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912443 | Irrationality of [imath]\sqrt{3}[/imath]
No doubt an easy question: I'm trying to follow Wikipedia's (second) proof of the irrationality of [imath]\sqrt{3}[/imath] and it relies on the notion that since [imath]3n^2 = m^2[/imath] is divisible by 3 then so is [imath]m[/imath]. Why is this so and can you help me prove it? | 2883 | Show that [imath]3p^2=q^2[/imath] implies [imath]3|p[/imath] and [imath]3|q[/imath]
This is a problem from "Introduction to Mathematics - Algebra and Number Systems" (specifically, exercise set 2 #9), which is one of my math texts. Please note that this isn't homework, but I would still appreciate hints rather than a complete answer. The problem reads as follows: If 3p2 = q2, where [imath]p,q \in \mathbb{Z}[/imath], show that 3 is a common divisor of p and q. I am able to show that 3 divides q, simply by rearranging for p2 and showing that [imath]p^2 \in \mathbb{Z} \Rightarrow q^2/3 \in \mathbb{Z} \Rightarrow 3|q[/imath] However, I'm not sure how to show that 3 divides p. Edit: Moron left a comment below in which I was prompted to apply the solution to this question as a proof of [imath]\sqrt{3}[/imath]'s irrationality. Here's what I came up with... [incorrect solution...] ...is this correct? Edit: The correct solution is provided in the comments below by Bill Dubuque. |
912514 | System of linear equation matrix?
How would I do this question. Determine the value(s) of [imath]h[/imath] such that the matrix is augmented of a consistent linear system. My matrix \begin{bmatrix} 1&h&4\\ 3&6&8 \end{bmatrix} I am not sure how to format it very well in latex. But I do not get this question I am not sure what they are asking. I just know how to find x and y. | 533681 | Determine the value of h such that the matrix is the augmented matrix of a consistent linear system.
Determine the value of h such that the matrix is the augmented matrix of a consistent linear system. [imath] \begin{bmatrix} 1 && h && 4 \\ 3 && 6 && 8 \end{bmatrix} [/imath] I'm entirely unsure how to go about solving this. This how far I got: [imath] \sim\begin{bmatrix} 1 && h && 4 \\ 0 && (6-3h) && -4 \end{bmatrix} R_2' = R_2-3R_1 [/imath] [imath] 6-3h=-4 \\ 6+4=3h \\ 10=3h \\ h=3/10 \text{ ?} [/imath] New progress [imath] \sim\begin{bmatrix} 3 && 6 && 8 \\ 1 && h && 4 \end{bmatrix} R_1 \leftarrow\rightarrow R_2 [/imath] [imath] \sim\begin{bmatrix} 3 && 6 && 8 \\ 0 && (h-2) && 4/3 \end{bmatrix} R_2' = - \frac13 R_1 [/imath] [imath] \sim\begin{bmatrix} 3 && 6 && 8 \\ 0 && 3(h-2) && 4 \end{bmatrix} R_2' = 3R_2 [/imath] [imath] \sim\begin{bmatrix} 3 && 6 && 8 \\ \\ 0 && 1 && \frac{4}{3(h-2)} \end{bmatrix} R_2' = R_2 \div 3(h-2) [/imath] [imath] \sim\begin{bmatrix} -3 && 0 && z \\ \\ 0 && 1 && \frac{4}{3(h-2)} \end{bmatrix} R_1' = 6R_2 - R_1 \\ z = -8 + \frac{24}{3(h-2)} = \frac{-32(h-2) + 24}{3(h-2)} [/imath] [imath] \sim\begin{bmatrix} 1 && 0 && z \\ \\ 0 && 1 && \frac{4}{3(h-2)} \end{bmatrix} R_1' = -\frac13 R_1 \\ z = - \frac{-32(h-2) + 24}{(h-2)} [/imath] |
911818 | How to obtain [imath]f(x)[/imath], if it is known that [imath]f(f(x))=x^2+x[/imath]?
How to get [imath]f(x)[/imath], if we know that [imath]f(f(x))=x^2+x[/imath]? Is there an elementary function [imath]f(x)[/imath] that satisfies the equation? | 913132 | Solve the function from the composition
I have equations as follows [imath]f(f(x))=x^2+x[/imath] Then solve for [imath]f(x)[/imath]. Can anyone give some hints about this question? |
912951 | Infinite Series [imath]\sum_{k=1}^\infty\left(\zeta(2)-\sum_{n=1}^k\frac1{n^2}\right)^2[/imath]
Evaluate: [imath]\sum_{k=1}^\infty\left(\zeta(2)-\sum_{n=1}^k\frac1{n^2}\right)^2[/imath] Recognizing that [imath]\zeta(2)-\sum_{n=1}^k\frac1{n^2}[/imath] can be written as [imath]\psi_1(1+k)[/imath] where [imath]\psi_1(z)[/imath] is the trigamma function, What remains to be done is to evaluate: [imath]\sum_{k=1}^\infty\psi_1^2(k+1)[/imath] Mathematica could not evaluate it in a closed form but the source assures that it exists. If you liked this problem check out Hard Definite integral involving the Zeta function. | 882621 | Polygamma function series: [imath]\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2[/imath]
Applying the Copson's inequality, I found: [imath]S=\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2\lt\dfrac{2}{3}\pi^2[/imath] where [imath]\Psi^{(1)}(k)[/imath] is the polygamma function. Is it known any sharper bound for the sum [imath]S[/imath]? Thanks. |
912470 | Why must a continuous function be null if its definite integral is null?
Let [imath] f(x) = \begin{cases} f:[a,b] \rightarrow\mathbb R \\ \int_{a}^{b}f = 0 \end{cases}[/imath]. Prove: if [imath]f[/imath] is continuous, then [imath]f\equiv 0[/imath]. I'm still trying to get the intuition on the situation. For instance, if [imath]f(x) = sin(x), x \in {[0, 2\pi]}[/imath], it's true that [imath]\int_{0}^{2\pi}sin(x) = 0[/imath], but it does not imply [imath]sin(x) \equiv0[/imath]. What did I misinterpret here? After the understanding of the situation, I'd like to know how would a formal proof follow. I'm a freshman Pure Math student who barely started the course and has very few practice in writing proofs, although slightly less worse at reading them. I'd like a level of rigor in this line of thought. EDIT: as it is noticeable from the comments below, there is a condition missing in the statement: [imath]f \geq0[/imath]. | 491475 | integration of function equals zero
Let [imath]f[/imath] be a continues function in [imath][a,b][/imath] [imath]\forall x \in [a,b] \ \ \ f(x)\geq 0[/imath] [imath] \int_{a}^{b}f(x)dx \ = 0[/imath] Proof that [imath] \forall x \in [a,b] \ \ f(x) = 0 [/imath] So how do I do that ? What I know is: because [imath]f[/imath] is continues function in [imath][a,b][/imath] I know it bounded and because [imath]\forall x \in [a,b] \ \ \ f(x)\geq 0[/imath] I know that [imath] \exists M \in \mathbb{N} \ \forall x \in [a,b] \ \ \ M \geq f(x)\geq 0[/imath]. I also know that because [imath]f[/imath] is continues function in [imath][a,b][/imath] that [imath]f \in R[a,b][/imath] meaning that there exists [imath]F[/imath] so that [imath]F(a) = F(b)[/imath], but how can I show that [imath] \forall x \in [a,b] \ \ F(b) = F(a) = F(x)[/imath] ? So how do I continue from here ? |
444271 | Swap real and imaginary part of a complex number
Say I have some complex number [imath]z=a+ib[/imath] with [imath]a,b\in \mathbb{R}[/imath]. Now I want to "swap" its real and imaginary parts, i.e. I want to get [imath]\tilde{z}=b+ia[/imath]. While I figure that the appropriate mapping would be [imath]\tilde{z}=-i\overline{ z }[/imath], (where [imath]\overline{z}[/imath] denoting the complex conjugate of [imath]z[/imath]), I keep wondering whether there is some standard notation/function for this. All of [imath]z^\dagger, z^*,\hat{z}[/imath] and the like usually have some other meaning to my knowledge. | 912995 | What is the correct notation for flipping [imath]a[/imath] and [imath]b[/imath] values in a complex?
I'm currently doing some experiments on fractals and in one of my equation I need to flip the real and imaginary components of a complex number, such as : [imath] z = a + bi [/imath] Becomes : [imath] z = b + ai [/imath] What would be the correct mathematical term or notation for this ? flip ? |
205346 | How to prove these inequalities: [imath]\liminf(a_n + b_n) \leq \liminf(a_n) + \limsup(b_n) \leq \limsup(a_n + b_n)[/imath]
The inequalities are: [imath]\liminf(a_n + b_n) \leq \liminf(a_n) + \limsup(b_n) \leq \limsup(a_n + b_n)[/imath] | 408135 | Prove [imath]\limsup\limits_{n \to \infty} (a_n+b_n) \le \limsup\limits_{n \to \infty} a_n + \limsup\limits_{n \to \infty} b_n[/imath]
I am stuck with the following problem. Prove that [imath]\limsup_{n \to \infty} (a_n+b_n) \le \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n[/imath] I was thinking of using the triangle inequality saying [imath]|a_n + b_n| \le |a_n| + |b_n|[/imath] but the problem is not about absolute values of the sequence. Intuitively it's clear that this is true because [imath]a_n[/imath] and [imath]b_n[/imath] can "reduce each others magnitude" if they have opposite signs, but I cannot express that algebraically... Can someone help me out ? |
60766 | What is the result of [imath]\infty - \infty[/imath]?
I would say [imath]\infty - \infty=0[/imath] because even though [imath]\infty[/imath] is an undetermined number, [imath]\infty = \infty[/imath]. So [imath]\infty-\infty=0[/imath]. | 1168993 | Why it is so? Please explain please help me
Why [imath] \infty - \infty \neq 0 ?[/imath] Please explain |
823351 | About [imath]\mathbb Z_{p}[\sqrt{k}][/imath], when is it a field?
I give up. I'm new in the fields world, and I'm trying to give a sufficient and necessary condition for [imath]\mathbb{Z}_{p}[\sqrt{k}]=\{a+b\sqrt{k}:a,b\in \mathbb{Z}_{p}\}[/imath] to be a field ([imath]p[/imath] is a prime and [imath]k[/imath] is a positive integer). I claim that the condition is that [imath]p[/imath] doesn't divide [imath]k[/imath], but I don't know if this is true, I've tried to prove this but I've got stuck in the "sufficient" part. It would be great if some of you can help me, thanks!. | 854801 | If [imath]k>0[/imath] is a positive integer and [imath]p[/imath] is any prime, when is [imath]\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in\mathbb Z_p\}[/imath] a field.
If [imath]k>0[/imath] is a positive integer and [imath]p[/imath] is any prime, when is [imath]\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in\mathbb Z_p\}[/imath] a field? Find necessary and sufficient condition. Attempt: Since we know that a finite integral domain is a field and since [imath]\mathbb Z_p[\sqrt{k}][/imath] is finite, it suffices to find the condition when [imath]\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in \mathbb Z_p\}[/imath] forms an integral domain. The given set forms a commutative ring with unity. Hence, the only condition that needs to be satisfied is the absence of any zero divisors. [imath](a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k) = (a_1a_2+b_1b_2k)+(a_1b_2+a_2b_1)\sqrt k[/imath] Case [imath]1[/imath]: When [imath]k[/imath] is not a perfect square [imath](a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k)=0 \implies (a_1a_2+b_1b_2k) \bmod p=0~~ \& ~~ (a_1b_2+a_2b_1) \bmod p =0[/imath] Case [imath]2[/imath]: When [imath]k[/imath] is a perfect square [imath] = u^2[/imath] [imath](a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k) = 0 \implies (a_1a_2+b_1b_2 u^2 + (a_1b_2+a_2b_1)u ) \bmod p=0 [/imath] Both the above cases can have many cases and seems a bit complicated. Am I missing out on something? The book which I am reading (Gallian) hasn't introduced Quadratic Residues as of yet. Thank you for the help. |
915383 | Subgroups and cyclic groups
Suppose a group [imath]G[/imath] has no proper subgroups (that is, the only subgroup of [imath]G[/imath] is [imath]G[/imath] itself and the trivial subgroup [imath]\{e\}[/imath]. Show that [imath]G[/imath] is cyclic. | 332926 | A group with no proper non-trivial subgroups
There is a lemma that says if a group [imath]G[/imath] has no proper nontrivial subgroups, then [imath]G[/imath] is cyclic. And here is the proof of the lemma: Suppose [imath]G[/imath] has no proper nontrivial subgroups. Take an element [imath]a[/imath] in [imath]G[/imath] for which [imath]a[/imath] is not equal to [imath]e[/imath]. Consider the cyclic subgroup [imath]\langle a \rangle[/imath]. This subgroup contains at least [imath]e[/imath] and [imath]a[/imath], so it is not trivial. But [imath]G[/imath] has no proper subgroups, so it must be that [imath]\langle a \rangle = G[/imath]. Thus [imath]G[/imath] is cyclic, by definition of a cyclic group. But here i do not understand the following: Why must [imath]\langle a \rangle[/imath] be a subgroup of [imath]G[/imath]? For every single element [imath]a[/imath] in [imath]G[/imath], if [imath]\langle a \rangle[/imath] is a subgroup of [imath]G[/imath], then every group should have at least as many subgroups as the number of its elements. I would appreciate any help. Thanks |
914818 | Problem about circle tangents
Circles [imath]c_1[/imath] and [imath]c_2[/imath] with origins [imath]O_1[/imath] and [imath]O_2[/imath] are on plane. [imath]O_1Z[/imath] and [imath]O_1X[/imath]are tangents to Circle [imath]c_2[/imath].These tangents intersect [imath]c_1[/imath] in [imath]A[/imath] and [imath]B[/imath].[imath]O_2Y[/imath] and [imath]O_2T[/imath]are tangents to Circle [imath]c_1[/imath].These tangents intersect [imath]c_2[/imath] in [imath]C[/imath] and [imath]D[/imath]. Prove [imath]AB=CD[/imath] Figure Things I have done: I was able to Prove that quadrilateral [imath]JO_1LO_2[/imath] is Kite (Triangles [imath]O_2J O_1[/imath] and [imath]O_2L O_1[/imath] are congruent by Angle-Side-Angle). So [imath]JO_1=LO_1[/imath] and [imath]CO_2=DO_2[/imath]. [imath]O_1A=O_1B[/imath] because Both of them are radius.So [imath]JA=BL[/imath].Similar to this,[imath]CJ=DL[/imath]. So By Side-Angle-Side, [imath]CJA[/imath] and [imath]DLB[/imath] are congruent.So [imath]AC=BD[/imath]. So if I Prove that [imath]AC ||DB[/imath],then [imath]CABD[/imath] will be a Parallelogram So [imath]DC=AB[/imath] will be true.This where I stuck. I would Appreciate if someone could edit title to something better. | 455395 | Two Circles and Tangents from Their Centers Problem
Let [imath]\Gamma_1[/imath] and [imath]\Gamma_2[/imath] be two non overlapping circles with centers [imath]O_1[/imath] and [imath]O_2[/imath] respectively. From [imath]O_1[/imath], draw the two tangents to [imath]\Gamma_2[/imath] and let them intersect [imath]\Gamma_1[/imath] at points [imath]A[/imath] and [imath]B[/imath]. Similarly, from [imath]O_2[/imath], draw the two tangents to [imath]\Gamma_1[/imath] and let them intersect [imath]\Gamma_2[/imath] at points [imath]C[/imath] and [imath]D[/imath]. Prove that [imath]AB=CD[/imath]. I've done some extensive angle chasing on this but have been unable to make any real progress. Can't decide whether [imath]ABCD[/imath] is supposed to be rectangle (as in my diagrams), a parallelogram or even a trapezium. There is a homothety taking one circle to the other but as far as I can see this doesn't help as we don't have a clearly defined center to this. Any help/hints would be greatly appreciated. |
910503 | How prove this integral inequality[imath] \int_{0}^{+\infty}\frac{1}{x^x}{\rm d}x<2 [/imath]
Show that[imath] \int_{0}^{+\infty}\frac{1}{x^x}{\rm d}x<2 [/imath] | 909637 | Inequality of numerical integration [imath]\int _0^\infty x^{-x}\,dx[/imath].
There was a friend asking me how to prove [imath]\int_0^\infty x^{-x}\,dx<2[/imath] Mathematica showed that its approximate value is 1.99546, so I think it isn't easy to solve it, can you provide me some ideas about this question? |
41807 | Variation on the Monty Hall Problem
Many of us know the Monty Hall Problem But the other day I was asked a variation of this riddle. The answer of the original question is, of course, [imath] 66\% [/imath] in favor of changing doors, but this is based on the fact that the game show host knows where the prize is. Suppose he does not know where the prize is, and after you make your pick, he opens one of the other two doors and it happens to be a goat. Is it still better to change doors when he asks? I believe it is. (After all it still leaves us with two chances instead of one.) But some of my friends think otherwise. We are not mathematicians, just a couple of riddle-likers, so we are not sure of the correct answer. So I thought to post it here. Edit : I read this on wikipedia and if I understand it correctly it seems to support my answer: Morgan et al. (1991) and Gillman (1992) both show a more general solution where the car is (uniformly) randomly placed but the host is not constrained to pick uniformly randomly if the player has initially selected the car, which is how they both interpret the well known statement of the problem in Parade despite the author's disclaimers. Both changed the wording of the Parade version to emphasize that point when they restated the problem. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q, having a value between 0 and 1. If the host picks randomly q would be 1/2 and switching wins with probability 2/3 regardless of which door the host opens. If the player picks Door 1 and the host's preference for Door 3 is q, then in the case where the host opens Door 3 switching wins with probability 1/3 if the car is behind Door 2 and loses with probability (1/3)q if the car is behind Door 1. The conditional probability of winning by switching given the host opens Door 3 is therefore (1/3)/(1/3 + (1/3)q) which simplifies to 1/(1+q). Since q can vary between 0 and 1 this conditional probability can vary between 1/2 and 1. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. However, it is important to note that neither source suggests the player knows what the value of q is, so the player cannot attribute a probability other than the 2/3 that vos Savant assumed was implicit. What are your thoughts? | 2164483 | Interesting variant of Monty Hall
I was wondering how we could think about the following variant of the classic Monty hall theorem, Suppose now that the host actually does not remember what is behind any of the doors, once you choose one of the three doors, he will open at random one of the other two doors. If he reveals the prize the show ends, if he does not reveal the prize, should you still switch doors? I was trying to formulate it but I am just not sure if I am making correct assumptions. For example, say we choose door [imath]A[/imath], then the probability that he opens door [imath]B[/imath] is equal to the probability he chooses door [imath]C[/imath] is equal to [imath]0.5[/imath]. The probability that the prize is behind door A, B and C is 1/3 the probability that Monty ends up opening any given door is also [imath]1/3[/imath] So should one just condition on something else now? Looking forward to hearing any opinions on this, Thanks |
917505 | Number of subsets of a set [imath]S[/imath] of n elements , is [imath]2^n[/imath]
I know that Number of subsets of a set [imath]S[/imath] of size n , is given by the binomial sum [imath]\sum_{k=0} ^n \binom{n}{k}=2^n[/imath] , n=1,2,3,.. elements how we could conclude of prove this formula ? how to prove that this gives the number of subsets . where did this formula come from ? PS:- i know how to prove [imath]\sum_{k=0} ^n \binom{n}{k}=2^n[/imath] | 717266 | Count subsets of an i-set in two ways
I am trying to prove that [imath]\sum_{k=0}^i\binom{i}{k} = 2^i[/imath] by counting in two ways. i.e. count the subsets of an [imath]i-set[/imath] in two ways. So far I have that the total number of subsets of an [imath]i-set[/imath] is [imath]\sum_{k=0}^{i}\binom{i}{k}[/imath] now for the second part let [imath]R=\{a_1,a_2,...,a_i\}[/imath] be an [imath]i-set[/imath] and let [imath]A_k \subset R [/imath] so for [imath]a_j, j=1,2,...,i[/imath] either [imath]a_j \in A_k[/imath] or [imath]a_j \notin A_k[/imath] I'm trying to use this to show that this give two choices for [imath]a_j[/imath] in [imath]A_k[/imath] and then use this to show that choosing each element in [imath]A_k \forall \space k = 1,2,...,i[/imath] is equivalent to [imath]2^i[/imath] but my argument is not so clear. Any help would be much appreciated. |
918788 | How to do this integral [imath]\int_{-\infty}^{\infty}{\rm e}^{-x^{2}}\cos\left(\,kx\,\right)\,{\rm d}x[/imath]
How to do this integral [imath]\int_{-\infty}^{\infty}{\rm e}^{-x^{2}}\cos\left(\,kx\,\right)\,{\rm d}x[/imath] for any [imath]k > 0[/imath] ?. I tried to use gamma function, but sometimes the series doesn't converge. | 388356 | How to prove [imath]\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}[/imath]
I've been grappling for a few hours with this problem ([imath]b[/imath] is a parameter) [imath]\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}[/imath] Tried direct integration by parts, differentiation w.r.t. [imath]b[/imath], then integrating... all to no avail. I think the second approach is the way... Pls help |
270118 | A few improper integral
[imath]\displaystyle \begin{align*} & \int_{0}^{+\infty }{\frac{\text{d}x}{1+{{x}^{n}}}} \\ & \int_{-\infty }^{+\infty }{\frac{{{x}^{2m}}}{1+{{x}^{2n}}}\text{d}x} \\ & \int_{0}^{+\infty }{\frac{{{x}^{s-1}}}{1+x}\text{d}x} \\ \end{align*}[/imath] | 1099614 | How to integrate [imath]\int_0^\infty\frac{dx}{1+x^n}[/imath]
I was playing around with the function [imath]\dfrac{1}{1+x^2}[/imath], and knowing that the integral over [imath](0,\infty)[/imath] was [imath]\dfrac{\pi}{2}[/imath], I was hoping to see if there was some neat pattern to determining the value of the following generalized form: [imath]I=\int_0^\infty \frac{dx}{1+x^n}[/imath] for natural [imath]n\ge2[/imath]. I ran some computations in Mathematica and got a consistent result of [imath]\dfrac{\pi}{n}\csc\left(\dfrac{\pi}{n}\right)[/imath], which also happens to be equivalent to [imath]\dfrac{1}{n}\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(1-\dfrac{1}{n}\right)=\mathrm{B}\left(\dfrac{1}{n},1-\dfrac{1}{n}\right)[/imath]. Is there some neat trick I can use to get this result? I've considered integrating along a contour, but I'm somewhat rusty with complex methods. Also, the result was posted previously here. I tried something like this instead: [imath]\frac{1}{1+x^n}=\begin{cases}\displaystyle \sum_{k=0}^\infty(-1)^kx^{nk}&\text{for }|x|<1\\\\ \displaystyle\sum_{k=0}^\infty (-1)^k x^{-n(k+1)}&\text{for }|x|>1\end{cases}[/imath] then integrating along the respective intervals [imath](0,1)[/imath] and [imath](1,\infty)[/imath]. Here's my attempt: [imath]\begin{align*}I &=\left\{\int_0^1+\int_1^\infty\right\}\frac{dx}{1+x^n}\\\\ &=\sum_{k=0}^\infty \left\{\int_0^1 (-1)^kx^{nk}\,dx+\int_1^\infty (-1)^kx^{-n(k+1)}\,dx\right\}\\\\ &=\sum_{k=0}^\infty \left\{(-1)^k\left[\frac{x^{nk+1}}{nk+1}\right]_{0}^{1}+(-1)^k\left[\frac{x^{-n(k+1)+1}}{-n(k+1)+1}\right]_1^\infty\right\}\\\\ &=\sum_{k=0}^\infty \left\{\frac{(-1)^k}{nk+1}-\frac{(-1)^{k}}{1-n(k+1)}\right\}\\\\ &=\left(1-\frac{1}{1-n}\right)+\left(-\frac{1}{n+1}+\frac{1}{1-2n}\right)\\ &\quad\quad+\left(\frac{1}{2n+1}-\frac{1}{1-3n}\right)+\left(-\frac{1}{3n+1}+\frac{1}{1-4n}\right)+\cdots\\\\ &=1-\frac{2}{1-n^2}+\frac{2}{1-4n^2}-\frac{2}{1-9n^2}+\cdots\\\\ &=\sum_{k=0}^\infty \frac{(-2)^k}{1-(kn)^2}\end{align*}[/imath] However, this series diverges by the ratio test, so I'm not sure if I made a mistake in my calculation or used some flawed reasoning. |
918876 | Suppose that [imath]a_0 >a_1 >...>a_{2013} >0.[/imath] Prove that [imath]\sum_{n = 0}^{2013}a_nz^n \neq 0[/imath] when [imath]|z|<1[/imath]
Suppose that [imath]a_0 >a_1 >...>a_{2013} >0.[/imath] Prove that [imath]\sum_{n = 0}^{2013}a_nz^n \neq 0[/imath] when [imath]|z|<1[/imath] Not sure where to begin with this. Any suggestions? Thanks. | 438485 | Proving that [imath]\sum\limits_{n = 0}^{2013} a_n z^n \neq 0[/imath] if [imath]a_0 > a_1 > \dots > a_{2013} > 0[/imath] and [imath]|z| \leq 1[/imath]
I'm going to teach a preparation course for the complex analysis qualifying exam from my university (which basically consists of me doing some problems from past exams) and I'm trying to solve some questions from previous exams. One of the questions is the following: Prove that [imath]\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}[/imath] if the coefficients satisfy [imath]a_0 > a_1 > \dots > a_{2013} > 0[/imath] and [imath]|z| \leq 1[/imath]. I tried to approach the problem by using the reverse triangle inequality to try to isolate the leading term, which I thought should be the constant coefficient, as follows: \begin{array} . \left| \sum_{n = 0}^{2013} a_n z^n \right| &= \left| a_0 + \sum_{n = 1}^{2013} a_n z^n \right| \\ &\geq |a_0| - \left| \sum_{n = 1}^{2013} a_n z^n \right|\\ &\geq |a_0| - \sum_{n = 1}^{2013} |a_n z^n|\\ &\geq a_0 - \sum_{n = 1}^{2013} a_n |z^n|\\ &\geq a_0 - \sum_{n = 1}^{2013} a_0 |z^n|\\ &= a_0\left( 1 - \sum_{n = 1}^{2013} |z^n| \right) \end{array} Then for [imath]|z| < 1[/imath], we have that this last expression is equal to [imath] a_0\left( 1 - |z| \frac{|z|^{2013} - 1}{|z| - 1} \right) [/imath] and my hope was to prove that this was always positive, nevertheless, after failing to prove it, I plotted the expression inside the parenthesis as a function of [imath]|z|[/imath], and found out that it changes sign in the interval [imath](0, 1)[/imath]. Question Can my approach somehow be made to work? And if not, how can I prove that the sum [imath]\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}[/imath] under the given conditions? Thank you very much for any help. |
699956 | Consecutive zeros in decimal expansion of [imath]\pi[/imath]
Is it known whether or not there exist arbitrarily long sequences of consecutive zeros in the decimal expansion of [imath]\pi[/imath]? | 39546 | What bizarrities lurk within the decimal expansion of an irrational number?
i.e. can we write [imath]\pi = 3.14159\dots X\dots[/imath] where [imath]X[/imath] consists of (say) [imath]10^{100}[/imath] consecutive zeroes? [Originally asked on reddit without response :-( ] |
919653 | Pigeonhole proof of the existence of two numbers with given sum
Let [imath]|W|=m+1[/imath] and [imath]W[/imath] be a subset of [imath]X=\{1,2,3,\dots ,2m\}[/imath] ([imath]m[/imath] is any natural number). Prove there exists two numbers in [imath]W[/imath] whose sum is [imath]2m+1[/imath]. Can anyone give me a hint to prove this? I know I should be using the pigeonhole principle. | 314629 | Using the Pigeonhole Principle to show that [imath]2[/imath] of any [imath]n+1[/imath] numbers from [imath]\{1,2,\ldots,2n\}[/imath] sum to [imath]2n+1[/imath]
Let n be greater or to 1, and let S be an (n+1)-subset of [2n]. Prove that there exist two numbers in S whose sum is 2n+1. I know I have to use the pigeonhole principle - no idea how to start... |
919859 | Error of Riemann sum is [imath]a/n + o(1/n)[/imath]
A problem from an old qual: For [imath]f[/imath] of class [imath]C^2[/imath], find [imath]a[/imath] such that [imath]\int_0^1 f(t)dt-\frac1n\sum_{k=1}^{n-1}f\left( \frac {k}{n}\right)=\frac{a}{n}+o\left(\frac1n \right).[/imath] If we divide [imath][0,1][/imath] into [imath]n[/imath] equal subintervals [imath]\left[\frac{k}{n},\frac{k+1}{n}\right],\,\,0\leq k <n[/imath], then [imath]\frac1n \sum_{k=1}^{n-1}f\left( \frac {k}{n}\right)[/imath] is taking a Riemann sum over this partition, except it's missing one interval (is this perhaps a mistake?) Forgetting for the moment the missed interval, the error of this Riemann sum is [imath]\sum_{i=1}^n \int_{(i-1)/n}^{i/n} f(x) - f\left( \frac{i}{n} \right)dx \\ = \sum_{i=1}^n \int_{(i-1)/n}^{i/n} f'(\xi_x)\left(x-\frac{i-1}{n}\right) dx.[/imath] Let [imath]m, M \in \mathbb{R}[/imath] be such that [imath]m\leq f'\leq M[/imath] on [imath][0,1][/imath]. Then by estimating [imath]f'(\xi_x)[/imath], integrating, and summing, we get [imath]m\frac{1}{2n}\leq \text{ error } \leq M\frac{1}{2n}.[/imath] Now since we were taking left-hand Riemann sums, and we missed the first interval, we should add to the error [imath]f(0)/n[/imath]. Let [imath]K:= \max\{|M|, |m|\}[/imath]. We have [imath]|\text{error}|\leq \frac{K/2 + |f(0)|}{n}.[/imath] I can see why the missing interval contributes something which is [imath]O\left(\frac{1}{n^2}\right).[/imath] But why does the result from estimating the derivative contribute something that is [imath]o\left(\frac{1}{n}\right)[/imath]? Or do we have to do a more careful estimation? Perhaps we could use that [imath]f'[/imath] is Lipschitz continuous? | 569750 | Speed of convergence of Riemann sums
This question is inspired by a previous question. It was shown that, for all function [imath]f \in \mathcal{C} ([0, 1])[/imath], [imath] \lim_{n \to + \infty} \sum_{k=0}^{n} f \left( \frac{k}{n+1} \right) - \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) = \int_0^1 f (x) \ dx.[/imath] A stronger statement would be that there exists some constant [imath]a(f)[/imath] such that: [imath]\sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) = n \int_0^1 f (x) \ dx + a(f) + o(1),[/imath] or, in other words, that there is an asymptotic development at order [imath]1[/imath] of the Riemann sums: [imath]\frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) = \int_0^1 f (x) \ dx + \frac{a(f)}{n} + o(n^{-1}).[/imath] Given [imath]f[/imath], can we always find such a constant [imath]a(f)[/imath]? If this is false, can we find a counter-example? If this is true, can [imath]a(f)[/imath] be written explicitely? I have had a quick look at the litterature, but most asymptotics for the Riemann sums involve different meshes, which depend on the function [imath]f[/imath]. |
919810 | Evaluate [imath]\sqrt{1 + 2\sqrt{1 + 3 \sqrt{1 + \dots}}}[/imath]
I was asked to show that the answer is 3. I don't have any idea on how to proceed. Thanks! | 458740 | Nested Radical of Ramanujan
I think I have sort of a proof of the following nested radical expression due to Ramanujan for [imath]x\ge 0[/imath]. [imath]\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}[/imath] for [imath] x\ge -1[/imath] I just want to know if my proof is okay or there is a flaw, and if there is one I request to give some suggestions to eliminate them. Thank you. The proof is the following: Proof: Let us define [imath] a_n(x)=\underbrace{\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}}_{n \ \mbox{terms}}[/imath] for [imath]x\ge 0[/imath] so that [imath] a_1(x)=\sqrt{1+x},\ a_2(x)=\sqrt{1+x\sqrt{1+(x+1)}},\ a_3(x)=\cdots[/imath] and so on. Since [imath]x\ge 0[/imath] each one of the [imath]a_n[/imath] is defined (I am taking only the positive square root). Also, we note that [imath]a_{n+1}^2(x)=1+xa_{n}(x+1)[/imath] Now, we note that [imath]\{a_n(x)\}[/imath] is an increasing sequence and that [imath]a_n(x)<x+1[/imath] [imath]\forall n\ge 1[/imath], this is easy to prove by induction as below: For [imath]n=1[/imath], [imath]a_1(x)=\sqrt{1+x}<1+x[/imath] since [imath]x\ge 0\Rightarrow 1+x\ge 1[/imath]. SO it is true for [imath]n=1[/imath]. Similarly, the truth can be proved for [imath]n>1[/imath]. Then [imath]a_n(x)[/imath] converges to [imath]l(x)=\sup_{n}a_n(x)\le x+1[/imath] Now I make the following claim: Claim: [imath]l(x)=x+1\quad \forall i\ge 0[/imath] Proof: Fix [imath]x[/imath]. Let [imath]l(x)<x+1[/imath]. Then, [imath]l(x)=x+1-\epsilon[/imath] for some [imath]\epsilon>0[/imath]. Now, I claim that there must be a [imath]n[/imath] such that [imath]x+1-a_n(x)<\epsilon[/imath], and if that is true then [imath]a_n(x)>x+1-\epsilon=l(x)[/imath] which is a contradiction since [imath]l(x)=\sup_{n}a_n(x)[/imath] and then it implies that [imath]l(x)=x+1[/imath] To prove my claim it requires [imath]x+1-a_n(x) < \epsilon[/imath] Now, \begin{align} x+1-a_n(x) = & x+1-\sqrt{1+xa_{n-1}(x+1)} \\ \ =& \frac{(x+1)^2-({1+xa_{n-1}(x+1)})}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ =& x\frac{(x+1)+1-a_{n-1}(x+1)}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ <& \frac{x}{x+2}((x+1)+1-a_{n-1}(x+1))\\ \ <& \frac{x}{x+2}\cdot\frac{x+1}{x+3}((x+2)+1-a_{n-2}(x+2))\\ \ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n} ((x+n-1)+1-a_{1}(x+n-1))\\ \ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n}(x+n-\sqrt{x+n})\\ \ <& \frac{x(x+1)}{x+n-1} \end{align} Now, if one is able to find [imath]n[/imath] such that [imath]\frac{x(x+1)}{x+n-1}<\epsilon \Rightarrow x< \frac{-(1-\epsilon)+\sqrt{(1-\epsilon)^2+4\epsilon(n-1)}}{2}[/imath] then we're done. Now from the upper bound it seems that there always exists some [imath]n[/imath] that satisfies this requirement. Hence the claim is proved. |
920750 | How many ways so that [imath]x+y+z\leq n[/imath]?
Can anyone help me out finding the number of solutions i.e. [imath](x,y,z)[/imath] for the inequality [imath]x+y+z\leq n[/imath] where, [imath]n[/imath] is a constant positive value and [imath]x,y,z[/imath] are non-negative? | 910809 | How to use stars and bars?
How to use the stars and bars method? Say I want to find number of combinations I can get with [imath]x_1+x_2+x_3+x_4=22[/imath], where [imath]x_i\in\mathbb{N}[/imath]. Is this the correct time to apply the method? This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions and here: List of abstract duplicates |
920782 | [imath]2x_1 + 2x_2 + \cdots + 2x_6 + x_7 = N[/imath]
How do I find the number of integral solutions to the equation - [imath]2x_1 + 2x_2 + \cdots + 2x_6 + x_7 = N[/imath] [imath]x_1,x_2,\ldots,x_7 \ge 1[/imath] I just thought that I should reduce this a bit more, so I replace [imath]x_i[/imath] with [imath](y_i+1)[/imath], so we have: [imath]y_1 + y_2 + \cdots + y_6 = \tfrac{1}{2}(N + 13 - y_7)[/imath] [imath]y_1,y_2,\ldots,y_7 \ge 0[/imath] I will be solving this as a programming problem by looping over [imath]y_7[/imath] from [imath][0, N+13][/imath]. How do I find the number of solutions to this equation in each looping step? | 924546 | How many solution are possible for this multivariable equation?
[imath]2(a+b+c+d+e+f)+g=N[/imath] where [imath]a,b,c, \cdots ,N \in \mathbb{N}[/imath] Any lead will be appreciated. |
921540 | "Waiter's paradox" - what's wrong with this reasoning?
Here's a puzzle I just heard and while I know that this reasoning is fundamentally wrong, I can't explain why: Three people bought a dish for, say, 25\[imath] and paid 30\[/imath] The waiter didn't want to divide 5 by 3, so he took 2\[imath] for himself and gave 1[/imath] to each of the clients Each of them seems to have spent 9\[imath], but 9\[/imath] x 3 = 27\$, not 28. What happened to this one dollar? | 656047 | Where is the lost dollar?
Somebody explained me this problem, but I am not sure to understand what is wrong. Three people go to a bar to have a drink. To pay the bill everyone gives [imath]10. The barman gives back [/imath]5 coins. Every people take [imath]1 back and leave the [/imath]2 other dollar for the barman. If everybody give \[imath]10 and get \[/imath]1 back, they paid \[imath]9 right?[/imath] 3 * 9 + [imath]2 = [/imath]29 Where is the missing dollar? |
922185 | Limit of [imath]\frac{1}{x^2}-\frac{1}{\sin^2(x)}[/imath] as [imath]x[/imath] approaches [imath]0[/imath]
Evaluate [imath]\lim \limits_{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2(x)}\right)[/imath] I tried to combine the fractions [imath]\frac{1}{x^2}-\frac{1}{\sin^2(x)} = \frac{\sin^2(x)-x^2}{x^2\sin^2(x)}[/imath] and apply L'Hopitals which only made a mess. I feel like there is a simpler way of doing this but I'm not quite sure what to do | 400541 | Find the limit without use of L'Hôpital or Taylor series: [imath]\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)[/imath]
Find the limit without the use of L'Hôpital's rule or Taylor series [imath]\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)[/imath] |
922573 | An elementary algebraic inequality
How do I go about proving that [imath]\frac{x^2}{yz+2}+\frac{y^2}{xz+2}+\frac{z^2}{xy +2}\ge \frac{x+y+z}{3} [/imath] for [imath]1\le x,y,z\le 2[/imath] ? This is straightforward when [imath]x=y=z[/imath], but I don't see where to go from there. | 920655 | How to prove that [imath]\frac{x^2}{yz+2}+\frac{y^2}{zx+2}+\frac{z^2}{xy+2}\geq \frac{x+y+z}{3}[/imath] holds for any [imath](x,y,z)\in[1,2]^3[/imath]
Prove that for [imath]x,y,z\in [1,2][/imath] the following inequality holds: [imath]\frac{x^2}{yz+2}+\frac{y^2}{zx+2}+\frac{z^2}{xy+2}\geq \frac{x+y+z}{3}.[/imath] I tried to apply the Cauchy-Schwarz inequality or the power mean inequality, but my attempts were unsuccessfull, so I am asking for help. |
922547 | sigma algebras generated by a set
If [imath]M[/imath] is the sigma algebra generated by [imath]E[/imath], then [imath]M[/imath] is the union of the sigma algebras generated by [imath]F[/imath] as [imath]F[/imath] ranges over all countable subsets of [imath]E[/imath]. (Hint: Show that the latter object is a sigma algebra) | 61617 | If [imath]E \in \sigma(\mathcal{C})[/imath] then there exists a countable subset [imath]\mathcal{C}_0 \subseteq \mathcal{C}[/imath] with [imath]E \in \sigma(\mathcal{C}_0)[/imath]
Given a collection of sets [imath]\mathcal{C}[/imath] and [imath]E[/imath] an element in the [imath]\sigma[/imath]-algebra generated by [imath]\mathcal{C}[/imath], how do I show that [imath]\exists[/imath] a countable subcollection [imath]\mathcal{C_0} \subset \mathcal{C}[/imath] such that [imath]E[/imath] is an element of the [imath]\sigma[/imath]-algebra, [imath]\mathcal{A}[/imath] generated by [imath]\mathcal{C_0}[/imath]? The hint says to let [imath]H[/imath] be the union of all [imath]\sigma[/imath]-algebras generated by countable subsets of [imath]\mathcal{C}[/imath]....although I don't know why. |
697629 | I found this odd relationship, [imath]x^2 = \sum_\limits{k = 0}^{x-1} (2k + 1)[/imath].
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to [imath]2x + 1[/imath] should equal [imath]x^2[/imath]. | 1033083 | Why is there a pattern for making orders of perfect squares (first one, second one, third one) by simply adding two to the next adding each time?
For example, if I had a perfect square of [imath]16[/imath], which is the fourth perfect square, I would add nine to get to the fifth perfect square, [imath]25[/imath]. This is probably how it goes:[imath]1+3=4+5=9+7=16+9=25+11=36+13=49+15=64+17=81+19=100[/imath]and it goes on forever. It looks like the amount being added each time gets added by two each time as they move on to the next and the next and the next and the next. |
923091 | Factorial Summation Problem
[imath]\sum_{j=0}^n j\cdot j![/imath] I got [imath](n+1)!-1[/imath] as the answer but I'm not sure if that's right or how I even got to that answer exactly. (my paper is a mess of random work and I can't make it out). Can somebody tell me if that's correct and explain the best and condensed way to get to the answer? | 910466 | Finding a formula for [imath]1+\sum_{j=1}^n(j!)\cdot j[/imath] using induction
I need help with finding the formula and proving it by induction. Am stuck, but the professor says we should know this by now. Find the formula for [imath]1 + \sum \limits _{j=1} ^n j! j[/imath], and show work proving the formula is correct using induction. |
128201 | Proving mathematical induction with arbitrary base using (weak) induction
I attempted a proof of mathematical induction using an arbitrary base case, but was unsuccessful (and hence this question). Below is what I was trying to do and along with my thinking; if anyone can point me in the right direction I'd appreciate it. The induction I am using: Let [imath]P[/imath] be a property about the natural numbers [imath]\mathbb{N}[/imath] with [imath]0\in\mathbb{N}[/imath], and let [imath]P(n)[/imath] denote the statement that the property [imath]P[/imath] holds for [imath]n\in\mathbb{N}[/imath]. Suppose [imath]P(0)[/imath]. Furthermore suppose that for each natural number [imath]k[/imath], [imath]P(k)[/imath] implies [imath]P(k+1)[/imath]. Then [imath]\forall nP(n)[/imath]. What I am trying to prove: Let [imath]P[/imath] be a property about the natural numbers [imath]\mathbb{N}[/imath] with [imath]0\in\mathbb{N}[/imath], and let [imath]P(n)[/imath] denote the statement that the property [imath]P[/imath] holds for [imath]n\in\mathbb{N}[/imath]. Suppose for [imath]n_0\in\mathbb{N}[/imath], [imath]P(n_0)[/imath]. Furthermore suppose that for each natural number [imath]k\geq n_0[/imath], [imath]P(k)[/imath] implies [imath]P(k+1)[/imath]. Then [imath]\forall n\geq n_0 P(n)[/imath]. My attempted proof: define [imath]Q(n)[/imath] to be [imath]n\geq n_0 \to P(n)[/imath]. Then we wish to prove [imath]\forall nQ(n)[/imath]. We induct on [imath]n[/imath]. Base case (for ordinary induction): [imath]Q(0)[/imath] is [imath]0\geq n_0\to P(0)[/imath]. Since [imath]n_0\in\mathbb{N}[/imath], [imath]0\geq n_0[/imath] implies that [imath]n_0=0[/imath]. Since [imath]P(n_0)[/imath], [imath]P(0)[/imath], which proves the base case. Inductive step: we want to show [imath]\forall n(Q(n)\to Q(n+1))[/imath]. To do this, we assume [imath]k\geq n_0 \to P(k)[/imath] and try to show [imath]k+1\geq n_0 \to P(k+1)[/imath]. Since [imath]k\geq n_0 \to P(k)[/imath], we first prove the case where [imath]k\geq n_0[/imath] and [imath]P(k)[/imath]. By the hypothesis of the proof, we see that [imath]P(k+1)[/imath], which proves the case for [imath]k+1[/imath]. This is where I am having trouble: For [imath]k<n_0[/imath], I am unable to show the implication for [imath]k+1[/imath]. So my questions would be: (1) is the overall approach for the proof correct? (2) If so, how might I go on to prove the case when [imath]k<n_0[/imath]? Thanks in advance. (This is not homework, by the way.) | 2877083 | In the principle of Mathematical Induction, why do we take the base case as [imath]P(1)[/imath] only?
I kinda understand the logic and motivation behind the proof, but what bothers me is the fact, why is the base case (the first statement that we write) is always [imath]P(1)[/imath] when we are proving a proposition is true for all [imath]\mathbb{N} [/imath]? Why can't it be anything other than [imath]1[/imath]? What about [imath]P(5)[/imath] or something? Does it change anything? |
923607 | Prove that [imath]\sum a_i e^{\alpha_ix}=0[/imath] has at most [imath]n-1[/imath] real solutions.
We have to prove that [imath]\sum_{i=1}^n a_i e^{\alpha_ix}=0,a_i\neq0,\alpha_i\neq\alpha_j\forall i\neq j[/imath] has at most [imath]n-1[/imath] real solutions. I tried differentiating and other stuff but was unable to do anything... Any suggestions will be helpful. | 921226 | Show that the equation [imath]a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0[/imath] has at most [imath]n - 1[/imath] real roots.
For non-zero [imath]a_1, a_2, \ldots , a_n[/imath] and for [imath]\alpha_1, \alpha_2, \ldots , \alpha_n[/imath] such that [imath]\alpha_i \neq \alpha_j[/imath] for [imath]i \neq j[/imath], show that the equation [imath]a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx} = 0[/imath] has at most [imath]n - 1[/imath] real roots. I thought of applying Rolle's theorem to the function [imath]f(x) = a_1e^{\alpha_1x} + a_2e^{\alpha_2x} + \cdots + a_ne^{\alpha_nx}[/imath] and reach a contradiction, but I can't find a way to use it. |
923726 | Show that [imath]\overline{x}\in\Bbb Z/(n\Bbb Z)[/imath] is invertible iff [imath]\gcd(x,n)=1[/imath]
Be [imath]n[/imath] greater than 2 integer and is [imath]\bar x\in\mathbb{Z}_n-\{\bar0,\bar1,\bar2,\ldots,\overline{n-1}\},[/imath] [imath]\;0\leq x<n[/imath]. Show that [imath]\exists\; \bar y\in\mathbb{Z}_n[/imath], such that [imath]\bar x\cdot \bar y=\bar y\cdot \bar x=\bar1\Longleftrightarrow gcd\{x,n\}=1[/imath] (i.e., the elements [imath]\bar x[/imath], [imath]0\leq x<n[/imath], invertible in [imath]\mathbb{Z}_n[/imath] são aqueles tais que [imath]gcd\{x,n\}=1[/imath]) | 348636 | How to compute [imath]\mathbb{Z}_n^*[/imath], the unit group of the integers modulo [imath]n[/imath]?
I am trying to wrap my head around calculating residue classes, for example [imath]\mathbb{Z}^*_{12} = [1]...[/imath] Would anyone mind making up an example with a decent explanation. I just wrap my head around things easier seeing the process. Thanks |
918990 | [imath]√2|z|≥|Rez|+|Imz|[/imath]
I just came across this simple inequality which I am finding it difficult to prove. For any complex number z I need to prove that the following inequality holds [imath]√2|z|≥|Rez|+|Imz|[/imath] I would much appreciate if someone could give me a hint to solve this. Thanks | 918416 | Verify that [imath]\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|[/imath]
Verify that [imath]\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|.[/imath] I started off noting that [imath]z=x+iy[/imath] and that [imath]Re(z)=x[/imath] and [imath]Im(z)=y[/imath] Then I know that I have to square both sides, giving me [imath]2(x^2 +y^2) \ge x^2+2|xy|+y^2[/imath] I'm not sure what to do after this. |
924303 | Exam exercise on sequence [imath]a_n = \sin(n)[/imath]
Prove that the sequence [imath]a_n = \sin(n)[/imath] cannot converge when [imath]n \rightarrow \infty [/imath] I tried to find two subsequences that converge to different values but I am having trouble with the fact that [imath]n \in N[/imath] and so I can't pick values in the whole domain of the function. Maybe there is an other way. | 27218 | Prove that the limit of [imath]\sin n[/imath] as [imath]n \rightarrow \infty[/imath] does not exist
Using only the delta definition of a limit, how can we prove that the sequence [imath]\{a_n\}[/imath], where [imath]a_n = \sin n[/imath], as [imath]n[/imath] tends to infinity does not have a limit? Thanks! |
924313 | Prove [imath]99^{100}>100^{99}[/imath] using binomial theorem
Prove [imath]0<(1+\frac{1}{n})<3[/imath] and hence prove [imath]99^{100}>100^{99}[/imath]. I did the first part and showed [imath]0<\frac{1}{n^{n-1}}\le3[/imath] and hence [imath]0<(1+\frac{1}{n})<3[/imath]. But for the second bit, I don't know how to incorporate the first bit to help me prove th inequality. | 103431 | Which of the numbers [imath]99^{100}[/imath] and [imath]100^{99}[/imath] is the larger one?
Which of the numbers [imath]99^{100}[/imath] & [imath]100^{99}[/imath] is the larger? Solve without using logarithms. |
924393 | Prove for all whole numbers n, [imath](n+1)(n+2)...(2n-1)(2n)=2^n(1)(3)(5)...(2n-1)[/imath]
Prove for all whole numbers n, [imath](n+1)(n+2)...(2n-1)(2n)=2^n(1)(3)(5)...(2n-1)[/imath]. I got upto [imath](2n)(2n-2)(2n-4)...=2^n[/imath], after which I'm stuck. | 69162 | Proving formula for product of first n odd numbers
I have this formula which seems to work for the product of the first n odd numbers (I have tested it for all numbers from [imath]1[/imath] to [imath]100[/imath]): [imath]\prod_{i = 1}^{n} (2i - 1) = \frac{(2n)!}{2^{n} n!}[/imath] How can I prove that it holds (or find a counter-example)? |
923738 | Number of bases of [imath]V= \mathbb{F}_p^2[/imath]
Let [imath]\mathbb{F}_p[/imath] be a prime field, and let [imath]V= \mathbb{F}_p^2[/imath] I need to prove that the number of bases of [imath]V[/imath] is equal to the order of the general linear group [imath]GL_2(\mathbb{F}_p)[/imath] In order to do so, I wanted to show that a pair [imath](v_1,v_2)[/imath] of column vectors forms a basis of [imath]F^2[/imath] if and only if the matrix whose columns are the two vectors is invertible. How can I prove that above statement ? | 79356 | Using the Determinant to verify Linear Independence, Span and Basis
Can the determinant (assuming it's non-zero) be used to determine that the vectors given are linearly independent, span the subspace and are a basis of that subspace? (In other words assuming I have a set which I can make into a square matrix, can I use the determinant to determine these three properties?) Here are two examples: Span Does the following set of vectors span [imath]\mathbb R^4[/imath]: [imath][1,1,0,0],[1,2,-1,1],[0,0,1,1],[2,1,2,-1][/imath]? Now the determinant here is [imath]1[/imath], so the set of vectors span [imath]\mathbb R^4[/imath]. Linear Independence Given the following augmented matrix: [imath]\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 2 & 0 \end{array}\right], [/imath] where again the determinant is non-zero ([imath]-2[/imath]) so this set S is linearly independent. Of course I am in trouble if you can't make a square matrix - I figure for spans you can just rref it, and I suppose so for linear independence and basis? |
924081 | What's the result? [imath]1/i=?[/imath], where [imath]i=\sqrt{-1}[/imath]
I just had my first math class in the university, and I understood everything pretty well, but I think I have misread this one because I read that the result is [imath]-1[/imath]. Thanks for your answers! | 682638 | How to show [imath]i^{-1} = -i[/imath]?
How can I show [imath]i^{-1} = -i[/imath], where [imath]i[/imath] is the imaginary unit? Here's what I've tried: [imath]i^{-1} = (-1)^{-1/2} = \dots ?[/imath] |
556662 | Show that [imath] \{\lnot,\leftrightarrow\} [/imath] is not functional complete
I have to prove that this set of logical operators is not functional complete - [imath] \{\lnot,\leftrightarrow\} [/imath] i've tried implement this set by [imath] \{\rightarrow,\lor\} [/imath] which is not functional complete too, but didn't success.. thanks ! | 1649423 | Can you represent other logical operations using only [imath]\neg[/imath] and [imath]\Leftrightarrow[/imath] (not and equivalence)?
I can check on WolframAlpha for how to represent some logical operations using others, but I don't see anything for [imath]\Leftrightarrow[/imath]. Is it because it is impossible? http://www.wolframalpha.com/input/?i=a+xor+b |
925900 | "Heine–Borel" for the Sorgenfrey line
The Heine–Borel theorem perfectly characterizes the compact subsets of the real line [imath]\mathbb{R}[/imath] (with the usual metric/order topology): Heine–Borel Theorem. A subset [imath]A \subseteq \mathbb R[/imath] is compact if and only if it is closed and bounded. What characterisations are there for the compact subsets of the Sorgenfrey line, [imath]\mathbb{S}[/imath]? Since the topology on [imath]\mathbb S[/imath] is finer than the topology on [imath]\mathbb R[/imath], every compact subset of [imath]\mathbb{S}[/imath] is also a comapct subset of [imath]\mathbb{R}[/imath], and so they must be closed (as subsets of [imath]\mathbb{R}[/imath]) and bounded. But more is needed: [imath][0,1][/imath] is not compact, since it has an infinite disjoint open cover: [imath] [0,\tfrac 12) , [\tfrac 12, \tfrac 34 ) , [ \tfrac 34 , \tfrac 78 ) , \ldots , [ \tfrac{2^n-1}{2^n}, \tfrac{2^{n+1}-1}{2^{n-1}}), \ldots , [1 , 2 ).[/imath] | 51251 | Connectedness problem and countability of compact subsets of the Sorgenfrey line
1) Stuck with this problem, can you please help? Show that if [imath]X[/imath] is connected and for a connected subspace [imath]A[/imath] of [imath]X[/imath] we have [imath]X \setminus A = U \cup V[/imath] where [imath]U,V[/imath] are open in [imath]X \setminus A[/imath] and disjoint, then the sets [imath]A \cup U[/imath] and [imath]A \cup V[/imath] are connected. 2) The following result appears in Encyclopaedia of mathematics by Hazewinkel page [imath]375[/imath]: Any compact subset of the Sorgenfrey line is countable. I tried to prove this without any luck. Why is this true? |
518487 | How is it, that [imath]\sqrt{x^2}[/imath] is not [imath] x[/imath], but [imath]|x|[/imath]?
As far as I see, [imath]\sqrt{x^2}[/imath] is not [imath]x[/imath], but [imath]|x|[/imath], meaning the "absolute". I totally get this, because [imath]x^2[/imath] is positive, if [imath]x[/imath] is negative, so [imath]\sqrt{y}[/imath], whether [imath]y = 10^2[/imath] or [imath]y = -10^2[/imath]: [imath]y[/imath] is positive. But then I remember that [imath]\sqrt(x)[/imath] is the same as [imath]x^{1/2}[/imath] and thus, [imath]\sqrt{x^2}[/imath] is the same as [imath]x^{1}[/imath]. So, as far as I get it, [imath]\sqrt{x^2} = x^{2/2}[/imath]. But then, I can cancel [imath]\frac{2}{2}[/imath] to [imath]1[/imath]. So: [imath]\sqrt{x^2} = x^1 = x[/imath]. Where does the absolute value in this derivation of the calculation come from? I get it why its there: Because [imath](-x)^2 = x^2[/imath], so the radicand is always positive. But from [imath]\sqrt{x^n} = x^{n/2}[/imath] I dont understand how it gets there. | 2068883 | Why does [imath] \sqrt{x^2} = |x| [/imath]
This equality keeps coming up in my algebra course and I don't understand why this is true. [imath] \sqrt{x^2} = |x| [/imath] shouldn't it just be [imath] \sqrt{x^2} = x [/imath]? |
926269 | How do I solve this tricky definite integral ?!
[imath]I=\int\limits_0^1\dfrac{x^2-1}{\ln x}\mathrm dx[/imath] I tried numerous substitutions but nothing seems to work.. any ideas ???! | 566475 | What is [imath]\int_0^1\frac{x^7-1}{\log(x)}\mathrm dx[/imath]?
/A problem from the 2012 MIT Integration Bee is [imath] \int_0^1\frac{x^7-1}{\log(x)}\mathrm dx [/imath] The answer is [imath]\log(8)[/imath]. Wolfram Alpha gives an indefinite form in terms of the logarithmic integral function, but times out doing the computation. Is there a way to do it by hand? |
913675 | Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertices
Triangle [imath]ABC[/imath] is an equilateral triangle and [imath]P[/imath] is an arbitrary point inside it. The distance from [imath]P[/imath] to [imath]A[/imath] is [imath]4[/imath] and the distance from [imath]P[/imath] to [imath]B[/imath] is [imath]6[/imath] and the distance from [imath]P[/imath] to [imath]C[/imath] is [imath]5[/imath]. How to find the side of an equilateral triangle from this information? | 922270 | Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertice
Triangle ABC is an equilateral triangle and P is an arbitrary point inside it. The distance from P to A is 4 and the distance from P to B is 6 and the distance from P to C is 5. How to find the side of an equilateral triangle from this information? [imath]\frac{a^2+x^2-y^2}{2xa} = \frac{\sqrt{3}}{2}\frac{a^2+x^2-z^2}{2xa}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+x^2-z^2}{2xa}\Big)^2}[/imath] @Aditya based from that equation where did you get that square root of 3 over 2 ....up to ...... plus one half then square root of 1 minus something and so on.. pls explain further |
926434 | Probability density function of sum of random variables
Assume [imath]X_i[/imath] probability density function is : [imath]f(x,\lambda)=\Bbb{I}_{(0,\infty)}(x)\lambda \exp(-\lambda x)[/imath] how to find the probability density function of [imath]\sum X_i[/imath] ? The result is [imath]\Bbb{I}_{(0,\infty)}(x)\frac{1}{(n-1)!}\lambda^nx^{n-1} \exp(-\lambda x)[/imath] But I don't know how to find it. where [imath]\Bbb{I}_{A}(x)=\begin{cases} 1, & x \in A \\ 0, & x\not\in A \end{cases}[/imath] | 655302 | Gamma Distribution out of sum of exponential random variables
I have a sequence [imath]T_1,T_2,\ldots[/imath] of independent exponential random variables with paramter [imath]\lambda[/imath]. I take the sum [imath]S=\sum_{i=1}^n T_i[/imath] and now I would like to calculate the probability density function. Well, I know that [imath]P(T_i>t)=e^{-\lambda t}[/imath] and therefore [imath]f_{T_i}(t)=\lambda e^{-\lambda t}[/imath] so I need to find [imath]P(T_1+\cdots+T_n>t)[/imath] and take the derivative. But I cannot expand the probability term, you have any ideas? |
924601 | What is [imath]\cdots ((((1/2)/(3/4))/((5/6)/(7/8)))/(((9/10)/(11/12))/((13/14)/(15/16))))/\cdots[/imath]?
What does this number equal if it goes on forever? [imath]\frac{\frac{\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac56}{\frac78}}}{\frac{\frac{\frac{9}{10}}{\frac{11}{12}}}{\frac{\frac{13}{14}}{\frac{15}{16}}}}}{\frac{\frac{\frac{\frac{17}{18}}{\frac{19}{20}}}{\frac{\frac{21}{22}}{\frac{23}{24}}}}{\frac{\frac{\frac{25}{26}}{\frac{27}{28}}}{\frac{\frac{29}{30}}{\frac{31}{32}}}}}[/imath] [imath].[/imath][imath].[/imath][imath].[/imath] Treat the fraction in blocks, one divides what is under it, then that pair is divided by the next pair, those four are divided by the next four. ETC Edit. I'm reformatting as I can't read the original... aged eyes I guess ;-( but in deference to almagest's comment I am leaving the original too. [imath]\frac{\frac{\frac{1}{2}\Big/\frac{3}{4}}{\frac{5}{6}\Big/\frac{7}{8}}\Bigg/ \frac{\frac{9}{10}\Big/\frac{11}{12}}{\frac{13}{14}\Big/\frac{15}{16}}} {\frac{\frac{17}{18}\Big/\frac{19}{20}}{\frac{21}{22}\Big/\frac{23}{24}}\Bigg/ \frac{\frac{25}{26}\Big/\frac{27}{28}}{\frac{29}{30}\Big/\frac{31}{32}}} \cdots[/imath] | 29234 | Tall fraction puzzle
I was given this problem 30 years ago by a coworker, posted it 15 years ago to rec.puzzles, and got a solution from Barry Wolk, but have never seen it again. Consider the series: [imath]1, \frac{1}{2},\frac{\displaystyle\frac{1}{2}}{\displaystyle\frac{3}{4}},\frac{\displaystyle\frac{\displaystyle\frac{1}{2}}{\displaystyle\frac{3}{4}}}{\displaystyle\frac{\displaystyle\frac{5}{6}}{\displaystyle\frac{7}{8}}},\cdots[/imath] Each fraction keeps its large bars while being put atop a similar structure. This can also be represented as [imath]\frac{1\cdot 4 \cdot 6 \cdot 7 \cdot\cdots}{2 \cdot 3 \cdot 5 \cdot 8 \cdot\cdots}[/imath] terminating at [imath]2^n[/imath] for some [imath]n[/imath], where it is much closer to the limit than elsewhere. The challenge: Find the limit, not too hard by experiment In the last expression, find a simple, nonrecursive, expression to say whether [imath]n[/imath] is in the numerator or denominator Prove the limit is correct-this is the hard one. |
918422 | Between [imath]n[/imath] and [imath]2n[/imath] there is always a prime number.
Between [imath]n[/imath] and [imath]2n[/imath] there is always a prime number. I was thinking of this and looked it up on the google to find that this is true. Now, I am wondering what is the proof for it? Does any elementary proof exist for it? Thank you. | 38917 | Are there any Combinatoric proofs of Bertrand's postulate?
I feel like there must exist a combinatoric proof of a theorem like: There is a prime between [imath]n[/imath] and [imath]2n[/imath], or [imath]p[/imath] and [imath]p^2[/imath] or anything similar to this stronger than there is a prime between [imath]p[/imath] and [imath](\prod_p p) + 1[/imath] (Euclid's theorem). I was trying to prove one by the Sieve on this grid 1 2 3 4 ... p p+1 p+2 p+3 p+4 ... 2p 2p+1 2p+2 2p+3 2p+4 ... 3p ........................... ........................... ........................... p^2 p^2+1 p^2+2 p^2+3 p^2+4 ... but it did not work. Do any good arguments like this exist? I don't expect anything as strong as Prime Number Theorem or even Bertrand, but surely a direct combination proof can prove that there are lots of primes? |
927593 | A question on left cosets of distinct subgroups and index
Let [imath]H_1 , H_2 , ... , H_k [/imath] be subgroups of [imath]G[/imath] and [imath]x_1,x_2,... ,x_k[/imath] be elements of [imath]G[/imath] such that [imath]G=\cup_{i=1}^k x_iH_i[/imath] , then how do we prove that some subgroup [imath]H_i[/imath] has finite index in [imath]G[/imath] ? | 536479 | Group covered by finitely many cosets
This question appears in my textbook's exercises, who can help me prove it? If a group [imath]G[/imath] is the set-theoretic union of finitely-many cosets, [imath]G=x_1S_1\cup\cdots\cup x_nS_n[/imath] prove that at least one of the subgroups [imath]S_i[/imath] has finite index in [imath]G[/imath]. I think that the intersection of these cosets is either empty or a coset of the intersection of all the [imath]S_i[/imath]. I want to start from this point to prove it. So I suppose none of these [imath]S_i[/imath] has finite index, but I don't know how to continue? |
927716 | Show that [imath]{n \choose 1} + {n \choose 3} +\cdots = {n \choose 0} + {n \choose 2}+\cdots[/imath]
Show [imath]{n \choose 1} + {n \choose 3} +\cdots = {n \choose 0} + {n \choose 2}+\cdots[/imath] A hint is given to consider the expansion [imath](x-y)^n[/imath] However, when I plug in a number for [imath]n[/imath], I don't get an equality. [imath]n=5[/imath], for instance, I get [imath]5+10 = 1 +10[/imath]. How is this equality possible? | 120231 | Evaluating even binomial coefficients
Can someone give me a hint how to evaluate [imath] \binom{n}{0}+\binom{n}{2}+\cdots+\binom{n}{o(n)},[/imath] where [imath]o(n)[/imath] is [imath]n[/imath] if [imath]n[/imath] is even and [imath]n-1[/imath] otherwise ? |
927709 | If [imath]X[/imath] generates [imath]\Bbb{Q}[/imath] then [imath]X\setminus\{x\}[/imath] also generates [imath]\Bbb{Q}[/imath]
If [imath]X[/imath] is a generator subset of [imath]\Bbb{Q}[/imath] then for [imath]x\in X[/imath], [imath]X\setminus\{x\}[/imath] also generates [imath]\Bbb{Q}[/imath]. Clearly if I can express [imath]x[/imath] as a combination of the remaining generators we are done. After struggling for an hour I took a different approach: tried to show that if [imath]\langle X\setminus\{x\}\rangle\not=\Bbb{Q}[/imath], [imath]\langle X\setminus\{x\}\rangle[/imath] is a maximal subgroup of [imath]\Bbb{Q}[/imath]. That is, if [imath]G\leq\Bbb{Q}[/imath] and [imath]\langle X\setminus\{x\}\rangle\subset G[/imath], then [imath]G=\Bbb{Q}[/imath]. For this purpose it is enough to show that [imath]x\in G[/imath]. But I doubt that this is true. Any ideas? | 487820 | Additive group of rationals has no minimal generating set
In a comment to Arturo Magidin's answer to this question, Jack Schmidt says that the additive group of the rationals has no minimal generating set. Why does [imath](\mathbb{Q},+)[/imath] have no minimal generating set? |
928176 | prove that the sum to n terms of the sequence is [imath]n(n+1)/2(2n+1)[/imath]
Prove that the sum to n terms of the Sequence: [imath]1^2/(1×3),2^2/(3×5),3^2/(5×7),...[/imath] is [imath] n(n+1)/2(2n+1).[/imath] Im having trouble with this question, firstly ive begun by stating that p(n) denotes the statement. Then to prove that p(1) holds where [imath]n = 1[/imath] [imath]n(n+1)2(2n+1).[/imath] = 1/3 and [imath]1^2/1*3[/imath] = [imath]1/3[/imath] therefore true Now i know I have to show [imath]n=k[/imath] and therefore [imath]n=k+1[/imath] but how can i do this? I'm very tired so clear instruction would be great. | 927335 | Prove the sum to n terms of the series
Prove that the sum to [imath]n[/imath] terms of a sequence: [imath]\frac{1^2}{1\times 3}+\frac{2^2}{3\times 5}+\frac{3^2}{5\times 7}+\cdots [/imath] is [imath] \frac{n(n+1)}{2(2n+1)} [/imath] |
928228 | Induction -- n to n+1
I'm trying to understand an induction proof that aims to prove some function is in [imath]O(n\log{ n})[/imath]. It's on page 5 of this PDF: https://courses.engr.illinois.edu/cs573/fa2010/notes/99-recurrences.pdf The function is [imath]T(n)=\sqrt{n}\, T(\sqrt{n}) +n[/imath] We want to prove that [imath]T(n) \leq an \log{ n}[/imath] for sufficiently large [imath]n[/imath] and some constant [imath]a[/imath] to be determined later. The author uses induction as follows: \begin{align*} T(n) &= \sqrt{n}\,T(\sqrt{n}) +n \\ &\le \sqrt{n}\,a\sqrt{n}\, \log{\sqrt{n}} +n \qquad \text{[induction hypothesis]} \\ &= (a/2)n \log{ n} +n \\ &\le an \log{n } \end{align*} First, this is an induction proof, so there should be a base case, but i understand that one needn't prove the base case first. That can always be done later, as the author acknowledges further down the page. Second, and this is what i don't understand, how is this induction if the proof isn't going from some [imath]n[/imath] to an [imath]n+1[/imath]? Is the jump from [imath]\sqrt{n}[/imath] to [imath]n[/imath] equivalent to the jump from [imath]n[/imath] to [imath]n+1[/imath]? Or is something else going on. | 928433 | Induction for recurrence
I'm trying to understand an induction proof that aims to prove some function is in [imath]O(n\log{ n})[/imath]. It's on page 5 of this PDF: https://courses.engr.illinois.edu/cs573/fa2010/notes/99-recurrences.pdf The function is [imath]T(n)=\sqrt{n}\, T(\sqrt{n}) +n[/imath] We want to prove that [imath]T(n) \leq an \log{ n}[/imath] for sufficiently large [imath]n[/imath] and some constant [imath]a[/imath] to be determined later. The author uses induction as follows: \begin{align*} T(n) &= \sqrt{n}\,T(\sqrt{n}) +n \\ &\le \sqrt{n}\,a\sqrt{n}\, \log{\sqrt{n}} +n \qquad \text{[induction hypothesis]} \\ &= (a/2)n \log{ n} +n \\ &\le an \log{n } \end{align*} First, this is an induction proof, so there should be a base case, but i understand that one needn't prove the base case first. That can always be done later, as the author acknowledges further down the page. Second, and this is what i don't understand, how is this induction if the proof isn't going from some [imath]n[/imath] to an [imath]n+1[/imath]? Is the jump from [imath]\sqrt{n}[/imath] to [imath]n[/imath] equivalent to the jump from [imath]n[/imath] to [imath]n+1[/imath]? Or is something else going on. btw, i asked this question at maths stackexchange, and i got no good answer. That's why I am reposing here. |
928417 | Prove that an inverse mapping for a surjective function exists.
Let [imath]f : A \to B [/imath] be a surjective function. Prove that there exists [imath]g : B \to A[/imath] such that [imath]\forall b \in B[/imath], [imath]f(g(b)) = b[/imath]. This does make sense. Because for each element of [imath]B[/imath], we can reverse the arrows in the mapping from [imath]A[/imath] to [imath]B[/imath] and that will take us to corresponding element [imath]a[/imath] in [imath]A[/imath] but again [imath]f(a)[/imath] will make it come back to [imath]b[/imath]. But I'm not sure how to go towards proving this. Any hint would be really helpful! | 440832 | Right Inverse for Surjective Function
Prove that if [imath]f:X\to Y[/imath] is a surjective function between sets, then there must exist a function [imath]g:Y\rightarrow X[/imath] such that [imath]f\circ g=1_Y[/imath]. I know that the identity function is onto, and if [imath]f[/imath] has right inverse, then [imath]f[/imath] must be onto; although, I haven't seen a proof of this. A good hint would be nice. |
928655 | Find solution to matrix sandwich product
For any two [imath]n \times n[/imath] real symmetric and positive definite matrices [imath]B[/imath] and [imath]C[/imath], is it always possible to find a third real symmetric and positive definite matrix [imath]A[/imath] such that [imath]ABA=C[/imath]? If not, give a counter example in a [imath]2 \times 2[/imath] case? | 284890 | Find [imath]C[/imath], if [imath]A=CBC[/imath], where [imath]A[/imath],[imath]B[/imath],[imath]C[/imath] are symmetric matrices.
If [imath]A=CBC[/imath], where [imath]A[/imath],[imath]B[/imath],[imath]C[/imath] are symmetric matrices and [imath]A[/imath],[imath]B[/imath] are given find [imath]C[/imath]. [imath]A[/imath],[imath]B[/imath],[imath]C[/imath] are assumed to be real valued and [imath]B[/imath] is positive definite matrix. Does the unique solution always exist ? What additional assumptions are needed to obtain unique solution. |
928964 | [imath]\sqrt 2+ \sqrt{3} \approx 3.14[/imath]
The square [imath]\sqrt 2+ \sqrt{3} \approx 3.14[/imath] Is this a coincidence or is their some mathemtical significance? | 701822 | Geometric explanation of [imath]\sqrt 2 + \sqrt 3 \approx \pi[/imath]
Just curious, is there a geometry picture explanation to show that [imath]\sqrt 2 + \sqrt 3 [/imath] is close to [imath] \pi [/imath]? |
928408 | How to show that every nonempty [imath]X \subset \mathbb N[/imath] has an [imath]\in[/imath]-minimal element
I am trying to prove the following: Every nonempty [imath]X \subset \mathbb N[/imath] has an [imath]\in[/imath]-minimal element. Proof: Take some [imath]n \in X[/imath]. Then [imath]\min \{ n \cap X \} = \min \{ X \}[/imath]. Is this ok or should it be more detailed? | 358979 | Proving the so-called "Well Ordering Principle"
Is there anything wrong with the following proof? Theorem. Every non-empty subset [imath]B \subset\mathbb{N}[/imath] has a least member. Proof. Assume not. Then, of necessity, we'd have to have [imath]B=\varnothing[/imath], for if [imath]B[/imath] contained even one [imath]b\in\mathbb{N}[/imath], then that [imath]b[/imath] would satisfy [imath]b=min(B). [/imath] Contradiction, end proof. I get the sense there is something wrong here, but I can't seem to define exactly what. Also, is there a context where one can simply take this principle as an axiom, and not have to prove it? After all, it is extremely intuitive. Thanks. |
928430 | Zero divisors of [imath]C[0,1][/imath]
Find the zero divisors of the ring [imath]R=C[0,1][/imath] the continuous functions [imath]f:[0,1] \to [0,1][/imath]. I could thought of a set [imath]S[/imath] that I think is included in the set of zero divisors, but I am not sure if [imath]S[/imath] contains all the zero divisors. [imath]S=\{f:[0,1] \to [0,1]: \text{there is} \space \epsilon>0, (x_0-\epsilon,x_0+\epsilon) \subset [0,1], f((x_0-\epsilon,x_0+\epsilon))=0\}[/imath] I would like some help to find the zero divisors. Is the set [imath]S[/imath] a subset of the set of zero divisors? How can I show that? | 155938 | What are the zero divisors of [imath]C[0,1][/imath]?
Suppose you have a ring [imath](C[0,1],+,\cdot,0,1)[/imath] of continuous real valued functions on [imath][0,1][/imath], with addition defined as [imath](f+g)(x)=f(x)+g(x)[/imath] and multiplication defined as [imath](fg)(x)=f(x)g(x)[/imath]. I'm curious what the zero divisors are. My hunch is that the zero divisors are precisely the functions whose zero set contains an open interval. My thinking is that if [imath]f[/imath] is a function which is at least zero on an open interval [imath](a,b)[/imath], then there exists some function which is nonzero on [imath](a,b)[/imath], but zero everywhere else on [imath][0,1]\setminus(a,b)[/imath]. Conversely, if [imath]f[/imath] is not zero on any open interval, then every zero is isolated in a sense. But if [imath]fg=0[/imath] for some [imath]g[/imath], then [imath]g[/imath] is zero everywhere except these isolated points, but continuity would imply that it is also zero at the zeros of [imath]f[/imath], but then [imath]g=0[/imath], so [imath]f[/imath] is not a zero divisor. I have a hard time stating this formally though, since I'm only studying algebra, and not analysis. Is this intuition correct, and if so, how could it be rigorously expressed? |
929572 | Question about the value of the Infinite series
I tried all day long to get the value of the following infinite series.. [imath]\displaystyle \sum_{k=1}^{\infty} \dfrac{2^{k}}{{2k+1}\choose {k}}[/imath] which seems to be [imath]\dfrac{\pi}{2}[/imath] by my matlab iteration. Can anyone help me to get this value? | 928968 | Summation of series [imath]\sum_{k=0}^\infty 2^k/\binom{2k+1}{k}[/imath]
How to find the sum of this series? [imath]\sum_{k=0}^{\infty}\cfrac{{2}^{k}}{\binom{2k+1}{k}}[/imath] It seems very easy. But I still can not work it out, can anyone help? |
929378 | Left ideals of [imath]M_n(K)[/imath]
Let [imath]K[/imath] be a field and [imath]n \in \mathbb N[/imath]. Show the following: (i) Let [imath]V \subset K^n[/imath] be a subspace and [imath]I_V[/imath] the subset of [imath]M_n(K)[/imath] consisting of all the matrices whose rows belong to [imath]V[/imath]. Prove that [imath]I_V[/imath] is a left ideal of [imath]M_n(K)[/imath]. (ii) Show that every left ideal of [imath]M_n(K)[/imath] is of the form defined in (i). I think I could show (i). I am having problems with (ii) For (i), take [imath]M \in M_n(K)[/imath] and [imath]N \in I_V[/imath]. Let [imath]P=MN[/imath], I want to show that [imath]P \in I_V[/imath]. Let [imath]P_i=(P_{i1} ... P_{in})[/imath] be the i-th row of [imath]P[/imath]. Then, by definition of matrix multiplication, [imath]P_i=(\sum_{k=1}^n M_{ik}N_{k1} ... \sum_{k=1}^n M_{ik}N_{kn})[/imath] [imath]=\sum_{k=1}^n M_{ik} (N_{k1} ... N_{kn})[/imath] [imath]=M_{i1}(N_{11} ... N_{1n})+...+M_{in}(N_{n1} ... N_{nn})[/imath] This means that the [imath]i-th[/imath] row of [imath]P[/imath] is a linear combination of the rows of [imath]N[/imath]. From here it follows [imath]P=MN \in I_V[/imath]. This proves [imath]I_V[/imath] is a left ideal of [imath]M_n(K)[/imath]. I don't know how to show (ii), I would appreciate some help with that part. | 2186644 | Finding Left Ideals of [imath]M_2( \mathbb{R})[/imath]
I can't figure out how to find all left ideals of the ring [imath]M_2(\mathbb R)= \left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\quad |\quad a,b,c,d \in \mathbb{R} \right\}.[/imath] Thanks in advance. |
162134 | Is this proof that [imath]\sqrt 2[/imath] is irrational correct?
Suppose [imath]\sqrt 2[/imath] were rational. Then we would have integers [imath]a[/imath] and [imath]b[/imath] with [imath]\sqrt 2 = \frac ab[/imath] and [imath]a[/imath] and [imath]b[/imath] relatively prime. Since [imath]\gcd(a,b)=1[/imath], we have [imath]\gcd(a^2, b^2)=1[/imath], and the fraction [imath]\frac{a^2}{b^2}[/imath] is also in lowest terms. Squaring both sides, [imath]2 = \frac 21 = \frac{a^2}{b^2}[/imath]. Lowest terms representations of rational numbers are unique, so we have [imath]a^2 = 2[/imath] and [imath]b^2=1[/imath]. But there is no such integer [imath]a[/imath], and therefore we have a contradiction and [imath]\sqrt 2[/imath] is irrational. I am not interested in the pedagogical value of this purported proof; I am only interested in whether the logic is sound. | 2068936 | Proving there is no rational number such that square of it is equal to 2.
I start the proof by assuming there is [imath]m/n \in Q[/imath], expressed in lowest terms, such that [imath](m/n)^2=2[/imath] My textbook tells me to derive a contradiction by showing [imath]m,n[/imath] are both even. Here's how I did it. [imath](m/n)^2=2[/imath] [imath]m^2/n^2=2[/imath] [imath]m^2=2n^2[/imath] Hence, [imath]m^2[/imath] is even and [imath]m[/imath] must be even. Then, [imath]m^2=2k[/imath] for some integer [imath]k.[/imath] Plugging [imath]m^2[/imath] back to the equation, [imath](2k)^2/n^2=2[/imath] [imath]4k^2=2n^2[/imath] [imath]n^2=2k^2[/imath] Hence, [imath]n^2[/imath] is even so as [imath]n.[/imath] Therefore, it contradicts [imath]m/n[/imath] are in lowest terms. Is this right way to prove this? I feel I am doing wrong because I am plugging back the equation I got from the equation to the same equation. |
919477 | does [imath]\sum_{n=1}^\infty \frac {1}{n^{1.7 +\sin n}}[/imath] converge?
[imath]\sum_{n=1}^\infty \frac {1}{n^{1.7 + \sin(n)}}[/imath] i was trying resolve it , by any method convergence , but I could not show if the series converge or diverge | 6818 | Testing the series [imath]\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}[/imath]
We know that the "Harmonic Series" [imath] \sum \frac{1}{n}[/imath] diverges. And for [imath]p >1[/imath] we have the result that the series converges [imath]\sum \frac{1}{n^{p}}[/imath] converges. One can then ask the question of testing the convergence the following 2 Series: [imath]\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}, \quad \sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \sin{n}}}[/imath] where [imath] k \in (0,2)[/imath]. Only thing which i have as tool for this problem is the inequality [imath]| \sin{n} | \leq 1[/imath], which i am not sure whether would applicable or not. |
930302 | Isomorphism of a set
We know that [imath]\operatorname {Aut}(G) \over \operatorname {Inn}(G)[/imath] [imath]\cong \operatorname {Out}(G)[/imath]. Is it true that [imath]\operatorname {Aut}(G) \cong \operatorname {Inn}(G) \rtimes \operatorname {Out}(G)? [/imath] If not, could you give me a counterexample? | 456073 | Splitting of Automorphism Group
It is well-known that for any group [imath]G[/imath] there is an exact sequence [imath]0 \rightarrow \text{Inn}(G) \rightarrow \text{Aut}(G) \rightarrow \text{Out}(G) \rightarrow 0[/imath]. Does this sequence always split, i.e. is it always true that [imath]\text{Aut}(G)[/imath] is a semidirect product of [imath]\text{Inn}(G)[/imath] and [imath]\text{Out}(G)[/imath]? I suspect not, but have not yet come across a counterexample. Thanks in advance! |
928578 | Is this a Markov chain?
Let [imath]\{\xi_n \}_{n \geq 1}[/imath] be i.i.d random variables taking values on [imath]\mathbb{Z}[/imath]. Let [imath]\xi_0 = 0[/imath]. [imath]S_n = \sum\limits_{i=1}^{n} \xi_i,[/imath] where [imath]S_0=0[/imath] [imath]Y_n = \sum\limits_{i=0}^{n} S_i[/imath]. My question is whether [imath]Y_n[/imath] is a Markov chain. I know it has to satisfy [imath]P(Y_{n+1} = y_{n+1}| Y_{n} = y_{n}, \ldots, y_0=y_{0}) = P(Y_{n+1} - y_{n+1}| Y_n = {y_n})[/imath]. Intuitively, I know it is not a Markov chain since by knowing the full information of the previous states, I can predict the next state [imath]Y_{n+1}[/imath] better but I am having trouble formulating a mathematical proof. | 925511 | [imath](S_0+\ldots + S_n)_{n\geq 0}[/imath] not a Markov chain
Assume that [imath]Y_0,\ldots , Y_n[/imath] are independent random variables with the following identical distribution: [imath]Y_i=1[/imath] with propability [imath]p[/imath] and [imath]Y_i=0[/imath] with propability [imath]1-p[/imath]. Also set [imath]S_0=0[/imath] and [imath]S_n=Y_0+\ldots + Y_n[/imath]. It's quite trivial that [imath](Y_n)_{n\geq 0}[/imath] is a Markov chain, and I managed to show that [imath](X_n)=(S_n)_{n\geq 0}[/imath] is also a Markov chain, by showing that [imath]\mathbb{P}(X_{n+1}=i_{n+1}\,|\,X_0=i_0,\ldots,X_n=i_n)=\mathbb{P}(X_{n+1}-X_n=i_{n+1}-X_n\,|\,X_0=i_0,\ldots,X_n=i_n)=\mathbb{P}(Y_{n+1}=i_{n+1}-X_n\,|\,X_0=i_0,\ldots,X_n=i_n)=\mathbb{P}(Y_{n+1}=i_{n+1}-i_n)=\mathbb{P}(Y_{n+1}+i_n=i_{n+1})=\mathbb{P}(X_{n+1}=i_{n+1}\,|\,X_n=i_n)[/imath] However I am now running into the problem that I have to find whether or not [imath](X_n)=(S_0+\ldots + S_n)_{n\geq 0}[/imath] is a Markov chain. I suspect that it isn't, but i'm having trouble finding a counterexample. I also need to find if [imath](X_n)=(S_n, S_0+\ldots+S_n)[/imath] is a Markov chain, but I think this actually might be one, since [imath]X_n[/imath] gives "more information" (so to speak) to [imath]X_(n+1)[/imath]. Again, having trouble actually proving that. I appreciate any help, tips etc you can give me. |
357693 | Inequality concerning limsup and liminf of Cesaro mean of a sequence
Let [imath]\{x_n \}[/imath] be a sequence of real numbers and let [imath]y_n = \frac{(x_1 + x_2 + ... + x_n)}{n}[/imath]. (a) Prove that [imath]\liminf x_n \le \liminf y_n \le \limsup y_n \le \limsup x_n[/imath] (b) Give an example of a sequence [imath]\{x_n\}[/imath] for which all inequalities of part (a) are strict. I honestly have no idea where to start on this. I can observe some of the easier things such as [imath]\lim \inf x_n \le \lim \sup x_n[/imath] Any hints would be appreciated. | 2232672 | Theorem by Cesàro: how to prove?
The following was introduced to me in class as «an old theorem, almost forgotten, by an old mathematician, Cesàro, 1960». Theorem Suppose [imath]\{a_n\}[/imath] is a real sequence. Set: [imath]A_n=\frac{1}{n+1}\sum_{k=1}^na_k.[/imath] Then: [imath]\liminf_na_n\leq\liminf_NA_N\leq\limsup_NA_N\leq\limsup_na_n.[/imath] I tried googling for it, but was unable to find it. I cannot seem to be able to prove it myself. How would I go about this? |
928008 | Find the area where dog can roam
A dog is tied to circular pillar by a rope. Radius of this pillar is [imath]1m[/imath] and length of rope is [imath]\pi m[/imath]. What is an area where dog can roam? I tried to find the area of all semicircles and then to find its sum. It is easy to find an area at front side of the pillar. It is [imath]\displaystyle\frac12\pi^2\pi=\frac{\pi^3}{2}[/imath]. Problem is how to find remaining area. I tried to write this area using compass and straightedge, but I couldn't. Then I wrote this in AutoCAD and it looks like this: Is it possible to find the exact value of this area? | 865178 | Area of the field that the cow can graze.
How do we find the area that the cow can graze? The question goes as follows-- There is a circular barn house surrounded by a huge grazing field. A cow is tied to the rope ([imath]AB[/imath]) at the end [imath]A[/imath] as shown. The length of the rope is half the circumference of the barn. Find the area that the cow can graze. The left side of the area is obvious but i cannot get a hang of what is happening on the right side..all i can say is that the rope starts to wrap around the barn when the cow goes to the right. The length of the rope is [imath]16\pi[/imath] units. |
928955 | How to show that [imath]\prod_{d/n} d = n^{\frac{\tau(n)}{2}}[/imath]
set [imath] n, n \in \mathbb{N}[/imath] and prove that [imath]\prod_{d/n} d = n^{\frac{\tau(n)}{2}}[/imath] ¨I have tried this¨ If [imath]n > 1[/imath] then [imath]n = p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}}[/imath] so [imath]n^{\frac{\tau(n)}{2}}=(p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}})^{\frac{(\alpha_{1}+1)\cdot(\alpha_{2}+1)\cdots(\alpha_{k}+1)}{2}}[/imath] but i dont know how stablish a relation with [imath]\prod_{d/n} d[/imath] | 337022 | How to prove [imath] \prod_{d|n} d= n^{\frac{\tau (n)}{2}}[/imath]
how to prove: [imath] \prod_{d|n} d= n^{\frac{\tau (n)}{2}}[/imath] [imath]\prod_{d|n} d[/imath] is product of all of distinct positive divisor of [imath]n[/imath], [imath]\tau (n)[/imath] is number (count)of all of positive divisor of [imath]n[/imath] |
930979 | On the existence of a non-constant sequence whose differentiable image converges
Let [imath]f: [a,b] \to \mathbb R[/imath] be a function differentiable in [imath](a,b)[/imath] , then is it true that there is a non-constant sequence [imath](x_n)[/imath] in [imath](a,b)[/imath] such that the sequence [imath]\big(f(x_n)\big)[/imath] is convergent ? | 930780 | Is a differentiable function always continuous?
Continuous Functions are not Always Differentiable. But can we safely say that if a function f(x) is differentiable within range [imath](a,b)[/imath] then it is continuous in the interval [imath][a,b][/imath] . If so , what is the logic behind it ? |
930902 | [imath]G[/imath] a finite group [imath]n[/imath]-abelian goup and g.c.d.[imath]\big(|G|,n(n-1)\big)=1[/imath] , then to show [imath]G[/imath] is abelian
Let [imath]G[/imath] be a finite group and [imath]n[/imath] be a given positive integer such that [imath](ab)^n=a^nb^n , \forall a,b \in G[/imath] and g.c.d.[imath]\big(|G|,n(n-1)\big)=1[/imath] , then how to prove that [imath]G[/imath] is abelian ? If I can show that [imath](ab)^{n(n-1)}=a^{n(n-1)}b^{n(n-1)} , \forall a,b \in G[/imath] then I would be done , so I'm also asking , is this identity true ? | 928772 | If[imath](ab)^n=a^nb^n[/imath] & [imath](|G|, n(n-1))=1[/imath] then [imath]G[/imath] is abelian
Let [imath]G[/imath] be a group. If [imath](ab)^n=a^nb^n[/imath] [imath]\forall a,b \in G[/imath] and [imath](|G|, n(n-1))=1[/imath] then prove that [imath]G[/imath] is abelian. What I have proved that If [imath]G[/imath] is a group such that [imath](ab)^i = a^ib^i[/imath] for three consecutive integers [imath]i[/imath] for all [imath]a, b\in G[/imath], then [imath]G[/imath] is abelian. |
931466 | Holder continuity, brwonian motion
Let [imath]B[/imath] stand for a brownian motion on a finite interval [imath][0,1][/imath]. If i am not wrong, i think that there exists a positive constant [imath]c[/imath], such that almost surely, for h small enough , for all [imath]0< t < 1- h[/imath] \begin{align} |B(t+h)-B(t)| < c\sqrt{h\log(1/h)} \end{align} or something like this. As a result \begin{align} \bigg|\frac{B(t+h)-B(t)}{h}\bigg| < K(h) \end{align} Am i correct ? | 929573 | Brownian motion - Hölder continuity
Let [imath]B[/imath] stand for a Brownian motion on a finite interval [imath][0,1][/imath]. If I am not wrong, I think that there exists a positive constant [imath]c[/imath], such that almost surely, for [imath]h[/imath] small enough , for all [imath]0< t < 1- h[/imath] \begin{align} |B(t+h)-B(t)| < c\sqrt{h\log(1/h)} \end{align} or something like this. As a result \begin{align} \bigg|\frac{B(t+h)-B(t)}{h}\bigg| < K(h) \end{align} Am I correct ? |
930163 | Transitive action on two sets!
Suppose [imath]G[/imath] is a finite group and G acts transitively on sets [imath]X[/imath] and [imath]Y[/imath]. Let [imath]a[/imath] and [imath]b[/imath] belongs to [imath]X[/imath] and [imath]Y[/imath] respectively and [imath]G_{a}[/imath] be stabilizer of [imath]a[/imath] in [imath]X[/imath] and [imath]G_{b}[/imath] be stabilizers of [imath]b[/imath] in [imath]Y[/imath].Let $G[imath]=[/imath]G_{a}$.[imath]G_{b}[/imath] then show that $G[imath]=[/imath]G_{x}$.[imath]G_{y}[/imath] [imath]\forall x \in X[/imath], [imath]\forall y \in Y[/imath] Edit:[imath]G_{x}[/imath] denotes stabilizer of [imath]x[/imath] in [imath]X[/imath] and [imath]G_{y}[/imath] denotes stabilizers of [imath]y[/imath] in [imath]Y[/imath] | 905931 | Prove a result on transitive group actions.
Let [imath]G[/imath] be a group and [imath]A[/imath] & [imath]B[/imath] be two sets s.t. [imath]G[/imath] acts transitively on each of [imath]A[/imath] & [imath]B[/imath]. Choose some [imath]\alpha[/imath] and [imath]\beta[/imath] in [imath]A[/imath] & [imath]B[/imath] respectively then prove that if [imath]G=G_\alpha G_\beta[/imath] then [imath]G=G_x G_y[/imath] for all [imath]x[/imath] [imath]\in A[/imath] & [imath]y[/imath] [imath]\in B[/imath] where [imath]G_x[/imath] is stabilizer of [imath]x[/imath] in [imath]G[/imath] and similarly [imath]G_\alpha[/imath] be the stabilizer of [imath]\alpha[/imath] and [imath]G_y[/imath] is the stabilizer of [imath]y[/imath]. I know we have to choose some elements and apply some manipulative trick. it is hardly 2 liner but just can't think of right elements. help! |
931480 | how i could show that [imath]f[/imath] is a constant if it has intermediate value and local extremum properties?
let [imath] f\colon R\to R [/imath] be a function with intermediate value property . if [imath]f[/imath] has a local extremum at every point [imath]x\in R [/imath] . my question is how i could show that [imath]f[/imath] is constant ? I would be interest for any replies or any comments | 886113 | Can non-constant functions have the IVP and have local extremum everywhere?
Let [imath]f:\mathbb R \to \mathbb R[/imath] has Intermediate value property. If f has local extremum at every point of [imath]\mathbb R[/imath], can we say f is constant? We know [imath]f(x)=\begin{cases}1 & x \in \Bbb{Q} \\ 0 & x \not \in \Bbb{Q}\end{cases}[/imath] has I.V.P and every point of [imath]\mathbb R[/imath] is local extremum but f is not constant. We know If f is continuous and have local extremum at every point then we can conclude f is constant if all extremum are maximum or minimum then proof is easy and is as follow: If f is continuous and has a local maximum everywhere, then, let [imath]a \in \mathbb R[/imath]. By continuity, [imath]\{x:f(x)≤f(a)\}[/imath] is closed, and by the hypothesis, [imath]\{x:f(x)≤f(a)\}[/imath] is open. The set [imath]\{x:f(x)≤f(a)\}[/imath] is nonempty, open and closed set of [imath]\mathbb R[/imath] so it is all of [imath]\mathbb R[/imath] by connectedness of [imath]\mathbb R[/imath]. Therefore, for all [imath]x[/imath] and [imath]y[/imath] in [imath]\mathbb R[/imath], [imath]f(b)≤f(a)[/imath] and similarly [imath]f(a)≤f(b)[/imath]. |
213285 | Prove: If a sequence converges, then every subsequence converges to the same limit.
I need some help understanding this proof: Prove: If a sequence converges, then every subsequence converges to the same limit. Proof: Let [imath]s_{n_k}[/imath] denote a subsequence of [imath]s_n[/imath]. Note that [imath]n_k \geq k[/imath] for all [imath]k[/imath]. This easy to prove by induction: in fact, [imath]n_1 \geq 1[/imath] and [imath]n_k \geq k[/imath] implies [imath]n_{k+1} > n_k \geq k[/imath] and hence [imath]n_{k+1} \geq k+1[/imath]. Let [imath]\lim s_n = s[/imath] and let [imath]\epsilon > 0[/imath]. There exists [imath]N[/imath] so that [imath]n>N[/imath] implies [imath]|s_n - s| < \epsilon[/imath]. Now [imath]k > N \implies n_k > N \implies |s_{n_k} - s| < \epsilon[/imath]. Therefore: [imath]\lim_{k \to \infty} s_{n_k} = s[/imath]. What is the intuition that each subsequence will converge to the same limit I do not understand the induction that claims [imath]n_k \geq k[/imath] | 2907024 | If [imath]{a_n}[/imath] converges, then [imath]{a_{2n}}[/imath] converges proof question
I am trying to prove that if a sequence [imath]\{a_n\}[/imath] converges, then the sequence [imath]\{a_{2n}\}[/imath] converges as well using the definition of convergence. What I have so far is if [imath]\{a_n\}[/imath] converges, say to [imath]L[/imath], then for any given [imath]\epsilon >0[/imath], there exists an [imath]n^* \in \mathbb{N}[/imath] such that if [imath]n > n^*[/imath], then [imath]\mid(a_n - L)\mid < \epsilon[/imath]. I think I want to choose an [imath]n_1^*[/imath] such that this is true for [imath]\{a_{2n}\}[/imath]. I'm just not sure exactly how to choose the [imath]n_1^*[/imath]. I thought maybe [imath]\frac{n^*}{2}[/imath], but I am not sure this is right and wouldn't know how to implement it. Thanks in advance! |
931798 | Is a horizontal line considered periodic?
Given the following definition of a periodic function: [imath]\exists P, P > 0, f(x + P) = f(x)[/imath] It is possible to argue that [imath]f(x)=k[/imath] ([imath]k[/imath] being a constant) is a periodic function, since you can define [imath]P[/imath] to be any given constant within the real numbers and the definition would be valid. ([imath]f(x + P) = f(x)[/imath] will always be true if [imath]P\in \mathbb R[/imath]). So my question is, can a horizontal line be considered periodic, even when its period is ambiguous? | 907041 | Is [imath]f(x)=10[/imath] a periodic function?
I am not getting satisficatory explanation for this. Clearly [imath]f(x+T) = f(x)[/imath] for all values of [imath]T[/imath]. If we assume it is periodic, does this mean period = [imath]0[/imath]? |
931776 | Distance between points in the plane
I have this problem and I honestly don't even have a clue of how to start, would someone help me please? Let [imath]A[/imath] = {[imath]v_1[/imath],[imath]v_2[/imath], . . . ,[imath]v_n[/imath]} be a set of points in the plane such that the distance between any two points is at least one. Show that there are at most [imath]3n[/imath] pairs of points at a distance exactly one. I found this hint but doesn't help me much: Define a graph on S in which [imath]v_i[/imath] and [imath]v_j[/imath] are adjacent if and only if they are at distance one. Show that in this graph each vertex has degree at most six. | 34458 | Pairs of points exactly [imath]1[/imath] unit apart in the plane
This is a problem I found in a graph theory text, but I can't figure it out. Let [imath]S[/imath] be a set of [imath]n[/imath] points in a plane, the distance between any two of which is at least one. Show that there are at most [imath]3n[/imath] pairs of points of [imath]S[/imath] at distance exactly one. Experimenting, I figured the way to get the most points of distance exactly 1 would be to lay out the points in a grid made out of equilateral triangles. While building up this grid, it seems that when adding a new vertex, I can connect it to 2 or 3 other points to have distance exactly 1, which implies that I can only add at most 3 new pairs of points 1 unit apart for every point I add, which suggests the result. Is there a nonhandwavey way to show this? |
247910 | Fundamental group of [imath]\mathbb{R}^3[/imath] \ finite number of lines passing through origin.
I want to confirm my answer for this question: Calculate the fundamental group of [imath]X=\mathbb{R}^3\setminus \{\text{union of n lines passing through origin}\}[/imath]. My idea is that [imath]X[/imath] deformation retracts onto [imath]S^2\setminus \{\text{union of [/imath]2n[imath] number of points}\}[/imath], which will be homeomorphic to [imath]\mathbb{R}^2\setminus\{\text{union of [/imath](2n-1)[imath] number of points}\}[/imath]. Hence fundamental group of [imath]X[/imath] is same as fundamental group of wedge of [imath](2n-1)[/imath]'s circles, which is free group on [imath](2n-1)[/imath]'s symbols. Please confirm it. | 129380 | Fundamental group of complement of [imath]n[/imath] lines through the origin in [imath]\mathbb{R}^3[/imath]
Just a quick question to verify whether I'm right. Claim: The fundamental group of the complement of [imath]n[/imath] lines through the origin in [imath]\mathbb{R}^3[/imath] is [imath]F_n[/imath], the free group on [imath]n[/imath] generators. Proof: remove a line from [imath]\mathbb{R}^3[/imath]. We may deformation retract the remaining space onto a cylinder radius [imath]\epsilon[/imath] about the line, and thence to a circle [imath]S^1[/imath]. There is no trouble repeating this process with a second distinct line, except that then we will be a wedge union [imath]S^1 \vee S^1[/imath]. Continue inductively, and recall that the wedge union of [imath]n[/imath] circles has the stated fundamental group. I'm only just starting to really get my head around this stuff, so any feedback would be really useful! Thanks! |
897764 | Characterize the natural numbers [imath]n[/imath] such that there is a surjective group homomorphism from [imath]S_n[/imath] to [imath]S_{n-1}[/imath].
This has always been one of my most favorite exercises from Group Theory, and I was surprised to see that this hasn't been asked before. To repeat: Characterize the natural numbers [imath]n[/imath] such that there is a surjective group homomorphism from [imath]S_n[/imath] to [imath]S_{n-1}[/imath]. I have a solution which I will post in a couple days (if someone doesn't recreate it), but I am more interested in seeing how other people would approach this problem. I am very interested in alternate proofs of this characterization. | 1699505 | Is it true that when [imath]n \geq 5[/imath] there is a surjective homomorphism from the symmetric group [imath]S_n[/imath] to [imath]S_{n-1}[/imath]?
Is it true that when [imath]n \geq 5[/imath] there is a surjective homomorphism from the symmetric group [imath]S_n[/imath] to [imath]S_{n-1}[/imath]? How come this is so? Does it have to deal with the subgroup [imath]A_n[/imath] being simple? |
932377 | Intuition for high school students regarding square roots and logarithms
These are some common mistakes high schoolers make: [imath] \sqrt{a + b} = \sqrt{a} + \sqrt{b} [/imath] [imath] \log(a+b) = \log (a) + \log(b)[/imath] So I can obviously show numeric examples to say why these are wrong, but I want to show why in general these are wrong. What are some intuitive arguments to show these are wrong? For example, for [imath](a+b)^2[/imath] there are some nice visual (geometric) illustrations which show why it equals [imath]a^2 + b^2 + 2ab[/imath], and I'd like some similar examples for the more difficult square roots and logarithms. | 630339 | Pedagogy: How to cure students of the "law of universal linearity"?
One of the commonest mistakes made by students, appearing at every level of maths education up to about early undergraduate, is the so-called “Law of Universal Linearity”: [imath] \frac{1}{a+b} \mathrel{\text{“=”}} \frac{1}{a} + \frac{1}{b} [/imath] [imath] 2^{-3} \mathrel{\text{“=”}} -2^3 [/imath] [imath] \sin (5x + 3y) \mathrel{\text{“=”}} \sin 5x + \sin 3y[/imath] and so on. Slightly more precisely, I’d call it the tendency to commute or distribute operations through each other. They don't notice that they’re doing anything, except for operations where they’ve specifically learned not to do so. Does anyone have a good cure for this — a particularly clear and memorable explanation that will stick with students? I’ve tried explaining it several ways, but never found an approach that I was really happy with, from a pedagogical point of view. |
932597 | Fibonacci sequence: Prove the formula [imath]f_{2n+1}=f_{n+1}^2 + f_n^2[/imath]
I can't seem to figure out this proof. I'm using weak induction and always get stuck during the inductive step. Prove for n > 0: [imath]f_{2n+1} = f_{n+1}^2 + f_n^2[/imath] | 300345 | Induction Proof: Fibonacci Numbers Identity with Sum of Two Squares
Using induction, how can I show the following identity about the fibonacci numbers? I'm having trouble with simplification when doing the induction step. Identity: [imath]f_n^2 + f_{n+1}^2 = f_{2n+1}[/imath] I get to: [imath]f_{n+1}^2 + f_{n+2}^2[/imath] Should I replace [imath]f_{n+2}[/imath] using the recursion? When I do that, I end up with the product of terms, and that just doesn't seem right. Any guidance on how to get manipulate during the induction step? Thanks! |
625167 | Determine the convergence of [imath] \sum_{n=1}^{\infty}\left[1-\cos\left(1 \over n\right)\right] [/imath]
I'm having trouble determining the convergence of the series: [imath] \sum_{n=1}^{\infty}\left[1-\cos\left(1 \over n\right)\right]. [/imath] I have tried the root test: [imath]\lim_{n\rightarrow\infty}\sqrt[n]{1-\cos\frac{1}{n}}=\lim_{n\rightarrow\infty}\left(1-\cos\frac{1}{n}\right)^{1/n}=\lim_{n\rightarrow\infty}\mathrm{e}^{\frac{\log(1-\cos\frac{1}{n})}{n}}=\mathrm{e}^{\lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n})}{n}}[/imath] Now by applying the Stolz–Cesàro theorem, that upper limit is equal to: \begin{align} \lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})}{(n+1)-n}&=\lim_{n\rightarrow\infty}\left(\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})\right) \\&=\lim_{n\rightarrow\infty}\log{\frac{1-\cos{\frac{1}{n+1}}}{1-\cos{\frac{1}{n}}}} \end{align} Now I'm totally stuck, unless that quotient is actually 1, in which case the limit would be 0, the Root test result would be [imath]\mathrm{e}^0=1[/imath] and all this would have been to no avail. I'm not sure this method was the best idea, the series sure seems way simpler than that, so probably another method is more appropriate? | 1590889 | Using Maclaurin series to study the convergence of [imath]\sum \limits _{k=1}^\infty (1- \cos\frac{1}{n})[/imath]
I know that the series [imath]\sum \limits _{n=1}^\infty (1- \cos\frac{1}{n})[/imath] converges. One way is to use trigonometric identity which says [imath](1- \cos\frac{1}{n})[/imath] = [imath]2\sin^2(\frac{1}{2n})[/imath] and from here I need to show that [imath]\sum \limits _{n=1}^\infty 2 \left(\frac{1}{2n}\right)^2 [/imath] converges, which is pretty simple because the denominator's [imath]n^p[/imath] and [imath] p > 1 [/imath]. But I want to try another approach. I know that the beginning of [imath]\sin(\frac{1}{n})[/imath] Maclaurin series is ([imath]x_0 = 0[/imath])[imath] 1 - (\frac{1}{n})^2 \ 2! [/imath]. which means I need to check: [imath]\sum \limits _{n=1}^\infty (1- (1 - (\frac{1}{n})^2 \ 2!))[/imath] = [imath]\sum \limits _{n=1}^\infty (\frac{1}{n})^2 \ 2![/imath] and this is clearly converges because it is harmonic sequence where [imath]n^p[/imath] and [imath] p > 1 [/imath]. Does it make sense what I just wrote? What I struggle with: Is my Maclaurin evaluation true? |
933261 | Give an example of A continuous function from X onto Y where X=[0,1] ;Y=[0,1]×[0,1]
Give an example of A continuous onto function [imath]f:X\to Y[/imath] where [imath]X=[0,1][/imath] ;[imath]Y=[0,1] \times[0,1][/imath].Why can't this function be one to one on [0,1]? As far as an example is concerned the only way to construct such a function is to use a space-filling curve http://en.wikipedia.org/wiki/Space-filling_curve. And as far as how to prove why cant it be one-one it seems it will somehow contradict ontoness,but i am not able to proceed as to how to do it. | 101000 | Is there a one-one onto continuous function [imath]f:[0,1]\rightarrow[0,1]^2[/imath]?
Is there a one-one onto continuous function [imath]f:[0,1]\rightarrow[0,1]^2[/imath]? I was trying to prove that there is no such function, but failed. Any suggestions? |
933297 | As[imath]\ n \to \infty[/imath], how does a product over the primes less than[imath]\ p_n[/imath] equal the same product over the primes less than[imath]\ n[/imath]?
How is[imath]\ \lim_{x\to \infty} \log \log x \prod_{i< \log x} \frac{p_i -1}{p_i}= \\ \lim_{x\to \infty} \log \log x \prod_{p < \log x}_{p prime} \frac{p-1}{p}[/imath]? | 929691 | Let[imath]\ p_n[/imath] be the[imath]\ n[/imath]-th prime. Is[imath]\ \lim_{n\to\infty} \log \log n \prod_{i=1}^{\lfloor \log n \rfloor} \frac{p_i-1}{p_i}>0[/imath]?
I'm less than a novice in analysis, I don't even know how to approach this. Thanks in advance. |
818749 | Find the value of [imath]\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}[/imath]
Find the value of [imath]\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}[/imath] My approach : I used [imath]\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 [/imath] [imath]\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(8\pi/7) [/imath] [imath]= 2\sin(3\pi)/7\times\cos(\pi/7)+\sin(\pi + \pi/7) [/imath] [imath]= 2\sin(3\pi)/7\times\cos(\pi/7)-\sin(\pi/7) [/imath] I am not getting any clue how to proceed further or whether it is correct or not. Please help thanks.. | 1745060 | Proving a trigonometric identity:
Prove that [imath]\sin \frac{{2\pi }}{7} + \sin \frac{{4\pi }}{7} + \sin \frac{{8\pi }}{7} = \frac{{\sqrt 7 }}{2}[/imath]. I have tried to square both side and got [imath]{\sin ^2}\frac{{2\pi }}{7} + {\sin ^2}\frac{{4\pi }}{7} + {\sin ^2}\frac{{8\pi }}{7} = \frac{7}{4}[/imath]. But I cannot proceed further. Any help would be appreciated. |
933605 | Properties of [imath]L^p[/imath] spaces
Let [imath]\Omega[/imath] [imath]\subseteq[/imath] [imath]\Bbb R^{n}[/imath] be bounded and [imath]u[/imath] be a measurable function on [imath]\Omega[/imath] Such that [imath]|u|^{p} \in {L^{1}_{\mathrm{loc}}(\Omega)}[/imath] for some [imath]p \in \Bbb R[/imath]. Then how to show the following? [imath]\lim_{p\rightarrow0} \left(\frac{1}{|\Omega|}\int_{\Omega} |u|^{p}\,\mathrm{d}x\right)^{\frac{1}{p}} = \exp \left(\frac{1}{|\Omega|}\int_{\Omega}\log|u|\,\mathrm{d}x\right)[/imath] I attempted as follows: [imath]\lim_{p\rightarrow0} \left(\frac{1}{|\Omega|}\int_{\Omega} |u|^{p}\,\mathrm{d}x\right)^{\frac1p} = \lim_{p\rightarrow0} \frac{1}{|\Omega|^\frac{1}{p}}\left(\int_{\Omega}|u|^{p}\,\mathrm{d}x\right)^{\frac1p} = \exp \lim_{p\rightarrow0} \frac{1}{|\Omega|^\frac1p} \log\left(\int_{\Omega} |u|^{p}\,\mathrm{d}x\right)^{\frac1p}= \exp\left(\frac{1}{|\Omega|}\int_{\Omega}\log|u|\,\mathrm{d}x\right)[/imath] However;I think some steps especially the last two are not satisfactory for me.If you see that you're well come! | 289494 | How compute [imath]\lim_{p\rightarrow 0} \|f\|_p[/imath] in a probability space?
I not solve the follow limit [imath]\lim_{p\rightarrow 0} \bigg[\int_{\Omega} |f|^p d\mu \bigg]^{1/p} = \exp\bigg[ \int_{\Omega} \log|f|d\mu \bigg],[/imath] where [imath](\Omega, \mathcal{F}, \mu)[/imath] is a probability space and [imath]f,\log |f| \in L^1(\Omega).[/imath] Can someone help me? Thank you! |
933450 | A limit ordinal [imath]\gamma[/imath] is indecomposable if and only if [imath]\alpha + \gamma = \gamma[/imath] if and only if [imath]\gamma = \omega^{\alpha}[/imath] for some [imath]\alpha[/imath].
I am trying to prove ex 2.13 from Jech's book on set theory: Ex 2.13: A limit ordinal [imath]\gamma[/imath] is indecomposable if and only if [imath]\alpha + \gamma = \gamma[/imath] if and only if [imath]\gamma = \omega^{\alpha}[/imath] for some [imath]\alpha[/imath]. I think this is equivalent to proving that the 3 following statement are equivalent: [imath]\gamma[/imath] is indecomposable (i.e. there are no [imath]\alpha,\beta < \gamma[/imath] s.t. [imath]\gamma = \alpha + \beta[/imath]). For every [imath]\alpha < \gamma[/imath], [imath]\alpha + \gamma = \gamma[/imath]. There exists [imath]\alpha[/imath] s.t. [imath]\gamma = \omega^{\alpha}[/imath]. I think that I can prove [imath](1) \rightarrow (3)[/imath] since it seems almost straight forward from Cantor's normal form theorem. I can also prove that [imath](1) \Rightarrow (2)[/imath] since, for every [imath]\alpha < \gamma[/imath], there exists a unigue [imath]\delta[/imath] s.t. [imath]\alpha + \delta = \gamma[/imath]. Given, (1), [imath]\delta \geq \gamma[/imath]. But [imath]\delta > \gamma \Rightarrow \alpha + \delta > \gamma[/imath] and we are left with [imath]\delta = \gamma[/imath]. My problem is with the other directions. Any help? | 349204 | Indecomposable limit ordinals
A limit ordinal [imath]\gamma>0[/imath] is said to be indecomposable iff [imath]\nexists\alpha,\beta<\gamma[/imath] such that [imath]\alpha+\beta=\gamma[/imath]. In view of this definition, I’m trying to prove the equivalence of the following statements for limit ordinals [imath]\gamma[/imath]: (1) [imath]\gamma[/imath] is indecomposable. (2) [imath]\alpha+\gamma=\gamma[/imath] [imath]\forall\alpha<\gamma[/imath]. (3) [imath]\gamma=\omega^\alpha[/imath] for some [imath]\alpha\in\mathrm{ON}[/imath]. I’ve been able to get [imath](1)\implies(2)[/imath] and [imath](3)\implies(1)[/imath], but I have nothing more than scratch work for [imath](2)\implies(3)[/imath]. Any suggestions would be appreciated. |
933681 | Some problem similar to Dido's problem
The question is : "Let [imath]A[/imath] and [imath]B[/imath] be two fixed points in [imath]\mathbb{R}^{2}[/imath]. Given [imath]L>[/imath] length of [imath]AB[/imath]. Show that the curve [imath]\alpha[/imath] joining A and B, with length [imath]L[/imath], which together with AB forms a Jordan curve (i.e. a simple closed curve), bounds the largest possible area is an arc of a circle passing through [imath]A[/imath] and [imath]B[/imath]." This is an homework question from a Differential Geometry Course. It seems that it is related to isoperimetric inequality but I don't have any idea to prove it. | 22584 | The Dido problem with an arclength constraint
It is well known that the solution to the classical Dido problem is a semicircle, and that the solution to the classical isoperimetric problem is a circle. It's also reasonably obvious that the solution to the following variant is a circular arc: Let [imath]A[/imath] and [imath]B[/imath] be fixed points on a plane, and let [imath]l[/imath] be a length greater than [imath]\overline{AB}[/imath]. Which (smooth) curve through [imath]A[/imath] and [imath]B[/imath], of length [imath]l[/imath], maximises the area between itself and the line [imath]AB[/imath]? It's a straightforward exercise to verify extremality using the calculus of variations, but are there alternative proofs that do not invoke e.g. the Euler-Lagrange equations? This was originally a homework problem with the isoperimetric inequality given as a hint, and I'm just wondering what the intended solution was... |
892651 | Divisor function asymptotics
Define [imath]\tau_{r}(n) = \sum_{d_1...d_r = n}1[/imath]. One exercise in a book on sieve theory asked for an elementary proof by induction of the fact that [imath]\sum_{n\le x}\tau_r(n) = \frac{1}{(r - 1)!}x(\ln x)^{r - 1} + O\left(x(\ln x)^{r - 2}\right)[/imath] The base case [imath]r = 2[/imath] is easy with reversing the order of summation. The only other progress that I made was the fact that [imath]\sum_{n\le x}\tau_{r}(n) = \sum_{d\le x}\left\lfloor\frac{x}{d}\right\rfloor\tau_{r- 1}(d)[/imath], but I don't know how to proceed. | 1223827 | Divisor number asymptotic?
I have got an interesting task, but I can't solve it: We use [imath]d(n)[/imath] as the number of divisors for the positive [imath]n[/imath] integer. We have: [imath]a(n)=\sum_{i=1}^n d(i)[/imath] How much is [imath]a(n)[/imath] asymptotic? [imath]a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 8,...[/imath] I tried something with Euler's Totient Function, since it counts the numbers which are relative prime to [imath]n[/imath], but it is not really exact, since if one number is not relative prime to the other, it is not in all case a divisor. Any other ideas, will the formula contain logarithm, square root, or anything? Thanks for any help! |
933911 | If [imath]\sum a_n b_n[/imath] converges for all [imath](b_n)[/imath] such that [imath]b_n \to 0[/imath], then [imath]\sum |a_n|[/imath] converges.
Let [imath]a_n[/imath] be a sequence of real numbers. Suppose that for any sequence of real numbers [imath](b_n)[/imath], [imath]b_n \to 0 \implies \sum a_nb_n \;\; \text{converges}[/imath] Prove that [imath]\sum|a_n|[/imath] converges. I attempted a proof by contradiction. If [imath]\sum|a_n|[/imath] does not converge, then it diverges to [imath]\infty[/imath]. Therefore, [imath]\displaystyle \frac{1}{\sum_{k=1}^n |a_k|}\to 0[/imath] as [imath]n[/imath] goes to [imath]\infty[/imath]. As a result, [imath]\displaystyle\sum_{n\geq 1} a_n\frac{1}{\sum_{k=1}^n |a_k|}[/imath] converges. How is that supposed to be a contradiction? I must be on the wrong track... | 577036 | If [imath]\sum a_nb_n[/imath] converges for each null sequence [imath](b_n)[/imath] then [imath]\sum a_n[/imath] is absolutely convergent
A friend of mine gave me the following problem recently: Let [imath](a_n)[/imath] be a sequence of real numbers. Suppose that [imath]\sum_{n=1}^\infty a_nb_n[/imath] converges for any sequence [imath](b_n)[/imath] such that [imath]\lim\limits_{n\to\infty} b_n=0[/imath]. Show that [imath]\sum |a_n|<\infty[/imath]. Hint he gave me was to use Banach-Steinhaus theorem (a.k.a. uniform boundedness principle). I would like to know whether there are different solutions to this, in particular, whether there is an elementary solution. (Using just calculus, without resorting to more advanced results from functional analysis or other areas.) The argument using Banach-Steinhaus theorem goes as follows: Let [imath]a=(a_n)[/imath] be a given (fixed) sequence such that [imath]\sum a_nb_n[/imath] converges for each [imath]b=(b_n)\in c_0[/imath]. For every [imath]n[/imath] we can define [imath]f_n\colon c_0\to{\mathbb R}[/imath] as [imath]\sum_{k=1}^n a_nb_n[/imath]. We consider [imath]c_0[/imath] with the usual sup-norm, which makes [imath]c_0[/imath] into a Banach space. Each function [imath]f_n[/imath] is obviously linear. For any [imath]b\in c_0[/imath] such that [imath]\|b\|=\sup\limits_{n\in\mathbb N} |b_n| \le 1[/imath] we have [imath]|f_n(b)| \le \sum_{k=0}^n |a_k|.[/imath] This shows that all functions [imath]f_n[/imath] are bounded linear functionals. Now if we fix some [imath]b=(b_k)\in c_0[/imath] we have [imath]\left|\sum_{k=0}^n a_kb_k\right| \le \sum_{k=0}^\infty |a_k|\cdot |b_k|.[/imath] The sum on the RHS is finite, since it can be expressed as [imath]\sum a_kc_k[/imath], where [imath]c_k=\pm b_k[/imath]. (The choice of sign here depends on the signs of [imath]b_k[/imath] and [imath]a_k[/imath].) So we have shown that at each point [imath]b\in c_0[/imath] the values [imath]f_n(b)[/imath] are bounded by the same constant (depending only on [imath]b[/imath]). By Banach-Steinhaus theorem we have that there exists a constant [imath]M[/imath] such that [imath]\|f_n\|<M[/imath] for each [imath]n\in\mathbb N[/imath]. Now it only remains to show that [imath]\|f_n\|=\sum_{k=0}^n |a_k|[/imath]. We already know that [imath]\|f_n\|\le\sum_{k=0}^n |a_k|[/imath]. The other inequality can be obtained by choosing [imath]b[/imath] in unit ball, such that [imath]b_k[/imath] is either [imath]1[/imath] or [imath]-1[/imath], depending on the sign of [imath]a_k[/imath] for [imath]k=1,2,\dots,n[/imath]; and [imath]b_k=0[/imath] for [imath]k>n[/imath]. Now we have that [imath]\sum_{k=0}^n |a_k|<M[/imath] for each [imath]n[/imath], which implies [imath]\sum_{k=0}^n |a_k|\le M[/imath]. |
933846 | Solve for Theta: [imath]a = b\tan(\theta) - \frac{c}{\cos(\theta)}[/imath]
The title pretty much sums it up. How do I solve for [imath]\theta[/imath] given the following equation. [imath]a = b\tan(\theta) - \frac{c}{\cos(\theta)}[/imath] I am not a student and this is not homework. It's been quite a while since I've done any significant trig and I'm out of time to figure this out. | 910269 | Solve for [imath]\theta[/imath]: [imath]a = b\tan\theta - \frac{c}{\cos\theta}[/imath]
This question was initially posted on SO (Link). I'm not sure the answer given there was correct. I cannot get the results from those expressions to match my CAD model. The title pretty much sums it up. How do I solve for theta given the following equation. [imath]a = b\tan\theta - \frac{c}{\cos\theta}[/imath] I am not a student and this is not homework. It's been quite a while since I've done any significant trig and I'm out of time to figure this out. |
934107 | If a series diverges, dividing it by the sequence of partial sums preserves divergence
Let [imath]x_n [/imath] be positive, [imath]s_n = \sum_{i=1}^{n} x_i [/imath] and the series [imath]\sum x_n[/imath] diverges. I'm trying to see how we can prove the following two propositions withouth using anything related to Cauchy ( Cauchy criteria for series and notion of cauchy sequences ) or too advanced but i can't seem to suceed in finding a way. The propositions are : a) [imath]\sum_{i=1}^{k} \frac{x_{n+i}}{s_{n +i}} \geq 1 - \frac{s_n}{s_{n+k}}[/imath] b) The series [imath]\sum_{n=1}^{\infty} \frac{x_n}{s_n} [/imath] diverges Thanks in advance | 411817 | Rudin: Problem Chp3.11 and need advice.
I am working on the following problems and I have a couple of questions. Suppose [imath]a_n>0, s_n = \sum_{i = 1}^{n}[/imath] and [imath]\Sigma a_n[/imath] diverges. RTP (a) [imath]\Sigma \frac{a_n}{1+a_n}[/imath] diverges. (b) By showing [imath]\sum_{i=1}^{k} \left[ \frac{a_{N+i}}{s_{N+i}} \right] \ge 1 - \frac{s_N}{s_{N+k}}[/imath] prove that [imath]\sum \frac{a_n}{s_n}[/imath] diverges. (c) By showing [imath]\frac{a_n}{s_n^2} \le \frac{1}{s_{n-1}}-\frac{1}{s_n}[/imath] and deduce that [imath]\sum \frac{a_n}{s_n^2}[/imath] converges. (d) What can be said about [imath]\Sigma \frac{a_n}{n a_n +1} \text{ and } \Sigma \frac{a_n}{a+n^2a_n}[/imath] I understand (a). But I am generally very iffy with proving these kind of problems because I usually get stuck with showing some inequalities being true. 1), How are the rest of the problem solved ? 2), Is there a general way to solve most of these inequalities ? Especially with those that have summations ? 3), Are these problem considered easy ? I am teaching myself analysis and it's such a struggle for me. |
934227 | Calculate exact value of and infinite sum
Im trying to find the exact value of the infinite sum : 3 + 1/3 + 1/27 + 1/243 + 1/2187 + ... I can see that to generate new terms we take the previous term and divide by 9 or multiply by 9. Not exactly sure if one way or the other makes it any easier. This is what i have come up with to solve the series is this correct and where do i go from here? [imath]\sum_{k=3}^\infty= K_0 + K_n(1/9)\\ [/imath] | 934215 | Closed formula for the sum of the following series
How can I find the closed formula for the sum [imath]1/3 + 1/27 + 1/243 + ...?[/imath] |
934263 | Preimage of maximal ideal is maximal
I've just started a commutative algebra course and I'm stuck on the very first homework problem: Let [imath]A \not= \{0\}[/imath] be a commutative ring. Let [imath]\Phi : A \longrightarrow B[/imath] be a ring homomorphism where [imath]B[/imath] has finitely many elements. Show that if [imath]I[/imath] is a maximal ideal of [imath]B[/imath], then [imath]\Phi^{-1}(I)[/imath] is a maximal ideal of [imath]A[/imath]. It's easy enough to prove that if [imath]I[/imath] is an ideal then [imath]\Phi^{-1}(I)[/imath] is also an ideal, but I'm stuck on the maximality. My idea was to assume, for a contradiction, that there exists some proper ideal [imath]J \subset A[/imath] that strictly contains [imath]\Phi^{-1}(I)[/imath]; then its image would perhaps be a proper ideal in [imath]B[/imath] that strictly contains [imath]I[/imath], which would contradict the maximality of [imath]I[/imath]. However, this doesn't seem to work, since I can't prove that the image of [imath]J[/imath] is a proper ideal. It is not given that [imath]\Phi[/imath] is surjective. I'm also not sure how to exploit the fact that [imath]B[/imath] has finitely many elements. Any hints on this problem (no complete solutions please)? | 933204 | A question on ring homomorphisms and maximal ideals.
Let [imath]A,B[/imath] be commutative rings, and let [imath]\phi: A \to B[/imath] be a ring homomorphism where [imath]B[/imath] has finitely many elements. Prove that if [imath]I \subset B[/imath] is a maximal ideal then [imath]\phi^{-1}(I)[/imath] is also a maximal ideal in [imath]A[/imath] I've tried to construct a proof by contradiction but with no success, im not looking for answers but rather hints. |
681728 | Solving the recurrence [imath]T (n) = \sqrt{n} T(\sqrt{n}) + O (n)[/imath]
I want to show that the requrrence [imath]T (n) = \sqrt{n} T(\sqrt{n}) + O (n)[/imath] is in [imath]O(n \log \log n)[/imath] Here's my attempt: If we expand the recursion tree, at a level [imath]i[/imath], there are [imath]n^{1/2^k}[/imath] subproblems each requiring work equal to [imath]n^{1/2^k}[/imath]. There are [imath]\log \log n[/imath] levels in this tree. So the summation is: [imath]\sum_{i = 0}^{\log \log n} n^{1/2^{(k-1)}}[/imath] Is this summation correct, and how can I show it is in [imath]O(n \log \log n)[/imath]? | 239402 | Solve the recurrence relation:[imath] T(n) = \sqrt{n} T \left(\sqrt n \right) + n[/imath]
[imath]T(n) = \sqrt{n} T \left(\sqrt n \right) + n[/imath] Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable. The answer is [imath]Θ (n \log \log n)[/imath]. Can anyone arrive at the solution. |
932942 | Why are continuous functions the "right" morphisms between topological spaces?
Recently, someone mentioned to me that given a function [imath]f: X \to Y[/imath] there are two natural functions between the powersets [imath]P(X)[/imath] and [imath]P(Y)[/imath]. Namely [imath]f: U \subset X \mapsto f(U)[/imath] and [imath]f^{-1}: V \subset Y \mapsto f^{-1}(V)[/imath]. Then if we consider maps between [imath]P(P(X)), P(P(Y))[/imath], each of the above maps induce two more, so there are four natural maps. Thus, it seems on the face of it like there are four natural choices for morphisms of topological spaces (since a topology on [imath]X[/imath] is an element of [imath]P(P(X))[/imath]). Why is it that continuous functions are the morphisms we choose and not one of the other four maps? I understand that the theory we get from taking continuous functions as morphisms is incredibly rich and so this alone provides adequate justification. However, I am looking for a different sort of justification along the lines of "is there some property of continuous functions that immediately suggests they are the 'right' choice of morphisms between topological spaces?". | 1857830 | Motivation for the definition of continuous maps on topological spaces
In any category where the objects are sets equipped with certain relations and operations, the notion of "morphism" arises perfectly naturally. (Generally, a morphism between objects is one that commutes with the operations, and maps tuples in a relation to tuples in a relation.) On the other hand, the definition of a morphism from topological space [imath](X, \tau_X)[/imath] to [imath](Y, \tau_Y)[/imath] is of course a continuous map [imath]f[/imath], i.e. a map such that the inverse image of [imath]f[/imath], [imath] f^{-1} : \mathcal{P}(Y) \to \mathcal{P}(X) [/imath] preserves open sets, rather than [imath]f[/imath] itself preserving open sets. Therefore my (very basic!) question is: in what sense is taking the morphisms to be the continuous maps natural or canonical? I'm assuming that there is some a priori reason that we might expect the continuous maps to be a better definition of morphism than, say, functions such that the image of an open set isn't an open set. Perhaps an answer comes from pointless topology, or from a better understanding of the correspondence between a function [imath]f: X \to Y[/imath] and its inverse image [imath]f^{-1}: \mathcal{P}(Y) \to \mathcal{P}(X)[/imath]. |
924299 | Evaluating [imath] \lim_{x\to0} \frac{\tan(2x)}{\sin x}[/imath]
[imath]\lim_{x\to0} \frac{\tan(2x)}{\sin x}[/imath] How would I evaluate that? I was thinking changing the tan to sin/cos, but when I tried that, it did not work. | 467780 | Can someone explain this trigonometric limit?
I have [imath]\lim \limits_{x\to 0} \frac {\tan(2x)}{\sin(x)}[/imath] and in my case the result is [imath]\frac{2}{1}[/imath] =2 not whether it is right. This is my procedure. [imath]\lim \limits_{x\to 0} \frac{\frac {\sin(2x)}{\cos(2x)}}{\frac{\sin(x)}{1}}= \lim \limits_{x\to 0} {\dfrac {\sin(2x)}{(\cos(2x))(\sin(x))}}=\dfrac{2x\frac {\sin(2x)}{2x}}{\cos(2x)\frac{x\sin(x)}{x}}[/imath] I separate the limit. [imath]\frac{\left(\lim \limits_{x\to 0}2x\right) \cdot \left(\lim \limits_{x\to 0}\frac {\sin(2x)}{2x}\right)}{\lim \limits_{x\to 0}\left(\cos(2x)\right)\cdot\left(\lim \limits_{x\to 0}\frac{x\sin(x)}{x}\right)} = \lim \limits_{x\to 0} \dfrac{2x}{x}=\frac{2}{1} =2[/imath] |
934927 | What is the generalized definition of '<' and '>' for complex numbers?
I'd expect this question to be asked here before, but I've not been able to find it. The generalized definition of the multiplication operator for complex numbers is simple: The product of the lengths and the sum of the angles (relatively to the [imath]X[/imath]-axis). But what is the generalized definition of the [imath]<[/imath] and [imath]>[/imath] comparators for complex numbers? | 310931 | Comparing complex numbers
If [imath]a+ib[/imath], [imath]c+id[/imath], [imath]e+if[/imath] are three complex numbers, than can we tell which one is greater or smaller between them? If yes, then how and if no then why not? Can somebody give explanation on this.... I will be grateful to him. |
392871 | Is the Converse of the Cayley-Hamilton Theorem true or not?
I was thinking about taking the Cayley-Hamilton Theorem in the opposite direction. Does this hold : If for [imath]F\in[/imath] End[imath](V)[/imath] [imath]\exists P\in K[t][/imath] such that [imath]P(F)=0[/imath] then [imath]P[/imath] is the characteristic polynomial of [imath]F[/imath] ? If not could you give a counterexample? | 48123 | Is the converse of Cayley-Hamilton Theorem true?
The question is motivated from the following problem: Let [imath]I\neq A\neq -I[/imath], where [imath]I[/imath] is the identity matrix and [imath]A[/imath] is a real [imath]2\times 2[/imath] matrix. If [imath]A=A^{-1}[/imath], then the trace of [imath]A[/imath] is [imath] (A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2[/imath] Since [imath]A=A^{-1}[/imath], [imath]A^2=I[/imath]. If the converse of Cayley-Hamilton Theorem is true, then [imath]\lambda^2=1[/imath] and thus [imath]\lambda=\pm1[/imath]. And then [imath]\rm{trace}(A)=1+(-1)=0[/imath]. Here are my questions: Is [imath]C[/imath] the answer to the quoted problem? Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix [imath]A[/imath], if [imath]p(A)=0[/imath], then [imath]p(\lambda)[/imath] is the characteristic polynomial of the matrix [imath]A[/imath]" true? If it is not, then what's the right method to solve the problem above? |
935198 | Solve the following integration
[imath]\int \sqrt{cot\theta} d\theta[/imath] I tried to set [imath]t=\sqrt{cot\theta},t^2=cot\theta[/imath] and substitute into the original integration and get[imath]-\int\frac{2t^2}{1+t^4}dt[/imath], but then what can I do? | 425603 | Integral [imath]\int\!\sqrt{\cot x}\,dx [/imath]
Find the integral [imath]\int\!\sqrt{\cot x}\,dx [/imath] How can one solve this using substitution? Can this be solved by complex methods? |
935245 | Solving a question by mathematical induction
Question : Prove that [imath] \sum_{k=1}^n\frac{1}{\sqrt{k}}\le 2\sqrt{n}-1 [/imath] for all positive integers [imath]n[/imath]. I've been thinking a solution for this question for hours but still can't solve it. | 880818 | Prove that [imath]\sqrt{n} \le \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1[/imath] is true for [imath]n \in \mathbb{N}^{\ge 1}[/imath]
I'm trying to solve these induction exercises proposed by the department of mathematics of Oxford University. I don't know how to give a valid proof for the third one which says the following: Prove that for [imath]n \in \mathbb{N}^{\ge 1}[/imath]: [imath] \sqrt{n} \le \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1 [/imath] My first approach has been to convert the summation into a function in terms of [imath]n[/imath] so that I could find the maximum and minimum values in the interval. However, there doesn't seem to be an easy way to express the summation as a function. Thanks in advance! |
91457 | Combinatorial interpretation for the identity [imath]\sum\limits_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}[/imath]?
A known identity of binomial coefficients is that [imath] \sum_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}. [/imath] Is there a combinatorial proof/explanation of why it holds? Thanks. | 963164 | Verify the following combinatorial identity: [imath]\sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}[/imath]
[imath]\sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}[/imath] Nice, so I've proven some combinatorial identities before via induction, other more simple ones by committee selection models.... But this one is weird, induction doesn't even seem feasible here without things getting nasty, and the summation on the left is not making things easier. Can anyone help? |
935955 | Compute the flux of [imath](z \sin x, yz \cos x, x^2 + y^2)[/imath] through a paraboloid.
Given the vector field [imath]F(x, y, z) = \langle z \sin x, yz \cos x, x^2 + y^2 \rangle,[/imath] calculate the flux [imath]\int_S F \cdot \hat{n} \; dS[/imath] through the paraboloid [imath]S = \{(x,y,z) : z = -3(x^2 + y^2) + 3, x^2 + y^2 \leq 1\},[/imath] where [imath]\hat{n}[/imath] is the unit normal pointing in the positive [imath]z[/imath]-direction. My question: does this have a solution that can be done by hand? Stokes' theorem doesn't work since it's not divergence free. I tried to compute the integral explicitly but it seems that there is no clean way to do so by hand. This is an exam question, and I don't know if it's written correctly (I suspect a sign change is missing from either the first or second component). I am asking because if there is no reasonable solution, then I have good support that the problem was not stated correctly. I just wanted to make sure that this was the case before bringing it up. | 935756 | Compute the flux of [imath](z \sin x, yz \cos x, x^2 + y^2)[/imath] through the paraboloid.
Given the vector field [imath]F(x, y, z) = \langle z \sin x, yz \cos x, x^2 + y^2 \rangle,[/imath] calculate the flux [imath]\int_S F \cdot \hat{n} \; dS[/imath] through the paraboloid [imath]S = \{(x,y,z) : z = -3(x^2 + y^2) + 3, x^2 + y^2 \leq 1\},[/imath] where [imath]\hat{n}[/imath] is the unit normal pointing in the positive [imath]z[/imath]-direction. Question: What does the integral evaluate to? Can this be computed by hand? This is an exam question, and I don't know if it's written correctly. Stokes' theorem doesn't work since it's not divergence free. I tried to compute the integral explicitly by both the divergence theorem and explicitly taking the dot product with the normal, but it seems that there is no way to evaluate the integral by hand. If one of the [imath]x[/imath] or [imath]y[/imath] components has a sign change, then the flux integral becomes easy to evaluate, so I don't know if that's what they meant. I have no way to contact the exam writer directly. I am asking this question to confirm that the problem may be incorrect before bringing it up. |
57686 | Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces
The following is a well-known result in functional analysis: If the vector space [imath]X[/imath] is finite dimensional, all norms are equivalent. Here is the standard proof in one textbook. First, pick a norm for [imath]X[/imath], say [imath]\|x\|_1=\sum_{i=1}^n|\alpha_i|[/imath] where [imath]x=\sum_{i=1}^n\alpha_ix_i[/imath], and [imath](x_i)_{i=1}^n[/imath] is a basis for [imath]X[/imath]. Then show that every norm for [imath]X[/imath] is equivalent to [imath]\|\cdot\|_1[/imath], i.e., [imath]c\|x\|\leq\|x\|_1\leq C\|x\|.[/imath] For the first inequality, one can easily get [imath]c[/imath] by triangle inequality for the norm. For the second inequality, instead of constructing [imath]C[/imath], the Bolzano-Weierstrass theorem is applied to construct a contradiction. The strategies for proving these two inequalities are so different. Here is my question, Can one prove this theorem without Bolzano-Weierstrass theorem? UPDATE: Is the converse of the theorem true? In other words, if all norms for a vector space [imath]X[/imath] are equivalent, then can one conclude that [imath]X[/imath] is of finite dimension? | 2854517 | All norms on [imath]\mathbb{R}[/imath] are equivalent.
Let [imath]X[/imath] be a vector space. Two norms [imath]\|\cdot\|,\|\cdot\|':X\to\mathbb{R}[/imath] are equivalent, if there are constants [imath]\alpha,\beta >0[/imath], such that for all [imath]x\in X[/imath] holds: [imath]\alpha\|x\|\leq \|x\|'\leq \beta\|x\|[/imath] Show that all norms on [imath]\mathbb{R}[/imath] are equivalent. A detailed hint is given: It is enough to show, that an arbitrary norm [imath]\|\cdot\|'[/imath] is equivalent to the Euclidean norm [imath]\|\cdot\|[/imath]. [Why is that the case?] Consider the set [imath]M:=\{x\in\mathbb{R}:\|x\|'\leq 1\}[/imath] and show, that it exists an [imath]r\in (0,\infty)[/imath], such that [imath]M=[-r,r][/imath]. It is possible to show, that [imath]r\|x\|'=\|x\|[/imath] for all [imath]x\in\mathbb{R}[/imath]. My first question is, why it is enough to show, that an arbitrary norm is equivalent to the Euclidean norm? Then I want to follow the hint and find [imath]r[/imath] with [imath]M=[-r,r][/imath], but I am stuck and do not know how to start here... I would appreciate a hint to get me started. Thanks in advance. |
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