t+1}>\cdots>\frac{a}{n}[/imath] [imath]\frac{a^n}{n!}<\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}[/imath] since [imath]\frac{a}{t+1}<1[/imath], [imath]\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0[/imath] Hence, [imath]\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0[/imath] And by using sandwich theorem, [imath]y=0[/imath]. Is this correct?
|
2013261
|
What is the mistake?
[imath]1=1[/imath] [imath]\Rightarrow\frac{-1}{1}=\frac{1}{-1}[/imath] [imath]\Rightarrow \sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}[/imath] [imath]\Rightarrow\frac{i}{1}=\frac{1}{i}[/imath] [imath]\Rightarrow\frac{i}{2}=\frac{1}{2i}[/imath] [imath]\Rightarrow\frac{i}{2}+\frac{3}{2i} = \frac{1}{2i} +\frac{3}{2i}[/imath] [imath]\Rightarrow i(\frac{i}{2}+\frac{3}{2i} ) = i(\frac{1}{2i} +\frac{3}{2i})[/imath] [imath]\Rightarrow\frac{-1}{2}=\frac{1}{2}[/imath] [imath]\Rightarrow1=2[/imath] What is wrong in this?
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2264960
|
What's wrong in this proof of [imath]i^2=1[/imath]
Complex number [imath]i=\sqrt{-1}[/imath] Now i consider [imath]\frac{1}{i}=\frac{1}{\sqrt{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}=i[/imath] so [imath]i^2=1[/imath]
|
2013157
|
Progression where nth term repeats n times.
How to deal with sequences like [imath]1,2,2,3,3,3,4,4,4,4,........[/imath] Now they don't come under any specific sequence (A.P., G.P. or H.P.). Can we deduce any simple formula or working method to deduce their [imath]n^{th}[/imath] term and the sum upto [imath]n^{th}[/imath] Any kind of hint would certainly work.
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455511
|
Formula for the [imath]n[/imath]th term of [imath]1, 2, 2, 3, 3, 3, 4, 4 ,4, 4, 5, ...[/imath]
There exists a formula for the [imath]n[/imath]th term of this sequence A002024 from the OEIS ("[imath]n[/imath] appears [imath]n[/imath] times") [imath]1, 2, 2, 3, 3, 3, 4, 4 ,4, 4, 5...[/imath] which is [imath]\left \lfloor \frac {1+\sqrt{1+8n}}{2} \right \rfloor.[/imath] Is there a better formula for the [imath]n[/imath]th term? A formula that isn't a 'magic' one that just happens to work but one that comes from a well-founded observation. Preferably one not using the floor function. I also have suspicions about the sustainability of the above formula as I feel with very [imath]n[/imath] it may not function properly.
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