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2544721 | If [imath]G[/imath] has no nontrivial subgroups, show that [imath]G[/imath] must be finite of prime order
We know the following fact from gorup theory: If [imath]G[/imath] is a group of prime order then it has no nontrivial subgroups. Lets try to prove the converse statement: If [imath]G[/imath] has no nontrivial subgroups, show that [imath]G[/imath] must be finite of prime order. Proof: Suppose by contradiction: If [imath]G[/imath] has no nontrivial subgroups [imath]\Rightarrow[/imath] [imath]G[/imath] is infiinite or [imath]|G|\neq p[/imath]. The case when [imath]G[/imath] is infinite we can rule out using that topic. Suppose that [imath]|G|=n[/imath] where [imath]n[/imath] is composite [imath]\Rightarrow[/imath] [imath]n=pm[/imath] where [imath]p[/imath] - prime and [imath]m\geqslant 2[/imath]. Let [imath]a\in G[/imath] such [imath]a\neq e[/imath] then [imath]a^{pm}=e[/imath]. If [imath]a^p\neq e[/imath] then considering the cyclic group of G, namely [imath]H=\left \langle a^p\right \rangle[/imath] [imath]\Rightarrow[/imath] [imath]1<|H|\leqslant m<pm[/imath] this is a contradiciton. If [imath]a^p=e[/imath] then we know that [imath]a\neq e[/imath] and considering the cyclic subgroup of [imath]G[/imath], namely [imath]H=\left \langle a\right \rangle[/imath] [imath]\Rightarrow[/imath] [imath]1<|H|\leqslant p<pm[/imath] So we get two contradiction and it follows that [imath]G[/imath] is finite and its order is prime. P.S. I think that my post is not duplicate because my solution somewhat is different from given duplicates. And this solution was created by me and it was important for me to understand is it correct or not. | 226365 | Group with more than one element and with no proper, nontrivial sub groups must have prime order.
I want to show that if [imath]G[/imath] is a group with more than one element, and that [imath]G[/imath] has no proper non-trivial subgroups. Prove that [imath]|G|[/imath] is prime. (Do not assume at the outset that |G| is finite). My question is not that how to prove it. I am saying that suppose [imath]|G|\geq 2[/imath] possibly [imath]|G|=\infty.[/imath] By assumption the only subgroups of [imath]G[/imath] are [imath]\{e\}[/imath] and [imath]G[/imath], i.e., the trivial groups. Let [imath]a[/imath] be non-identity element in [imath]G[/imath]. Consider [imath]\langle a\rangle[/imath]. Then [imath]\langle a\rangle=G.[/imath] So [imath]G[/imath] is cyclic. My question is, why can I say that [imath]G=\langle a\rangle[/imath]. I know there are only two subgroups and [imath]\langle a\rangle\neq e[/imath] because [imath]a\neq e[/imath]. Therefore we must have [imath]G=\langle a\rangle[/imath]. But my problem is why cant I say that consider [imath]a,b\in G[/imath] and then we look at [imath]\langle a,b\rangle[/imath]. And then I say [imath]G=\langle a,b\rangle[/imath] and then I cannot say that [imath]G[/imath] is cyclic, and then I will have problem proving question. |
2545062 | worksheet problem...Prove the set of all subsets of N is countably infinite
Prove that the set of all finite subsets of [imath]\mathbb{N}[/imath] is countably infinite. I understand what countably infinite means but I don't know how to arrange the subsets according to their sum and I can't figure out how to form the proof Please help thanks | 2537255 | Proof of a countable set
I want to prove that [imath] E = \{ A \subset \mathbb{N} | \hspace{0.1cm} |A| < ∞ \} [/imath] is countable. Additionally I got the hint to first prove, that [imath] E_{k} := \{ A \subset \mathbb{N} | \hspace{0.1cm} |A| < k \} [/imath] is countable; to show that I should consider [imath] \mathbb{N}^{k} = \prod_{i=1}^{k}\mathbb{N} [/imath] So far, I know that [imath] \prod_{i=1}^{k}A_{i} [/imath] is countable by definition, where [imath](A_{1},..., A_{n})[/imath] is a finite family of countable sets. Furthermore, I know, per definition, that [imath]\mathbb{N}[/imath] is also a countable set. Therefore, I assume that [imath] \prod_{i=1}^{k}\mathbb{N} = \mathbb{N}^{k} [/imath] is countable as well. Unfortunately I don't know how to continue at this point as I do not have a clear understanding how the last hint can be used to show that E is actually countable. Any help is greatly appreciated! |
2544214 | Logarithm of a Dirichlet series
I am currently struggling to understand Section 11.9 in Tom Apostol's Introduction to Analytic Number Theory, where he talks about taking the logarithm of a Dirichlet series. The idea is that we have some Dirichlet series [imath]F(s) = \sum_{n=1}^\infty f_nn^{-s}[/imath] with [imath]f(1)\neq 0[/imath] and [imath]F(s)[/imath] is absolutely convergent [imath]F(s) \neq 0[/imath] for any [imath]s[/imath] with [imath]\mathfrak{R}(s)>\sigma_0[/imath]. Apostol says that we can then always find some Dirichlet series [imath]G(s)[/imath] that is also absolutely convergent for [imath]\mathfrak{R}(s)>\sigma_0[/imath] and fulfils [imath]e^{G(s)} = F(s)[/imath] for any such [imath]s[/imath], and that [imath]G(s) = \ln f(1) + \sum_{n=2}^\infty \frac{(f' * f^{-1})(n)}{\ln n} n^{-s}[/imath] where [imath]f^{-1}[/imath] is the Dirichlet inverse of [imath]f[/imath] and [imath]f'(n) = f(n) \ln n[/imath], i.e. they correspond to the multiplicative inverse and the negative derivative of [imath]F[/imath]. However, if I am not missing anything here, I think all of this only works if [imath]\sum_{n=1}^\infty f'(n) n^{-s}[/imath] and [imath]\sum_{n=1}^\infty f^{-1}(n) n^{-s}[/imath] converge absolutely, but I don't see why they should, in general. | 2369463 | Theorem 11.14, Apostol, pg 238 - need explanation
An excerpt from Introduction to Analytic Number Theory by Tom M. Apostol. I have three main concerns regarding this proof: What is the abscissa of convergence for [imath]\frac{1}{F(s)}[/imath] - why can we take the product of the two series. I only know this is true if the two series are absolutely convergent. Why can we do term by term integration? I believe its suffices to show that the series converges locally uniformly - BUT is this true? How does the proof show that [imath]G(s)[/imath] converges uniformly? |
1860405 | Find the coordinates of P and Q.
A line is drawn through the point [imath]A(1,2)[/imath] to cut the line [imath]2y=3x-5[/imath] in [imath]P[/imath] and the line [imath]x+y=12[/imath] in [imath]Q[/imath]. If [imath]AQ=2AP[/imath], find the coordinates of [imath]P[/imath] and [imath]Q[/imath]. From: Mathematics, The Core Course for A-level, Bostock and Chandler. Chapter 4 Q15. The answer is [imath](4,3.5)(7,5)[/imath]or[imath](0.4,-1.9)(2.2,9.8)[/imath] I have no idea where to begin with this question. I've tried letting [imath]P=(x,12-x)[/imath] and [imath]Q=(x,\frac{3}{2}x-\frac{5}{2})[/imath] and then used the distance formula to try and equate [imath]|AQ|=2|AP|[/imath] but didn't get anywhere. How should I approach this problem? | 1613664 | A line is drawn through the point [imath]A(1,2)[/imath] to cut the line [imath]2y=3x-5[/imath] in [imath]P[/imath] and the line [imath]x+y=12[/imath] in [imath]Q[/imath]. If [imath]AQ=2AP[/imath], find [imath]P[/imath] and [imath]Q[/imath].
A line is drawn through the point [imath]A(1,2)[/imath] to cut the line [imath]2y=3x-5[/imath] in [imath]P[/imath] and the line [imath]x+y=12[/imath] in [imath]Q[/imath]. If [imath]AQ=2AP[/imath], find the coordinates of [imath]P[/imath] and [imath]Q[/imath]. I found the lengths of the lines [imath]AQ[/imath] and [imath]AP[/imath] in terms of [imath]x[/imath] and [imath]y[/imath] and used [imath]AQ=2AP[/imath] to get this equation [imath]11x^2-40x-16=0[/imath] Working: [imath]AP=\sqrt{(x-1)^2+(\frac{3}{2}x-\frac{9}{2})^2}=\sqrt{\frac{1}{4}(13x^2-62x+85)}[/imath] [imath]AQ=\sqrt{(x-1)^2+(-x+10)^2}=\sqrt{2x^2-22x+101}[/imath] [imath]AQ=2AP\Rightarrow \sqrt{2x^2-22x+101}=2\sqrt{\frac{1}{4}(13x^2-62x+85)}[/imath] [imath]\Rightarrow 2x^2-22x+101=13x^2-62x+85\Rightarrow 11x^2-40x-16=0[/imath] Using the quadratic formula I found [imath]x=4,x=-\frac{4}{11}[/imath] I then substituted these values into the two equations in the question to get the coordinates [imath](4,\frac{7}{2}),(-\frac{4}{11},-\frac{15}{11})[/imath] or [imath](4,8),(-\frac{4}{11},\frac{136}{11})[/imath] But the answers in the book are [imath](4,\frac{7}{2}),(7,5)[/imath] or [imath](\frac{2}{5},-\frac{19}{10}),(\frac{11}{5},\frac{49}{5})[/imath] I don't know where I went wrong in my method. Edit: Found some errors in my method and calculation. Using [imath]x=4[/imath] in [imath]2y=3x-5[/imath] gives [imath]P(4,\frac{7}{2})[/imath] Line with points [imath]AP[/imath] has equation [imath]y-2=\frac{\frac{7}{2}-2}{3}(x-1)\Rightarrow y=\frac{1}{2}x+\frac{3}{2}[/imath] [imath]Q[/imath] lies on the same line. Therefore, the intersection between [imath]y=\frac{1}{2}x+\frac{3}{2}[/imath] and [imath]y+x=12[/imath] is the point [imath]Q[/imath]. Working: [imath]\frac{1}{2}x+\frac{3}{2}=-x+12\Rightarrow x=7[/imath] Therefore, [imath]y=-7+12=5[/imath]. Point Q has coordinates [imath](7,5)[/imath]. But [imath]x=-\frac{4}{11}[/imath] is not the other [imath]x[/imath] coordinate of [imath]P[/imath] according to the answers in the book. I have no idea where I went wrong at this point. |
2545710 | Prob. 6, Sec. 24, in Munkres' TOPOLOGY, 2nd ed: For a well-ordered set [imath]X[/imath], [imath]X\times[0,1)[/imath] in the dictionary order is a linear continuum
Here is Prob. 6, Sec. 24, in the book Topology by James R. Munkres, 2nd edition: Show that if [imath]X[/imath] is a well-ordered set, then [imath]X \times [0, 1)[/imath] in the dictionary order is a linear continuum. Here is the definition of well-ordered set, from Sec. 10 in Munkres: A set [imath]A[/imath] with an order relation [imath]<[/imath] is said to be well-ordered if every non-empty subset of [imath]A[/imath] has a smallest element. And, here is the definition of linear continuum: A simply ordered set [imath]L[/imath] having more than one element is called a linear continuum if the following hold: (1) [imath]L[/imath] has the least upper bound property. (2) If [imath]x < y[/imath], there exists [imath]z[/imath] such that [imath]x < z < y[/imath]. My Attempt: Of course, [imath]X \times [0, 1)[/imath] has more than one (in fact uncountably many) elements, even if [imath]X[/imath] is a singleton set. Suppose [imath]x_1 \times r_1[/imath] and [imath]x_2 \times r_2[/imath] are any two elements of [imath]X \times [0, 1)[/imath] such that [imath]x_1 \times r_1 \prec x_2 \times r_2[/imath]. Then either [imath]x_1 \prec_X x_2[/imath], or [imath]x_1 = x_2[/imath] and [imath]r_1 < r_2[/imath]. If [imath]x_1 \prec_X x_2[/imath], then the ordered pair [imath]x_1 \times \frac{r_1+1}{2}[/imath], for instance, is in [imath]X \times [0, 1)[/imath], and also [imath] x_1 \times r_1 \prec x_1 \times \frac{r_1+1}{2} \prec x_2 \times r_2. [/imath] Let us put [imath]x \times r \colon= x_1 \times \frac{r_1+1}{2}.[/imath] On the other hand, if [imath]x_1 = x_2[/imath] and [imath]r_1 < r_2[/imath], then the ordered pair [imath]x_1 \times \frac{r_1+r_2}{2}[/imath], for instance, is in [imath]X \times [0, 1)[/imath], and also [imath] x_1 \times r_1 \prec x_1 \times \frac{ r_1 + r_2 }{2} \prec x_2 \times r_2. [/imath] Let us put [imath]x \times r \colon= x_1 \times \frac{r_1+r_2}{2}.[/imath] In either case, we can find an element [imath]x \times r \in X \times [0, 1)[/imath] such that [imath] x_1 \times r_1 \prec x \times r \prec x_2 \times r_2, [/imath] whenever [imath]x_1 \times r_1[/imath] and [imath]x_2 \times r_2[/imath] are any elements of [imath]X \times [0, 1)[/imath] such that [imath]x_1 \times r_1 \prec x_2 \times r_2[/imath]. Now let [imath]A[/imath] be a non-empty subset of [imath]X \times [0, 1)[/imath] such that [imath]A[/imath] is bounded from above in [imath]X \times [0, 1)[/imath]. Then there is an ordered pair [imath]x \times r[/imath] in [imath]X \times [0, 1)[/imath] such that, for every ordered pair [imath]y \times s \in A[/imath], we have [imath] y \times s \preceq x \times r, [/imath] that is, either [imath]y \prec_X x[/imath], or [imath]y = x[/imath] and [imath]s \leq r[/imath]. Let [imath]\pi_1 \colon X \times [0, 1) \to X[/imath], [imath]z \times t \mapsto z[/imath], be the projection map of [imath]X \times [0, 1)[/imath] onto [imath]X[/imath]. Then the set [imath] \pi_1 (A) \colon= \left\{ \ \pi_1 (y \times s) \ \colon \ y \times s \in A \ \right\} = \left\{ \ y \in X \ \colon \ y \times s \in A \mbox{ for some } s \in [0, 1) \ \right\} [/imath] is bounded from above in [imath]X[/imath], the element [imath]x[/imath] being an upper bound of this set. Thus the set of all the upper bounds in [imath]X[/imath] of the set [imath]\pi_1(A)[/imath] is a non-empty subset of the well-ordered set [imath]X[/imath] and so has a smallest element [imath]x_0[/imath], say. If [imath]x_0 \not\in \pi_1(A)[/imath], then, for every element [imath]y \times s \in A[/imath], we have [imath] y \prec_X x_0[/imath]. So [imath]x_0 \times 0[/imath] is an upper bound of [imath]A[/imath] in [imath]X[/imath]. Moreover, if [imath]x \times r[/imath] is any upper bound of [imath]A[/imath] in [imath]X[/imath], then, as we have seen above, [imath]x[/imath] would then be an upper bound of the set [imath]\pi_1(A)[/imath], and so [imath]x_0 \preceq_X x[/imath] must hold (because [imath]x_0[/imath] is the least upper bound of [imath]\pi_1(A)[/imath]). Therefore, [imath]x_0 \times 0 \preceq x \times r[/imath], thus showing that [imath]x_0 \times 0[/imath] is the least upper bound of [imath]A[/imath] in [imath]X \times [0, 1)[/imath]. On the other hand, if [imath]x_0 \in \pi_1(A)[/imath], then the set [imath] \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A = \left\{ \ x_0 \times s \in A \ \colon \ s \in [0, 1) \ \right\} [/imath] is a non-empty subset of [imath]A[/imath], and this subset has the order type of [imath][0, 1)[/imath] (i.e. there is a bijective, order-preserving mapping of this set with [imath][0, 1)[/imath]). Now let [imath]\pi_2 \colon X \times [0, 1) \to [0, 1)[/imath], [imath]z \times t \mapsto t[/imath], be the projection of [imath]X \times [0, 1)[/imath] onto [imath][0, 1)[/imath]. Then the set [imath] \pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right) = \left\{ \ \pi_2 \left( x_0 \times s \right) \ \colon \ x_0 \times s \in A \ \right\} = \left\{ \ s \in [0, 1) \ \colon \ x_0 \times s \in A \ \right\} [/imath] is a non-empty subset of [imath][0, 1)[/imath]. If the set [imath]\pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right)[/imath] is bounded above by some element [imath]r[/imath] in [imath][0, 1)[/imath], then this set is also bounded above in [imath]\mathbb{R}[/imath] and so has a least upper bound [imath]r_0[/imath], say. Then [imath]r_0 \leq r < 1[/imath]. But since [imath]\pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right) [/imath] is a non-empty subset of [imath][0, 1)[/imath] which is bounded above by [imath]r_0[/imath], we can also conclude that [imath]0 \leq r_0[/imath]. Therefore [imath]r_0 \in [0, 1)[/imath]. Then [imath]x_0 \times r_0[/imath] is the least upper bound of [imath]A[/imath] in [imath]X \times [0, 1)[/imath]. So let us assume that the set [imath]\pi_2 \left( \left( \left\{ x_0 \right\} \times [0, 1) \right) \cap A \right) [/imath] is not bounded above in [imath][0, 1)[/imath]. Now if [imath]x_0[/imath] were the largest element of [imath]X[/imath], then set [imath]A[/imath] would not be bounded from above in [imath]X \times [0, 1)[/imath]. But since [imath]A[/imath] by our supposition is bounded from above in [imath]X \times [0, 1)[/imath], we can conclude that [imath]x_0[/imath] cannot be the largest element of [imath]X[/imath] and thus the subset [imath] \left\{ \ x \in X \ \colon \ x_0 \prec_X x \ \right\}[/imath] of the well-ordered set [imath]X[/imath] must be non-empty; let [imath]x_1[/imath] be the smallest element of this set. Then the subset [imath] \left( x_0, x_1 \right)_X = \left\{ \ x \in X \colon \ x_0 \prec_X x \prec_X x_1 \ \right\}[/imath] of [imath]X[/imath] is empty. Then [imath]x_1 \times 0[/imath] is the least upper bound of [imath]A[/imath] in [imath]X \times [0, 1)[/imath]. Is each and every step in the logic of my proof correct? If so, then is my presentation clear and accessible enough? If not, then where am I lacking? | 2527288 | Example 2, Sec. 24, in Munkres' TOPOLOGY, 2nd ed: If [imath]X[/imath] is a well ordered set, then [imath]X \times [0, 1)[/imath] is a linear continuum
Let [imath]X[/imath] be a well ordered set. Then how to show that the set [imath]X \times [0, 1)[/imath] is a linear continuum in the dictionary order? Here [imath] [0, 1) \colon= \{ \ x \in \mathbb{R} \ \colon \ 0 \leq x < 1 \ \}.[/imath] Definition of Linear Continuum: A simply ordered set [imath]L[/imath] having more than one element is called a linear continuum if the following hold: (1) [imath]L[/imath] has the least upper bound property. (2) If [imath]x < y[/imath], there exists [imath]z[/imath] such that [imath]x < z < y[/imath]. My Attempt: In what follows, [imath]x \times r[/imath] would denote an order pair, and [imath]r < s[/imath] would denote the usual less than relation between real numbers [imath]r[/imath] and [imath]s[/imath]. Suppose [imath]x_1 \times r_1[/imath] and [imath]x_2 \times r_2[/imath] are two arbitrary elements of [imath]X \times [0, 1)[/imath] such that [imath]x_1 \times r_1 \prec x_2 \times r_2[/imath]. This means that either [imath]x_1 \prec_X x_2[/imath], or [imath]x_1 = x_2[/imath] and [imath]r_1 < r_2[/imath]. Case 1. Suppose [imath]x_1 \prec_X x_2[/imath]. Then [imath]x_1 \times \frac{r_1+1}{2}[/imath] is an element of [imath]X \times [0, 1)[/imath] such that [imath] x_1 \times r_1 \prec x_1 \times \frac{r_1+1}{2} \prec x_2 \times r_2.[/imath] Case 2. Suppose [imath]x_1 = x_2[/imath] and [imath]r_1 < r_2[/imath]. Then [imath]x_1 \times \frac{r_1+r_2}{2}[/imath] is an element of [imath]X \times [0, 1)[/imath] such that [imath] x_1 \times r_1 \prec x_1 \times \frac{r_1+r_2}{2} \prec x_2\times r_2.[/imath] Let [imath]\pi_1 \colon X \times [0, 1) \to X[/imath] and [imath]\pi_2 \colon X \times [0, 1) \to [0, 1)[/imath] be the maps [imath]x \times r \mapsto x[/imath] and [imath]x \times r \mapsto r[/imath], respectively. Now let [imath]S[/imath] be a non-empty subset of [imath]X \times [0, 1)[/imath] such that [imath]S[/imath] is bounded from above in [imath]X \times [0, 1)[/imath]. Let [imath]a \times u[/imath] be an upper bound in [imath]X \times [0, 1)[/imath] of set [imath]S[/imath]. Then, for every element [imath]x \times r \in S[/imath], we have [imath] x \times r \preceq a \times u, [/imath] which is tantamount to saying that, either [imath]x \prec_X a[/imath], or [imath]x = a[/imath] and [imath]r \leq u[/imath]. In either case, we have [imath]x \prec_X a[/imath] or [imath]x = a[/imath] for every element [imath]x \in \pi_1(S)[/imath]. Thus the set [imath]\pi_1(S)[/imath] is bounded from above in [imath]X[/imath] (by the element [imath]a \in X[/imath], for example). Let [imath]a_0[/imath] be the supremum in [imath]X[/imath] of set [imath]\pi_1(S)[/imath]. [The existence of this supremum follows from the fact that the set of all the upper bound in [imath]X[/imath] of the set [imath]\pi_1(S)[/imath] is non-empty and thus has a smallest element, because [imath]X[/imath] is well ordered.] Case 1. If [imath]a_0 \not\in \pi_1(S)[/imath], then, for every element [imath]x \times r \in S[/imath], we have [imath]x \prec_X a_0[/imath] and hence [imath]x \times r \prec a_0 \times 0[/imath]. Moreover, if there were some element [imath]a \times u \in X \times [0, 1)[/imath] such that [imath]a \times u \prec a_0 \times 0[/imath], then [imath]a \prec_X a_0[/imath] and so [imath]a[/imath] could not be an upper bound for [imath]\pi_1(S)[/imath], which would imply the existence of an element [imath]x \times r \in S[/imath] for which [imath]a \prec_X x = \pi_1(x \times r) [/imath]. Thus the element [imath]a_0 \times 0 \in X \times [0, 1)[/imath] is the supremum of set [imath]S[/imath]. Case 2. If [imath]a_0 \in \pi_1(S)[/imath], then there is some [imath]r \in [0, 1)[/imath] for which [imath]a_0 \times r \in S[/imath]. Thus the set [imath]S \cap \left( a_0 \times [0, 1) \right)[/imath] is non-empty. Is what I have done so far correct? If so, then what next? If not, then where have I gone wrong? PS: The set [imath]a_0 \times [0, 1)[/imath] has the order type of [imath][0, 1)[/imath] and so has the least upper bound property. If [imath]S \cap \left( a_0 \times [0, 1) \right)[/imath] has an upper bound in [imath]a_0 \times [0, 1)[/imath], then the set [imath]\pi_2 \left( \ S \cap \left( a_0 \times [0, 1) \right) \ \right)[/imath] is bounded above in [imath][0, 1)[/imath]; let [imath]r_0[/imath] be the supremum of this set. Then [imath]a_0 \times r_0[/imath] is the supremum of [imath]S[/imath] in [imath]X \times [0, 1)[/imath]. So we suppose that [imath]S \cap \left( a_0 \times [0, 1) \right)[/imath] has no upper bound in [imath]a_0 \times [0, 1)[/imath]. But as the set [imath]S[/imath] is bounded above in [imath]X \times [0, 1)[/imath], so [imath]a_0[/imath] cannot be the largest element of [imath]X[/imath], because otherwise the set [imath]S \cap \left( a_0 \times [0, 1) \right)[/imath] and hence the set [imath]S[/imath] would be unbounded in [imath]X \times [0, 1)[/imath]. And, as [imath]a_0[/imath] is not the largest element of [imath]X[/imath], so the set [imath] \{ \ x \in X \ \colon \ a_0 \prec_X x \ \}[/imath] is a non-empty subset of [imath]X[/imath] and so has a smallest element, say [imath]a_1[/imath]. Then the open interval [imath]\left( a_0, a_1 \right) \colon= \left\{ \ x \in X \ \colon \ a_0 \prec_X x \prec_X a_1 \ \right\}[/imath] is an empty subset of [imath]X[/imath]. Therefore, the element [imath]a_1 \times 0[/imath] is the supremum of [imath]S[/imath]. Is my logic correct? If so, then does this PS part complete the proof satisfactorily enough? |
2545415 | [imath]\lim_{N\to\infty}1/N\sum_{n=1}^Nf(n\alpha)=\int_0^1f(t)dt[/imath]
Suppose [imath]f[/imath] is a continuous function on [imath]R^1[/imath] with period 1. Prove that [imath]\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^Nf(n\alpha)=\int_0^1f(t)dt[/imath] for every irrational number [imath]\alpha[/imath]. Hint: Do it first for [imath]f(t)=e^{2πikt}[/imath] How are these two thing connected? Intuitively the left hand side is acting like a Riemann sum but it isn't a Riemann sum. [imath]n\alpha[/imath] is fairly ambiguous and hard to understand. In the long run I understand that [imath](n\alpha)[/imath] (the decimal value of [imath]n\alpha[/imath]) populates the entire interval [imath](0,1[/imath] fairly uniformly. I don't understand how that may help. I don't know any assistance | 2033021 | proof of [imath]\lim_\limits{m\to\infty}\frac1m\sum_\limits{n=1}^m f(an)=\int_\limits{0}^1fdx[/imath]
I have taken a compact seminar about Fourier analysis and someone mentioned the following: Let [imath]f\in C(\mathbb R)[/imath] with period 1. Let [imath]a[/imath] be an arbitrary irrational number. Then [imath] \lim_{m\to\infty}\frac1m\sum_{n=1}^m f(an)=\int_\limits{0}^1fdx [/imath] Since it wasn't mentioned any proof, I've tried to verify the equation by showing it first for [imath]f(x)=\exp(2\pi ikx)[/imath] but didn't get far. Anybody could help using Fourier? |
2545829 | On direct product of groups
By definition, A group [imath]G[/imath] is a direct product of the subgroups [imath]N_1, N_2, ... , N_k[/imath] if an only if [imath]i.[/imath] [imath]G=N_1N_2...N_k[/imath] [imath]ii.[/imath] For each [imath]j[/imath], [imath]N_j \cap(N_1N_2...N_{j-1}N_{j+1}...N_k)=e[/imath], and [imath]iii.[/imath] Each of [imath]N_1, N_2, ..., N_k [/imath] is normal in [imath]G[/imath]. My question is... Is the second condition the same as saying [imath](a)[/imath] [imath]H_i \cap H_j =e[/imath] for [imath]i \neq j[/imath] ? I am proving something. Now, I'm having difficulty in showing [imath](ii.)[/imath]. However, it will be easier if [imath](ii.)[/imath] is the same as [imath](a)[/imath]. Thank you. | 2368718 | Example of a group [imath]G[/imath] such that [imath]G = N_1\cdots N_n[/imath] and [imath]N_i \cap N_j = \{e\}[/imath] for all [imath]i \neq j[/imath] but [imath]G[/imath] is not the internal direct product of them.
I was trying to find an example of a group [imath]G[/imath] and normal subgroups [imath]N_1 , \ldots , N_n[/imath] such that [imath]G = N_1\cdots N_n[/imath] and [imath]N_i \cap N_j = \{e\}[/imath] for all [imath]i \neq j[/imath] and yet [imath]G[/imath] is not the internal direct product of [imath]N_1, \ldots , N_n[/imath] . It is easy to show that G is the internal direct product of normal subgroups [imath]N_1 , \ldots , N_n[/imath] iff (i) [imath]G = N_1\cdots N_n[/imath] (ii) [imath]N_i\cap (N_1 \cdots N_{i-1}\cdots N_{i+1} \cdots N_n) = \{e\}[/imath] for [imath]i = 1, 2, \ldots, n[/imath]. So I guess the question boils down to produce an example of a group [imath]G[/imath] and its normal subgroups [imath]N_1 , \ldots , N_n[/imath] such that (1) [imath]G = N_1\cdots N_n[/imath] (2) [imath]N_i \cap N_j = \{e\}[/imath] for all [imath]i \neq j[/imath] but (3) [imath]N_i \cap (N_1 \cdots N_{i-1} N_{i+1} \cdots N_n) \neq \{e\}[/imath] for at least one [imath]1 \le i \le n.[/imath] Thanks in advance for help.... |
2545836 | Every Infinite-dimensional normed vector space containes a non continous linear functional
Given an infinite-dimensional normed vector space M, there exists a linear functional [imath] f: M \rightarrow K [/imath], where [imath]K[/imath] is the filed of scalars. I am trying to prove it following the strategy of dealing with finite dimensional basis. Can somebody say how one should procede ? Thanks. | 99206 | Discontinuous linear functional
I'm trying to find a discontinuous linear functional into [imath]\mathbb{R}[/imath] as a prep question for a test. I know that I need an infinite-dimensional Vector Space. Since [imath]\ell_2[/imath] is infinite-dimensional, there must exist a linear functional from [imath]\ell_2[/imath] into [imath]\mathbb{R}[/imath]. However, I'm having trouble actually coming up with it. I believe I'm supposed to find an unbounded function (although I'm not sure why an unbounded function is necessarily not continuous; some light in that regard would be appreciated too), so I thought of using the vectors [imath]e^i[/imath], which have all entries equal to zero, except for the [imath]i[/imath]-th one. Then, you can define [imath]f(e^i)=i[/imath]. That'd be unbounded, but I'm not sure if it'd be linear, and even if it is, I'm not sure how to define it for all the other vectors in [imath]\ell_2[/imath]. A friend mentioned that at some point the question of whether the set [imath]E=\{e^i:i\in\mathbb{Z}^+\}[/imath] is a basis would come up, but I'm not sure what a basis has to do with continuity of [imath]f[/imath]. I'm just learning this topic for the first time, so bear with me please. The space of sequences that are eventually zero (suggested by a few people) turned out to be exactly what I needed. It also helped to cement the notions of Hamel basis, not continuous, etc. |
2544838 | Existence of a linear continuous function
I found the following problem: Let [imath](H,(\cdot,\cdot))[/imath] be a Hilbert space and B a continuous Bilinear form on H. Then there is a [imath]A\in L(H)[/imath], with [imath]L(H)[/imath] being the linear, continuous functions [imath]H\rightarrow H[/imath], with [imath]B(x,y)=(Ax,y) \forall x,y\in H[/imath] I tried to use the Lax-Milgram theorem, which can be used on continuous koerziv sesquilinear forms. (A is a koerziv Sesequilinear form if there is an [imath]a>0[/imath] so that for all [imath]u\in H[/imath]: [imath]Re(A(u,u))\ge a||u||^2[/imath]). Let [imath]a:H\times H\rightarrow\mathbb{K}[/imath] be a continuous, koerziv sesquilinear form on a [imath]\mathbb{K}[/imath]-Hilbert space. Then there is exactly one bijective function [imath]A\in L(H)[/imath], so that for all [imath]u,v\in H[/imath] is true: [imath]a(v,u)=(v,Au)[/imath] In the real numbers every sesquilinear form is a bilinear form which is koerziv. And since the dot product is symmetric, this would solve my problem for the real numbers. Do I have to specify it also for the complex numbers? I don't know how I could do this | 1175721 | Question about bilinear form on Hilbert space
I am trying to verify the following Let [imath]H[/imath] be a Hilbert space, and let [imath]a(\cdot,\cdot)[/imath] be a real continuous bilinear form on [imath]H[/imath] Then, define the operator [imath]A:H-> H'[/imath] as [imath]Au(v) :=a(u,v), v\in H[/imath] Also, define the operator [imath]B:H->H'[/imath] as [imath]Bu(v) :=a(v,u), v\in H[/imath]. Then, I want to show that A is surjective if and only if B is injective. Here is my attempt: Assume A is surjective. Let [imath]w,z \in H[/imath] such that [imath]Bw(v) = Bz(v), v\in H[/imath]. That is, [imath]a(\cdot,w) = a(\cdot,z)[/imath]. By using surjectivity of A, then I can find [imath]s[/imath] and [imath]t[/imath] in [imath]H[/imath] such that [imath]a(s,\cdot) = a(\cdot,w)[/imath] and [imath]a(t,\cdot) = a(\cdot,z)[/imath]. Then, I somehow want to conclude that z = w, but I am not sure how what to do next. How should I proceed? |
2546027 | Showing that [imath]2^{50}<3^{33}[/imath]
I'd never encountered a problem like "show that [imath]2^{50}<3^{33}[/imath]" but I think I ended up solving it after doing some weird stuff with logs and Maclaurin series: [imath] \begin{align*} 2^{50}&<3^{33}\\ \ln2^{50}&<\ln3^{33}\\ 50\ln2&<33\ln3\\ \frac{50}{33}&<\frac{\ln3}{\ln2}\\ \end{align*} [/imath] When I got to this point I thought I might be able to use the result [imath]\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots[/imath] to rewrite the right-hand side into something like a fraction. [imath] \begin{align*} \frac{50}{33}&<\frac{\ln3}{\ln2}\\ &<\frac{\ln(1+2)}{\ln(1+1)}\\ &<\frac{2-\frac{2^2}2+\frac{2^3}3-\cdots}{1-\frac12+\frac13-\cdots}\\ \end{align*} [/imath] Here I noticed that actually, the bottom is clearly greater than a half, because the sum of all the terms is greater than the sum of the first two, [imath]1-\frac12+\frac13-\cdots>1-\frac12[/imath]. So I actually went back a step: [imath] \begin{align*} \frac{50}{33}&<\frac{\ln3}{\frac12}\\ \frac{50}{33}&<2\ln3\\ \end{align*} [/imath] Here I think I've proved that [imath]2^{50}<3^{33}[/imath], because the above inequality ([imath]\frac{50}{33}<2\ln3[/imath]) is clearly true: [imath] \ln3>1\implies 2\ln3>2.\\ \frac{50}{33}<2\implies \frac{50}{33}<2\ln3. [/imath] However, I'm not sure if I've done this the "right" way. I'm pretty sure there must be alternative, simpler, "more elegant" ways to prove [imath]2^{50}<3^{33}[/imath]. Can people show me how they would prove this result? | 2276227 | How would you prove that [imath]2^{50} < 3^{33}[/imath] without directly calculating the values
Could you generalise the question and get something along the lines of [imath]n^{50} < (n+1)^{33}[/imath] ? |
2546528 | Calculate x when n = x^x
Is it possible when you know some n calculate the x if [imath]n = x^x[/imath] ? Sorry, I did't even know how this equation is called, any help appreciate. | 1569750 | How to solve x^x = const?
How to solve following equation: [imath]x^x = C[/imath] as [imath]x, C \in \mathbb{R}_{>0}[/imath]. I tried to use exponent and such that: [imath]x\ln x = \ln C[/imath]. |
2546549 | If [imath]|z| < 1 [/imath] , calculate [imath] s= \sum_{n=1}^{\infty} n^2 \cdot z^n [/imath]
As far as I understand we should try to find a geometric series [imath]g(z)[/imath] with a known sum and then get the sum [imath]s[/imath] from manipulating [imath]g(z)[/imath]. I have a hard time finding [imath]g(z)[/imath] though. Any help appreciated! | 1010403 | Derive an explicit formula for a power series [imath]\sum_{n=1}^\infty n^2x^n[/imath]
Could anyone help me find an explicit formula for: [imath] \sum_{n=1}^\infty n^2x^n [/imath] We're supposed to use: [imath]\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2} \qquad |x| <1 [/imath] |
2546395 | Question on Linear Alegbra.
Let [imath]A[/imath] be [imath]2 \times 2[/imath] matrix with trace 2 and determinant -3. Let [imath]T:M_2(\mathbb{R})\to M_2(\mathbb{R})[/imath] defined by[imath]T(B)=AB[/imath]. Then How to show that [imath]T[/imath] is diagnalizable? Could you please give me some hints? | 401697 | [imath]A[/imath] be a [imath]2\times 2[/imath] real matrix with trace [imath]2[/imath] and determinant [imath]-3[/imath]
[imath]A[/imath] be a [imath]2\times 2[/imath] real matrix with trace [imath]2[/imath] and determinant [imath]-3[/imath], consider the linear map [imath]T:M_2(\mathbb{R})\to M_2(\mathbb{R}):=B\to AB[/imath] Then which of the following are true? [imath]T[/imath] is diagonalizable [imath]T[/imath] is invertible [imath]2[/imath] is an eigen value of [imath]T[/imath] [imath]T(B)=B[/imath] for some [imath]0\ne B\in M_2(\mathbb{R})[/imath] if [imath] A=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}[/imath] Then I have calculated that matrix of [imath]T[/imath] will be \begin{pmatrix}a_{11}&a_{12}&0&0\\a_{21}&a_{22}&0&0\\0&0&a_{11}&a_{12}\\0&0&a_{21}&a_{22}\end{pmatrix} so [imath]T[/imath] is invertible I can say as [imath]\det T=(\det A)^2=9[/imath], could any one help to find out others as true/false? |
2534855 | On the way of evaluate [imath]\int_{0}^{\pi}\frac{dx}{1+2\sin^{2}x}[/imath].
Evaluate [imath]\int_{0}^{\pi}\frac{dx}{1+2\sin^2x}[/imath] My approach [imath]\Longrightarrow\int_{0}^{\pi}\frac{dx}{1+2\sin^{2}x}=2\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}x}{1+3\tan^{2}x}dx=\frac{2}{\sqrt{3}}\left[\tan^{-1}\left(\sqrt{3}\tan x\right)\right]_{0}^{\frac{\pi}{2}}[/imath] [imath]\tan x[/imath] is undefined at [imath]\dfrac{\pi}{2}[/imath]. So I need to solve [imath]\lim_{x\rightarrow\frac{\pi}{2}}[/imath][imath]\left[\tan^{-1}\left(\sqrt{3}\tan x\right)\right][/imath]. I don't know how to solve. | 958920 | Evaluating [imath]\int\limits_0^{\pi} \frac{dx}{1+2\sin^2x}[/imath]
After making u substitution two times, I am getting indefinite integral as [imath]\int\limits\dfrac{dx}{1+2\sin^2x} = \dfrac{\arctan(\sqrt{3}\tan(x))}{\sqrt{3}}+ C[/imath] I am stuck at working the bounds because I am getting [imath]0[/imath] but the answer needs to be [imath]\pi/\sqrt{3}[/imath]. I think [imath]\arctan(x)[/imath] is defined in [imath]-\pi/2 \lt x\lt\pi/2[/imath] while making u substitution. However I don't see how to use this observation to my advantage. Can anybody help in getting the answer and understanding whats going on. Thanks ! |
2546458 | Showing [imath](1+x)^r<(1+rx)[/imath] (MVT)
So the full question goes as following: Let [imath]0<r<1[/imath]. If [imath]x>0[/imath] or [imath]-1\le x<0[/imath], show that [imath](1+x)^r < (1+rx)[/imath] I'm having a real hard time getting the idea of the usefullness of MVT. It seems every approach I take is faulty. I don't know how to think. If I take [imath]f'(c)[/imath] for [imath]0<c<x[/imath] I get [imath](1+c)r -1 / c[/imath] which I can conclude is larger than [imath]0[/imath], which leads me nowhere. Help is very appriciated! | 556266 | Using Bernoulli's Inequality to prove an inequality
Using Bernoulli's Inequality show that [imath]\left(1+\frac{1}{k+1}\right)\left(1-\frac{1}{(k+1)^2}\right)^k\geq 1[/imath]for all [imath]k\in\mathbb{Z^+}[/imath]. My initial thought was to start be noting that \begin{align*}1-\frac{1}{(k+1)^2}&\geq 1-\frac14 \text{ for all natural [imath]k[/imath]}\\\implies\left(1-\frac{1}{(k+1)^2}\right)^k&\geq\left(1-\frac14\right)^k\geq1-\frac k4,\end{align*}by Bernoulli's Inequality. Multiplying both sides by [imath](1+1/(k+1))[/imath] gives the required expression on the left hand side, but the expression on the right hand side is not always greater than or equal to one. So my approach is clearly flawed. However, if I use Bernoulli's Inequality as follows [imath]\left(1-\frac{1}{(k+1)^2}\right)^k\geq1-\frac{k}{(k+1)^2}[/imath] and multiply both sides by [imath](1+1/(k+1))[/imath] I get an expression greater than or equal to one. However, this seems to me completely false; isn't Bernoulli's inequality [imath](1+x)^k\geq1+xk[/imath] supposed to work for a variable real [imath]x>-1[/imath] and a fixed natural [imath]n[/imath]? Any help will be appreciated. |
2545960 | Weaker form of inverse Galois problem
I do not know if this correct. So I ask. Given a number [imath]n\geq 2[/imath], can we find a Galois extension of [imath]\mathbb Q[/imath] such that the group has order [imath]n[/imath]? Similarly, given [imath]n\in \mathbb N[/imath] can we find a totally imaginary number field that is a Galois extension of [imath]\mathbb Q[/imath] and for which the order of its Galois group is [imath]2n[/imath]? | 2204162 | Real Cyclic Extensions of [imath]\mathbb Q[/imath] of given degree
An old qual problem asks us to Show that for every positive integer [imath]n[/imath], there exists a cyclic extension of [imath]\mathbb{Q}[/imath] of degree [imath]n[/imath] which is contained in [imath]\mathbb{R}[/imath]. A first thought might be towards Kummer theory: we could adjoin an [imath]n^\text{th}[/imath] root of, say, a prime number. But when [imath]n>2[/imath], [imath]\mathbb{Q}[/imath] lacks the full cohort of roots of unity that would make this work. If [imath]n[/imath] is a power of [imath]2[/imath] we can get what we want by adjoining ([imath]\mathbb{Q}[/imath]-linearly independent) square roots to [imath]\mathbb{Q}[/imath], and I think some casus irreducibilis things can be done in other degrees ( at least [imath]n=3[/imath] and [imath]n=5[/imath]) but a more general [imath]n[/imath] has me stumped. Could I get a nudge in the right direction on this problem? |
2547086 | [imath]\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}\right)=0=\sum_{k=0}^{n-1}\sin\left(\frac{2\pi k}{n}\right)[/imath]
How would you prove [imath]\sum_{k=0}^{n-1}\cos\left(\frac{2\pi k}{n}\right)=0=\sum_{k=0}^{n-1}\sin\left(\frac{2\pi k}{n}\right)[/imath] We tried induction first, but it doesnt seem to work. Any help would be appreciated! | 809089 | Trigonometric series sum proof
So, how do I prove that this trigonometric sum equals to zero? [imath]\sum_{k = 0}^{n - 1} \cos \left(\frac{2\pi k}{n}\right) = 0[/imath] |
2547176 | Is a continuous interval countable or uncountable infinite?
I wondered if a continuous interval like [imath][0,1]^{\mathbb{R}}[/imath] has countable or uncountable infinite elements. So i looked for a bijective mapping and found following: [imath]f:\mathbb{N} \to ]0,1[^{\mathbb{R}}[/imath] [imath]f:x \mapsto \frac{x}{10^{\lfloor\log_{10}x\rfloor+1}}[/imath] [imath]f(1234) = 0.1234[/imath] Therefore [imath][0,1]^{\mathbb{R}}[/imath] should have countable infinite elements, shouldn't it? | 555517 | What's wrong with this argument that [imath][0,1][/imath] is countable?
Every real number in [imath][0,1][/imath] has a decimal expansion [imath]0.d_1d_2d_3...[/imath], so construct an infinite tree rooted at 0 where each node has branches leading to [imath]\{0,1,2,3,4,5,6,7,8,9\} [/imath], and let each path through the tree represent the successive decimal digits of a real number. To enumerate them, go through the tree breadth-first i.e. layer-by-layer. |
2547699 | What is the maximum total displacement needed for sorting a list?
I imagined a system of [imath]n[/imath] "things" that are ordered: each "thing" has a specific place. Also, the [imath]n[/imath]th thing is the last thing in the system, meaning that there is always a "thing" to which number [imath]1[/imath] is assigned. Given such a system, we can "mess it up" — put things in wrong order. Now we must put them back. Like so (an example): We see that in this case, we have a system [imath]n = 5[/imath]. If all the "things" are in their spots, then the output of the function I wanted to deduce (but failed) is [imath]0[/imath]. But if they are not, my "function" adds [imath]1[/imath] to its output for each place a misplaced "thing" must travel to get to its original location. In the above case, the output of function in order to "clean up the mess" is [imath]8[/imath]. The question is: for some system [imath]n[/imath] (where [imath]n \in \mathbb{N}[/imath]), what is the maximum of the function [imath]f[/imath]? I already found [imath]f_{\text{max}}[/imath] for [imath]n = 1[/imath] ([imath]f_{\text{max}}(1) = 0[/imath]), [imath]n = 2[/imath] ([imath]f_{\text{max}}(2) = 2[/imath]) and [imath]n = 3[/imath] ([imath]f_{\text{max}}(3) = 4[/imath]). But it is quite clear that this problem needs a different approach; bruteforce won't do any good. So, my main interest is to know what a formula for [imath]f_{\text{max}}(n)[/imath]. I am very eager to see the answer. I tackled a few hard problems, but this one (and the strange thing is that I made it up myself) is one tough cookie, at least for me. | 623516 | What's the name of this quantity?
For each permutation [imath]\sigma[/imath] of [imath] \left\{ 1, 2, \dots, n \right\}[/imath] define [imath]\operatorname{dist}(\sigma)=\sum_{i=1}^{n}\left| \sigma (i)-i \right|[/imath] For each [imath]n\in\mathbb{N}[/imath], I'm interested in finding the maximum value this function can score, namely [imath]M_n=\max_{\sigma\in D_n}\left\{\operatorname{dist}(\sigma)\right\}[/imath] Does [imath]M_n[/imath] has a conventional name? Uses? Is there a formula to find it? I've noticed that [imath]M_n[/imath] grows "in pairs", meaning that for each odd [imath]a\in \mathbb{N}[/imath] it holds that [imath]M_{a+1}-M_a=M_{a+2}-M_{a+1}[/imath] I guess it can represent some kind of measure for hardness to sort. |
2547722 | Prove that [imath]na_n \rightarrow 0[/imath] as [imath]n \rightarrow \infty[/imath].
Let [imath]\{a_n \}[/imath] be a decreasing sequence of positive real numbers. Prove that if the series [imath]\sum_{n=0}^{\infty} a_n \sin (nx)[/imath] converges uniformly on [imath]\mathbb R[/imath] then [imath]na_n \rightarrow 0[/imath]. I find difficulty to prove this. Please help me. Thank you in advance. | 976970 | [imath] \sum a_n \sin (nx)[/imath] converges uniformly iff [imath]na_n\to 0[/imath] as [imath]n \to \infty[/imath]
Let [imath]\{a_n\}[/imath] be decreasing sequence of positive terms then prove that [imath]\displaystyle \sum a_n \sin (nx)[/imath] converges uniformly on [imath]\Bbb{R}[/imath] iff [imath]na_n \to 0[/imath] as [imath]n\to \infty[/imath]. I proved that convergence of [imath] \sum a_n \sin (nx)[/imath] implies [imath]na_n \to 0[/imath] as [imath]n\to \infty[/imath] . I got stuck while prove the converse, I tried ussing Dirichlet's test but the problem here is that partial sums of [imath]\displaystyle \sum^{n}_{k=1} \sin (kx)[/imath] are bounded by [imath]\displaystyle \frac{1}{|\sin (\frac{t}{2})|}[/imath] , so as per Dirichlet's test requirement I'm not getting uniform bound. |
2140495 | [imath]a_0 = 5/2[/imath]. [imath]a_k = (a_{k-1})^{2} - 2[/imath] for [imath]k\geq1. \prod_{k=0}^{\infty}{\left(1-1/a_k\right)}.[/imath]
Some hints for the evaluation of this infinite product would be appreciated. [imath]a_0 = 5/2[/imath]. [imath]a_k = (a_{k-1})^{2} - 2[/imath] for [imath]k\geq1.[/imath] Evaluate : [imath]\prod_{k=0}^{\infty}{\left(1-1/a_k\right)}.[/imath] I tried finding [imath]a_1[/imath] [imath]a_2[/imath] and so on, and then multiplying them to see if I could find an easier sequence to solve but I could no proceed any further, any help would be appreciated. | 2548319 | Computing an infinite product
Let [imath]a_0=5/2[/imath] and [imath]a_k=a^2_{k-1}-2[/imath] for all [imath]k\geq 1[/imath].The question is to compute [imath]\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)[/imath] I tried to calculate few terms.[imath]a_0=5/2[/imath], [imath]a_1=17/4,a_2=257/16[/imath] it seems that [imath]a_k[/imath] is of the form [imath]2^{2^k}+2^{-2^{k}}[/imath] however I am not sure about it.How to proceed without doing much guesswork.Any ideas? |
2547895 | What is a better explanation for why $\sum\limits_{n=2}^{\infty}\frac1{n\ln(n)^{1.01}}$ converges?
So I want to show that [imath]\sum_{n=2}^{\infty}\displaystyle\frac{1}{n\ln(n)^{1.01}}[/imath] converges. I suppose if [imath]\ln(n)>n^{0.01}[/imath]then [imath]n\ln(n)>n^{1.01}[/imath] and so [imath]\displaystyle\frac{1}{n\ln(n)^{1.01}}<\displaystyle\frac{1}{n^{1.01}},[/imath] meaning that [imath]\int\limits_{2}^{\infty}\displaystyle\frac{1}{n\ln(n)^{1.01}}\mathrm dn<\int\limits_{2}^{\infty}\displaystyle\frac{1}{n^{1.01}}\mathrm dn.[/imath] And since the RHS of the inequality converges by the [imath]p[/imath] test, the right hand term also converges, and so the series converges. Is this correct? If it is, is there a better (e.g. more straightforward/obvious/elegant...etc.) way of showing that the series converges? | 2069399 | series convergence with comparison test [imath]\frac{1}{n\log(n)^p}[/imath]
I came across a question about convergence tests and I'm not sure how to show that [imath]\sum_{n=2}^{\infty}\frac{1}{n \log(n)^p}[/imath] converges for [imath]p>1[/imath]. I found that the integral test works but I haven't learnt it yet. I have learnt the comparison test, so I tried that but couldn't find any series that converges to which I can compare. So my question is both how can I show convergence here with comparison test and also: Is there some list of convergent series that I can compare to? PS: I also found out that this is called a ln seriesAlso, this is not a duplicate of this question. (The power is on the [imath]\log[/imath] here) |
2489364 | Understanding the Convergence of Series (Rudin theorem)
In Rudin's Principles of Mathematical Analysis, there's a theorem that reads: [imath]\sum a_n[/imath] converges if and only if for every [imath]\epsilon > 0[/imath], there is an integer [imath]N[/imath] such that [imath]\lvert \sum_{k = n}^m a_k \rvert \le \epsilon[/imath] if [imath]m \ge n \ge N[/imath]. (3.22) I'm having a bit of trouble understanding this theorem. Is it saying that every sum of [imath]a_k[/imath] from one point to another (both greater than or equal to [imath]N[/imath]) must be less than some arbitrary epsilon? If so, what exactly is it "saying" and what's the significance? If someone could provide me with a simplified/"dumbed down" explanation of this theorem, it'd be tremendously helpful. Thank you. | 1786273 | Criteria for convergent sequence (Baby Rudin Theorem 3.22)
Theorem 3.22 of Rudin's Principles of Mathematical Analysis says: The series [imath]\sum a_{n}[/imath] of (real or) complex numbers converges iff for every [imath]\varepsilon > 0[/imath], there is an integer [imath]N[/imath] such that [imath] \left| \sum_{k=n}^m a_{k} \right| \leq \varepsilon [/imath] if [imath]m\geq n \geq N.[/imath] My doubt is: The condition is that the sequence of partial sums {[imath]s_{n}[/imath]} is Cauchy, i.e for every [imath]\varepsilon>0[/imath] there is an integer [imath]N[/imath] such that [imath] \left|s_{m}-s_{n} \right|<\varepsilon [/imath] if [imath]m \geq n \geq N.[/imath] But [imath]\left|s_{m}-s_{n} \right|[/imath] would be [imath]\left|\sum_{k=n+1}^m a_{k}\right|[/imath], isn't it? How come the summation starts from [imath]n[/imath] in the theorem? How are the two summations equivalent? |
2548541 | How can [imath]\sum_{1}^\infty 1/(n^2) = \pi^2/6[/imath] when [imath]\int_1^\infty 1/(x^2) =1[/imath]?
How can [imath]\sum_{1}^\infty 1/(n^2) = \pi^2/6[/imath] when [imath]\int_1^\infty 1/(x^2) =1[/imath] ? | 37911 | Why is [imath]\int_1^{\infty}(1/x^2)\,dx[/imath] less than [imath]\sum_{n=1}^{\infty} 1/(n^2)[/imath]?
The value of [imath]\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx[/imath] equals 1. The series [imath]\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}[/imath] equals [imath]\pi^2/6 \approx 1.66493[/imath]. Shouldn't the area under the curve of the function be larger than the sum of the terms of the sequence? |
2548537 | How to prove [imath]\sqrt{1+x} \le 1+\dfrac{x}{2}[/imath], [imath]x > -1[/imath] using MVT
How to prove this question?(picture) I don't know the way and sorry for weird typing (i'm korean...) | 602292 | prove inequality through mean value theorem
I want to prove that if [imath]x[/imath] is a real number and [imath]x>-1[/imath], [imath]\sqrt{1+x} \leq 1+\dfrac{x}{2}[/imath] I'm not sure which one should I choose for [imath]f(x)[/imath]. I also don't understand why we need the mean value theorem to prove this. |
2548438 | Integral asked on MIT bee qualifier 2012 [imath]\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx[/imath]
I have been trying to solve this one but I have no clue, I put it into wolfram and the result is absurd, considering this one is taken from MIT bee qualifier 2012, this is the integral: [imath]\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \, dx[/imath] | 1403717 | [imath]\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}[/imath]
[imath]\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}[/imath] My Attempt: [imath]\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}[/imath] Replacing [imath]x[/imath] by [imath]1-x[/imath],we get [imath]\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}[/imath] Then I got stuck. Please help. |
2548707 | Derivative of an indicator function
I was wondering what the derivative is of an indicator function. So we have the function: [imath]f(Y, a) = \Bbb1(Y \le a).[/imath] I am trying to differentiate this function to [imath]a[/imath]. Is this equal to [imath]0[/imath]? Thank you | 1993827 | Derivative of unit step function?
How would I find the derivative of a unit step function? I understand that the unit impulse function will be used but I'm not sure how to use it. I am trying to find the derivative of this: [imath]v(t) = u(t+1) - 2u(t) + u(t-1)[/imath] [imath]u(t) = 0[/imath] when [imath]t < 0[/imath] [imath]u(t) = 1[/imath] when [imath]t > 0[/imath] The relationship between unit step function and impulse function: δ(n) = u(n) - u(n-1) [imath] δ(t)=du(t)/dt [/imath] |
403505 | Given [imath]x,y\in\mathbb{C}^n[/imath] s.t. [imath]f(x,y)=\sup_{\theta,\phi}\{\|e^{i\theta}x-e^{i\phi}y\|^2,\theta,\phi\in\mathbb{R}\}[/imath]
Given[imath]x,y\in\mathbb{C}^n,\quad f(x,y)=\sup_{\theta,\phi}\{\|e^{i\theta}x-e^{i\phi}y\|^2,\theta,\phi\in\mathbb{R}\}[/imath] Then which is/are the following are true? [imath]1.\ f(x,y)\le \|x\|^2+\|y\|^2-2Re|\langle x,y\rangle|[/imath] [imath]2.\ f(x,y)\le \|x\|^2+\|y\|^2+2Re|\langle x,y\rangle|[/imath] [imath]3.\ f(x,y)= \|x\|^2+\|y\|^2+2Re\langle x,y\rangle[/imath] [imath]4.\ f(x,y)\ge \|x\|^2+\|y\|^2-2Re\langle x,y\rangle[/imath] I have no idea how to start with or how to solve, could any one help me? | 257738 | Consider [imath]f(x,y)=\sup_{θ,φ}\{||e^{i θ }x+ e^{i φ }y||^2: θ,φ∈R\}[/imath] in which [imath]x,y\in\mathbb C^n[/imath]
Assuming [imath]x,y\in\mathbb C^n[/imath],consider: [imath]f(x,y)=\sup_{θ,φ}\{||e^{i θ }x+ e^{i φ }y||^2: θ,φ\in\mathbb R\}[/imath] Which of the following is/are correct? 1.[imath] f(x,y)[/imath]≤[imath]||x||^2+||y||^2+2|(x,y)|[/imath] 2. [imath]f(x,y)[/imath]= [imath]||x||^2+||y||^2+2Re(x,y)[/imath] 3. [imath]f(x,y)[/imath]= [imath]||x||^2+||y||^2+2|(x,y)|[/imath] 4. [imath]f(x,y)[/imath]> [imath]||x||^2+||y||^2+2|(x,y)|[/imath] How can I solve this problem ,I am completely stuck on it . can anyone help me?thanks |
2549492 | inequality problem tough - putnam level
Suppose we have a necklace of n beads. Each bead is labeled with an integer and the sum of all these labels is [imath]n − 1[/imath]. Prove that we can cut the necklace to form a string whose consecutive labels [imath]x_1, x_2,\dots, x_n[/imath] satisfy [imath]\sum_{i=1}^k x_i \le k −1[/imath] for [imath]k = 1,2,\dots,n[/imath] | 2523484 | Interesting property of finite integer sequence with sum 1
I ran into nice problem which I proved and would be thankful if anyone assess it. Given a sequence of integers [imath]x_1, x_2, \dots, x_n[/imath] whose sum is 1? prove that exactly one of the cyclic shifts [imath]x_1,x_2,\dots, x_n; \quad x_2,\dots, x_n,x_1; \cdots \quad x_n, x_1\dots, x_{n-1}[/imath] has all of his partial sums positive. (By a partial sum we mean the sum of the first [imath]k[/imath] terms, [imath]k\leqslant n[/imath].) My solution: For [imath]n=1[/imath] we have a single-term sequence and it is obvious. Assume that our proposition is true for [imath]n[/imath]. We'll try to prove it for [imath]n+1[/imath]. Let we have the sequence [imath](x_1, x_2, \dots, x_n,x_{n+1})[/imath] then [imath]\exists \ i[/imath] such that [imath]x_i>0[/imath]. If [imath]1\leqslant i <n+1[/imath] then consider the following sequence [imath](x_1, \dots,x_i+x_{i+1},\dots, x_{n+1})[/imath] which has [imath]n[/imath] elements. Then using the assumption of induction we have that there is exactly one cyclic shift with desired property, And if we write [imath]x_i, x_{i+1}[/imath] instead of [imath]x_i+x_{i+1}[/imath] and since [imath]x_i>0[/imath] we get our desired sequence. If [imath]i=n+1[/imath], i.e. [imath]x_{n+1}>0[/imath] then replace the sequence [imath](x_1, x_2, \dots, x_n,x_{n+1})[/imath] by [imath](x_2, \dots, x_n,x_{n+1}+x_1)[/imath] which consists of [imath]n[/imath] terms. Then using the assumption of induction we get exactly one sequence with desired properties and separating [imath]x_{n+1}+x_1[/imath] to [imath]x_{n+1}, x_1[/imath] we get the sequence with [imath]n+1[/imath] elements since [imath]x_{n+1}>0[/imath]. |
2548665 | Show that [imath]\left|\varphi\right|^{2}+\left|\eta\right |^{2}=\frac{1}{2}\left (\left|\varphi+\eta\right|^{2}+\left|\varphi -\eta\right|^{2}\right ) [/imath]
I want to show that [imath]\forall_{\varphi, \eta \in \mathbb{C}} \ \left | \varphi \right |^{2} + \left | \eta \right |^{2} = \frac{1}{2}\left ( \left | \varphi + \eta \right |^{2} + \left | \varphi -\eta \right |^{2} \right ) [/imath] How can I generally prove this? Any help for a starting point of the proof would be greatly appreciated. | 959586 | Absolute value of a complex number proof
Ok, so I have the following proof. Let [imath]z[/imath] and [imath]w[/imath] be complex numbers. Prove [imath]\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2[\lvert z \rvert^2 + \vert w \rvert^2][/imath]. Using [imath]\vert z \rvert^2=z\bar{z}[/imath], I have said [imath]\lvert z+w \rvert ^2 + \lvert z-w \rvert^2 = 2(\lvert z \rvert^2) + \vert w \rvert^2)[/imath] [imath](z+w)\bar{(z+w)}+(z-w)(\bar{(z-w)}[/imath] [imath](z+w)(z-w)+(z-w)(z+w)[/imath] [imath]z^2 - w^2 + z^2 - w^2[/imath] [imath]z^2 + z^2 + -w^2 + -w^2[/imath] [imath]2(z^2) + 2(-w^2)[/imath] [imath]2[z^2 + (-w)^2][/imath] [imath]2[\lvert z \rvert^2 + \vert w \rvert^2][/imath] I guess my question is, is my logic correct? I am not so skilled with proof writing and though I feel that this is correct, I have been known to be wrong in the past. Just need some insight. Thank you. |
2550109 | axiom of choice is equivalent to "the product of nonempty collection of nonempty sets is nonempty"
Axiom of choice:\ If {[imath]A_\alpha \vert \alpha \in A[/imath] } is an indexed family of nonempty pairwise-disjoint sets, there is a function [imath]f[/imath] from [imath]A[/imath] to [imath]\cup A_\alpha [/imath] such that [imath]f(\alpha )\in A_\alpha[/imath], for each [imath]\alpha \in A[/imath] From this definition I need to obtain an equivanlence to the product of nonempty collection of nonempty sets is nonempty, but I have a problem in the part of pairwise disjoint sets. Help! | 1876846 | Proving equivalencies of two versions of the Axiom of Choice
I am reading through some course notes to prepare for an upcoming Real Analysis course I have to take in the fall, and found two definitions of the Axiom of Choice. [AC] If [imath]\Lambda \neq \emptyset[/imath] and for each [imath]\lambda \in \Lambda[/imath], [imath]X_{\lambda}[/imath] is a non-empty set, then [imath]\prod_{\lambda \in \Lambda} X_{\lambda} \neq \emptyset[/imath]. [ACD] Suppose that [imath]\Lambda \neq \emptyset[/imath] and that [imath]\forall \ \lambda \in \Lambda[/imath], [imath]X_{\lambda}[/imath] is a non-empty set, and [imath]X_{\lambda} \cap X_{\beta} = \emptyset[/imath] if [imath]\lambda \neq \beta \in \Lambda[/imath] Then [imath]\prod_{\lambda \in \Lambda}X_{\lambda} \neq \emptyset[/imath]. It wasn't immediately apparent to me that they were equivalent, so I started trying to prove it. I think that it is trivial to show that [imath]1 \implies 2[/imath], more or less just by saying that if every non-empty set has a choice function then every pairwise non-empty set will also have a choice function. But I am stuck on trying to prove that [imath] 2 \implies 1 [/imath] ... I can't seem to construct any meaningful way to use the disjoint version to prove the more general version. Any tips? |
2548799 | Limit point of set [imath]A=\{1/n:n\in\mathbb{N}\}[/imath]
Consider set [imath]A=\{1/n:n\in\mathbb{N}\}[/imath] show that 0 is only limit point. Though I prove that but to prove 1 is not a limit point. For this consider neighborhood [imath](1/2,3/2)[/imath] of 1, so if we delete 1 from it we get points like 0.6 , 0.7, 0. 8 but the set [imath]A[/imath] also contains it as it is between 0 to 1. Please let me know where I am incorrect. | 2522897 | Limit point of the set [imath]G= \{1/n: n \in \mathbb N\}[/imath]?
In [imath]\mathbb{R}[/imath] , what is the limit point of the set [imath]G= \{1/n: n \in \mathbb N\}[/imath]? Definition: A point [imath]p \in X[/imath] is a limit of of E if every neighborhood of [imath]p [/imath] contains a point [imath]q \in E[/imath] [imath]q\neq p[/imath] I do understand [imath]0[/imath] is a limit point but shouldn't it be the case that all the points in [0,1] are limit points by the definition? |
418961 | [imath]\epsilon[/imath]-[imath]\delta[/imath] proof that [imath]\lim\limits_{x \to 1} \frac{1}{x} = 1[/imath].
I'm starting Spivak's Calculus and finally decided to learn how to write epsilon-delta proofs. I have been working on chapter 5, number 3(ii). The problem, in essence, asks to prove that [imath]\lim\limits_{x \to 1} \frac{1}{x} = 1.[/imath] Here's how I started my proof, [imath]\left| f(x)-l \right|=\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| < \epsilon \implies \left| x-1 \right| < \epsilon |x|[/imath] I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach? | 2807989 | Epsilon delta proof fraction
Prove using the epsilon-delta definition? [imath]\lim \limits_{x \to 2}\frac{1}{x} = \frac{1}{2}[/imath] [imath]|f(x)-L|<\epsilon[/imath] [imath]|\frac{1}{x}-\frac{1}{2}|<\epsilon[/imath] [imath]-\epsilon<\frac{1}{x}-\frac{1}{2}<\epsilon[/imath] [imath]-\epsilon+\frac{1}{2}<\frac{1}{x}<\epsilon + \frac{1}{2}[/imath] Any tips on how do I continue from here? |
2550720 | Integral of point-wise product of 2 functions greater than product of 2 integrals
Let [imath]f[/imath] and [imath]g[/imath] be continuous increasing functions defined on [imath][0,1][/imath]. Prove that [imath]\left(\int_0^1 f(x)\,dx\right)\left(\int_0^1 g(x)\,dx\right) \le \int_0^1 f(x)g(x)\,dx[/imath] This looks like Cauchy-Schwarz, using [imath]L^1[/imath](?) but that makes little sense to me. Perhaps there is a more elementary way of proving this inequality? Note: This question is from a undergraduate single variable calculus course, but feel free to use any suitable theorems from analysis. | 1163512 | Given two increasing continuous functions [imath]f,g[/imath] prove that [imath](b-a) \int^b_a f(x)g(x) dx > \int^b_a f(x) dx \int^b_a g(x) dx[/imath]
Given two monotonically increasing continuous functions [imath]f,g[/imath] prove that [imath](b-a) \int^b_a f(x)g(x) dx > \int^b_a f(x) dx \int^b_a g(x) dx,\; b>a[/imath] what I have tried: Let [imath]h(x) = (x-a)\int^x_a f(t)g(t) dt - \int^x_a f(t)dt\int^x_a g(t)dt [/imath] and trying to figure out that [imath]h'>0[/imath] using MVT to reform [imath]h'(x)[/imath] as [imath]h'(x) = (x-a)(-f(x)g(\xi_1) - g(x)f(\xi_2) + f(x)g(x) + f(\xi_3)g(\xi_3))[/imath] where [imath]\xi_{1,2,3} \in (a,x)[/imath] (That method does not seem to work) |
2549380 | Is there a simple proof that if [imath](b-a)(b+a) = ab - 1[/imath], then [imath]a, b[/imath] must be Fibonacci numbers?
Consider the identity [imath](b-a)(b+a) = ab - 1[/imath], where [imath]a, b[/imath] are nonnegative integers. We can also express this identity as [imath]a^2 + ab - b^2 = 1[/imath]. This identity is clearly true when [imath]a = F_{2i-1}[/imath] and [imath]b = F_{2i}[/imath], where [imath]F_i[/imath] is the [imath]i^{th}[/imath] term of the Fibonacci sequence. This is equivalent to one case of Cassini's identity, [imath](F_{2i-1})(F_{2i+1}) = F_{2i}^2 + 1[/imath], and is easily proven by induction or several other simple elementary means. My question is this: Is there a simple elementary proof that these Fibonacci numbers are the only solutions of this identity? By simple elementary proof, ideally I mean a proof using methods and steps that a mathematically gifted high school student could follow and understand. Alternatively, I could define it as a proof using methods that would have been known to mathematicians in Cassini's time, in the late 17th century. In other words, I am looking for a proof that does not rely on more advanced methods such as quadratic number fields or generalized solutions of Pell equations. | 2058057 | Prove the inequality [imath]\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|<\frac{1}{mn}[/imath]
Prove that the inequality [imath]\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|<\frac{1}{mn}[/imath] holds for positive integers [imath]m, n[/imath] if and only if [imath]m[/imath] and [imath]n[/imath] where [imath]m > n[/imath] are two successive terms of the Fibonacci sequence. I thought about using the explicit formula for the Fibonacci sequence, which is [imath]F_n = \dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right),[/imath] but this seems to get computational. Is there an easier way to think about this? |
2551769 | Is [imath]L^p[/imath] space separable?
If ([imath]X[/imath],[imath]\Sigma[/imath],[imath]\mu[/imath]) is [imath]\sigma[/imath]-finite, is [imath]L^p[/imath] space separable with 1[imath]\leq[/imath]p<[imath]\infty[/imath]? If it is correct, please tell me a proof. | 1447055 | Space [imath]\mathcal{L}^p(X, \Sigma, \mu)[/imath] is separable iff [imath](\Sigma, \rho_\Delta)[/imath] is separable
Let's consider the space [imath]\mathcal{L}^p(X, \Sigma, \mu)[/imath] of all functions [imath]f\colon X \to \mathbb{R}[/imath] (or [imath]\mathbb{C}[/imath]) for which: [imath] \int\limits_X|f|^p \mu(dx) < \infty. [/imath] Here [imath]X[/imath] is a metric space, [imath]\Sigma\subset 2^X[/imath] — sigma-algebra on [imath]X[/imath] and [imath]\mu[/imath] — is the measure on [imath]\Sigma[/imath]. As usual, functions that are equal almost everywhere considered to be equivalent, so technically we are dealing with classes [imath][f]=\{g\in \mathcal{L}^\infty(X,\Sigma, \mu) \colon \mu \{x\in X\colon f(x)\neq g(x)\}=0\}.[/imath] This space is endowed with integral metric [imath]\rho_p(f,g)=\left(\int\limits_X|f-g|^p \mu(dx)\right)^{1/p}[/imath] The statement is that space [imath]\mathcal{L}^p(X, \Sigma, \mu)[/imath] is separable iff [imath](\Sigma, \rho_\Delta)[/imath] is separable, where [imath]\forall A,B\in\Sigma[/imath] [imath]\rho_\Delta(A,B)=\mu(A\Delta B)[/imath] Here, again, we consider [imath]A\in \Sigma[/imath] as a class [imath][A]=\{A'\in\Sigma\colon \mu(A\Delta A')=0\}[/imath] to have the identity of indiscernibles. How do one prove the statement? |
2547901 | Consider the surface [imath]S[/imath] in [imath]\Bbb R^3[/imath] given by the graph of [imath]f(x,y)[/imath]. Find the intersection point of the tangent planes.
Consider the surface [imath]S[/imath] in [imath]\Bbb R^3[/imath] given by the graph of [imath]f(x,y) = xy^2 − 2y^2 + e^x[/imath]. Calculate the intersection point of the tangent planes to [imath]S[/imath] above [imath](x, y) = (0, 0)[/imath], [imath](0, 1)[/imath], and [imath](0, 2)[/imath]. I am confused about what this question is asking. Would I have to choose a tangent plane above that point and see where it intersects [imath]S[/imath]? Thanks | 2548228 | Consider the surface [imath]S[/imath] in [imath]\mathbb{R}^ 3[/imath] given by the graph of [imath]f(x, y) = xy^2 − 2y^2 + e^x [/imath].
Consider the surface [imath]S[/imath] in [imath]\mathbb{R}^3[/imath] given by the graph of [imath]f(x, y) = xy^2 − 2y^2 + e^x[/imath] . Calculate the intersection point of the tangent planes to [imath]S[/imath] above [imath](x, y) = (0, 0), (0, 1)[/imath], and [imath](0, 2)[/imath]. Can someone guide me through solving this problem? I am confused on where to start because we didn't cover this in class. Thanks so much |
2552127 | Continuity of Lebesgue measure w.r.t Hausdorff distance?
If one considers the metric space [imath]\mathcal{K}[/imath] of all compact subsets of [imath]\mathbb{R}^2[/imath] endowed with the Hausdorff distance (that is [imath]\Delta(A,B)=\inf \{ \delta: A\subset B^{\delta},B \subset A^{\delta} \}[/imath]). Is it true (and how can it be shown in that case) that the function [imath]\mu: \mathcal{K} \to \mathbb{R}^2[/imath] ([imath]\mu[/imath] is the 2-dimensional lebesgue measure) defined by [imath]A \mapsto \mu{(A)}[/imath] is continuous w.r.r the metric [imath]\Delta[/imath]? | 1339102 | Interplay of Hausdorff metric and Lebesgue measure
Consider the space [imath]\mathcal{K}(\mathbb{R}^{n})[/imath] of compact subsets of [imath]\mathbb{R}^{n}[/imath] endowed with the Hausdorff metric [imath]\rho[/imath], and let [imath]\lambda[/imath] denote the [imath]n[/imath]-dimensional Lebesgue measure on [imath]\mathbb{R}^{n}[/imath]. Now, I understand that [imath]\lambda[/imath] is not continuous with respect to [imath]\rho[/imath], i.e. that [imath][ \lim_{k \to \infty} \rho(K, K_{k}) = 0 ] \not \Rightarrow [ \lambda(K) = \lim_{k \to \infty} \lambda(K_{k}) ][/imath]. But my question is: Suppose [imath]\lambda(K_{i}) = \lambda(K_{j}) = C[/imath] for every [imath]i, j \in \mathbb{N}[/imath], i.e. the sequence [imath]k \mapsto \lambda(K_{k})[/imath] is constant; does this imply that [imath]\lambda(\lim_{k \to \infty} K_{k}) = C[/imath]? If not, I'm interested to see a counterexample. Thanks. |
2552519 | Prove that a simple group of order 168 is isomorphic to a subgroup of A_8
Let [imath]G[/imath] be a simple group of order [imath]168=2^3 \cdot 3 \cdot 7[/imath]. By simplicity we get that [imath]n_7=8[/imath]. Let [imath]G[/imath] act on the set of [imath]7[/imath]-Sylow P subgroups (of which there are 8). This induces an injective homomorphism [imath]\psi:G \to S_8[/imath], and by the first isomorphism theorem [imath]G/\ker(\psi) = G \cong \psi(G)[/imath] How do I show that [imath]\psi(x)[/imath] is necessarily even to complete this proof? | 1050268 | A simple finite group [imath]G[/imath] with [imath]n[/imath] p-Sylows is isomorphic to a subgroup of [imath]\mathbb A_n[/imath]
I am trying to solve this problem: Let [imath]G[/imath] be a finite and simple group, and let [imath]p[/imath] be a prime number such that [imath]p[/imath] divides [imath]|G|[/imath]. If [imath]n_p(G)=n[/imath] for [imath]n>1[/imath] (n_p denotes the number of p-Sylows) then [imath]G[/imath] is isomorphic to a subgroup of [imath]\mathbb A_n[/imath]. What I thought of was the following: I consider the action of [imath]G[/imath] on [imath]X[/imath]={H subgroup of G, H p-Sylow} by conjugation. Then there is a morphism [imath]\psi:G \to S(X)[/imath] [imath]g \to gSg^{-1} \space \forall S \in X,[/imath] where [imath]S(X)[/imath] is the set of all bijective functions [imath]f:X \to X[/imath]. First of all, I would like to affirm [imath]S(X) \cong \mathbb S_n[/imath], but I don't know if this is true. Second, note that [imath]Ker(\psi) \neq G[/imath] because if not [imath]n_p=1[/imath] and [imath]Ker(\psi)[/imath] can't be a proper subgroup because if that was the case [imath]Ker(\psi) \lhd G[/imath], then we have [imath]Ker(\psi)=\{e\}[/imath]. By the first isomorphism theorem, [imath]G \cong G/Ker(\psi) \cong Im(\psi)[/imath]. Now, if I could show that [imath]Im(\psi)[/imath] is isomorphic to a subgroup of [imath]\mathbb A_n[/imath], then I would be done. I would appreciate any hints or suggestions. |
173590 | Sufficient condition for convergence of a real sequence
Let [imath](x_n)[/imath] be a sequence of real numbers. Prove that if there exists [imath]x[/imath] such that every subsequence [imath](x_{n_k})[/imath] of [imath](x_n)[/imath] has a convergent (sub-)subsequence [imath](x_{n_{k_l}})[/imath] to [imath]x[/imath], then the original sequence [imath](x_n)[/imath] itself converges to [imath]x[/imath] . Thanks for any help. | 741297 | Show sequence converges to L
If a sequence [imath]\{x_n\}[/imath] has the property that [imath]\lim_{n\rightarrow\infty} x_{2n} = \lim_{n\rightarrow\infty}x_{2n+1} = L[/imath] Show that the sequence [imath]\{x_n\}[/imath] convergences to [imath]L[/imath]. I tried a proof by contradiction against the given information, then I considered saying that if all the subsequences of the already given convergent subsequences converge to [imath]L[/imath], then [imath]\{x_n\}[/imath] must converge to [imath]L[/imath]. What is the proper way to prove this? |
2552672 | Let G be non abelian and |G| = [imath]p^3[/imath] , where p is a prime. Prove that Z(G) = G′ . What is the order of Z(G)?
Let G be non-abelian and |G| = [imath]p^3[/imath] , where p is a prime. Prove that Z(G) = G′ . What is the order of Z(G)? | 1264974 | Let [imath]p[/imath] be a prime number and [imath]G[/imath] a non abelian group or order [imath]p^3[/imath]. Prove that [imath]Z(G) = [G,G][/imath].
Let [imath]p[/imath] be a prime number and [imath]G[/imath] a non abelian group or order [imath]p^3[/imath]. Prove that [imath]Z(G) = [G,G][/imath]. I have already figured out that [imath]|Z(G)| = p[/imath], and that [imath]G'=[G,G] \lhd G[/imath]. Also, [imath]|G'| = p[/imath] or [imath]p²[/imath]. I suppose I'd have to prove that [imath]G' \subset Z(G)[/imath], but I've been trying and I have no idea how. Any hints? Thanks! |
2552671 | gcd of two polynomials using euclidean algorithm
Using the Euclidean algorithm, I need to find the gcd. I believe it needs to be a linear combination, but I honestly don't understand how to find it. So ([imath]x^4+x^3+2x^2+x-1[/imath])-(something)[imath]\cdot (x^3+1[/imath]) I believe but I don't know how to find the something | 352079 | Euclidean Algorithm for GCD of polynomials
I am struggling to use the Euclidean algorithm for polynomials. Given something like [imath]GCD(x^5+1, x^3+1)[/imath] I can easily use it as follows: [imath]x^5+1 = x^2(x^3+1) -x^2 +1 \\ x^3+1 = -x(-x^2+1) + x +1 \\ -x^2+1 = (x+1)(-x+1)[/imath] GCD = x+1 But for something like [imath]GCD(2x^2+6x+3, 2x+1)[/imath] I cannot figure out how to do it using the same method. I start like this: [imath]2x^2+6x+3 = x(2x+1) + 5x+3\\ 2x+1 = \frac{2}{5}(5x+3) - \frac{1}{5}\\[/imath] and I am not sure how to continue. Any help would be appreciated, sorry for my poor attempt at using LaTeX. |
187525 | Sufficiency to prove the convergence of a sequence using even and odd terms
Given a sequence [imath]a_{n}[/imath], if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms: [imath]\lim_{n\rightarrow\infty} a_{2n}=\lim_{n\to\infty} a_{2n-1}=L[/imath] Is this sufficient to prove that the [imath]\lim_{n\to\infty}a_{n}=L[/imath]? If so, how can I make this more rigorous? Is there a theorem I can state that covers this case? | 3038788 | True or false: a sequence [imath]\{a_{n}\}[/imath] is convergent if and only if the subsequences [imath]a_{2n}[/imath] and [imath]c_{2n + 1}[/imath] both converge to the same number
True or false: a sequence [imath]\{a_{n}\}[/imath] is convergent if and only if the subsequences [imath]a_{2n}[/imath] and [imath]c_{2n + 1}[/imath] both converge to the same number I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample. |
2552747 | Remembering the PI angles
What is a good way to remember the essential angles from the unit circle? For instance [imath]\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \text{ and } \sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}[/imath] | 628644 | How to memorize the families that are [imath]\sin[/imath], [imath]\cos[/imath], and [imath]\tan[/imath] of [imath]\pi[/imath] over something?
Is there any way to easily memorize these? Such as [imath]\sin \frac{\pi}{6} = 1/2[/imath]. Any help needed!!! |
2553012 | Isometric embedding [imath]e: (E, \Vert \cdot \Vert) \hookrightarrow (C_{\mathbb{K}}(X), \Vert \cdot \Vert_{\infty})[/imath]
In a course of functional analysis, I came across the following exercise which asks me to prove the existence of a specific space under which an isometric embedding exists. I do not seem to get which space it is. The exact exercise is as follows. Let [imath](E, \Vert \cdot \Vert)[/imath] be a normed vector space over [imath]\mathbb{K}[/imath]. Prove that there exists a compact topological space [imath]X[/imath] which also possesses the Hausdorff property and satisfies a linear isometric embedding [imath]e: (E, \Vert \cdot \Vert) \hookrightarrow (C_{\mathbb{K}}(X), \Vert \cdot \Vert_{\infty}).[/imath] Also show that if [imath]E[/imath] is complete, then [imath]E[/imath] is isometrically isomorph to a closed subspace of [imath]C_{\mathbb{K}}(X)[/imath]. Since I do not know which [imath]X[/imath] is meant I also cannot figure out the second partial task. Any suggestions or hints are appreciated. | 542545 | Every Banach space is isometric to a subspace of [imath]C(X)[/imath] for some compact space [imath]X[/imath]
The book of Douglas says on page 12: Theorem (Banach): Every Banach space [imath]B[/imath] is isometrically isomorphic to a closed subspace of [imath]C(X)[/imath] for some compact Hausdorff space [imath]X[/imath]. Proof: Let [imath]X[/imath] be [imath](B^*)_1[/imath] with the w*-topology and define [imath]\beta[/imath]: [imath]B \rightarrow C(X)[/imath] by [imath](\beta f)(\phi) = \phi(f)[/imath] with [imath]\phi \in (B^*)_1[/imath] and [imath]f \in B[/imath]. Next he prove the linearity of [imath]\beta[/imath] and the fact that [imath]\beta[/imath] is an isomorphism using the Hahn-Banach theorem My question is: What is [imath]\beta[/imath]? I don't understand much of its definition. Isn't supposed to be a function from [imath]B \rightarrow C(X)[/imath]? Here it seems to me a function [imath]B \rightarrow X[/imath] because it's equal to [imath]\phi(f)[/imath] that is a function in [imath]X = (B^*)_1[/imath]. |
1300866 | Suppose that [imath]P[/imath] is a monic polynomial of degree [imath]n[/imath] in one variable with real coefficients and [imath]K[/imath] a real number.Then which is /are true?
Suppose that [imath]P[/imath] is a monic polynomial of degree [imath]n[/imath] in one variable with real coefficients and [imath]K[/imath] a real number.Then which is /are true? If [imath]n[/imath] is even and [imath]K>0[/imath] , then there exists [imath]x_0 \in \mathbb{R}[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. If [imath]n[/imath] is odd and [imath]K<0[/imath] , then there exists [imath]x_0 \in \mathbb{R}[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. For any natural number [imath]n[/imath], and [imath]0<K<1[/imath] then there exists [imath]x_0 \in \mathbb{R}[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. If [imath]n[/imath] is odd and [imath]K \in \mathbb{R}[/imath] , then there exists [imath]x_0 \in \mathbb{R}[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. I feel that Rolles Theorem will be helpful but cant apply i. Can i get some help please. | 1410008 | [imath]P[/imath] is a monic polynomial of degree [imath]n[/imath] , then which are correct?
Suppose that [imath]P[/imath] is a monic polynomial of degree [imath]n[/imath] in one variable with real coefficients and [imath]K[/imath] is a real number. Then which of the following statements are necessarily correct ? If [imath]n[/imath] is even and [imath]K>0[/imath] then there exists [imath]x_0\in \mathbb R[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. If [imath]n[/imath] is odd and [imath]K<0[/imath] , then there exists [imath]x_0 \in \mathbb{R}[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. For any natural number [imath]n[/imath], and [imath]0<K<1[/imath] then there exists [imath]x_0 \in \mathbb{R}[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. If [imath]n[/imath] is odd and [imath]K \in \mathbb R[/imath] , then there exists [imath]x_0 \in \mathbb{R}[/imath] such that [imath]P(x_0)=Ke^{x_0}[/imath]. For (3), consider, [imath]P(x)=x^2+5[/imath]. Then , there does not exist [imath]x_0\in \mathbb R[/imath] such that [imath]0<e^{-x_0}P(x_0)<1[/imath]. So it is FALSE. But what about the others ? |
2551737 | No meromorphic function dominates another meromorphic function?
It is well known that "genuinely different" entire functions cannot dominate each other. More precisely, let [imath]f[/imath] and [imath]g[/imath] be entire functions in [imath]\mathbb{C}[/imath] satisfying [imath]|f(z)|\le |g(z)|[/imath] for all [imath]z \in \mathbb{C}.[/imath] Then [imath]f=Cg[/imath] for some [imath]C\in \mathbb{C}.[/imath] I'd like to know the result for meromorphic functions similar to the above result for entire functions. Please let me know if you have any comment about this question. Thanks in advance! | 902256 | [imath]f(z)[/imath] and [imath]g(z)[/imath] are Meromorphic functions such [imath]|f(z)|\le|g(z)|[/imath] for all [imath]z\in\mathbb{C} [/imath] then [imath] f=ag[/imath]
We know that if [imath]f(z)[/imath] and [imath]g(z)[/imath] are entire functions such that [imath]g(z)\ne0[/imath] and [imath]|f(z)|\le|g(z)|[/imath] for all [imath]z\in\mathbb{C} [/imath] then by Liouville's theorem [imath] f=ag[/imath] for some constant [imath]a\in \mathbb{C} [/imath] . Now my question is this that similar to above argument, if [imath]f(z)[/imath] and [imath]g(z)[/imath] are Meromorphic functions such [imath]|f(z)|\le|g(z)|[/imath] for all [imath]z\in\mathbb{C} [/imath] then I want to show [imath] f=ag[/imath] for some constant [imath]a\in \mathbb{C} [/imath] . I am thinking in this way that because because poles and zeros of Meromorphic functions are isolated , by the Riemann's theorem on removable singularities and By using analytic continuation to eliminate removable singularities also we can have [imath] f=ag[/imath] for some constant [imath]a\in \mathbb{C} [/imath] . Is this true way? |
2551914 | Using Riemann integrals to find limits
Evaluate [imath]\lim_{n\to\infty}\frac{\sum_{k=1}^n\sqrt{k}}{\sum_{k=1}^n\sqrt{n+k}}[/imath] I can't seem to remove the [imath]\frac{1}{n}[/imath] term in order to use the riemann integral any help? | 2013189 | Use Riemann integral to evaluate the limit [imath]\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^n \sqrt{k}}{\sum_{k=1}^n \sqrt{n+k}}[/imath]
Use Riemann integral to evaluate the limit [imath]\lim_{n\to\infty}\frac{\sum_{k=1}^n \sqrt k}{\sum_{k=1}^n\sqrt{n+k}}[/imath] |
2550403 | convergence of [imath]\sum_{n=1}^{\infty} \frac{a^n}{n!}[/imath]
How can i check convergence of [imath]\sum_{n=1}^{\infty} \frac{a^n}{n!}[/imath] ? I tried some of the tests for checking if it's convergent , but it's does not work. | 2136683 | Convergence of the [imath]e^x[/imath] Taylor series
I have figured out a way to show that [imath]e=\sum_{i=0}^\infty {1\over i!}[/imath] I am wanting to formally show that [imath]e^x = \sum_{i=0}^\infty {x^i\over i!}[/imath] I have been looking at power series/Taylor series for a long period of time (absolute convergence) and have seen multiple proofs that I look past because something seems illegitimate with radius of convergence. If someone explains the general proof behind Taylor series/power series absolutely converging so there is no gray area, that would work as well. I will probably ask some questions in the comment area if this is the case. There must be something I am missing. |
2550850 | poinwise limit and limit in convergence is same if they are continuous
let [imath]{f_k}[/imath] be a sequence of integrable functions on [imath][a,b][/imath] that converges in pointwise to [imath]f[/imath] and converges in mean to [imath]g[/imath]. If [imath]f[/imath] and [imath]g[/imath] are continuous, then f = g. The 'convergence in mean' means, [imath]\lim_{k\to \infty}\int_{a}^{b} [f_k(x) - g(x)]^2 dx = 0[/imath] . I tried to show that, if there exists some point [imath]x_0[/imath] s.t [imath]f(x_0) \neq g(x_0)[/imath], then on some neighborhood of [imath]x_0[/imath], [imath]f_k-g[/imath] is being uniformly nonzero, so that the value of integration can't be 0, so that [imath]f_k[/imath] can't converge in mean to [imath]g[/imath]. However, I couldn't prove this way, since and [imath]f_k[/imath] is integrable, but not provided that [imath]f_k[/imath] is continuous. Am I trying in a correct way? give me any idea please. (This question is from elementary analysis textbook, so please assume that properties about [imath]L_2[/imath] convergence or almost everywhere convergence is not known) | 2549933 | Pointwise Convergence and Convergence in the mean of Riemann-integrable functions
Let [imath](f_n)_{n\in\mathbb{N}}[/imath] be a sequence of Riemann-integrable functions on [imath][a, b][/imath]. Assume that [imath](f_n)_{n\in\mathbb{N}}[/imath] converges pointwise to [imath]f[/imath] on [imath][a, b][/imath] and converges in the mean to [imath]g[/imath] on [imath][a, b][/imath], where [imath]f[/imath] and [imath]g[/imath] are both continuous. Show that [imath]f = g[/imath]. ([imath](f_n)_{n\in\mathbb{N}}[/imath] converges in the mean to g on [imath][a , b][/imath] means that [imath]\lim_{n\to\infty} d(f_n, g) = 0[/imath], where [imath]d[/imath] is the [imath]L_2[/imath] metric.) I am studying real analysis now and want to prove this statement. But I don't know how to use condition "[imath](f_n)_{n\in\mathbb{N}}[/imath] converges in the mean to g on [imath][a , b][/imath]." How can I prove this? |
810825 | Which of the statements are true (CSIR)?
Question : Let [imath]f[/imath] be real valued function on [imath]R^3[/imath] satisfying (for a fixed [imath]\alpha \in \mathbb R[/imath]) [imath]f(rx) = r^{\alpha}f(x)[/imath] for any [imath]r>0[/imath] and [imath]x \in \mathbb R^3[/imath]. Then which of the following statements are true. If [imath]f(x) = f(y)[/imath] whenever [imath]||x|| = ||y|| = \beta[/imath] for a [imath]\beta >0[/imath], then [imath]f(x) = \beta ||x||^{\alpha}[/imath] If [imath]f(x) = f(y)[/imath] whenever [imath]||x|| = ||y|| = 1[/imath] for a [imath]\beta >0[/imath], then [imath]f(x) = ||x||^{\alpha}[/imath] If [imath]f(x) = f(y)[/imath] whenever [imath]||x|| = ||y|| = \beta[/imath] for a [imath]\beta >0[/imath], then [imath]f(x) = c||x||^{\alpha}[/imath] for a some constant c. If [imath]f(x) = f(y)[/imath] whenever [imath]||x|| = ||y|| [/imath] , then [imath]f[/imath] must be constant function. Please help me to see this question more clearly and help me to answer this. | 363612 | is [imath]f[/imath] constant or dilation
[imath]f:\mathbb{R}^3\to\mathbb{R}, f(rx)=r^{\alpha}f(x)[/imath] for some [imath]\alpha>0[/imath], [imath]\forall x\in\mathbb{R}[/imath], for any [imath]r\in\mathbb{R}[/imath], could any one tell me which of the following is true? [imath]1.[/imath] If [imath]f(x)=f(y)[/imath] whenever [imath]||x||=||y||=\beta>0[/imath], then [imath]f(x)=\beta ||x||^{\alpha}[/imath] [imath]2.[/imath] If [imath]f(x)=f(y)[/imath] whenever [imath]||x||=||y||=1[/imath], then [imath]f(x)= ||x||^{\alpha}[/imath] [imath]3.[/imath] If [imath]f(x)=f(y)[/imath] whenever [imath]||x||=||y||=1[/imath], then [imath]f(x)=c||x||^{\alpha}[/imath] for some constant [imath]c[/imath] [imath]4.[/imath] If [imath]f(x)=f(y)[/imath] whenever [imath]||x||=||y||[/imath], then [imath]f(x)[/imath] must be a constant function. Thank you. |
2552499 | Show that [imath]\mathbb{R}^{n}[/imath] ([imath]n>2[/imath]) minus a countable number of lines is still path-connected
Define [imath]C[/imath] to be a set of countable number of lines. Let [imath]p,q[/imath] be a couple of points in [imath]\mathbb{R}^{n}\setminus C[/imath]. Then if the line segment [imath]\bar{pq}\cap C=\emptyset[/imath], we're done. If [imath]\bar{pq}\cap C\neq\emptyset[/imath], the intersection is a set of countable number of points. Since [imath]n>2[/imath], at each point in the intersection, there exists an arc that goes around the point. Thus, there exists a path connecting [imath]p[/imath] and [imath]q[/imath] that goes along the line segment [imath]\bar{pq}[/imath] with arcs replacing the points in [imath]\bar{pq}\cap C[/imath]. Therefore, [imath]\mathbb{R}^{n}\setminus C[/imath] is path-connected. It's pointed out to me that this proof is not exactly correct. I don't know where it went wrong. | 2465578 | Let [imath]X[/imath] be obtained by removing finitely many lines from [imath]\mathbb{R}^3[/imath]. How do I prove [imath]X[/imath] is connected?
Let [imath]X[/imath] be obtained by removing finitely many lines from [imath]\mathbb{R}^3[/imath]. How do I prove [imath]X[/imath] is connected? I think the way to go about this is to show [imath]X[/imath] is path-connected but I'm not quite sure. |
2553590 | Prove that [imath]\tan 2A = \frac{2\tan A}{1- \tan^2 A} [/imath]
Prove that [imath]\tan 2A = \frac{2 \tan A}{1- \tan^2 A} [/imath] using identities of [imath]\sin 2A[/imath] and [imath]\cos 2A[/imath] Can I get a hint on how do I start this ? Identities from [imath]\sin 2A[/imath] and [imath]\cos 2A[/imath] doesn't give me a [imath]\tan A[/imath] so how do I prove it ? | 787255 | use the fact that [imath]\tan 2x=\sin2x \ /\cos2x[/imath] to prove that [imath]\tan 2x=2\tan x/(1-\tan^2x)[/imath]
I need to use the fact that [imath]\tan 2x=\sin2x \ /\cos2x[/imath] to prove that: [imath]\tan 2x=\frac{2\tan x}{1-\tan^2x}[/imath] I don't know where to start. Please help or hint. Thanks in advance. |
2555201 | Solve the differential equation [imath](2x^2+3y^2-7)x\mathrm{d}x - (3x^2+2y^2-8)y\mathrm{d}y = 0[/imath]
Problem: Solve the differential equation [imath](2x^2+3y^2-7)x\mathrm{d}x - (3x^2+2y^2-8)y\mathrm{d}y = 0[/imath] My attempt: [imath](2x^2+3y^2-7)x\mathrm{d}x - (3x^2+2y^2-8)y\mathrm{d}y = 0[/imath] Let [imath]u = x^2 +y^2 -3[/imath] [imath]2u = 2x^2+2y^2-6[/imath] so, [imath]2x^2+3y^2-6= 2u +(y^2-1)[/imath] and [imath]3x^2+2y^2-8 = 3u-(y^2-1)[/imath] so, [imath]\frac{x}{y} \frac{\mathrm{d}x}{\mathrm{d}y} = \frac{3u-(y^2-1)}{2u+(y^2-1)}[/imath] To convert the equation into variables [imath]u[/imath] and [imath]y[/imath], [imath]\frac{\mathrm{d}(x^2+y^2-3)}{\mathrm{d}y} = \frac{\mathrm{d}u}{\mathrm{d}y}= 2x \frac{\mathrm{d}x}{\mathrm{d}y}+2y[/imath] Solving for [imath]\frac{\mathrm{d}x}{\mathrm{d}y}[/imath] [imath]\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{2x}\biggr(\frac{\mathrm{d}u} {\mathrm{d}y}-2y\biggr)[/imath] Plugging in the problem [imath]\frac{x}{y}\cdot\frac{1}{2x}\biggr(\frac{\mathrm{d}u} {\mathrm{d}y}-2y\biggr)= \frac{3u-(y^2-1)}{2u+(y^2-1)}[/imath] Upon solving, [imath]\frac{1}{2y}\frac{\mathrm{d}u} {\mathrm{d}y}= \frac{5u}{2u+(y^2-1)}[/imath] Moreover, let [imath]t = y^2-1[/imath] [imath]\mathrm{d}t= 2y\mathrm{d}y[/imath] Plug in the above equation (ambiguous step?), [imath]\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{5u}{2u+t}[/imath] Even now, I can't separate the variables. All suggestions are welcome. Please attempt to provide alternate solutions. Thanks! | 2103851 | [imath](2x^2+3y^2-7)xdx-(3x^2+2y^2-8)ydy=0[/imath]
The question is basically to find out the solution of the differential equation [imath](2x^2+3y^2-7)xdx-(3x^2+2y^2-8)ydy=0[/imath]. Since the given differential equation is non homogeneous I tried to find out the point of intersection of [imath]2x^2+3y^2-7=0[/imath] and [imath]3x^2+2y^2-8=0[/imath].I got the point of intersection and then I tried to shift the coordinate system to that point.this removed the constant terms.However I was not able to proceed further.Please help me in this regard.thanks. |
2555257 | Show that the following equation has exactly [imath]n[/imath] real roots.
If [imath]0<a_1< \cdots < a_n[/imath], show that the following equation has exactly [imath]n[/imath] real roots. [imath] \frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+ \frac{a_3}{a_3-x}+ \cdots + \frac {a_n}{a_n - x} = 2015[/imath] My working: The LHS easily becomes [imath]a_1(a_2-x)...(a_n-x)+a_2(a_1-x)(a_3-x)...(a_nx)+.....+a_n(a_1-x)(a_2-x)...(a_{n-1} -x)[/imath] This shows it has [imath]n[/imath] roots But what about exactly and real? | 2275845 | Proving it has all real roots
Let [imath]0<a_1<a_2<\cdots<a_n[/imath] be real numbers. Show that the equation has exactly [imath]n[/imath] real roots. [imath]\displaystyle \sum_{i=1}^n \dfrac{a_i} {a_i-x} =2015[/imath] I multiplied out denominators and that was very long, but I don't think descarte's rule will be helpful. Please any help is appreciated. |
2555326 | What is the maximum probablility of survival?
Suppose you are a prisoner. The king gives you [imath]100[/imath] marbles, [imath]50[/imath] black and [imath]50[/imath] white. You have to put all the marbles into [imath]2[/imath] urns such that none of the urns is empty. The king will then choose an urn at random and from that a marble at random. If the marble is white, you will be set free, otherwise you will be sentenced to death. So you have to distribute the marbles in such a way so as to maximize the chance of survival. Basically you have to maximize the probability that the king chooses white marble. My attempt : Let there be [imath]b[/imath] black and [imath]w[/imath] white marbles in the 1st urn . Therefore, there will be [imath]50-b[/imath] black and [imath]50-w[/imath] white marbles in the 2nd urn. Therefore P(Survival) = P(Choosing white marble) = [imath]\frac{1}{2}\cdot\frac{w}{w+b} +\frac{1}{2}\cdot\frac{50-w}{100-w-b}[/imath] . Now I need to maximize the above equation but I don't know how. Any ideas? | 2022884 | Probability: A flaw in logic? The emperor's proposition with marbles and two urns
I've tried searching for this question but couldn't find it on stackexchange. This is a common type of interview question; I ran into it doing brain teasers on a probability puzzles app, and if you fine people agree with my logic, I will inform the app developer that his/her answers are incorrect. The problem is essentially this: You are sentenced to death for thievery. The King is magnanimous and decides to put your fate in the hands of chance. You are given [imath]100[/imath] white marbles and 100 black marbles, and [imath]2[/imath] urns. The king will choose an urn at random and pull out a single marble at random; if the marble is white, you live, if its black, you die. If you place the marbles in the best way possible, what is your probability of survival? I started with the base case: [imath]100[/imath] white marbles in one urn, [imath]100[/imath] black marbles in the other. This comes down to a [imath]50[/imath]-[imath]50[/imath] chance of survival. I then worked my way to deciding that placing [imath]1[/imath] white marble in one urn and [imath]99[/imath] white marbles + [imath]100[/imath] black marbles in the other urn would be the "best way possible", which yields the following: [imath]P(\text{Survival}) = \frac{1}{2}(1+\frac{99}{199}) \approx .749[/imath] Selecting [imath]1[/imath] of [imath]2[/imath] urns at random gives [imath]\frac{1}{2}[/imath], the urn containing [imath]1[/imath] marble gives [imath]1[/imath], and the other that contains [imath]99[/imath] white marbles and [imath]100[/imath] black marbles gives [imath]\frac{99}{199}[/imath] because there are [imath]99[/imath] possible white marbles to select out of [imath]199[/imath] total marbles. The app claims that the correct answer is [imath]\frac{1}{2}(1+\frac{99}{200}) \approx .748[/imath] I see where the [imath]200[/imath] comes from, but I do not think it is right to say that there are [imath]200[/imath] marbles in the other urn. Who is correct? |
2554072 | Express the negation of the statement limit
Express the negation of the statement: [imath] \forall \epsilon >0 , \exists \delta>0 , \forall x , 0 <|x-a| <\delta \Rightarrow |f(x)-a| <\epsilon \ [/imath] Answer: The negation is- [imath] \exists \epsilon >0 , \forall \delta>0 , \exists x , 0 <|x-a| <\delta \Rightarrow |f(x)-a| > \epsilon \ [/imath] Am I right? | 1473090 | Negation of the Definition of Limit of a Function
Question Suppose we are dealing with real valued functions [imath]f(x)[/imath] of one real variable [imath]x[/imath]. We say that the limit of [imath]f(x)[/imath] at point [imath]a[/imath] exists and equals to [imath]L[/imath] if and only if [imath]\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)[/imath] and show this by the symbolism [imath]\mathop {\lim f(x)}\limits_{x \to a} = L[/imath] Now, what do we say when we want to state that the limit of function [imath]f(x)[/imath] does not exist at [imath]a[/imath]? In fact, what is the negation of the above statement? I am interested to obtain the negation with a step by step approach using tautologies in logic. To be specific, I want to start from [imath]\neg \left[ {\exists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)} \right][/imath] and then go through it to get the final form of negation (See the example below). My Thought I just wrote down the two following negations without going through a step by step approach. [imath]\forall L,\exists \varepsilon > 0:\left( {\forall \delta > 0,\exists x:0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| \ge \varepsilon } \right)[/imath] or [imath]\nexists L:\left( {\forall \varepsilon > 0,\exists \delta > 0:\left( {\forall x,0 < \left| {x - a} \right| < \delta \implies \left| {f(x) - L} \right| < \varepsilon } \right)} \right)[/imath] I want to know weather these are true or not. Example I will give an example of what I mean by a step by step approach. Consider the following statement [imath]{P}\implies R[/imath] I want to take a step by step approach to obtain the negation of the above statement. Here is what they usually do in logic \begin{align} \, \neg \left( {P \implies R} \right) &\iff \neg \left( {\neg P \vee R} \right) & \text{Conditional Disjunction} \\ \qquad \qquad \quad &\iff \neg \neg P \wedge \neg R & \text{Demorgan's Law} \\ \qquad \qquad \quad &\iff P \wedge \neg R & \text{Double Negation} \end{align} |
2554265 | Can the result for [imath]\prod_{p\ge 3}\left(1-\frac 2p\right)[/imath] be extended for general [imath]\prod_{p\ge n+1}\left(1-\frac np\right)[/imath]?
In the excellent answer at https://math.stackexchange.com/a/22435/86846, the identity [imath]\log\left(1-\frac{2}{p}+\frac{1}{p^2}\right) +\log\left(1- \frac{1}{(p-1)^2} \right)=\log\left(1- \frac{2}{p} \right)[/imath] was put forward as a starting point. I thought I would try a similar thing for estimating [imath]\prod_{3\lt p\le x}\left(1-\frac 3p\right)[/imath] and thus use some identity which might be along the lines of [imath]\log\left(1-\frac 3p\right)=\log\left(1-\frac 1p\right)+\log\left(1-\frac 2p\right)+\log\left(1-\frac 2{p-1}-\frac 2{p-2}\right)[/imath] but as I work with these quantities I realize that no identity will completely remove the factor [imath]p-3[/imath] from the results entirely. In my search results I keep finding the same "Euler Product" and "Mertens' Theorem" pages on Wikipedia and Mathworld, but none of those pages has a [imath]p-3[/imath] factor listed that I have noticed, nor any suggestion as to how to approach the broader question. I am not especially worried about the accuracy of the error term, as long as the error is not asymptotically as big as the rest of the result. Are there known results that my primitive searches haven't found yet? Are there any reference materials I could look at that might help me work out some of these bounds myself, or perhaps some simple observation that has a relatively large error term but simplifies the approach in general? | 741633 | Derivation of a generalization of Mertens' Third Theorem.
One of Mertens' Theorems states [imath]\prod_{p\le x}(1-\frac{1}{p})\sim \frac{e^{-\gamma}}{\ln(x)}.[/imath] I have seen a generalized version that states [imath]\prod_{m<p\le x}(1-\frac{m}{p})\sim \frac{c(m)}{(\ln(x))^m},[/imath] where [imath]m[/imath] is a positive integer and [imath]c(m)[/imath] is a real number that depends on [imath]m[/imath]. How can I derive this result? Is there a way to find what the values of [imath]c(m)[/imath] are? Thank you very much. |
2555389 | A ring is never a direct sum of infinitely many non-zero left ideals?
Is it true or false that a ring is never a direct sum of infinitely many non-zero left ideals? It should be false because we can make a direct product of infinite copies of [imath]\Bbb{R}[/imath], and every copy is a left ideal. But it was written in a book I was doing. I don't understand the reasoning behind it. | 1093631 | Ring With Unity As a Direct Sum of Non-Zero Ideals
Let [imath]R[/imath] be a ring with unity. Let [imath] (a_i),i\in I [/imath] be a family of non-zero ideals of [imath]R[/imath]. Suppose that [imath]R[/imath] is a direct sum of the family [imath] (a_i),i\in I [/imath] (i.e the additive group [imath]R[/imath] is a direct sum of the subgroups [imath] a_i,i\in I [/imath]). I wish to prove that in this case [imath]I[/imath] must be a finite set. I will explain the argument which I gave until the point of my question: Take the identity element [imath] 1\in R [/imath]. Since [imath]R[/imath] is the direct sum of the [imath]a_i[/imath], there exists a finite subset [imath]I_0 \subseteq I[/imath] and elements [imath]e_i \in a_i \setminus 0 , i \in I_0[/imath] such that [imath]1 = \sum_{i \in I_0} e_i [/imath]. Then, for every [imath]x \in R[/imath] we have [imath]x = x.1 = \sum_{i \in I_0} xe_i \in \sum_{i \in I_0} a_i [/imath] since each [imath]a_i[/imath] is an ideal, and so [imath]R = \sum_{i \in I_0} a_i [/imath]. My idea now is to prove that one must have [imath]I = I_0[/imath], but I have a feeling that I'm missing something trivial. Can anyone help me with this final step if possible? Thank you in advance. On a final note, I'm assuming that each [imath]a_i[/imath] is an ideal, but as far as I was informed, this is also true for left ideals (or right ideals). |
2556371 | Basic equinumerous examples
If [imath]a < b[/imath] are real numbers, then [imath](a,b) \sim (0,1)[/imath] [imath][a,b] \sim (0,1)[/imath] For the first one, it is quite simple. Pf (1): Let [imath]f:(a,b)\rightarrow (0,1)[/imath] be defined by [imath]f(x)=\frac{x-a}{b-a}[/imath]. Clearly, [imath]f[/imath] is a bijection. Pf(2): Let [imath]g:[a,b]\rightarrow (0,1)[/imath]... (here I want to say something along the lines of) be defined by [imath]g(x)=\frac{x-a}{2(b-a)}, a<x<b[/imath] which clears space for a and b as they are included.. Is this reasoning correct? How do I formally write up the a and b part? Looking for an explanation. | 205392 | Intervals and cardinality
Let [imath]x,y,a,b[/imath] be real numbers. For [imath]x<y[/imath] and [imath]a<b[/imath], show the cardinality of [imath][a,b][/imath] equals the cardinality of [imath][x,y][/imath]. I did the above problem, by defining a linear function as a bijection between the intervals. The next part of the problem states, Extend your result (from above) to open and half-open intervals. I'm just not sure what exactly I'm supposed to do next, the question seems vague. Am I supposed to show [imath](a,b)\sim (x,y)[/imath]? [imath][a,b]\sim (x,y)[/imath]? [imath][a,b)\sim (x,y][/imath]? |
2554915 | Calculate [imath]Pr(X \ge 0, Y \ge 0)[/imath] where [imath]X, Y \sim N(0,1)[/imath] and their correlation is 1/2
I've been working on this problem and got stuck. Assume that X and Y have joint normal distribution, that each [imath]X, Y \sim N(0,1)[/imath], and that their correlation is [imath]\frac{1}{2}[/imath]. Calculate [imath]Pr(X \ge 0, Y \ge 0)[/imath]. I know that I can rewrite (X,Y) in terms of independent normal variables (W,T) as follows [imath] \begin{bmatrix} X\\ Y\\ \end{bmatrix} = \begin{bmatrix} 1 & 0\\ \rho & \sqrt{1-\rho^2}\\ \end{bmatrix} \begin{bmatrix} X\\ Z\\ \end{bmatrix} [/imath] where [imath]\rho[/imath] is the correlation. This has left me with [imath] X=X\\ Y=\frac{1}{2}X + Z\sqrt{\frac{3}{4}} [/imath] And so I have [imath] Pr(X \ge 0, Y \ge 0) = Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0) [/imath] So I just realized that the T variable I had before was superfluous, so I removed it. I am wondering from here if it would make sense to put this: [imath] Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0)=Pr(X \ge 0, Z\sqrt{\frac{3}{4}} \ge -\frac{1}{2}X) [/imath] Still not sure how to proceed from here. Maybe simplify to [imath] Pr(X \ge 0, \frac{1}{2}X + Z\sqrt{\frac{3}{4}} \ge 0)=Pr(X \ge 0, Z\sqrt{\frac{3}{4}} \ge 0) [/imath] Since we are looking for [imath]Pr(X \ge 0)[/imath] anyway? | 255368 | [imath]P(X>0,Y>0)[/imath] for a bivariate normal distribution with correlation [imath]\rho[/imath]
[imath]X[/imath] and [imath]Y[/imath] have a bivariate normal distribution with [imath]\rho[/imath] as covariance. [imath]X[/imath] and [imath]Y[/imath] are standard normal variables. I showed that [imath]X[/imath] and [imath]Z= \dfrac{Y-\rho X}{\sqrt{1-\rho^2}}[/imath] are independent standard normal variables. Using this I need to show that [imath]P(X >0,Y>0) = \frac14 + \frac{1}{2\pi} \cdot\arcsin(\rho).[/imath] |
2556143 | Proof Verification: Order/Cardinality of a coset
For G a group and H a subgroup of G, I'm required to show that [imath]|aH| = |bH|[/imath] for all [imath]a,b\epsilon G[/imath] Proof: Define a function [imath]f:aH\to bH[/imath] such that [imath]f(ah) = bh[/imath] for some [imath]h\epsilon H[/imath] Then we show, f is 1-1 and onto, i.e. f is a bijection 1) f is 1-1: let [imath]f(ah)=f(h')[/imath] then [imath]bh = bh'[/imath] multiplying by [imath]b^{-1}[/imath] to the left we get [imath] h= h'[/imath] Hence, f is injective 2) I have to show that f is onto but I don't know how. Can anyone let me know as well as verify the aforementioned proof for injectivity of f? | 187572 | Clarification on proof: Order of left cosets equal
There is a lemma that says that all left cosets [imath]aH[/imath] of a subgroup [imath]H[/imath] of a group [imath]G[/imath] have the same order. The proof given is as follows... The multiplication by [imath]a \in G[/imath] defines the map [imath]H \rightarrow aH[/imath] that sends [imath]h\mapsto ah[/imath]. This map is bijective because its inverse is multiplication by [imath]a^{-1}[/imath]. I don't quite understand the proof. Why does having a bijective map mean that all sets of left cosets have the same order? Thank you |
2556048 | Determine the convergence of [imath]\sum\limits_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} [/imath]
I'm having trouble determining the convergence of the series: [imath] \sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} [/imath] I am thinking it doesn't converge and since neither the root test or [imath]|\frac{a_{k+1}}{a_{k}}|[/imath] seemed to work for me I would have to use a comparing test Keep in mind I am not allowed to actually calculate what it converges to. | 2124581 | Find the infinite sum [imath]\sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} [/imath]
I am facing a problem, where I have to find the partial sum of a sequence/sum and with that, the infinite sum of the sequence. [imath]\sum_{k=1}^{\infty}\sqrt k - 2\sqrt {k + 1} + \sqrt {k + 2} [/imath] The problem here is that I don't know how to proceed. I would be thankful if someone would steer me in the right direction. |
2556178 | What is the proper way to notate the anti-derivative of x^-1
just a quick calculus question here. I'm a bit new to indefinite integrals and I was presented with this problem. Find [imath]f(x)[/imath] if [imath]f''(x) = x^{-2}[/imath]. Naturally it is easy to realize that the first derivative [imath]f'(x)[/imath] is simply [imath]-x^{-1}+C_1[/imath], however, when calculating the anti-derivative of [imath]-x^{-1}[/imath] to find [imath]f[/imath], what is the proper way to denote [imath]f(x)[/imath]? Because the natural answer is [imath]-x^0[/imath] due to the power rule but that obviously doesn't make any sense. | 2118082 | Why isn't [imath]\int \frac{1}{x}~dx = \frac{x^0}{0}[/imath]?
I know that [imath]\int \frac{1}{x}~dx = \ln|x| + C[/imath] and I know the antiderivative method works for all powers of [imath]x[/imath] except [imath]-1[/imath]. But why is that the case? I am still in high school and teachers aren't really helpful with these questions. Edit: I do realize dividing by zero is meaningless. That's not the question. Just because dividing by zero is meaningless doesn't mean mathematicians just choose to ignore it and work just make up a new answer. There has to be an explanation, and that's what I am asking for. The explanation. |
2555648 | Real projective space homeomorphic to sphere
Consider the unit sphere [imath]S^n[/imath] with the equivalence relation that identifies antipodal points of [imath]S^n[/imath]. Show that the quotient space [imath]S^n/\sim[/imath] is homeomorphic to the projective space [imath]\mathbb{RP}^n[/imath]. I know that [imath]\mathbb{RP}^n=\overline{B^n}/\sim[/imath], so if I can define a homeomorphism from [imath]\overline{B^n}[/imath] to [imath]S^n[/imath], I think that should be sufficient. My thought was to define a function on the ball so that every point on a line that goes through the origin is mapped to the outermost point, a point on the sphere. But I don't know how to write this explicitly. I have seen questions like this stated, with the explanation I came up with given, but there is no explicit definition of the homeomorphism. | 29730 | Why are these two descriptions of real n-dimensional projective space the same?
I saw this in Basic Topology by M.A.Armstrong. It gives three descriptions of real n-dimensional projective space [imath]P^n[/imath]. Two of them are: (a) Begin with the unit sphere [imath]S^n[/imath] in [imath]E^{n+1}[/imath] and identify its antipodal points. (c) Begin with the unit ball [imath]B^n[/imath] and identify antipodal points of its boundary sphere. I find it hard to imagine why these two descriptions lead to the same space. Can you please help? Thank you. |
2556192 | Number of ways to select [imath]k[/imath] books from a shelf with [imath]n[/imath] books so that no two adjacent books are selected
There are [imath]n[/imath] books in a line on a shelf and we want to choose [imath]k[/imath] of them in such a way that no two adjacent books are selected. Denote this number by [imath]{n \brack k}[/imath]. Show that [imath]{n \brack k}=\binom{n-k+1}{k}[/imath] for [imath]n \geq 1[/imath] whenever [imath]0 \leq k <\frac{n}{2}+1[/imath]. I want to show this by forming a bijection from the set of selections of [imath]k[/imath] books to the set of compositions of [imath]n-k+2[/imath] of length [imath]k+1[/imath]. The reason being that I know that this number is [imath]\binom{n+k-1}{k}[/imath]. Say the books are labeled [imath]b_1, \ldots, b_n[/imath]. Whenever I select the [imath]k[/imath] books, say [imath]b_{j_1}, \ldots, b_{j_k}[/imath], I know that [imath]|j_i-j_\ell|\geq 2[/imath] for each [imath]i \neq \ell[/imath]. I'm just not sure how to form a composition from this. **Recall that a composition of [imath]n-k+2[/imath] of length [imath]k+1[/imath] is a finite sequence of positive integers [imath](a_1, \ldots, a_{k+1})[/imath] such that [imath]\sum_{i=1}^k a_i=n-k+2[/imath]. | 2518827 | Counting the Lotto number draws which contain consecutive numbers
Witness a draw of lotto numbers, say '[imath]6[/imath] out of [imath]45[/imath]'—or [imath]49[/imath], or [imath]59[/imath], this may vary by locality—and watch out for two (or more) consecutive numbers in the draw, e.g. [imath]\,\lbrace 3,15,20,26,27,36\rbrace[/imath]. This appears to happen fairly often. How many draws are there, if a draw consists of [imath]\,k\,[/imath] numbers out of [imath]\,\lbrace 1,2,\dots,n\rbrace\:[/imath]? Assume that [imath]\,n\,[/imath] and [imath]\,1\,[/imath] are not adjacent, that [imath]k>1[/imath], and that [imath]k\leqslant\frac n2\,[/imath] if [imath]\,n\,[/imath] is even, or [imath]k\leqslant\frac{n+1}2\,[/imath] for odd [imath]\,n[/imath]. The preceding bounds for [imath]\,k\,[/imath] are reasonable upon regarding the draws [imath]\,\lbrace 2,4,\ldots,n\rbrace[/imath] or [imath]\,\lbrace 1,3,\ldots,n\rbrace[/imath]. My first approach was to choose the consecutive pair first, having [imath]n-1[/imath] possibilities, then drawing [imath]k-2\,[/imath] out of the remaining [imath]\,n-2\,[/imath] numbers, yielding [imath]\,(n-1)\binom{n-2}{k-2}\,[/imath] draws. But the associated probability [imath]{(n-1)\binom{n-2}{k-2}\over\binom{n}{k}}\:=\:\frac{k\,(k-1)}n[/imath] shows this cannot be correct. There must be some overcounting when the secondly drawn [imath]k-2[/imath] numbers are adjacent to the already fixed pair. A more weird approach was to consider the random variable [imath] X(d_1,\dots,d_k):=\sum_{i=1}^k(d_{i+1}-d_i)^2\;\text{ where }d_{k+1}=d_1[/imath] on the set of all draws [imath](d_1,\dots,d_k)[/imath] and somehow exploit that at least one summand equals [imath]1[/imath] when evaluated at a draw of interest. (It's then expected to be smaller than [imath]E[X][/imath].) |
1843654 | Every separable metric space has cardinality less than or equal to the cardinality of the continuum.
Every separable metric space has cardinality less than or equal to the cardinality of the continuum. This was stated in my book in a series of questions using the post office metric on [imath]\Bbb R^2[/imath], but I can't think of a way to prove this. | 2306597 | Cardinality of separable metric spaces
Show that [imath]\mathbb R[/imath] is cardinally larger than any separable metric space S. I have been trying to solve this on my own. My idea was to start by mapping the open balls of positive rational radius around the points in the dense set of S to the corresponding in [imath]\mathbb R[/imath]. Now any [imath]{x}\subset S[/imath] can be written as a limit of an appropriate sequence of open balls, which would correspond to a similar limit in [imath]\mathbb R[/imath], in case the limit exists. Then I would need to show that different limits are obtained in [imath]\mathbb R[/imath] for different [imath]x\in S[/imath]. This is a step I am not sure how to show. Alternatively, one could show that all possible sequences of rationals have no greater cardinality than reals. What would be a good way to show this? I know that all reals can be written as some convergent sequence of rationals, but I am looking for something slightly different here. I did not find this useful as it does not fill in the details: Every separable metric space has cardinality less than or equal to the cardinality of the continuum. |
2555456 | Conditional expected value of die roll
I need some help in solving the following problem. We roll a fair die repeatedly until see the number [imath]6[/imath]. Let [imath]X[/imath] be a random variable is a number of times we roll the die. Let [imath]A[/imath] be the event "The results of all rolls are even numbers". What is the value of [imath]\mathbb{E}(X|A)[/imath]? | 2463768 | Understanding The Math Behind Elchanan Mossel’s Dice Paradox
So earlier today I came across Elchanan Mossel's Dice Paradox, and I am having some trouble understanding the solution. The question is as follows: You throw a fair six-sided die until you get 6. What is the expected number of throws (including the throw giving 6) conditioned on the event that all throws gave even numbers? Quoted from Jimmy Jin in "Elchanan Mossel’s dice problem" In the paper it goes on to state why a common wrong answer is [imath]3[/imath]. Then afterwards explains that this problem has the same answer to, "What is the expected number of times you can roll only [imath]2[/imath]’s or [imath]4[/imath]’s until you roll any other number?" I don't understand why this is the case. If the original problem is asking for specifically a [imath]6[/imath], shouldn't that limit many of the possible sequences? I also attempted to solve the problem using another method, but got an answer different from both [imath]3[/imath] and the correct answer of [imath]1.5[/imath]. I saw that possible sequences could have been something like: [imath]\{6\}[/imath] [imath]\{2,6\}, \{4,6\}[/imath] [imath]\{2,2,6\}, \{2,4,6\}, \{4,2,6\}, \{4,4,6\}[/imath] [imath]\vdots[/imath] To which I set up the following summation and solved using Wolfram Alpha: [imath]\text{Expected Value} =\sum_{n=1}^\infty n\left( {\frac{1}{6}} \right)^n 2^{n-1} = 0.375[/imath] Obviously this is different and probably incorrect, but I can't figure out where the error in the thought process is. Any help on understanding this would be greatly appreciated. A blog post discussing the problem can be found here. |
2556835 | Extension degree of a splitting field
I want to show: If [imath]f(x) \in F[x][/imath] is a poly of degree n, then the splitting field of [imath]f(x)[/imath] over [imath]F[/imath] in [imath]\bar F[/imath] has an extension degree less or equal to [imath]n![/imath]. I think [imath]n![/imath] has something to do with permutation but I cannot proceed further. | 2318587 | Degree of a splitting field is no greater than [imath]n![/imath]
Is it possible to prove that if [imath]P[/imath] is a polynomial of degree [imath]n[/imath] then the degree of its splitting field is no greater than [imath]n![/imath] without using notion of Galois group? |
2555792 | Source: If [imath]f \in L^2[/imath] , then [imath]\hat{f} \in L^2[/imath] and [imath]2\pi ||f||_{L^2} = ||\hat{f}||_{L^2}[/imath], where [imath]\hat{f}[/imath] is the fourier transform
This claim turned up in one of my lectures, but there was no proof or source for it given (We don't have a textbook for this course). Does anyone have anymore information on this theorem, or an idea where I can learn more about it? EDIT: Looking at the Wiki article, they say that the following holds [imath]{\displaystyle \int _{-\infty }^{\infty }\|f(x)\|^{2}\,dx=\int _{-\infty }^{\infty }\|{\widehat {f}}(\xi )\|^{2}\,d\xi }[/imath] but then go on to say A more precise formulation is that if a function is in both L1(R) and L2(R), then its Fourier transform is in L2(R) What do they mean, "both in [imath]L^2[/imath] and [imath]L^1[/imath]" ? The above formula clearly states that if the left side is finite, ie [imath]f[/imath] is in [imath]L^2[/imath], then the right side is finite, ie [imath]\hat{f}[/imath] is in [imath]L^2 [/imath]? EDIT2: I think I get it, it is becuse they have defined the Fourier transform to only be applicable to [imath]L_1[/imath] functions, so they RHS might not exist if [imath]f[/imath] is not in [imath]L_1[/imath] | 1912143 | Step in the Stein's proof of Plancherel Theorem
I am reading Stein's Fourier Analysis, and I am confused by the proof of Plancherel's theorem. If [imath]f \in \mathcal S(\mathbb R)[/imath], then [imath]||f|| = ||\hat{f}||[/imath]. The norm here is defined in the [imath]L^2[/imath] sense. The proof goes like this. If [imath]f \in \mathcal S(\mathbb R)[/imath], define [imath]g(x) = \overline{f(-x)}[/imath]. Then [imath]\hat g(\xi) = \overline{\hat f(\xi)}[/imath]. Let [imath]h = f*g[/imath]. Then [imath]\hat h = \hat f \hat g = |\hat f|^2[/imath], and [imath]\displaystyle h(0) = \int f(-t)g(t) dt = \int f(-t)\overline{f(-t)} dt = \int|f|^2[/imath]. Apply Fourier inversion formula, and get [imath]\displaystyle h(0) = \int \hat h(\xi) d\xi[/imath]. Thus [imath]\displaystyle\int |\hat f|^2 = \int \hat h = h(0) = \int |f|^2[/imath]. I understand every step in this proof except for the setence [imath]\hat g(\xi) = \overline{\hat f(\xi)}[/imath]. I try to expand this thing, and get [imath]\displaystyle \hat g(\xi) = \int g(x)e^{-2i\pi x\xi} dx = \int \overline{f(-x)}e^{-2i\pi x\xi} dx = -\int \overline{f(u)}e^{2i\pi u\xi}du = -\overline{\hat f(\xi)}[/imath], which would make everything after this off by a minus sign. Could someone tell me what went wrong with my argument here? |
2557260 | Irreducibility of polynomial [imath]x^k-x^{k-1}-\ldots-x-1[/imath]
Can anybody give us a hint how to prove the irreducibility of polynomial [imath]x^k-x^{k-1}-\ldots-x-1[/imath] over [imath]\mathbb{Q}[/imath] ? Thanks in advance. Dusan | 334051 | Irreducibility of an infinite sequence of polynomials
Prove that [imath]P_n(X) = X^n - X^{n-1} - X^{n-2} - ... - X - 1[/imath] is irreducible over [imath]\mathbb{Z}[/imath] for all [imath]n[/imath]. I was able to prove the result for [imath]n=2^k-1[/imath] by applying Eisenstein's criterion to [imath]P_n(X+1)[/imath]. But for other values of [imath]n[/imath], I'm stuck. Has anyone an idea on this ? |
2557765 | Show that [imath]\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/imath]
Show that [imath]\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/imath] I have to show this using results I have found earlier. I started with [imath]0 = \frac{512}{10}+\sum_{n=1}^{\infty} 2048(\pi^2n^2-6)\frac{(-1)^n}{π^4n^4}[/imath] which was obtained from a fourier series. And have got to [imath]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4 }=\frac{ -7\pi^4}{720} [/imath] But I have no idea how to proceed with getting rid of the [imath](-1)^n.[/imath] Any help/tips would be greatly appreciated. Thanks in advance. | 1775879 | [imath] \sum_{n = 1}^{\infty} \frac{1}{n^4}[/imath]?
Using Fourier series I have managed to show that [imath] \frac{x^4}{12} = \frac{\pi^2 x^2}{6} + 4 \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^4}(1-\cos(nx)) , x \in [-\pi,\pi][/imath] From here apparently one need just one step to find an expression for [imath] \sum_{n = 1}^{\infty} \frac{1}{n^4}[/imath]. I need a hint what this step might be. |
2556988 | Prove the inequality [imath]f'(0) \ge -\sqrt {2}[/imath]
Let [imath]f[/imath] be a twice differentiable function on the open interval [imath](-1,1)[/imath] such that [imath]f(0)=1[/imath]. Suppose [imath]f[/imath] also satisfies [imath]f(x)\geq 0[/imath], [imath]f'(x)\leq 0[/imath], [imath]f''(x)\leq f(x)[/imath] for all [imath]x\geq0[/imath] Show that [imath]f′(0)\geq−{\sqrt[]{2}}[/imath]. All I am to see in this question is that [imath]f(x)[/imath] is decreasing in [imath](0,1)[/imath] but I don't know how to prove this inequality can somebody give me a hint on how to proceed with it. | 1072941 | If [imath]f[/imath] s twice differentiable and satisfies the following constraints, prove [imath]f'(0)>-\sqrt 2[/imath]
Let [imath]f[/imath] be a twice differentiable function on the open interval [imath](-1,1) [/imath]such that [imath]f(0)=1[/imath]. Suppose [imath]f[/imath] also satisfies [imath]f(x) \ge 0, f'(x) \le 0 [/imath]and [imath]f''(x) \le f(x)[/imath], for all [imath] x\ge 0[/imath]. Show that [imath]f'(0) \ge -\sqrt2.[/imath] My attempt [imath] \rightarrow[/imath] From Taylor's we see that [imath]0 \leq f(x)=f(0)+f'(0)x+\frac{f''(\zeta)x^2}{2} [/imath] for some [imath]\zeta \in (0,x) .[/imath] [imath]f'(x)\leq 0[/imath] and [imath] f'(x) \le 0 \implies[/imath] [imath] f''(\zeta) \leq f(\zeta)\leq 1[/imath] Thus [imath]1+f'(0)x+\frac{x^2}{2} \geq 0 [/imath]. I'm stuck upto this,I can see that if the discriminant of the quadratic is less than [imath] 0[/imath] we're done but how do I conclude that the discriminant would be [imath]0[/imath] ? |
2557944 | If the integral of f is 0 on every subinterval, show that the integral of |f| is 0
I'm trying to prove the following result: If [imath]f : [0,1] \to \mathbb{R}[/imath] is integrable and the integral of [imath]f[/imath] is [imath]0[/imath] on every subinterval [imath]I[/imath] of [imath][0,1][/imath], then the integral of [imath]\int_0^1|f|[/imath] is [imath]0[/imath]. I have no idea how to proceed. Can you please suggest a method of attack or provide a proof? Thank you. | 1659993 | Zero integral of measurable [imath]f[/imath] on every interval implies [imath]f=0[/imath]?
Let [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] be a Borel measurable function, such that [imath]\displaystyle \int_{a}^{b}f(x)dx=0[/imath] for every [imath]a<b[/imath] in [imath]\mathbb{R}.[/imath] Is it true that [imath]f(x)=0[/imath] for every [imath]x\in \mathbb{R}[/imath]? Comments: the starting point of this is to prove that the (Borel measurable) density function [imath]f[/imath] (if exists) of a symmetric random variable [imath]X[/imath] (meaning that [imath]X,~-X[/imath] have common cdf's) is even. If my question has a positive answer, it follows easily. Thanks a lot for any comments! |
2557940 | Let [imath]f:[0,\infty[ \rightarrow \mathbb{R}[/imath] be differentiable on [imath](0,\infty)[/imath] such that [imath]f'(x) \to b \; as \; x \to \infty[/imath]
Let [imath]f:[0,\infty[ \rightarrow \mathbb{R}[/imath] be differentiable on [imath](0,\infty)[/imath] such that [imath]f'(x) \to b \; as \; x \to \infty[/imath] (a) Show that for any [imath]h>0[/imath] we have [imath]\lim_{x \to \infty} \frac{f(x+h) - f(x)}{h}=b[/imath] (b) Show that if [imath]f(x) \to a [/imath] as [imath]x \to \infty[/imath] then [imath]b = 0[/imath] (c) Show that [imath]\lim_{x\to \infty} \frac{f(x)}{x}=b[/imath] I have no clue on how to approach this problem... I tried to show [imath]h \to 0[/imath] but I could not. | 2548583 | How to prove that [imath]\lim\limits_{x\to\infty} f(x)/x=L[/imath]
Let [imath]f\colon(0,\infty)\to\mathbb R[/imath] be differentiable on [imath](0,\infty)[/imath] and suppose that [imath]\lim\limits_{x\to\infty} f'(x)=L[/imath]. a. Show that for any [imath]h>0[/imath], [imath]\lim\limits_{x\to\infty} \frac{f(x+h)-f(x)}h=L[/imath]. b. Show that [imath]\lim\limits_{x\to\infty} \frac{f(x)} {x} =L[/imath]. this is a differentiation question. i don't know how to solve this question Because it is Korean, it is difficult to deal with this cite. So for me, it is not easy to write. thank you |
2521462 | Proving [imath](x-c)|(p(x)-p(c))[/imath]
Let [imath]F[/imath] be a field. [imath](x-c)|(p(x)-p(c))[/imath]. So if [imath]x-c|p(x)=q(x)-\frac{r}{x-c}[/imath] so for [imath]x-c|p(x)-p(c)=s(x-c)-\frac{r}{x-c}[/imath]. Is that correct? Or is there a different way to go about proving this? I have proved that [imath]x-c|p(x)[/imath] has a remainder of [imath]p(c)[/imath] so It wuld make sense that [imath]p(x)-p(c)[/imath] would be [imath]x-c[/imath]. | 2563966 | Why is [imath]x - a[/imath] a factor of [imath]p(x) - p(a)[/imath]?
I'm reading a proof in a linear algebra book. It mentions [imath]p(x) -p(c)= (x - c) h(x),[/imath] where [imath]c[/imath] is a constant, and [imath]p(x)[/imath] and [imath]h(x)[/imath] are polynomials. Can we always factor [imath]p(x) - p(c)[/imath] in this way? Please give a proof. |
2559009 | Show that [imath]\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}?[/imath]
[imath]\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}\tag1[/imath] How can we show that [imath](1)=\ln{3}[/imath] [imath]u=\sin^3{x}[/imath] then [imath]du=3\sin^2{x}\cos{x}dx[/imath] [imath]{1\over 3}\int{u\mathrm du\over \cos{x}+\cos^4{x}}\tag2[/imath] This is not a good substitution | 2198685 | How to show that [imath]\int_{0}^{\pi}{\sin^5 x\over 1+\cos^3 x}\mathrm dx=\ln 3?[/imath]
Given the integral [imath](1)[/imath] [imath]\int_{0}^{\pi}{\sin^5 x\over 1+\cos^3 x}\mathrm dx=\ln 3\tag1[/imath] This integral seems to be quite tough, I can't even know where to start. I have try substitution [imath]u=1+\cos^3 x[/imath] and [imath]u=\cos^3 x[/imath], can seem to remove the x variables. We have the following identities [imath]\sin^5a={5\over 8}\sin a-{5\over 16}\sin 3a+{1\over 16}\sin 5a\tag2[/imath] [imath]\cos^3 a={3\over 4}\cos a+{1\over 4}\cos 3a\tag3[/imath] We may write [imath](1)[/imath] as [imath]\int_{0}^{\pi}{\sin x(1-\cos^2 x)^2\over 1+\cos x(1-\sin ^2 x)}\mathrm dx\tag4[/imath] [imath]u=1-\sin^2 x\implies \mathrm du=-2\sin x\cos x\mathrm dx[/imath] then [imath](4)[/imath] becomes [imath]{1\over 2}\int_{0}^{1}{(1-u)^2\over \sqrt{u}+u^2}\mathrm du\tag5[/imath] [imath]u=v^2\implies du=2vdv[/imath] then [imath](5)[/imath] becomes [imath]\int_{0}^{1}{(1-v^2)^2\over v+v^4}\cdot{v\mathrm dv}\tag6[/imath] Simplify to [imath]\int_{0}^{1}{(1-v)(1-v^2)\over 1-v+v^2}\mathrm dv\tag7[/imath] I guess using partial decomposition of fraction and integrate term by term, seem lengthy process How else can we prove [imath](1)[/imath]? |
2559397 | What is the sum of all positive integers [imath]n[/imath] such that [imath]n^2+n+34[/imath] is a perfect square?
What is the sum of all positive integers [imath]n[/imath] such that [imath]n^2+n+34[/imath] is a perfect square? I tried to form a perfect square by factorising [imath]n^2+n+34[/imath] but was unsuccessful, and began to compute values whilst also using a graphing software. I found three number that were successful: [imath]5, 10, 33[/imath]. What is a simpler approach, as I am certain there is one. Thanks | 2152757 | Determine all integer values [imath]$ for which [/imath]^2 + 9 + 14$ is a perfect square.
Calculate the integer value(s) of [imath]n[/imath] for which [imath]^2 + 9 + 14[/imath] is a perfect square. Suppose [imath]^2 + 9 + 14 = k^2[/imath] where [imath]k\in\mathbb{Z}[/imath], reducing to the form [imath](2(n+k)+9) (2(n-k)+9) = 25[/imath]. My question is how can I calculate for such integer values when the values I've gotten aren't integers Thanks |
2559039 | Complex analysis and entire functions
Be [imath]f : \mathbb{C} \Rightarrow \mathbb{C}[/imath] a entire function. Supose that is exist [imath]a, b, c \in \mathbb{R}[/imath] such as [imath] a Re(f(z)) + b Im(f(z)) \leq c[/imath] for all [imath]z \in \mathbb{C}[/imath]. Show that [imath]f[/imath] is constant. I want prove that [imath]f[/imath] is bounded and apply the Liouville's Theorem, but i don't know show that [imath]f[/imath] is bounded. | 1703999 | Prove that the entire function is constant over certain conditions.
Let [imath]f[/imath] a entire function such that there exists [imath]3[/imath] real numbers not all zero [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] which [imath]a \Re(f(z))+ b\Im(f(z)) \leq c[/imath], [imath]z \in \mathbb{C}[/imath]. Show that [imath]f[/imath] is constant. I think we could consider the entire function [imath]e^{\alpha f(z)}[/imath] with [imath]\alpha=a+bi[/imath]. I know that [imath]|e^{\alpha f(z)}|= |e^{\Re({\alpha f(z)})+i \Im({\alpha f(z)})}|=e^{\Re({\alpha f(z)})}[/imath], but I don't know how to finish it. Is anyone could help at this point? Maybe we have to use Liouville theorem. |
2560272 | Prove that the following polynomial has at least one zero in the interval [imath](0,1)[/imath] by using Rolle's theorem
Suppose that: [imath]\frac{a_0}{n+1}+\frac{a_1}{n}+\cdots+a_n=0[/imath] Prove that the polynomial: [imath]a_ox^{n}+a_1x^{n-1}+\cdots+a_n[/imath] has at least one zero in the interval [imath](0,1)[/imath] using Rolle's theorem. | 2497775 | Prove that [imath]\sum_{k=0}^n a_k x^k = 0[/imath] has at least [imath]1[/imath] real root if [imath]\sum_{k=0}^n \frac{a_k}{k+1} = 0[/imath]
Knowing that [imath] \frac{a_0}{1} + \frac{a_1}{2} + \frac{a_2}{3} +\cdots + \frac{a_n}{n+1} =0[/imath] Prove that [imath] a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = 0[/imath] has at least one real solution. I suspect that it's proven with the intermediate value theorem. But can't find two numbers that satisfy it. |
2560250 | Let [imath]A[/imath] be a set of all continuous functions [imath]f:[0,1] \to \mathbb R^+ \cup \{0\}[/imath] that satisfies the following condition.
Let [imath]A[/imath] be a set of all continuous functions [imath]f:[0,1] \to \mathbb R^+ \cup \{0\}[/imath] that satisfies the following condition: [imath]\int_{0}^{x}f(t)dt \ge f(x), \forall x\in [0,1].[/imath] Which of the following statements is true? (A) [imath]A[/imath] has cardinality [imath]1[/imath]. (B) [imath]A[/imath] has cardinality [imath]2[/imath]. (C) [imath]A[/imath] is infinite. (D) [imath]A=\phi[/imath] Case 1 [imath]\int_{0}^{x}f(t)dt =f(x)[/imath], differentiating on both the sides. We get, [imath]f'(x)=f(x)\implies f(x)=Ke^{x}[/imath]. hence option (C) is correct. Am I correct? Please check my steps. how it is wrong? | 2536232 | Does [imath]\varphi(x)\le\int_0^x\varphi(t)dt[/imath] for all [imath]x\in[0,\infty)[/imath] imply [imath]\varphi\equiv0[/imath].
Let [imath]\varphi[/imath] be a nonnegative and continuous function on [imath][0,\infty)[/imath] and such that [imath]\varphi(x)\le\int_0^x\varphi(t)dt[/imath] for all [imath]x\in[0,\infty)[/imath]. Can we infer from here that [imath]\varphi\equiv0[/imath]. By taking limit on both sides, I got [imath]\varphi(0)\le 0[/imath]. As [imath]\varphi[/imath] is nonnegative, I can say [imath]\varphi(0)=0[/imath]. But what then. I could have progressed if I could show that [imath]\varphi(x)\le \varphi'(x)[/imath], but it is not quite evident from the given condition. |
2560269 | Relations of numbers or geometric distances.
Does there exist a real number [imath]r[/imath] such that [imath]\pi^r[/imath] is a rational number? Clearly, [imath](\sqrt{2})^2=2[/imath] etc. Is there any real number such that my question is true? | 399478 | Can you raise [imath]\pi[/imath] to a real power to make it rational?
We're all familair with this beautiful proof whether or not an irrational number to an irrational power can be rational. It goes something like this: Take [imath](\sqrt{2})^{\sqrt{2}}[/imath] If it's rational, then you proved it, if it's irrational, take [imath]((\sqrt{2})^{\sqrt{2}} ){^\sqrt{2}} = 2[/imath] and you've proved it. I'm wondering if you can raise [imath]\pi[/imath] or [imath]e[/imath] to a certain non-trivial real power to make it rational? And if not, where is the proof that it can't be done? p.s. - I almost left out the real part, but then I realized that [imath]e^{i\pi} = -1[/imath]. |
1696962 | Infinite series with e and pi
Please, help me to prove that [imath] \sum_{n=1}^\infty \frac{\cos(\pi n-\sqrt{\pi^2n^2-9})}{n^2}=-\frac{\pi^2}{12 e^3} [/imath] I found this fact here https://en.wikipedia.org/wiki/List_of_representations_of_e but there's no reference. | 1128657 | Gosper's unusual formula connecting [imath]e[/imath] and [imath]\pi[/imath]
Wolfram MathWorld quotes (see equation [imath](26)[/imath]) Gosper gives the unusual equation connecting [imath]\pi[/imath] and [imath]e[/imath] [imath]\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\cos\left(\frac{9}{n\pi + \sqrt{n^{2}\pi^{2} - 9}}\right) = -\frac{\pi^{2}}{12e^{3}}\tag{1}[/imath] I did not find any references to a paper in which Gosper proved the above formula. I am perplexed by this cosine term in the formula and it does not seem to resemble any familiar series. A direct proof of this formula would be greatly appreciated. Or if someone knows Gosper's paper where this formula is established then do provide a link to that paper. Update: We can see that [imath]\frac{9}{n\pi + \sqrt{n^{2}\pi^{2} - 9}} = n\pi - \sqrt{n^{2}\pi^{2} - 9}\tag{2}[/imath] and noting that [imath]\cos(n\pi - \alpha) = (-1)^{n}\cos \alpha[/imath] we can see that the formula of Gosper can be written as [imath]\sum_{n = 1}^{\infty}\frac{(-1)^{n}}{n^{2}}\cos\sqrt{n^{2}\pi^{2} - 9} = -\frac{\pi^{2}}{12e^{3}}\tag{3}[/imath] I think that the formula given by Gosper is probably a special case of a more general formula for the sum [imath]\sum_{n = 1}^{\infty}\frac{(-1)^{n}}{n^{2}}\cos\sqrt{n^{2}\pi^{2} + a^{2}}\tag{4}[/imath] (the formula [imath](3)[/imath] corresponds to [imath]a = 3i[/imath]). Looks like Gosper is also fond of strange formulas like Ramanujan. |
2560650 | Closed form solution for the series [imath]\sum _{n=1} ^{\infty} \frac{1}{2^n+1}[/imath]
I'd like to start by thanking you for your time reading this question and your passion for mathematics. It would be a pleasant surprise to me if there were a closed form expression for the sum: [imath]\sum _{n=1} ^{\infty} \frac{1}{2^n+1}[/imath] Knowing well that it converges, as term by term, [imath]\frac{1}{2^n+1}<\frac{1}{2^n}[/imath], I searched for a taylor series expansion, I rewrote the expression as [imath]\frac{1}{2^n}-\frac{1}{2^n \cdot (2^n+1)}=\frac {2^{-n}}{1+2^{-n}}=1-\frac {1}{1+2^{-n}}[/imath] in hopes of something, telescoping, nth partial-sum-exposing, or otherwise, all to no avail. So I ask more accomplished and brilliant minds their thoughts. Thank goodness for MSE. | 662795 | Closed-Form Solution to Infinite Sum
Does the convergent infinite sum [imath] \sum_{n=0}^{\infty} \frac{1}{2^n + 1} [/imath] have a closed form solution? Quickly coding this up, the decimal approximation appears to be [imath]1.26449978\ldots[/imath] |
2560732 | How to prove this relation between Ramsey Numbers: [imath]R(s,t)≤R(s,t−1)+R(s−1,t)-1[/imath] for [imath]s,t>2[/imath] when [imath]R(s,t-1)[/imath] and [imath]R(s-1,t)[/imath] are even.
I am beginner in combinatorics and have been following this book PRINCIPLE AND TECHNIQUES IN COMBINATORICS by CHEN CHUAN-CHONG. In this book the authors improves the relation [imath]R(s,t)≤R(s,t−1)+R(s−1,t)[/imath] for [imath]s,t>2[/imath] to above relation by making a constraint that [imath]R(s,t-1)[/imath] and [imath]R(s-1,t)[/imath] are both even. I am unable to follow through the author's reasoning for this proof. Please help. | 1927889 | Strengthening of a lemma in Ramsey Theory: [imath]R(m, n) \le R(m-1, n) + R(m, n-1)-1[/imath] if both numbers are even
A cool lemma in Ramsey theory is the following: [imath]R(m, n) ≤ R(m-1, n) + R(m, n-1).[/imath] It is not hard to prove it. However, a friend of mine said that, if [imath]R(m-1, n)[/imath] and [imath]R(m, n-1)[/imath] are both even, then [imath]R(m, n) ≤ R(m-1, n) + R(m, n-1)-1.[/imath] How can I prove it? |
2560771 | prove if p is a prime having the form 4k+3
Hi I was having a little trouble with this question that prove if p is a prime having the form [imath]4k+3[/imath], and if [imath]m[/imath] is the number of quadratic residues less than [imath]p/2[/imath], then [imath]1.3.5...(p-2)≡(-1)^{m+k+1}\pmod p[/imath], and[imath]2.4.6...(p-1)≡(-1)^{m+k}\pmod p[/imath] I just dont understand the question. | 1946063 | [imath]1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p[/imath], and [imath]2\cdot4\cdot6\cdots(p - 1) = ( -1)^{m +k} \pmod p[/imath].
Prove that if [imath]p[/imath] is a prime having the form [imath]4k + 3[/imath], and if [imath]m[/imath] is the number of quadratic residues less than [imath]\frac p2[/imath], then we have [imath]1\cdot3\cdot5 \cdots (p - 2) = (-1)^{m+k+1} \pmod p, \text{ and } 2\cdot4\cdot6\cdots(p - 1) = ( -1)^{m +k} \pmod p.[/imath] I am stuck with the problem....Help Needed. |
2561080 | Lebesgue Outer Measure of [imath]\mathbb Q[/imath] from the Definition
Define [imath]\lambda^*:\wp\left(\mathbb R\right)\to\left[0,\infty\right][/imath] by [imath]\lambda^*\left(E\right)=\inf\left\{\sum_{n\in\mathbb N}\left(b_n-a_n\right):E\subseteq\bigcup_{n\in\mathbb N}\left(a_n,b_n\right)\right\}.\tag{$\star$}[/imath] In other words, [imath]\lambda^*[/imath] is the Lebesgue outer measure. Clearly, [imath]\lambda^*\left(\left\{q\right\}\right)=0[/imath] for every [imath]q\in\mathbb R[/imath]. Therefore, since outer measures are countably subadditive, [imath]\lambda^*\left(\mathbb Q\right)=0[/imath]; [imath]\lambda^*\left(\mathbb Q\right)=\lambda^*\left(\bigcup_{q\in\mathbb Q}\left\{q\right\}\right)\leq\sum_{q\in\mathbb Q}\lambda^*\left(\left\{q\right\}\right)=0.[/imath] However, how does this follow from [imath]\left(\star\right)[/imath] alone? I am having a hard time coming up with a sequence of collections of intervals whose unions contain [imath]\mathbb Q[/imath] and whose sums approach zero since [imath]\mathbb Q[/imath] is dense in [imath]\mathbb R[/imath]. | 1138305 | How come [imath]\lambda(\mathbb{Q})= \inf\left\{\sum_{n \ge 1} \lambda([a_n,b_n)) : \mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)\right\} = 0[/imath]?
I have an intuition problem regarding a certain claim. I've learned that since [imath]\mathbb{Q}[/imath] is measurable than [imath]\lambda(\mathbb{Q})= \inf\left\{\sum_{n \ge 1} \lambda([a_n,b_n)) : \mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)\right\}[/imath] This is because Lebesgue measure is the Caratheodory extension over the semiring whose elements are [imath][a,b)[/imath]. But it seems so false, since as we know [imath]\lambda(\mathbb{Q}) = 0[/imath], But I can't even imagine a countable family of sets such that [imath]\mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)[/imath] and [imath]\sum_{n \ge 1} \lambda([a_n,b_n)) < \infty[/imath]. How can the infimum of the above set be [imath]0[/imath]? |
1646734 | How to show that [imath]R^3[/imath] is the direct sum of [imath]W_1=[/imath]Span[imath](1,1,1)[/imath] and [imath]W_2=[/imath]Span[imath](\{1,0,0\}, \{1,1,0\})[/imath]?
How to show that [imath]R^3[/imath] is the direct sum of [imath]W_1=[/imath]Span[imath](1,1,1)[/imath] and [imath]W_2=[/imath]Span[imath](\{1,0,0\}, \{1,1,0\})[/imath]? So we write it as [imath]R^3=W_1+W_2[/imath] because every [imath](x_1, x_2, x_3)\in R^3[/imath] can be written as [imath]a((1,1,1)-(1,1,0))+b((1,1,0)-(1,0,0))+c(1,0,0)[/imath] Do I have the correct direction? | 1626743 | How to show that [imath]\mathbb{R}^3[/imath] is the direct sum of [imath]W_1={\rm span}\{(1,1,1)\}[/imath] and [imath]W_2={\rm span}\{(1,0,0),(1,1,0)\}[/imath]?
We write linear combination of two spans as [imath]a(1,1,1)+b(1,0,0)+c(1,1,0)[/imath]. Any vector [imath]v[/imath] in [imath]\mathbb{R}^3[/imath] can be expressed as [imath](a+b+c, a+b, a)[/imath]. Also linear combination of [imath]W_2[/imath] cannot form vector in [imath]W_1[/imath] other than [imath](0,0,0)[/imath], [imath]W_1\cap W_2=(0,0,0)[/imath], and therefore direct sum is [imath]\mathbb{R}^3[/imath]. I am not sure if my method is correct, so could anyone help? |
2561178 | How to prove that there are infinitely many irreducibles in [imath]F_q[T][/imath]
Can any one give a hint to prove that, there are infinitely many irreducibles in [imath]F_q[T][/imath], function field. | 430921 | Is the number of irreducibles in any number field infinite?
Are there infinitely many irreducibles in the ring of integers of any algebraic number field ? I tried to follow the same argument as we usually do for integers. Suppose there are finitely many irreducibles, say [imath]p_1,\ldots ,p_n[/imath] and let [imath]\alpha :=1+ p_1\cdots p_n[/imath]. Now if [imath]\alpha [/imath] is not a unit then it must have an irreducible [imath]p[/imath] such that [imath]p|\alpha[/imath] but then [imath]p[/imath] can not be any of the [imath]p_i[/imath]'s and we have a contradiction. [imath]\textit{What if [/imath]\alpha[imath] is an unit ? Is it possible for [/imath]\alpha[imath] to be an unit ?}[/imath] Of course, one can replace [imath]p_1\cdots p_n[/imath] by [imath]p_1^{k_1}\cdots p_n^{k_n}[/imath] for any [imath]k_1,\ldots ,k_n\in\mathbb{N}[/imath] and the same argument would go through. |
2560738 | How/why does this generating function approach work?
Suppose I wanted to find the number of solutions of [imath]e_1 + e_2 + e_3 = 17[/imath] where [imath]e_1, e_2,[/imath] and [imath]e_3[/imath] are nonnegative integers with [imath]2 \leq e_1 \leq 5, 3 \leq e_2 \leq 6, 4 \leq e_3 \leq 7[/imath]. The solution [apparently] lies in the fact that the number of solutions with the indicated constraints is the coefficient of [imath]x^{17}[/imath] in the expansion of [imath]\bigg(\sum_{k=2}^5 x^k \bigg)\bigg(\sum_{k=3}^6 x^k \bigg) \bigg(\sum_{k=4}^7 x^k\bigg) [/imath]Why is this the case? Why, on a fundamental level, do elementary generating function-based counting methods (such as this one) work? | 2550082 | Find the number of solutions of [imath]a_0+a_1+a_2=17[/imath] if [imath]2\le a_0\le 5[/imath], [imath]3\le a_1\le 6[/imath], [imath]4\le a_2\le 7[/imath].
Find the number of solutions of [imath]a_0+a_1+a_2=17[/imath] if [imath]2\le a_0\le 5[/imath], [imath]3\le a_1\le 6[/imath], [imath]4\le a_2\le 7[/imath]. This is an exmaple given by my professor, and his solution is: The number of solutions to this equation satisfying the given constraints is equal to the coefficient of [imath]x^{21}[/imath] in the expression [imath]\left(\sum_{n=2}^5x^n\right)\left(\sum_{n=3}^6x^n\right)\left(\sum_{n=4}^7x^n\right)=x^9\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right).[/imath] This must be 3, since the coefficient of [imath]x^8[/imath] in [imath](1+x+x^2+x^3)^3[/imath] is 3. I understand actually nothing here. The expression of summation just pops up, and then comes the answer. What I only know that the idea of generating function is used here. Can someone help explain how the steps in the solution come? |
2561419 | How to prove: [imath]\int\limits_0^1 \frac{|f''(x)|}{f(x)}dx>4,[/imath] using elementary math tools.
Let [imath]f\in C^2 [0,1][/imath] and [imath]f(0)=0=f(1)[/imath] and for all [imath]x\in (0,1)[/imath] we have [imath]f(x)>0[/imath]. Prove that [imath] \int_0^1 \dfrac{|f''(x)|}{f(x)}dx>4.[/imath] I want to prove this problem with elementary math tools. It is given to the first year university students as a calculus exercise. It is noticed that: Hint: Use average value theorem in suitable intervals. | 213459 | Integral Inequality [imath]|f''(x)/f(x)|[/imath]
Let [imath]f[/imath] be a [imath]C^2[/imath] function in [imath][0,1][/imath] such that [imath]f(0)=f(1)=0[/imath] and [imath]f(x)\neq 0\,\forall x\in(0,1).[/imath] Prove that [imath]\int_0^1 \left|\frac{f{''}(x)}{f(x)}\right|dx\ge4[/imath] |
2561532 | Integral [imath]\int_{-\infty}^{\infty} \frac{c(\theta)}{(1 + x^2)^\theta} dx [/imath]
I want to calculate this integral:[imath]\int_{-\infty}^{\infty} \frac{c(\theta)}{(1 + x^2)^\theta} dx [/imath] it's easy to do, when [imath]\theta[/imath] is given: For example when [imath]\theta[/imath] is 1, then integral is equal to [imath]\pi[/imath] But how to calulate integral, when \theta is not given? And [imath]c(\theta)[/imath] is a constant, that is dependent on [imath]\theta [/imath] | 2124263 | Exact Values of the integal [imath]\int_0^\infty \frac{r^{n-1}}{(1+r^2)^{\frac{s}{2}}}\,dr[/imath]
Does any one know the exact expression of the integral, [imath]E_n(s)=\int_0^\infty \frac{r^{n-1}}{(1+r^2)^{\frac{s}{2}}}\,dr~~~~s>n, n\in \mathbb{N}[/imath] or more generally, [imath]E_a(s)=\int_0^\infty \frac{r^{a-1}}{(1+r^2)^{\frac{s}{2}}}\,dr~~~~s>a, a\in \mathbb{R}[/imath] For the special case [imath]s=n, n+2[/imath] I find out by induction that [imath] E_{n-1}(n)=\frac{\omega_{n-1}}{2\omega_{n-2}}~~\text{and}~~E_{n-1}(n+2)=\frac{\omega_{n-1}}{2n\omega_{n-2}}. [/imath] where [imath]\omega_{n-1} = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}[/imath] is the surface measure of the n-dimensional sphere of [imath]\mathbb{R}^n[/imath]. Further result is welcome |
2562701 | Finite quotient ring over prime ideals
Let [imath]d[/imath] be a square free integer. Let [imath]R=\{a+b\sqrt{d}:a,b\in\mathbb{Z}\}[/imath]. Also [imath]P\subset R[/imath] is a prime ideal. I want to show that the factor ring [imath]R/P[/imath] is finite. (No group theory only ring theory) I know that if [imath]I[/imath] is a nonzero ideal, the [imath]I\cap\mathbb{Z}[/imath] is a nonzero ideal in [imath]Z[/imath]. Also [imath]I\cap\mathbb{Z}=(n)[/imath] for some positive integer [imath]n[/imath]. Also I know [imath]R/P[/imath] has no zero divisors. So I looked at some other post and I still don't understand the connection between [imath]R/(n)[/imath] (which is a finite set), [imath](R/(n))/P[/imath] (how does this look like), and [imath]R/P[/imath]. | 2559019 | Prime/Maximal Ideals in [imath]\mathbb{Z}[\sqrt d][/imath]
Let [imath]d \in \mathbb{Z}[/imath] be a square free integer. [imath]R=\mathbb{Z}[\sqrt d][/imath] =[imath]\{a+b\sqrt d | a,b \in \mathbb{Z} \}[/imath]. Overall, I'm trying to show that every prime ideal [imath]P \subset R[/imath] is a maximal ideal. So far I showed that [imath]I \subset R[/imath] is finitely generated. [imath]I=\{(x,s+y \sqrt d)\}[/imath] And now I'm trying to show that R/P, for some prime ideal is a finite ring with no zero divisors. From there it would follow that R/P is a field and any prime ideal is maximal. I know R could also be written as [imath]R= \mathbb{Z}[x]/(x^2-d)[/imath]. How could I show that R/P is a quotient of [imath]\mathbb{Z}/n\mathbb{Z}[x]/(x^2-d)[/imath]? |
2563198 | Integrate fractions
I know there is a lot of answers on how to integrate with fractions, but I think they are very case specific. I've tried integration with substitution, but I can't seem to get a useful answer. I simply want to integrate [imath]\int[/imath][imath]\frac{v}{g+\alpha\cdot v}dv[/imath] | 1279982 | How to integrate [imath]\int \frac{v}{g-kv}dv[/imath]
How to integrate [imath]$$\int \frac{v}{g-kv}\,\mathrm dv$$[/imath] it's supposed to be equal to [imath]$$ -\frac{v}{k} -\frac{g}{k^2}\ln(g-kv)$$[/imath] but I can't get that, I tried a substitution [imath]u=g-kv[/imath] and got close but no right? Any help. |
2563353 | Prove that [imath]\cos A + \cos B + \cos C \geq \frac{1}{4}(3 + \cos(A - B) + \cos(B - C) + \cos(C - A))[/imath]
I have simplified the inequality as these but, can't proceed further: [imath]3(\cos A + \cos B + \cos C) \geq 3 + 2(\cos A\cos B + \cos B\cos C + \cos C\cos A)[/imath] [imath](2\cos A-1)(\cos B-1)+(2\cos B-1)(\cos C-1)+(2\cos C-1)(\cos A-1) \leq 0[/imath] | 1589245 | Prove this inequality [imath]\sum \cos{A}\ge\frac{1}{4}(3+\sum\cos{(A-B)})[/imath]
Prove that in any triangle [imath]ABC[/imath] the following inequality holds [imath]\cos{A}+\cos{B}+\cos{C}\ge\dfrac{1}{4}(3+\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})[/imath] And I have gotten [imath]8(\cos{A}+\cos{B}+\cos{C})\ge 6+2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})[/imath] [imath]2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})+3=(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2[/imath] [imath]\Longleftrightarrow 8\sum_{cyc}\cos{A}\ge 3+(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2[/imath] then Any hints, ideas? Thanks in advance. |
772035 | How do I prove and expand Bayesian Networks?
Attempting to understand Exercise 20 (pdf page 44) in the paper: (Warning: large paper; small exercise) Bayesian Reasoning and Machine Learning The party animal problem corresponds to the network in g(3.14). The boss is angry and the worker has a headache - what is the probability the worker has been to a party? When set to 1 the statements are true: P = Been to Party, H = Got a Headache, D = Demotivated at work, U = Underperform at work, A =Boss Angry. Shaded variables (A, H) are observed in the true state. [imath]\begin{matrix} & & D \\ & & \downarrow \\ P & \rightarrow & U \\ \downarrow & & \downarrow \\ (H) & & (A)\end{matrix} [/imath] I would like to solve the following: Prove that p(P|H,A ) = a*p(P,H,A) where a is a constant. Expand p(P, H, A) by marginalizing over the variables U and D. how to compute 'a' Any help is greatly appreciated. Thank You. | 2545974 | Bayesian networks problem
Attempting to understand Exercise 20 (pdf page 44) in this book: The party animal problem corresponds to the network in g(3.14). The boss is angry and the worker has a headache - what is the probability the worker has been to a party? All variables are binary and [imath]p(U =tr|P =tr,D=tr)=0.999[/imath], [imath]p(U =tr|P =fa,D=tr)=0.9[/imath], [imath]p(U =tr|P =tr,D=fa)=0.9[/imath], [imath]p(U =tr|P =fa,D=fa)=0.01[/imath]. [imath]\begin{matrix} & & D \\ & & \downarrow \\ P & \rightarrow & U \\ \downarrow & & \downarrow \\ (H) & & (A)\end{matrix} [/imath] I'm unsure of how to solve this problem given only the above information, and help greatly appreciated. Can this even be solved with the information given in the question? |
2490947 | Problem about LFT carries one pair of concentric circles into another pair of concentric circles
While I tried to prove a problem, Suppose that a linear fractional transformation carries one pair of concentric circles into another pair of concentric circles. Prove that the ratios of the radii must be the same. Following is my try Suppose concentric circles [imath]z_1=\alpha+Re^{i\theta}[/imath] and [imath]z_2=\alpha+t Re^{i\theta}[/imath], where [imath]t>0,\ \theta\in [0,2\pi)[/imath], through LFT [imath]\frac{az+b}{cz+d}[/imath] mapping to [imath] \omega_1=\frac{a Re^{i\theta}+a\alpha+b}{cRe^{i\theta}+c\alpha +d};\\ \omega_2=\frac{a tRe^{i\theta}+a\alpha+b}{ctRe^{i\theta}+c\alpha +d}. [/imath] According to condition the mapping maps the concentric circles to concentric circles, so [imath]\omega_1[/imath] and [imath]\omega_2[/imath] are concentric. Suppose [imath]\omega_1=\beta_1+R_1e^{i\theta}[/imath] and [imath]\omega_2=\beta_2+R_2e^{i\theta}[/imath], solve the equations, we get the results as follows, [imath] \beta_1\to \frac{(a \alpha+b) (\alpha c+d)-a c R^2}{(\alpha c+d)^2-c^2 R^2},R_1\to \frac{R (a d-b c)}{(\alpha c+d)^2-c^2 R^2}; [/imath] [imath] \beta_2\to \frac{(a \alpha+b) (\alpha c+d)-a c R^2 t^2}{(\alpha c+d)^2-c^2 R^2 t^2},R_2\to \frac{R t (a d-b c)}{(\alpha c+d)^2-c^2 R^2 t^2}. [/imath] We simplify [imath]\beta_1-\beta_2=0[/imath], that is [imath] c(c\alpha+ d)=0. [/imath] (i)[imath]c=0[/imath], [imath]R_2/R_1=t[/imath]. (ii)[imath]c\alpha+d=0[/imath], [imath]R_2/R_1=1/t[/imath]. Now my question is the first situation is normal and second one seems like the original proposition is not right, what's wrong? Besides, I think my way to solve this question especially solving the equations are complicated, it there any other simpler solution? Thanks! | 23646 | Möbius transformations and concentric circles
Given a Möbius transformation that maps one pair of concentric circles to another pair of concentric circles, why is the ratio of the radii preserved through the map? I thought about how Möbius transformations are compositions of rotations, scaling, inversion, and translation, and that intuitively, these types of maps shouldn't change the ratio of radii between two circles. Would it be correct to just say that if [imath]\frac{r_1}{r_2}[/imath] is the ratio of radii between the two circles, then 1) The radii are invariant under translation, [imath]z \mapsto z+a[/imath], so [imath]\frac{r_1}{r_2}[/imath] stays the same 2) Under scaling by a factor [imath]z \mapsto az[/imath], [imath]\frac{ar_1}{ar_2} = \frac{r_1}{r_2}[/imath] 3) Under inversion, [imath]z \mapsto \frac{1}{z}[/imath], [imath]\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}[/imath] Or is there a different/better way to think about this problem? |
2563384 | Finding the area of the surface
FIind the area of the part of the cylinder [imath]x^2+y^2=2ay[/imath] that lies inside the sphere [imath]x^2+y^2+z^2=4a^2[/imath] The intersection of the cylinder with the sphere gives us the shape [imath]2ay+z^2=4a^2[/imath]. So if I were to find the surface area of this shape I am done. However it is almost impossible to express the surface area element since I get a [imath]0[/imath] in the denominator. Can somebody help me by using the suggested approach (calculate the surface area of [imath]2ay+z^2=4a^2[/imath] to get the answer or explain why that method is not possible? | 867961 | Surface area of sphere [imath]x^2 + y^2 + z^2 = a^2[/imath] cut by cylinder [imath]x^2 + y^2 = ay[/imath], [imath]a>0[/imath]
The cylinder is given by the equation [imath]x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2[/imath]. The region of the cylinder is given by the limits [imath]0 \le \theta \le \pi[/imath], [imath]0 \le r \le a\sin \theta[/imath] in polar coordinates. We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have: [imath]A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA[/imath] where [imath]F[/imath] is the equation of the sphere. Plugging in the expressions and simplifying ([imath]z \ge 0)[/imath], we get: [imath]A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy[/imath] Converting to polar coordinates, we have: [imath]A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta[/imath] Calculating this I get [imath]2\pi a^2[/imath]. The answer is [imath](2\pi - 4)a^2[/imath]. Where am I going wrong? |
2563598 | Let [imath]\{a_n\}[/imath] be a sequence of reals going to [imath]0[/imath] but whose series diverges. Prove that we can flip the signs of each term to get a convergent series
Suppose [imath]\{a_n\}[/imath] is a sequence of real numbers that converges to [imath]0[/imath] but whose series diverges. Show that for amy real number [imath]r[/imath] there is a sequence [imath]\sigma_n[/imath], consisting of [imath]1[/imath] and [imath]-1[/imath]'s, such that [imath]\sum_n \sigma_n a_n = r[/imath]. I am confused about how to show this. Considering the partial sums, we can construct a sequence of partial sums such that the sum of first few terms (by making all of them positive) go just above [imath]r[/imath], and make the 'second' frw terms all negative so that the partial sums up to that point goes just below [imath]r[/imath], and continue the process. But I wasn't sure how to make this mathematically rigorous. | 148663 | If [imath]a_n[/imath] goes to zero, can we find signs [imath]s_n[/imath] such that [imath]\sum s_n a_n[/imath] converges?
Let [imath]a_n[/imath] be a sequence of complex numbers that converge to zero. Can we always find [imath]s_n \in \{-1,1\}[/imath] such that [imath]\sum_{n=1}^{\infty} s_n a_n[/imath] converges? If the [imath]a_n[/imath] are real numbers, we can find such a sequence [imath]s_n[/imath]. If the partial sum of the first [imath]N[/imath] terms is positive we make sure the following terms are negative until the sum becomes less than zero. Then we switch to making the terms positive until the partial sum becomes greater than zero, and so on. It is easy to see that the partial sums will either tend monotonically towards zero, or oscillate around zero with decreasing amplitude. Edit: Actually there is no reason that the partials sums should go to zero in the monotone case. They will still converge however. |
2562549 | How to find the sum of this expression?
I want to find the sum of the following expression in terms of [imath]n[/imath]. I tried looking for it but couldn't come up with one. Any help will be greatly appreciated. [imath]\sum_{i=0}^n \frac{1}{4^i}\binom{2i}{i}[/imath] | 1989966 | Summation of Central Binomial Coefficients divided by even powers of [imath]2[/imath]
Whilst working out this problem the following summation emerged: [imath]\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m[/imath] The is equivalent to [imath]\begin{align} \sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\ &=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right) \end{align}[/imath] and terms are the same as coefficients in the expansion of [imath](1-x)^{-1/2}[/imath]. Once the solution [imath] \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}[/imath] is known, the telescoping sum can be easily derived, i.e. [imath]\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m[/imath] However, without knowing this a priori, how would we have approached this problem? |
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