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713568 | Entire [imath]f[/imath] and [imath]g[/imath] satisfying [imath]f^n+g^n=1[/imath] for [imath]n>3[/imath] must be constant.
Studying for a qualifying exam and came across Don Marshall's notes. This one has stumped me for some time. This question is related to another MSE question: [imath]\begin{align*}\text{If [/imath]f[imath] and [/imath]g[imath] are entire and } [f(z)]^n+[g(z)]^n=1\text{ for [/imath]n>3[imath] then [/imath]f,g[imath] are constant}. \end{align*}[/imath] The most I could deduce is that if [imath]f[/imath] were not constant then [imath][f(z)]^n[/imath] is necessarily surjective as missing one value for [imath][f(z)]^n[/imath] would force [imath]n[/imath] missed values for [imath]f[/imath]. That's about as far as I could get. | 470689 | Proving two entire functions are constant.
Let [imath]f[/imath] and [imath]g[/imath] be entire functions such that [imath]f^n+g^n=1[/imath], where [imath]n\geq 3[/imath] is an integer. Prove that [imath]f[/imath] and [imath]g[/imath] are constant. I suppose I should somehow prove that either [imath]f[/imath] or [imath]g[/imath] is bounded so that I can apply Liouville's Theorem, but I don't see how. I tried setting the derivative of the left hand side equal to zero and work with that but that did not seem to work. |
713969 | Computing partial sum
I'm trying to compute the partial sum [imath]\displaystyle\sum_{n=0}^m \frac{n}{n^4+n^2+1}[/imath]. Wolfram Alpha says that the answer is [imath]\displaystyle\frac{m^2+m}{2(m^2+m+1)}[/imath]. I have no idea how they got this answer. Can someone help me out? | 449510 | How to find the sum of the sequence [imath]\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....[/imath]
Problem : How to find the sum of the sequence [imath]\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....[/imath] I am unable to find out how to proceed in this problem.. this is a problem of arithmetic progression... Please suggest how to proceed...Thanks.. |
714276 | Calculating [imath]\lim\limits_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{x \cdot \tan x}[/imath]
I am trying to solve this limit without using Taylor expansions to explain it to a secondary school student [imath]\lim_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{x \cdot \tan x}.[/imath] | 695992 | Indeterminate form limits question
[imath]\lim_{x\to 0}\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x} [/imath] This is an indeterminate limit. I want help in solving this problem. Thanks in advance |
708688 | What choice of parameters makes this function uniformly continuous?
Given constants [imath]\alpha, \beta > 0[/imath] define [imath] f_{\alpha, \beta}(x) = x^{\alpha} \sin(x^{\beta})[/imath] for [imath]x > 0[/imath]. For which pairs [imath]\alpha, \beta[/imath] is [imath]f_{\alpha, \beta}[/imath] uniformly continuous? | 708762 | When is [imath]x^{\alpha}\sin(x^{\beta})[/imath] uniformly continuous for [imath]\alpha, \beta > 0[/imath]?
Consider a function [imath]f_{\alpha, \beta}\colon (0, \infty) \longrightarrow \mathbb{R}[/imath] defined in the following way: [imath]f_{\alpha, \beta} = x^{\alpha}\sin(x^{\beta}) \quad \alpha, \beta > 0[/imath] Then we can pose the questions: For which pairs [imath]\alpha, \beta[/imath] is this function uniformly continuous? For which sets [imath](\alpha, \beta)[/imath] in [imath](0, \infty)^2[/imath] is the family equicontinuous? I am baffled as to how to go about answering these questions in a clear and concise way. I think that it is possible to produce an answer by considering many cases and lots of tedious estimates. Is there a better way to approach the problem? Any help will be appreciated. |
715900 | Proving that a well-ordered set with the order topology is not homeomorphic to any initial segment of it
Let [imath]\langle X,\leq_X\rangle[/imath] be a well-ordered set. I want to show that [imath]X[/imath] with the topology on [imath]X[/imath] relative to [imath]\leq_X[/imath] (the order topology) is not homeomorphic to any initial segment of [imath]X[/imath]. And then I need to conclude that any two well-ordered sets [imath]\langle X,\leq_X\rangle[/imath] and [imath]\langle Y,\leq_Y\rangle[/imath] are homeomorphic iff they are isomorphic. I can feel both seem true, but while the first one I don't know how to proof, the second one I can prove, but not as a conclusion of the first. Any ideas on how can I show the second one as a conclusion of the first? And on how can I prove the first? Thanks! | 38993 | How do you find the smallest of homeomorphic ordinals?
I am trying to get a better feel for the topology of ordinals and just received a great answer to this question where the Cantor-Bendixson rank and degree turn out to be a complete homeomorphism invariant of countable compact spaces. Now I am curious about what is going on within homeomorphism classes of ordinals. In particular, I want to know how to characterize countable ordinals [imath]\alpha[/imath] which are minimal among all ordinals homeomorphic to [imath]\alpha[/imath]. Certainly finite ordinals and [imath]\omega[/imath] are easy. Next, if [imath]\alpha\cong \omega+1[/imath], then it seems like we would have to have [imath]\omega+1\leq \alpha<\omega +\omega[/imath] since there can only be one limit point in [imath]\alpha[/imath]. Things become a little fuzzier for me with ordinals that have infinitely many limit points. For instance, suppose [imath]\alpha[/imath] is an ordinal homeomorphic to [imath]\omega^{\omega}[/imath]. Is it necessarily true that [imath]\alpha\geq \omega^{\omega}[/imath]? Is there a general rule for knowing when an ordinal is the smallest among all ordinals homeomorphic to it? |
715823 | If [imath]x,y,z \in \mathbb{R^+}[/imath] such that [imath]x+y+z=3[/imath]. Prove the inequality [imath]\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx[/imath]
If [imath]x,y,z \in \mathbb{R^+}[/imath] such that [imath]x+y+z=3[/imath]. Prove the inequality [imath]\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx[/imath]. My work: We have [imath]3(x+y+z)=x^2+y^2+z^2+2(xy+yz+zx) \implies (xy+yz+zx)=\dfrac12(3x-x^2+3y-y^2+3z-z^2)[/imath] So, we have to prove, [imath]\sqrt x+\sqrt y+\sqrt z-\dfrac12(3x-x^2+3y-y^2+3z-z^2)\ge 0[/imath] Now, I cannot proceed further. Please help. | 336362 | Prove [imath]\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca[/imath]
Let [imath]a,b,c[/imath] are non-negative numbers, such that [imath]a+b+c = 3[/imath]. Prove that [imath]\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca[/imath] Here's my idea: [imath]\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca[/imath] [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)[/imath] [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0[/imath] [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0[/imath] [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0[/imath] [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2[/imath] [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9[/imath] And I'm stuck here. I need to prove that: [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2[/imath] or [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)[/imath], because [imath]a+b+c = 3[/imath] In the first case using Cauchy-Schwarz Inequality I prove that: [imath](a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2[/imath] [imath]3(a^2 + b^2 + c^2) \ge (a+b+c)^2[/imath] Now I need to prove that: [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)[/imath] [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)[/imath] [imath]\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2[/imath] I need I don't know how to continue. In the second case I tried proving: [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)[/imath] and [imath]a^2 + b^2 + c^2 \ge a+b+c[/imath] Using Cauchy-Schwarz Inequality I proved: [imath](a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2[/imath] [imath](a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2[/imath] [imath]a^2 + b^2 + c^2 \ge a+b+c[/imath] But I can't find a way to prove that [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)[/imath] So please help me with this problem. P.S My initial idea, which is proving: [imath]2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9[/imath] maybe isn't the right way to prove this inequality. |
716126 | Show that [imath]A \cong \mathbb{C}^n[/imath] with A a commutative algebra
Let A be a commutative algebra of finite dimension, and if [imath]A[/imath] has no nilpotent elements other than [imath]0[/imath], is true that [imath]A \cong \mathbb{C}^n[/imath] ? The question emerge to my mind, I thought that the finite dimension tell us that the scheme is Artinian (geometrically dimension 0). I think the pattern is just a meeting of [imath]n[/imath] points but I have not managed to prove it. Someone can enlighten me please ? Thanks | 191800 | Characterize finite dimensional algebras without nilpotent elements
Characterize all finite dimensional algebras (may not be commutative) over a field [imath]K[/imath] without nilpotent elements. My condition: Let [imath]A[/imath] be any algebra (may not be finite dimensional), then it's easy to prove that [imath]A[/imath] has no nilpotent elements iff the equation [imath]x^2=0[/imath] has a unique solution (the trivial solution). But is there a more explicit characterization of these finite dimensional algebras? |
322091 | on the factorization of maps between connected CW complexes
I'm working on problem 16 in section 4.1 of Hatcher's Algebraic Topology book. I really have no ideas so far: Show that a map [imath]f: X \to Y[/imath] between connected CW complexes factors as a composition [imath]X \to Z_n \to Y[/imath] where the first map induces isomorphisms on [imath]\pi_n[/imath] for [imath]i\le n[/imath] and the second map induces isomorphisms on [imath]\pi_n[/imath] for [imath]i\ge n+1[/imath]. Any help is appreciated. | 580104 | Factorization of a map between CW complexes
I've been working on problem 4.1.16 of Hatcher's Algebraic Topology and am at a complete impasse. The problem is as follows: Show that a map [imath]f:X→Y[/imath] between connected CW complexes factors as a composition [imath]X→Z_n→Y[/imath] where the first map induces isomorphisms on [imath]π_n[/imath] for [imath]i≤n[/imath] and the second map induces isomorphisms on [imath]π_n[/imath] for [imath]i≥n+1.[/imath] I applied Proposition 4.13 on the pair [imath](M_f,X)[/imath] to get an [imath]n[/imath]-connected CW model [imath](Z_n,X)[/imath]. Because [imath]M_f[/imath] deformation retracts to [imath]Y[/imath], this gives the desired isomorphisms for the second map. Moreover, because [imath](Z_n,X)[/imath] is [imath]n[/imath]-connected, the inclusion of [imath]X[/imath] in [imath]Z_n[/imath] gives the desired isomorphisms for the first map, with the exception of [imath]π_n(X)→π_n(Z_n)[/imath] (this map is, however, surjective). How do I prove injectivity? Any help would be appreciated. |
716635 | Showing that the sequence [imath]z^n[/imath] is normally but not uniformly convergent
I was able to show that [imath]z^n[/imath] is normally convergent on the unit disk centred at the origin, but I am not sure how to show that it is not uniformly convergent on the unit disk centred at the origin. I've read in a few places that this is true, but I have not seen a proof. My first idea was to fix an [imath]\epsilon[/imath] such that [imath]\epsilon=1/2[/imath]. Then [imath]|z^n|<1/2[/imath]. However, I haven't been able to get this to work out. | 185876 | Why is this sequence of functions not uniformly convergent?
For the following sequence of functions and its limit function, we can see that [imath]f_n(x)[/imath] is clearly pointwise convergent [imath]f_n(x) = x^n\text{ }\forall x\in[0,1]\text{ and }\forall n\in\mathbb N^*\\ f(x) = \begin{cases}0&\text{if } x\in[0,1)\\1&\text{if } x=1\end{cases}[/imath] However, I was wondering why this is not uniformly convergent. The condition for uniform convergence is: [imath]|f_n(x) - f(x)| < \epsilon,\ \ \ \forall x \text{ when } n > N[/imath] Now most sources present an argument along the lines of: assume that [imath]f_n(x)[/imath] is uniformly convergent and that [imath]0 < x < 1[/imath], this means that [imath]x^n<\epsilon[/imath] whenever [imath]n>N[/imath]. Specifically, this would mean [imath]x^{N+1}<\epsilon[/imath] for some fixed [imath]N[/imath]. But if we now pick [imath]x[/imath] such that [imath]1 > x > ε^{\frac{1}{N+1}}[/imath], then this would lead to a contradiction, therefore [imath]f_n(x)[/imath] is not uniformly convergent. However, I was wondering why couldn't we take [imath]n[/imath] to infinity. If [imath]0 < x < 1[/imath], then [imath]\lim_{n\rightarrow \infty} |f_n(x) - f(x)| = 0[/imath] (which is less than [imath]\epsilon[/imath]). Now since [imath]|f_n(x) - f(x)|[/imath] will always be [imath]0[/imath] if [imath]n > \infty[/imath], then wouldn't this be uniformly convergent? |
715418 | Solving an equality in 2 variables
I need to prove that [imath]\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}[/imath] if [imath]a+b = 1[/imath] and [imath]a b \le 1/4[/imath] I'd like a hint. Solve the equality first to [imath]a[/imath] or [imath]b[/imath], or stay in a and b as to get [imath]a b \le 4[/imath] in the inequality ? | 636893 | If [imath]a[/imath] and [imath]b[/imath] are positive real numbers such that [imath]a+b=1[/imath], prove that [imath](a+1/a)^2+(b+1/b)^2\ge 25/2[/imath]
If [imath]a[/imath] and [imath]b[/imath] are positive real numbers such that [imath]a+b=1[/imath], prove that [imath]\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.[/imath] My work: [imath]\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}[/imath] Now, we have [imath]a^2+\dfrac{1}{a^2}\ge 2[/imath] and [imath]b^2+\dfrac{1}{b^2}\ge 2[/imath]. Here, I am stuck, I cannot use the information provided, [imath]a+b=1[/imath] to any use. Please help! |
716789 | Proving divisibility by using induction: [imath]133 \mid (11^{n+2} + 12^{2n+1})[/imath]
If [imath]n > 0[/imath], then prove the following by using induction: [imath]133|(11^{n+2} + 12^{2n+1}).[/imath] | 150979 | Show that [imath]11^{n+1}+12^{2n-1}[/imath] is divisible by [imath]133[/imath].
Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck. Most likely, due to the late hour, I am missing something very trivial here. Prove that for any integer [imath]n[/imath], the number [imath]11^{n+1}+12^{2n-1}[/imath] is divisible by [imath]133[/imath]. I have tried multiplying through by [imath]12[/imath] and rearranging, but like I said with meager results. I arrived at [imath]11^{n+2}+12^{2n+1}[/imath] which satisfies the induction hypothesis for LHS, but for the RHS I got stuck at [imath]12 \times 133m-11^{n+1}-11^{n+3}[/imath] or [imath]12^2 \times 133m-12 \times11^{n+1}-11^{n+3}[/imath] and several other combinations none of which would let me factor out [imath]133[/imath]. |
716497 | Why does [imath]\sqrt{x^2}=|x|[/imath]?
By convention, we say that: [imath]\sqrt{x^2}=|x|[/imath] In fact, the above statement is how we define absolute value. We would not write [imath]\sqrt{4}=-2[/imath]. Although logically it is correct, by convention it is wrong. You have to say [imath]\sqrt{4}=2[/imath] unless the question specifically asks for negative numbers like this: [imath]-\sqrt{4}=-2[/imath] Why is this? I suspect it is because back then, square roots were used to calculate distances (e.g. with Pythagoras' theorem) and distances must be positive. Am I correct? Any other reasons why we only define square roots to be positive? Edit: This entire topic is confusing for me because for example, when you are finding the roots of the function [imath]f(x)=x^2-4[/imath], you would set [imath]f(x)=0[/imath], so now the equation is [imath]0=x^2-4[/imath]. This means that [imath]x^2=4[/imath], so [imath]x=\pm\sqrt{4}=\pm 2[/imath]. Therefore the roots are [imath]2, \ -2[/imath]. But normally we cannot say that [imath]\sqrt{4}=\pm 2[/imath]. Hope this clarifies things a bit. | 607568 | Is [imath]\sqrt{x^2}=|x|[/imath] or [imath]=x[/imath]? Isn't [imath](x^2)^\frac12=x?[/imath]
[imath]|x|=\sqrt{x^2}[/imath] as Wolfram|Alpha shows. But, as [imath](x^2)^\frac12=x[/imath], I can't understand where am I wrong interpreting Square-root. |
717148 | Show that [imath]\lim_{p \to \infty}||x||_p=||x||_M[/imath]
Let [imath]||x||_p=\left( \displaystyle \sum_{i=1}^n|x_i|^p\right)^{\frac{1}{p}}[/imath] and [imath]||x||_M=\max\{|x_1|,|x_2|,...,|x_n|\},[/imath] norms in [imath]\mathbb{R}^n[/imath]. Show that [imath]\lim_{p \to \infty}||x||_p=||x||_M, \ \forall x \in \mathbb{R}^n.[/imath] | 506777 | Trouble in proving that [imath]\|x\|_p = \max|x_j|[/imath]
We define p-norm in this way: [imath]\|x\|_p = \{\sum ^N_j=_1|x_j|^p\}^ {1\over p}[/imath] We know that It change to [imath]\|x\|_p = \max|x_j| [/imath] when [imath] p \to \infty [/imath] How can I prove this ? |
717185 | closed form for [imath]\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}[/imath]
closed form for [imath]\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}[/imath] I tried to solve it by : [imath]\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}=\sum_{k=0}^{n/3}\binom{n}{3k}[/imath] [imath]\sum_{k=0}^{n/3}\binom{n}{3k}=\sum_{k=0}^{n/3}\frac{1}{2\pi i}\int_{\left |z \right |=1}\frac{(1+z)^n}{z^{3k+1}}dz[/imath] then using geometric series but i got no result . what is your suggest to solve ? | 918 | How do I count the subsets of a set whose number of elements is divisible by 3? 4?
Let [imath]S[/imath] be a set of size [imath]n[/imath]. There is an easy way to count the number of subsets with an even number of elements. Algebraically, it comes from the fact that [imath]\displaystyle \sum_{k=0}^{n} {n \choose k} = (1 + 1)^n[/imath] while [imath]\displaystyle \sum_{k=0}^{n} (-1)^k {n \choose k} = (1 - 1)^n[/imath]. It follows that [imath]\displaystyle \sum_{k=0}^{n/2} {n \choose 2k} = 2^{n-1}[/imath]. A direct combinatorial proof is as follows: fix an element [imath]s \in S[/imath]. If a given subset has [imath]s[/imath] in it, add it in; otherwise, take it out. This defines a bijection between the number of subsets with an even number of elements and the number of subsets with an odd number of elements. The analogous formulas for the subsets with a number of elements divisible by [imath]3[/imath] or [imath]4[/imath] are more complicated, and divide into cases depending on the residue of [imath]n \bmod 6[/imath] and [imath]n \bmod 8[/imath], respectively. The algebraic derivations of these formulas are as follows (with [imath]\omega[/imath] a primitive third root of unity): observe that [imath]\displaystyle \sum_{k=0}^{n} \omega^k {n \choose k} = (1 + \omega)^n = (-\omega^2)^n[/imath] while [imath]\displaystyle \sum_{k=0}^{n} \omega^{2k} {n \choose k} = (1 + \omega^2)^n = (-\omega)^n[/imath] and that [imath]1 + \omega^k + \omega^{2k} = 0[/imath] if [imath]k[/imath] is not divisible by [imath]3[/imath] and equals [imath]3[/imath] otherwise. (This is a special case of the discrete Fourier transform.) It follows that [imath]\displaystyle \sum_{k=0}^{n/3} {n \choose 3k} = \frac{2^n + (-\omega)^n + (-\omega)^{2n}}{3}.[/imath] [imath]-\omega[/imath] and [imath]-\omega^2[/imath] are sixth roots of unity, so this formula splits into six cases (or maybe three). Similar observations about fourth roots of unity show that [imath]\displaystyle \sum_{k=0}^{n/4} {n \choose 4k} = \frac{2^n + (1+i)^n + (1-i)^n}{4}[/imath] where [imath]1+i = \sqrt{2} e^{ \frac{\pi i}{4} }[/imath] is a scalar multiple of an eighth root of unity, so this formula splits into eight cases (or maybe four). Question: Does anyone know a direct combinatorial proof of these identities? |
717428 | Why does variance divide by [imath]n-1[/imath]?
The variance is: [imath]\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}[/imath] I read that [imath]n-1[/imath] is used instead of just [imath]n[/imath] when we are measuring the variance of a sample taken from a bigger population. I don't understand why we're subtracting [imath]1[/imath]. Thank you. | 707272 | Why do statisticians like "[imath]n-1[/imath]" instead of "[imath]n[/imath]"?
Does anyone have an intuitive explanation (no formulas, just words! :D) about the "[imath]n-1[/imath]" instead of "[imath]n[/imath]" in the unbiased variance estimator [imath]S_n^2 = \dfrac{\sum\limits_{i = 1}^n \left(X_i-\bar{X}\right)^2}{n-1}?[/imath] |
717515 | How to prove that these two matrices have the same eigenvalues?
I need some help at the following exercise: "Prove that if [imath] A,B \in M_{n \times n} ( \mathbb{R})[/imath], then the matrices [imath]AB[/imath] and [imath]BA[/imath] have the same eigenvalues." Could you give some hint? I thought that I could use the fact that: If A and B are two square n×n matrices then characteristic polynomials of AB and BA coincide. So [imath]det(AB- λ I)=det(BA- λ I).[/imath] How could I continue? Or is this way wrong? | 520122 | Eigenvalues of [imath]AB[/imath] and [imath]BA[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be [imath]m \times n[/imath] and [imath]n \times m[/imath] real matrices. I proved that if [imath]\lambda[/imath] is a nonzero eigenvalue of the [imath]m \times m[/imath] matrix [imath]AB[/imath] then it is also an eigenvalue of the [imath]n \times n[/imath] matrix [imath]BA[/imath]. But I am having trouble finding an example showing that this need not be true if [imath]\lambda=0[/imath] Thank you in advance |
717616 | Consider the function h where [imath]h(x,y) = (x+y,x-y)[/imath], [imath]h : \mathbb N\times \mathbb N\to \mathbb N\times\mathbb N[/imath]
Is the function h onto and one to one? Prove this. Online bonus question on a recent proofs quiz on the topic of one-to-one and onto functions. Gave me a bit of grief (the mapping stuff). Also consider this is a proofs class, much help would be appreciated. I believe the answer would have to include something to do with x+y or x-y not being natural numbers? I could (most probably) be wrong though. Please help. | 713273 | Let [imath]g : \Bbb N \times \Bbb N \to\Bbb N \times \Bbb N[/imath] defined as [imath]g(m,n) = (m + n,m - n)[/imath]
Determine if [imath]g[/imath] is injective; surjective; bijective. Question on a recent test regarding one-to-one and onto functions. Was very difficult for me, could not even begin to answer either. This is a proofs class, much help would be appreciated. |
717674 | Naive calculations with infinite series
In the realm, where the sum of natural numbers is [imath]-1/12[/imath] : [imath]1+2+3+4+...=-1/12[/imath] Is this true?: [imath]2+4+6+8+...=2*(1+2+3+4+...)=-2/12[/imath] Can this kind of naive calculations always be done? -or are there dangers lurking in the infinity? | 180590 | When does (Riemann) regularization work?
I've seen the Wikipedia articles on how to sum [imath]1+1+1+1+\cdots=-1/2[/imath] or [imath]1+2+3+4+\cdots=-1/12[/imath]. Is there a theory behind it or is it a random trick? It basically uses analytic continuation of the Riemann zeta? The article says it may be used in physical applications. So I wonder, what are the conditions that this type of regularization will give me a useful and consistent result? There are other series which can be summed differently: [imath]1+2+4+8+\dots=-1[/imath]. Is it possible to obtain the same result with Riemann regularization? Are different type of regularizations giving the same results (if they exist)? If not, how can I know which one will work for my (physical?) calculation? |
718357 | [imath]\left\lfloor(\sqrt[3]{28}-3)^{-n}\right\rfloor[/imath] is not divisible by 6
let [imath]n[/imath] be a positive integer. Prove that the following expression: [imath]\left\lfloor(\sqrt[3]{28}-3)^{-n}\right\rfloor[/imath] is not divisible by 6. [imath]\lfloor x\rfloor[/imath] is the greatest integer less than or equal to [imath]x[/imath]. | 284112 | Prove [imath]6 \nmid [\left( \sqrt[3]{28} - 3 \right)^{-n}][/imath]
Prove that: [imath]6 \not\left|\ \left\lfloor\frac 1 {(\sqrt[3]{28} - 3)^{n}}\right\rfloor \ (n \in Z^+)\right.[/imath] ([imath]\lfloor x\rfloor[/imath] = largest integer not exceeding [imath]x[/imath]) I am very bad as English and number theory, please help me |
718254 | Is [imath]\widehat{\mathbb{R}/\mathbb{Z}} = \mathbb{Z}[/imath]?
Let [imath]\widehat{\mathbb{R}/\mathbb{Z}}[/imath] be the set of all homomorphisms from [imath]\mathbb{R}/\mathbb{Z}[/imath] to [imath]\mathbb{C}[/imath]. Is [imath]\widehat{\mathbb{R}/\mathbb{Z}} = \mathbb{Z}[/imath]? I think that [imath]\mathbb{R}/\mathbb{Z} = \{t + \mathbb{Z}: t \in \mathbb{R}\}[/imath]. We can define [imath]\varphi: \mathbb{R}/\mathbb{Z} \to \mathbb{C}[/imath] by letting [imath]\varphi(t+\mathbb{Z}) = e^{2\pi i n t}[/imath], [imath]n \in \mathbb{Z}[/imath]. It is easy to see that [imath]\varphi[/imath] is a homomorphism. Is it true that every element in [imath]\widehat{\mathbb{R}/\mathbb{Z}}[/imath] is of this form? Thank you very much. | 148155 | Why is that [imath]\widehat{\mathbb{R}/\mathbb{Z}}\cong\mathbb{Z}[/imath]?
[imath]\widehat{\mathbb{R}/\mathbb{Z}}\cong\mathbb{Z}[/imath], that is, every character of [imath]\mathbb{R}/\mathbb{Z}[/imath] is of the form [imath]x\mapsto e(mx)[/imath] for some integer [imath]m[/imath]. I was considering the dual of [imath]{\mathbb{R}/\mathbb{Z}}[/imath]. What confused me is why we can assign an integer [imath]m[/imath] to the character [imath]\chi.[/imath] Thanks in advance! Any comment is appreciated! |
718609 | Prove that if [imath]n[/imath] is composite, then [imath](n-1)! \equiv 0 \pmod n[/imath]
This theorem is the converse of Wilson's theorem: If [imath]n[/imath] is composite and [imath]n>4[/imath], then [imath](n-1)! \equiv 0 \pmod n[/imath] The question holds up for all the composites I have tried but I'm struggling to form a proof for all composites greater than [imath]4[/imath]. | 164852 | If [imath]n\ne 4[/imath] is composite, then [imath]n[/imath] divides [imath](n-1)![/imath].
I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you! Here is what I am asked to prove: If [imath]n[/imath] is composite then [imath](n-1)! \equiv 0 \pmod n[/imath]. Proof: [imath]n[/imath] is composite [imath]\implies n=ab[/imath] where [imath]a,b \in \mathbb{Z}[/imath] and [imath]0<a,b<n[/imath]. Case 1: If [imath]a=b[/imath] then [imath]n=a^{2}[/imath]. Now [imath]n \mid (n-1)! \implies a \mid (n-1)![/imath], so [imath]\begin{aligned} (n-1)! &\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &\equiv 0 \pmod n \end{aligned}[/imath] Case 2: [imath]0<a<b<n[/imath]. Then, since [imath]a \mid n[/imath], [imath]b \mid n[/imath] and [imath]n \mid (n-1)![/imath] we have that [imath]a \mid (n-1)![/imath] and [imath]b \mid (n-1)![/imath]. So this implies [imath](n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n[/imath], Q.E.D. |
718625 | Isomorphism problem
Problem Show that [imath]4 \mathbb{Z} / 8 \mathbb{Z} \simeq \mathbb{Z}_2[/imath] Can someone explain what the quotient group consists of in the most simple terms as humanly possible (freshman maths major here) and then explain to me how I'd go about showing that it is isomorphic to [imath] \mathbb{Z}_2[/imath] I'm having some real problems understand this concept. | 716993 | Seeing quotient groups
Can someone explain simply what a quotient group is? I've read a lot of convoluted and unnecessarily tough descriptions on it but it seems like a really simple idea. Question Show that [imath]8 \mathbb{Z} / 56 \mathbb{Z} \simeq \mathbb{Z}_7[/imath] I'm sure this is a really easy question if I can just realize what a quotient group is. If you got an easy way to describe isomorphism, that would be great too! |
718917 | Using [imath]S_n = \sum_{k=1}^{n}H_k[/imath] where [imath]H_k[/imath] are the harmonic numbers, show [imath]S_n = (n+1)H_n - n[/imath]
The question: Using [imath]S_n = \sum_{k=1}^{n}H_k[/imath] where [imath]H_k[/imath] are the harmonic numbers, show [imath]S_n = (n+1)H_n - n[/imath]. So far I have [imath]S_n = \sum_{k=1}^{n} H_k = \sum_{k=1}^{n} \sum_{j=1}^{k}\frac{1}{j} [/imath] is there perhaps some way to change the summation index? Or would the next step come from [imath]S_n = \sum_{k=1}^{n}[1+ \frac{1}{2} + \frac{1}{3} + ...+\frac{1}{k-1} + \frac{1}{k}][/imath] Any help would be appreciated. | 464957 | [imath]\sum_{k=1}^nH_k = (n+1)H_n-n[/imath]. Why?
This is motivated by my answer to this question. The Wikipedia entry on harmonic numbers gives the following identity: [imath] \sum_{k=1}^nH_k=(n+1)H_n-n [/imath] Why is this? Note that I don't just want a proof of this fact (It's very easily done by induction, for example). Instead, I want to know if anyone's got a really nice interpretation of this result: a very simple way to show not just that this relation is true, but why it is true. Has anyone got a way of showing that this identity is not just true, but obvious? |
693060 | [imath]\forall t \in \mathbb{R}, e^{i \alpha_n t} \rightarrow 1 \implies \alpha_n \rightarrow 0[/imath]?
[imath]\alpha_n \in \mathbb{R}[/imath] is a fixed sequence of real numbers, for which the following holds: [imath] \forall t \in \mathbb{R}, \lim_{n \rightarrow \infty}e^{i\alpha_nt} = 1 [/imath] Is it necessarily the case that [imath]\lim_{n \rightarrow \infty} \alpha_n = 0[/imath]? | 562506 | If [imath]e^{itx_n}[/imath] converges for every [imath]t\in\mathbb R[/imath], then does [imath]x_n[/imath] converge?
Question : Is the following true for any real number sequence [imath]x_n[/imath] ? "If [imath]e^{itx_n}[/imath] converges for every [imath]t\in\mathbb R[/imath], then [imath]x_n[/imath] converges." Note that [imath]i^2=-1[/imath]. Motivation : We know that the converse of the above proposition is true. However, I'm facing difficulty for the above proposition. Can anyone help? |
719304 | Convergence and Divergence
Suppose that the series [imath]\sum_{n=1}^\infty a_n[/imath] is conditionally convergent. Prove that the series [imath]\sum_{n=1}^\infty n^2a_n[/imath] is divergent. How should I start to prove this? I have absolutely no idea how to go about solving this problem. Thanks in advance! | 719181 | The series [imath]\sum a_n[/imath] is conditionally convergent. Prove that the series [imath]\sum n^2 a_n[/imath] is divergent.
Ratio and root tests won't help. And I can't use the comparison test because [imath] |a_n| [/imath] is not necessarily smaller than [imath] n^2a_n [/imath]. Can I use limits? We know: [imath]\lim\limits_{n \to \infty} a_n = 0 [/imath] [imath]\lim\limits_{n \to \infty} |a_n| \ne 0 [/imath] And we need to prove: [imath]\lim\limits_{n \to \infty} n^2a_n \ne 0 [/imath] Any ideas/hints? |
719451 | Show that [imath]3^n = 2^{O(n)}[/imath]
The formal description I have is that this is this: [imath]f(n) = n^{O(n)}[/imath] iff there exists some [imath]h(n) = O(n)[/imath] such that [imath]f(n) = n^{h(n)}[/imath]. I don't see how this can be applied to the problem to show that [imath]3^n = 2^{O(n)}[/imath]. What function [imath]h(n)[/imath] can we have such that [imath]3^n = n^{h(n)}[/imath]? Further, how is its existence shown, trial and error? | 719399 | Big O/little o true/false
These are all from Sipser's book, second edition. I was just hoping someone could verify/explain those that are more difficult for me. [imath]2n = O(n)[/imath]: true [imath]n^2 = O(n)[/imath]: false [imath]n^2 = O(n\log^2 n)[/imath]: I say false, because [imath]\log^2 n[/imath] grows more slowly than [imath]n[/imath] [imath]n\log n = O(n^2[/imath]): true because [imath]n[/imath] grows more quickly than [imath]\log^2n[/imath] [imath]3^n = 2^{O(n)}[/imath]: I wasn't sure what this notation meant. Any clarification here? [imath]2^{2^n} = O(2^{2^n})[/imath]: I would say true, but I'm not entirely sure why. [imath]n = o(2n)[/imath]: false, because [imath]o(2n) = o(n)[/imath] and [imath]n[/imath] grows as quickly as itself. [imath]2n = o(n^2)[/imath]: true, [imath]n^2[/imath] is always more than [imath]2n[/imath] for some constant. [imath]2^n = o(3^n)[/imath]: true [imath]1 = o(n)[/imath]: I'm not really sure. [imath]n = o(\log n)[/imath]: false, [imath]n[/imath] is always larger. [imath]1 = o(1/n)[/imath]: true, because 1 is always larger. |
719592 | Proofs by strong induction
I am trying to solve the following problem using strong induction, the problem is the following: For any positive integer [imath]n[/imath], let [imath]T_n[/imath] be the number [imath]1[/imath] if [imath]n<4[/imath] and the number [imath]T_{n − 1} +T_{n − 2} +T_{n − 3}[/imath] if [imath]n\ge 4[/imath]. We have [imath]T_1 =1, T_2 =1, T_3 =1, T_4 =T_3 +T_2+T_1=1+1+1=3[/imath], [imath]T_5 =T_4 +T_3 +T_2 =3+1+1=5[/imath], etc. Prove that: [imath]\forall n \in \Bbb{Z}^+[/imath], [imath]T_n <2^n[/imath] [imath]\;[/imath] Sadly, I don't even know where to start this question! | 719590 | Need help with Proof by Strong Induction question
So, here is the question: For any position integer [imath]n[/imath], let [imath]T(n)[/imath] be the number 1 if [imath]n<4[/imath] and the number [imath]T(n-1) + T(n-2) + T(n-3)[/imath] if [imath]n \geq 4[/imath]. We have [imath]T(1)=1, T(2)=2, T(3)=3[/imath] [imath]T(4)=T(3)+T(2)+T(1) = 1+1+1+1 = 3[/imath] [imath]T(5) = T(4)+T(3)+T(2) = 3+1+1 = 5[/imath] Prove that: (Universal n is an element of all positive integers), [imath]T(n)<2^n[/imath] Any suggestions? |
719474 | Equivalence relations--prove R/~ is homeomorphic to R
On [imath]\mathbb{R}[/imath] with the usual topology, generate an equivalence relation by [imath]x\sim x[/imath] for all [imath]x[/imath] in [imath]\mathbb{R}[/imath] and [imath]x\sim 0[/imath] for all [imath]x[/imath] in [imath][0,1][/imath]. Prove [imath]\mathbb{R}/\sim[/imath] is homeomorphic to [imath]\mathbb{R}[/imath]. I'm totally lost on this. I'm supposed to be proving that there is a homeomorphism but I don't understand how that is possibly true considering [imath]f:[0,1]\ \longrightarrow \{0\}[/imath] is not injective. | 719370 | Proof that [imath]\mathbb R/[0,1][/imath] and [imath]\mathbb R[/imath] are homeomorphic
On [imath]\mathbb{R}[/imath] with the usual topology, generate an equivalence relation by [imath]x\sim x[/imath] for all [imath]x[/imath] in [imath]\mathbb{R}[/imath] and [imath]x\sim 0[/imath] for all [imath]x[/imath] in [imath][0,1][/imath]. Prove [imath]\mathbb{R}/\sim[/imath] is homeomorphic to [imath]\mathbb{R}[/imath]. |
720783 | Simple inequality in euclidean [imath]n[/imath] space
Let [imath]x_1,...,x_n \in \mathbb{R}[/imath]. Then [imath] |x_1| + ... + |x_n| \leq \sqrt{n} \cdot \sqrt{ x_1^2 + ... + x_n^2} [/imath] Is this inequality true? I have proved it for [imath]n = 2[/imath]. For instance, it follows from the AM-GM inequality [imath] \frac{x^2 + y^2}{2} \geq xy \iff (x^2+y^2) \geq 2 xy \iff 2(x^2+y^2) \geq 2xy + x^2 + y^2 = (x+y)^2 \iff \sqrt{2} \sqrt{x^2 + y^2} \geq x + y[/imath] But, how can I prove this for the [imath]n[/imath] case? Induction does not seem to work well. Any help would be gratly appreciated. thanks | 678044 | Inequality $\left|\,x_1\,\right|+\left|\,x_2\,\right|+\cdots+\left|\,x_p\,\right|\leq\sqrt{p}\sqrt{x^2_1+x^2_2+\cdots+x^2_p}$
I could use some help with proving this inequality: [imath]$$\left|\,x_1\,\right|+\left|\,x_2\,\right|+\cdots+\left|\,x_p\,\right|\leq\sqrt{p}\sqrt{x^2_1+x^2_2+\cdots+x^2_p}$$[/imath] for all natural numbers p. Aside from demonstrating the truth of the statement itself, apparently [imath]\sqrt{p}[/imath] is the smallest possible value by which the right hand side square root expression must be multiplied by in order for the statement to be true. I've tried various ways of doing this, and I've tried to steer clear of induction because I'm not sure that's what the exercise was designed for (from Bartle's Elements of Real Analysis), but the best I've been able to come up with is proving that the statement is true when the right hand side square root expression is multiplied by p, which seems pretty obvious anyway. I feel like I'm staring directly at the answer and still can't see it. Any help would be appreciated. |
720813 | Do four dimensional vectors have a cross product property?
We know how to make cross product of three dimensional vectors. [imath] \vec A \times \vec B = \vec C[/imath] Where : [imath] \vec A = (A_i; A_j; A_k)[/imath] [imath] \vec B = (B_i; B_j; B_k)[/imath] [imath] \vec C = (C_i; C_j; C_k)[/imath] [imath]C_i = \left|\begin{matrix}A_j&A_k\\B_j&B_k\end{matrix}\right|[/imath] [imath]C_j = \left|\begin{matrix}A_k&A_i\\B_k&B_i\end{matrix}\right|[/imath] [imath]C_k = \left|\begin{matrix}A_i&A_j\\B_i&B_j\end{matrix}\right|[/imath] But what about if we have four dimensional vectors? Is it possible to make cross product of four dimensional vectors? If it is possible, then tell me when it can be possible? Let say we have two vectors: [imath] \vec A = (A_i; A_j; A_k; A_l)[/imath] [imath] \vec B = (B_i; B_j; B_k; B_l)[/imath] Then how to compute a cross product of this two vectors? Will it again vector? [imath] \vec A \times \vec B = \vec C[/imath] [imath] \vec C = (C_i; C_j; C_k; C_l)[/imath] Then how to compute those coordinates? We know that only square matrices have a determinant property! In this case it might not be correct if we will wright... [imath]\color{red} {\text { [/imath]C_i = \left|\begin{matrix}A_j&A_k&A_l\\B_j&B_k&B_l\end{matrix}\right|[imath]} C_j = \left|\begin{matrix}A_k&A_i&A_l\\B_k&B_i&B_l\end{matrix}\right| C_k = \left|\begin{matrix}A_i&A_j&A_l\\B_i&B_j&B_l\end{matrix}\right| C_l = \left|\begin{matrix}A_i&A_j&A_k\\B_i&B_j&B_k\end{matrix}\right|}[/imath] So tell me how to solve this problem? | 424482 | Cross product in [imath]\mathbb R^n[/imath]
I read that the cross product can't be generalized to [imath]\mathbb R^n[/imath]. Then I found that in [imath]n=7[/imath] there is a Cross product: https://en.wikipedia.org/wiki/Seven-dimensional_cross_product Why is it not possible to define a cross product for other dimensions [imath] \ge 4[/imath]? |
721173 | Proving [imath]\lim_{x \rightarrow 2} \sqrt[3]{x} = \sqrt[3]{2} [/imath] using [imath]\epsilon - \delta[/imath] definition
I'm trying to use the [imath]\epsilon[/imath] - [imath]\delta[/imath] definition to show that [imath]\lim_{x \rightarrow 2} \sqrt[3]{x} = \sqrt[3]{2}[/imath] . [imath]\forall \epsilon>0, \exists \delta >0[/imath] s.t. if [imath]0< | x-2 | < \delta[/imath] then [imath]| \sqrt[3]{x} - \sqrt[3]{2} | < \epsilon[/imath]. I'm not at all sure how to do this but... [imath]\sqrt[3]{2}-\epsilon < \sqrt[3]{x} < \epsilon+\sqrt[3]{2}[/imath] [imath]\Rightarrow[/imath] [imath](\sqrt[3]{2}-\epsilon)^3 <x< (\epsilon+\sqrt[3]{2})^3[/imath] and [imath]2-\delta < x < \delta + 2[/imath]. But now i'm stuck. If i say that [imath]2-\delta = (\sqrt[3]{2}-\epsilon)[/imath] and [imath]\delta + 2 = (\epsilon+\sqrt[3]{2})^3[/imath], then [imath]\delta[/imath] could be [imath]2-(\epsilon+\sqrt[3]{2})^3[/imath] or [imath](\epsilon+\sqrt[3]{2})^3 -2[/imath]...... How do i know which is the smaller of the two? Also this felt like a very complicated way to prove this. Is there any way to make [imath]| \sqrt[3]{x} - \sqrt[3]{2} |[/imath] seem more like | x-2 |??? | 687414 | Prove continuity for cubic root using epsilon-delta
I am trying to prove that a function is continuous at a point a using the [imath]\epsilon[/imath]-[imath]\delta[/imath] theorem. I managed to find a [imath]\delta[/imath] in this case [imath]|2x^2+1 - (2a^2+1)| < \epsilon[/imath]. But I have a hard time when the function under consideration is [imath]f(x) = \sqrt[3]{x}[/imath]. That is, I want to have if [imath]|x-a|<\delta[/imath], then [imath]|\sqrt[3]{x}-\sqrt[3]{a}| < \epsilon[/imath]. Any suggestions? |
721273 | Show that if [imath]\gcd(a,b)=1[/imath] and [imath]p[/imath] is an odd prime, then
Show that if [imath]\gcd(a,b)=1[/imath] and [imath]p[/imath] is an odd prime, then [imath]{\gcd(a+b,}\frac{a^p +b^p}{a+b}[/imath][imath]) = 1[/imath] or [imath]p[/imath] Sorry about the duplicate In another answer, however, the sum [imath]\sum\limits_{k=0}^{n-1} (-1)^{k}a^{n-1-k}b^{k}[/imath] was expressed as [imath]\left(\sum\limits_{k=0}^{n-2} (-1)^{k}(k+1)a^{n-2-k}b^{k}\right)(a+b) + (-1)^{n-1}nb^{n-1}[/imath] How was it done? | 340955 | [imath]\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1[/imath], or [imath]p[/imath]
Let [imath]p[/imath] be prime number ([imath]p\gt2[/imath]) and [imath]a,b\in\mathbb Z[/imath] ,[imath]a+b\neq0[/imath] ,[imath]\gcd(a,b)=1[/imath] how to prove that [imath]\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p[/imath] Thanks in advance . |
128021 | If [imath]n=2\phi(n)[/imath], then [imath]n=2^j[/imath].
I need to show that if [imath]n=2\phi(n)[/imath], then [imath]n=2^j[/imath], where [imath]n,j\in\mathbb{N}[/imath]. I have a strong feeling that this can only be shown by contradiction. Therefore, I assumed that both [imath]n=2\phi(n)[/imath] and [imath]n\neq2^j[/imath]. Then if we let, say, [imath]n=5[/imath], it follows that [imath]5=2\cdot\phi(5)=8[/imath], which is a contradiction. Hence, it must be the case that [imath]n=2^j[/imath]. Is this a valid proof? | 716805 | Find all n such that [imath]\phi(n) = n/2[/imath]
My idea for the solution is something like this: Since [imath]2 | n[/imath], [imath]n = 2^a p_1^{e1} p_2^{e2} \cdots p_t^{et}[/imath] where [imath]a \geq 1[/imath]. Then, [imath]n/2 = \phi(2^a) \phi(p_1^{e1}) \phi(p_2^{e2}) \cdots \phi(p_t^{et})[/imath]. Looking at the case that [imath]t = 0[/imath], [imath]n/2 = \phi(2^a) = 2^{a-1}[/imath], and therefore [imath]n = 2^a[/imath]. Otherwise, [imath]n = 2^a \phi(p_1^{e1}) \phi(p_2^{e2}) \cdots \phi(p_t^{et}) = n = 2^a p_1^{e1} p_2^{e2} \cdots p_t^{et}[/imath]. Since [imath]\phi(n) < n[/imath], this is a contradiction, and therefore, the only [imath]n[/imath]'s that apply are the powers of 2. Is this right? Also, we were asked to find [imath]n[/imath] such that [imath]\phi(n) = n/3[/imath]. How would I go about that one? |
528715 | For [imath]x>0[/imath], [imath]x + \frac1x \ge 2[/imath] and equality holds if and only if [imath]x=1[/imath]
Prove that for [imath]x>0[/imath], [imath]x + \frac1x \ge 2[/imath] and equality holds if and only if [imath]x=1[/imath]. I have proven that [imath]x+ \frac1x \ge 2[/imath] by re-writing it as [imath]x^2 -2x +1 \ge0[/imath] and factoring to [imath](x-1)^2\ge0[/imath] which is true because you cannot square something and it be negative. Now I am stuck on the part where I have to prove equality to hold if and only if [imath]x=1[/imath]. Any suggestions? | 705064 | How to prove this inequality [imath] x + \frac{1}{x} \geq 2 [/imath]
I was asked to prove that: [imath]x + \frac{1}{x}\geqslant 2[/imath] for all values of [imath] x > 0 [/imath] I tried substituting random numbers into [imath]x[/imath] and I did get the answer greater than [imath]2[/imath]. But I have a feeling that this is an unprofessional way of proving this. So how do I prove this inequality? |
722526 | Determinants and eigenvectors
Hello, I'm trying to work through this question. I define linearly independent as: [imath]a_1*v_1+a_2*v_2+...+a_n*v_n = 0[/imath] iff every [imath]a_i=0[/imath]. I also know that an eigenvector is a vector [imath]v[/imath] such that: [imath]T(v)=\lambda*v[/imath] where [imath]\lambda[/imath] is the eigenvalue. Thanks for your help. | 507641 | Show that the determinant of [imath]A[/imath] is equal to the product of its eigenvalues
Show that the determinant of a matrix [imath]A[/imath] is equal to the product of its eigenvalues [imath]\lambda_i[/imath]. So I'm having a tough time figuring this one out. I know that I have to work with the characteristic polynomial of the matrix [imath]\det(A-\lambda I)[/imath]. But, when considering an [imath]n \times n[/imath] matrix, I do not know how to work out the proof. Should I just use the determinant formula for any [imath]n \times n[/imath] matrix? I'm guessing not, because that is quite complicated. Any insights would be great. |
722869 | Sign of square root of a real number
May we write [imath]\sqrt{x^2}=\pm x[/imath]. Is [imath]\sqrt{x^2}=\sqrt{(\pm x)^2}=\pm{x}[/imath] true ? But we write [imath]\sqrt{x^2}=|x|=x[/imath] What is the actual logic? | 677123 | Values of square roots
Good-morning Math Exchange (and good evening to some!) I have a very basic question that is confusing me. At school I was told that [imath]\sqrt {a^2} = \pm a[/imath] However, does this mean that [imath]\sqrt {a^2} = +2[/imath] *and*[imath]-2[/imath] or does it mean: [imath]\sqrt {a^2} = +2[/imath] *OR*[imath]-2[/imath] Is it wrong to say 'and'? What are the implications of choosing 'and'/'or' Any help would be greatly appreciated. Thanks in advance and enjoy the rest of your day :) |
723455 | Is the value of the sum [imath]1-1+1-1+1-1+\cdots[/imath] does not exist?
Let [imath]a_k:=(-1)^k[/imath] where [imath]k\in\mathbb{N}[/imath]. [imath]\mathbb{N}[/imath] is the set of all non-negative integer. And we define the partial sum [imath]S_n:=\sum \limits_{k=0}^{n}a_k[/imath]. Notice that the sequence [imath]\{S_k\}[/imath] diverges which also implies that the infinite series [imath]\sum \limits_{k=0}^{\infty}a_k[/imath] cannot be defined. If I consider the infinite sum [imath]1-1+1-1+1-1+\cdots[/imath], then this statement is equivalent that I just defined the infinite series [imath]\sum \limits_{k=0}^{\infty}a_k[/imath], which is contradiction. But, suppose that the sum [imath]1-1+1-1+1-1+\cdots[/imath] exists and let the value of the sum be [imath]S[/imath]. Then, we can easily observe that [imath]S=1-S[/imath], therefore [imath]S=1/2[/imath]. The supposition of this proposition already proved as false, but if I ignore the definition of infinite series, it holds. Which one is right? Or are these propositions depends on what we define? | 635324 | Checking my understanding: [imath]1 - 1 + 1 - 1 + 1 - ... = \frac{1}{2}[/imath]
I've recently run into a proof that claims that [imath]1 - 1 + 1 - 1 + 1 - 1 ... = \frac{1}{2}[/imath] that proceeds as follows: Let [imath]S = 1 - 1 + 1 - 1 + 1 - 1 + ...[/imath]. Then [imath]S = 1 - (1 - 1 + 1 - 1 + 1 - 1 ...) = 1 - S[/imath] Therefore, [imath]2S = 1[/imath], so [imath]S = \frac{1}{2}[/imath]. QED I was under the impression that this sum doesn't converge, and therefore this step of the proof is invalid: Let [imath]S = 1 - 1 + 1 - 1 + 1 - 1 + ...[/imath] That is, it's not even valid to suppose that the sum is equal to some (implicitly real or complex) number [imath]S[/imath], and therefore the reasoning that follows is meaningless because [imath]S[/imath] doesn't exist and therefore has any properties we want it to have. Is the reasoning I've given above correct? That is, is it correct for me to claim that the proof fails because [imath]S[/imath] doesn't exist? (I've heard that there are techniques for evaluating divergent integrals by using complex analysis and Taylor series, so perhaps there's another way to prove that the summation is [imath]\frac{1}{2}[/imath]; I just wanted to see whether my reasoning is sufficient to explain why this particular proof is incorrect. If the proof actually is correct, then I stand corrected!) Thanks! |
723561 | How to solve this problem
Find the number of numbers between [imath]100[/imath] to [imath]400[/imath] which are divisible by either [imath]2,3,5,7[/imath] Please give some shortcut or some easy way | 504319 | How many positive integers [imath] n[/imath] with [imath]1 \le n \le 2500[/imath] are prime relative to [imath]3[/imath] and [imath]5[/imath]?
I am trying to understand this example from my study guide and am getting no where with it and need some help. Example: How many positive integers [imath]n[/imath] with [imath]1 \le n \le 2500[/imath] are prime relative to [imath]3[/imath] and [imath]5[/imath]? The example come from the chapter on Venn Diagrams. Let [imath]U = \{n \in \Bbb Z_+ \mid 1 \le n \le 2500\}[/imath]. We need to establish the number of positive integers in [imath]U[/imath] such that [imath]\gcd(3, n) = 1[/imath] and [imath]\gcd(5, n) = 1[/imath]. In this case, it is easier to count the integers that are relative primes with [imath]3[/imath] and [imath]5[/imath]. Note that an integer is not a relative prime with [imath]3[/imath] if it is a multiple of [imath]3[/imath]. Thus, [imath]A = \{n \in U\mid \gcd(3, n) = 1\} = \{3n \in U\mid 1 \le n \le 1249\}.[/imath] Can someone explain how [imath]249[/imath] is obtained? and [imath]B = \{n \in U\mid \gcd(5, n) = 1\} = \{5n \in U\mid 1 \le n \le 500\}.[/imath] Can someone explain how [imath]500[/imath] is obtained? Thus, [imath]A = \{n \in U\mid \gcd(3, n) = 1\}[/imath] and [imath]B = \{n \in U\mid \gcd(5, n) = 1\}[/imath]. We want to find [imath]|A \cap B| = |A \cup B| = |U| − |A \cup B|[/imath]. We have [imath]|A \cap B| = |\{15k \in U \mid 1 \le k \le 166\}| = 166[/imath]. Thus, [imath]|A \cap B| = |U| − |A| − |B| + |A \cap B| = 2500 − 1249 − 500 + 166 = 917[/imath] Thanks for any help. Tony |
216306 | [imath]\ell_p[/imath] is Hilbert if and only if [imath]p=2[/imath]
Can anybody please help me to prove this: Let [imath]p[/imath] be greater than or equal to [imath]1[/imath]. Show that for the space [imath]\ell_p=\{(u_n):\sum_{n=1}^\infty |u_n|^p<\infty\}[/imath] of all [imath]p[/imath]-summable sequences (with norm [imath]||u||_p=\sqrt[p]{\sum_{n=1}^\infty |u_n|^p}\ )[/imath], there is an inner product [imath]<\_\,|\,\_> [/imath] s.t. [imath]||u||^2=<u\,|\,u>[/imath] if and only if [imath]p=2[/imath]. | 1398003 | [imath]\ell^p\!,[/imath] for [imath]p\neq2[/imath], is not an inner product space.
Consider sequence spaces of the form, [imath]\ell^p=\Big\{x=\left(x_j\right)_{j=1}^\infty \mathrel{}\big|\mathrel{} \sum_{j=1}^\infty \left\lvert x_j \right\rvert ^p\lt\infty\Big\}[/imath] for [imath]1\le p\lt\infty[/imath], equipped with the norm, [imath]\left\|x\right\|_p:=\left(\sum_{j=1}^\infty\left\lvert x_j \right\rvert ^p\right)^{\frac{1}{p}}[/imath] for [imath]x\in\ell^p[/imath]. Now, in the case that [imath]n=2[/imath], the familiar norm on [imath]\ell^2[/imath] is obtained from the inner product, that is to say that the norm is induced by the inner product. And since [imath]\ell^2[/imath] is a infinite-dimensional Banach space, it follows that it is in fact an infinite dimensional Hilbert space. Why is it, that when [imath]p\neq2[/imath], the space [imath]\ell^p[/imath] is not an inner product space? |
722051 | Subgroup of [imath]S_4[/imath] with order 12
I need to find a subgroup of [imath]S_4[/imath] that has order 12 other than the subgroup of even permutations or anything isomorphic to it. What is the procedure to go about finding such a subgroup? | 722086 | Prove [imath]S_4[/imath] has only 1 subgroup of order 12
The subgroup in [imath]S_4[/imath] that I know has order 12 is the subgroup of all even permutations, otherwise known as the alternating group [imath]A_4[/imath]. However, I know this from a fact and not because I am able to show a subgroup of order 12 exists in [imath]S_4[/imath] in the first place. If I had not been told there existed a subgroup of order 12 in [imath]S_4[/imath], I would not have known there was one. So I guess this is actually a two-parter for me: 1) How do I go about showing that a subgroup of order 12 does indeed exist in [imath]S_4[/imath]? I know by Lagrange that if there exists a subgroup, then the order of that subgroup must divide the order of the group. However, this doesn't say anything about the existence of the subgroup. And Sylow only verifies subgroups of a prime to some power order, which 12 is not. I also know that there might be a cyclic subgroup of order 12, but without manually multiplying every possible permutation in [imath]S_4[/imath], I don't know any other way to check the existence of this or if this subgroup is isomorphic to [imath]A_4[/imath] because I don't know how to write these permutations as functions in order to check if there is a homomorphism (I know that a permutation is bijective by definition). And 2) How do I show that this group of order 12 is the only group of order 12 in [imath]S_4[/imath]? |
723769 | Groups of order 12 that aren't isomorphic
Give examples of four groups of order 12 no two of which are isomorphic. So far I've thought of [imath]Z_{12}[/imath] and [imath]D_6[/imath]. Thanks! | 683079 | Nonisomorphic groups of order 12.
I'm trying to find 4 groups of order 12, none of which are isomorphic to each other. Should I be trying external direct products? So far I have [imath]A_4, \mathbb Z_{12},\,[/imath] and [imath]\,\mathbb Z_6\times \mathbb Z_2.\,[/imath] How do I show all of these are non-isomorphic to each other? And how do I find a fourth? ` |
724115 | Question about elementary set theory
How can I explain that [imath]A\cup(B\cap C) = (A\cup B)\cap(A\cup C)[/imath] is true? | 660058 | Prove that [imath](X \cap Y) \cup Z = (X \cup Z) \cap (Y \cup Z)[/imath]
I've only the definition of union, intersection, subset, and complement available to me. [imath](X \cap Y) \cup Z = (X \cup Z) \cap (Y \cup Z)[/imath] [imath](X \cap Y) = \left\{a: a \in X, ~ a \in Y\right\}[/imath] [imath]\begin{eqnarray} (X \cap Y) \cup Z &=& \left\{a: a \in X ~ \text{or} ~ a \in Y\right\} \cup \left\{a: a \in Z\right\}\\ &=& \left\{a: a \in X ~ \text{or} ~ a \in Y, ~ \text{and} ~ a \in Z \right\} \tag1 \\ &=& \left\{a: a \in X~ \text{and} ~ a \in Z, ~ \text{or}, ~ a \in Y ~ \text{and} ~ a \in Z \right\} \tag2\\ &=&\left\{a: a \in X~ \text{and} ~ a \in Z\right\} \cap \left\{a: ~ a \in Y ~ \text{and} ~ a \in Z \right\}\tag3\\ &=& (X \cup Z) \cap (Y \cup Z)\tag4 \end{eqnarray}[/imath] I numbered those last few lines to make it easier to point out my blunders. I've never proven anything with sets before, so it probably doesn't make any sense. Many thanks in advance. |
724012 | Prove [imath]\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}[/imath] for [imath]0[/imath]
not sure how to approach the following [imath]\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}[/imath] for [imath]0<b<a<\pi/2[/imath]. Hints would be appreciated! | 719814 | Prove that [imath]\frac{\sin(a)}{\sin(b)} < \frac{a}{b} < \frac{\tan(a)}{\tan(b)}[/imath] where [imath]0 < b < a < \frac{\pi}{2}[/imath]
Prove the following: [imath]\frac{\sin(a)}{\sin(b)} < \frac{a}{b} < \frac{\tan(a)}{\tan(b)}[/imath] where [imath]0 < b < a < \frac{\pi}{2}[/imath] Hello everyone, I am trying to create some sort of function or maybe see an application of the mean value theorem, but I am just not getting it. Any help would be greatly appreciated! |
229312 | An entire function whose real part is bounded must be constant.
Greets This is exercise 15.d chapter 3 of Stein & Shakarchi's "Complex Analysis", they hint: "Use the maximum modulus principle", but I didn't see how to do the exercise with this hint rightaway, instead I knew how to do it with the Casorati-Weiestrass Theorem, here is my answer: Define [imath]g(z)=f(1/z)[/imath] for [imath]z\neq{0}[/imath],then by the hypothesis we must have that for any [imath]\epsilon>0[/imath] [imath]g(D_{\epsilon>0}(0)-\{0\})[/imath] is not dense in [imath]\mathbb{C}[/imath], then the singularity at [imath]0[/imath] of [imath]g[/imath] is not essential, this implies [imath]f[/imath] must be a polynomial, but if [imath]f[/imath] is a non-constant polynomial, it is easy to see that its real part must be unbounded, so [imath]f[/imath] must be constant. I would like to know an answer with the the maximum modulus principle. Thanks | 2022625 | A property of complex functions to be proven.
If a holomorphic on the whole complex plane funtion [imath]f[/imath] has a bounded real part then it is constant. Could I use here Cauchy-Riemann equations here? They seem to lead me nowhere. |
574379 | Proof for recurrence relation of laguerre polynomials
How can I prove the following recurrence relation for Laguerre polynomials. [imath] (n+1)L_{n+1}(x)=(2n+1-x)L_n(x)-nL_{n-1}(x) [/imath] please help me. thanks | 585770 | Laguerre polynomials recurrence relation
Laguerre polynomials has the recurrence relation [imath](n+1) L_{n+1} (x)=(2n+1-x) L_n (x)-nL_{n-1} (x)[/imath] In proving this, I differentiate the generating function of laguerre polynomials [imath]g(x,t)=\frac{e^{-\frac{xt}{1-t}}}{(1-t)}=\sum_{n=0}^\infty L_n (x)t^n [/imath] The process goes like this: \begin{align} \frac{\partial g}{\partial t} &=\frac{(1-t) D_t \left(e^{-\frac{xt}{1-t}}\right)-e^{-\frac{xt}{1-t}} D_t \left(1-t\right) }{(1-t)^2}\\ &=e^{-\frac{xt}{1-t}} \frac{1-t-x}{(1-t)^3} \end{align} [imath](1-t)^2 \frac{\partial g}{\partial t}=(1-t-x)g \tag 1[/imath] solving the right-hand side of eq. (1), we have: \begin{align} (1-t-x)g &=(1-t-x) \sum_{n=0}^\infty L_n (x)t^n\\ &=L_0 (x)-xL_0 (x)+\sum_{n=1}^\infty(L_n (x)-L_{n-1} (x)-xL_n (x) ) t^n \end{align} also, Solving the left-hand side of eq. (1), we have: [imath] \begin{align} (1-t)^2 \sum_{n=0}^\infty nL_n (x) t^{n-1}&=(1-2t+t^2)\sum_{n=0}^\infty nL_n (x) t^{n-1}\\ &=\sum_{n=0}^\infty nL_n (x)(t^{n-1}-2t^n+t^{n+1})\\ &=L_1 (x)+\sum_{n=1}^\infty t^n [(n+1) L_{n+1}(x)-2nL_n (x)+(n-1)L_{n-1} (x) ]. \end{align} [/imath] My question is: How did the derivative of the generating function with respect to [imath]t[/imath] in the left hand side equal to [imath]\sum_{n=0}^\infty nL_n (x) t^{n-1}[/imath] that is, [imath] \frac{\partial g}{\partial t}=\sum_{n=0}^\infty nL_n (x) t^{n-1} [/imath] Thanks! :) |
331556 | Missing dollar problem
This sounds silly but I saw this and I couldn't figure it out so I thought you could help. The below is what I saw. You see a top you want to buy for [imath]\[/imath]97[imath], but you don't have any money so you borrow [/imath]\[imath]50[/imath] from your mom and [imath]\[/imath]50[imath] from your dad. You buy the top and have [/imath]\[imath]3[/imath] change, you give your mom [imath]\[/imath]1[imath],your dad [/imath]\[imath]1[/imath], and keep [imath]\[/imath]1[imath] for yourself. You now owe your mom [/imath]\[imath]49[/imath] and your dad [imath]\[/imath]49[imath].[/imath] \[imath]49 + \[/imath]49 = \[imath]98[/imath] and you kept [imath]\[/imath]1[imath]. Where is the missing [/imath]\[imath]1[/imath]? | 1369374 | 3 men and a cold night
[imath]3[/imath] guys, each with [imath]\[/imath]10[imath] a piece, go to a hotel hoping to get a room to stay in for the night. A room costs [/imath]\[imath]60[/imath]. The men go in, and ask to rent a room, only having [imath]\[/imath]30[imath] between them. The mater deen says he can give them a broom closet to sleep in for the night. One of the men asks if they can keep a few dollars to get coffee and breakfast in the morning. Mater deen says no, and has a bellboy take them to the closet. Downstairs, the mater deen feels bad, and gives the bell boy [/imath]\[imath]5[/imath] to give back to them. The bellboy pockets [imath]2[/imath] of the [imath]5[/imath], and gives each guy a dollar back. IF the men, then, spent [imath]\[/imath]9[imath] each for the closet, and [/imath]3\times9 = 27[imath]. The bell boy pocketed [/imath]2[imath], so [/imath]27+2=29[imath]. what happened to the [/imath]30$th dollar? |
725208 | Finitely generated ideal in boolean ring
A boolean ring is a commutative ring where [imath]x^{2} = x[/imath] for every [imath]x[/imath]. Why in such a ring a finitely generated ideal is principal ? | 110329 | Finitely generated ideals in a Boolean ring are principal, why?
The classical book on commutative algebra Introduction to Commutative Algebra, by Atiyah and Macdonald, has the following as exercise I.11. A ring is Boolean if [imath]x^2=x[/imath] for any [imath]x[/imath] of [imath]A[/imath]. In a Boolean ring [imath]A[/imath], show that i) [imath]2x=0[/imath] for all [imath]x\in A[/imath]; ii) Every prime ideal of [imath]A[/imath] is maximal, and its residue field consists of two elements; iii) Every finitely generated ideal of [imath]A[/imath] is principal. I can but solve the first two of them; for the last one, I cannot even perceive the situation here, for it will not, generally speaking, be a Dedekind domain, whereof my knowledge is somewhat more. I have hitherto tried to find one element for each ideal of two generators which can stand the stead of them, and then proceed by induction. But even the first step fails to produce any result. Thanks for any hints and inspirations in advance. |
726465 | prove [imath]EO=OF[/imath] is there an euclidean solution?
[imath]PQ[/imath] is a chord in a circle such that [imath]PO=OQ[/imath] The chords [imath]AB,CD[/imath] are passing throw [imath]O[/imath]. The chords [imath]AD,BC[/imath] are cutting [imath]PQ[/imath] at [imath]E[/imath] and [imath]F[/imath] reciprocally. Need to prove: [imath]EO=OF[/imath]. If so i solved it in trigonometry but im asking if there is an euclidean solution. if so, i'd love to see! thanks. http://i57.tinypic.com/dexx0y.jpg EDIT: [imath]O[/imath] is not the center of the circle. | 2248 | How to prove the midpoint of a chord is also the midpoint of the line segment defined by the points of intersection of other two chords with it?
Bernhard Elsner, alias MathOMan, posted this exercise in plane Geometry, Theorem about a circle, three chords and a midpoint on January [imath]$29$[/imath]th, [imath]$2010.$[/imath] "Let [imath]\mathcal{C}[/imath] be a circle, [imath]A,B[/imath] two distinct points on [imath]\mathcal{C}[/imath] and [imath]M[/imath] be the midpoint of the chord [imath][AB][/imath]. Take two other chords,[imath][PQ][/imath] and [imath][SR][/imath], that pass through [imath]M[/imath]. Let [imath]C[/imath] (resp. [imath]D[/imath]) be the intersection of [imath][AB][/imath] with [imath][PS][/imath] (resp. [imath][RQ][/imath]). Prove that [imath]M[/imath] is the midpoint of the chord [imath][CD][/imath]." To prove it I've written the following (failed) argument, in the German version of this post (translation of mine): The figure is symmetric with respect to [imath]M[/imath]: [imath]\overline{AM}=\overline{MB}[/imath], [imath]\overline{PM}=\overline{MU}[/imath], [imath]\overline{RM}=\overline{MW}[/imath], [imath]\overline{QM}=\overline{MV}[/imath], [imath]\overline{QR}=\overline{VW}[/imath], [imath]\overline{SM}=\overline{MT}[/imath]. From [imath]\dfrac{\overline{SC}}{\overline{DT}}=\dfrac{\overline{CM}}{\overline{MD}}=1[/imath] follows that [imath]\overline{CM}=\overline{MD}[/imath]. Here is an extract of the author's reply (translation of mine): "It is not clear that [imath]\overline{QM}=\overline{MV}[/imath]. Is the point [imath]V[/imath] on the line [imath](QM)[/imath] defined by this equality or is [imath]V[/imath] defined as the intersection point of the lines [imath](QM)[/imath] and [imath](WC)[/imath]? Why are both definitions to give the same point? Let [imath]C^{\prime }[/imath] be the intersection of [imath](SP)[/imath] and [imath](VW)[/imath], and [imath]D^{\prime }[/imath] the intersection of [imath](TU)[/imath] and [imath](QR)[/imath]. (...) One still has to show that [imath]C^{\prime }=C[/imath] and [imath]D^{\prime }=D[/imath]." I have agreed with these objections. Until now no proof has been posted. The author considers that the "proof is not quite simple". Q. What is the theorem this exercise refers to? Or how does one prove it? |
726703 | Fibonacci General Formula - Is it obvious that the general term is an integer?
Given the recurrence relation for the Fibonacci numbers, [imath]F_{n+1}=F_{n}+F_{n-1}[/imath] with [imath]F_0=1[/imath] and [imath]F_1=1[/imath] it's obvious that [imath]F_n[/imath] is a positive integer for all [imath]n[/imath]. Suppose instead we were given [imath]F_n=\frac{1}{\sqrt{5}} \left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}} \left(\frac{1-\sqrt{5}}{2}\right)^n[/imath] Without spotting that [imath]F_{n+1}=F_{n}+F_{n-1}[/imath], how might one show that [imath]F_n[/imath] is an integer for all [imath]n[/imath]? | 672212 | How to prove that Fibonacci number is integer?
How to prove that formula for Fibonacci numbers are always integers, for all [imath]n[/imath]: [imath] F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}} [/imath] where, [imath]\varphi = \frac{1 + \sqrt{5}}{2}[/imath] and [imath]\psi = \frac{1 - \sqrt{5}}{2}[/imath]. I know how to prove that [imath]F_n[/imath] is rational, but what about integer? |
727211 | bounded monotone continuous function is uniform continuous
If [imath]f: \mathbb R \to \mathbb R[/imath] is bounded, increasing and continuous. Does [imath]f[/imath] have to uniform continuous? I know the answer is yes if [imath]f[/imath] has domain to be any open interval, say [imath](0,1)[/imath]. But I don't know how to prove this case. Any help is appreciated. | 105310 | Is an increasing, bounded and continuous function on [imath][a,+\infty)[/imath] uniformly continuous?
Supose [imath]f[/imath] is increasing, bounded and continuous on [imath][a,+\infty)[/imath]. Is [imath]f[/imath] uniformly continuous ? I think yes. how to prove that? My idea is to show there exists [imath]X[/imath] , [imath]f[/imath] is uniformly continuous on [imath][X,+\infty)[/imath]. How to fix such [imath]X[/imath]? |
75589 | Open maps which are not continuous
What is an example of an open map [imath](0,1) \to \mathbb{R}[/imath] which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map [imath]e^{1/z}[/imath] from [imath]\mathbb{C}[/imath] to [imath]\mathbb{C}[/imath] (filled in to be [imath]0[/imath] at [imath]0[/imath]). There must be a simpler example, using the usual Euclidean topology, right? | 3041740 | Finding a mapping that is not continuous but open
Find example of a mapping that is open but not continuous and a mapping that is closed but not continuous. I striving with these questions. I thought of using for the first case a map between [imath](\mathbb{Z},\tau)[/imath] where [imath]\tau[/imath] is the co-finite topology to [imath](\mathbb{R},\tau')[/imath] with the standard topology. I think the mapping would not be continuous and since any open set in [imath]\mathbb{R}[/imath] is infinite and uncountable and the inverse image would need to be countable. However I am not seeing how to prove this with more rigorous mathematical terminology and I do not see how this mapping might be open. Question: Can someone help me solve this question? Thanks in advance! |
727959 | Probability of rolling a six on seven die?
How would one calculate the probability of rolling a six (at least once) on seven dice? Would I be correct in proposing [imath]1 - P(\text{no sixes})[/imath] which is [imath]1 - (5/6) ^ 7[/imath]? | 473026 | Probability of 1x six from seven dice?
Could someone help me with how the following is calculated: What is the probability of rolling a die seven times and getting at least one six? My instinct told me it would be [imath]1/6+\cdots + 1/6[/imath] but this ends up being [imath]7/6[/imath]. |
727699 | Tangent bundle of a manifold
can anyone help me with this problem: Show that for a manifold [imath]M[/imath], the tangent bundle [imath]TM[/imath] also has the structure of a manifold. If [imath]M[/imath] is an n-manifold, what is the dimension of [imath]TM[/imath]? for the 1st part it is enough to show that it is locally euclidean. thanx. | 316065 | Tangent bundle : is a manifold
I have studied what a differentiable manifold is, and what a tangent space at a given point is, and read the proof that its dimension is equal to the dimension of the manifold. Here a tangent vector is a derivation on the germs of differentiable functions at a point. Now I am trying to see that the tangent bundle is indeed a differentiable manifold. I get that for an open set [imath]U[/imath] in Eucliden space, it takes the form [imath]U \times \mathbb R^n[/imath] with the vector [imath]\sum a_i x_i[/imath] giving the tangent vector [imath]\sum a_i \partial /\partial x_i[/imath]. Now the book says that it is "obvious" that these glue up and give a differentiable manifold. I may be stupid; but I will be grateful if somebody can hold my hand through this "obvious" statement. |
726029 | Does [imath]\dim V=\dim(\operatorname{Ker}(T))+\dim(\operatorname{Im}(T))[/imath] imply [imath]V=\operatorname{Ker}(T)+\operatorname{Im}(T)[/imath] iff [imath]T[/imath] is a homomorphism?
The will be linear transofmation [imath]T:V \rightarrow V[/imath] The formula says that [imath]\dim V=\dim(\operatorname{Ker}(T))+\dim(\operatorname{Im}(T))[/imath] and [imath]\operatorname{Ker}(T)\subseteq V[/imath] and [imath]\operatorname{Im}(T) \subseteq V[/imath] Does it implies that [imath]V=\operatorname{Ker}(T)\oplus \operatorname{Im}(T)[/imath] iff [imath]T[/imath] is a homomorphism? | 247911 | When does [imath]V = \ker(f) \oplus \operatorname{im}(f)[/imath]?
For a vector space [imath]V[/imath] and a linear operator [imath]f : V \to V[/imath], under what conditions does [imath]V = \ker(f) \oplus \operatorname{im}(f)[/imath]? Is it always true, or only in special cases? Edit: [imath]V[/imath] is finite dimensional. |
727835 | Different solutions to [imath]\int^\pi_{-\pi} \cos^3(x) \cos(ax)~dx[/imath]
Linked to my previous question, when solving the following integral ([imath]a[/imath] is an integer) I get: [imath]\int^\pi_{-\pi} \cos^3(x) \cos(ax)~dx = \frac{2a(a^2-7)\sin(\pi a)}{a^4 - 10a^2 + 9}[/imath] However, trivially, [imath]\sin(\pi a) = 0[/imath] for all integer values of [imath]a[/imath]. Therefore the integral is always equal to [imath]0[/imath]. Wolfram Alpha agrees with this solution. However, let us substitute [imath]a=1[/imath] and [imath]a=3[/imath] into the integral and then solve: [imath]\int^\pi_{-\pi} \cos^3(x) \cos(x)~dx = \frac{3\pi}{4}[/imath] and [imath]\int^\pi_{-\pi} \cos^3(x) \cos(3x)~dx = \frac{\pi}{4}[/imath] Why do these answers disagree? | 727664 | Evaluating [imath]\int^\pi_{-\pi} \cos^3(x) \cos(ax)~dx[/imath]
I need to evaluate [imath]I = \int^\pi_{-\pi} \cos^3(x) \cos(ax)~dx[/imath] where [imath]a[/imath] is some integer. I get: [imath]\dfrac{2a(a^2-7)\sin(\pi a)}{a^4 - 10a^2 + 9}[/imath]. However [imath]\sin(\pi a)[/imath] is [imath]0[/imath] for all [imath]a[/imath] so [imath]I=0[/imath]. But as noted by an answer, there are answers for [imath]a = 1,3,-1,-3[/imath]. |
636702 | The Limit of [imath]x\left(\sqrt[x]{a}-1\right)[/imath] as [imath]x\to\infty[/imath].
How to find the limit of: [imath] \lim_{x\to \infty }x\left(\sqrt[x]{a}-1\right)[/imath] Without using L'hôspital rule. I've tried to bound the term and use the squeeze theorem but I couldn't find the right upper bound. I've also tried to convert [imath]a^\frac{1}{x}[/imath] to [imath]e^{\frac{1}{x}\ln{a}}[/imath] but it didn't help me. Whats the right way to evaluate that limit? | 195446 | Finding [imath]\lim_{n \to \infty} n(\sqrt[n]{2} - 1)[/imath]
I am trying to find the limit of: [imath]\lim_{n \to \infty} n(\sqrt[n]{2} - 1)[/imath] I know it should be very simple but I don't seem to get it. Thanks in advance for any help! |
729010 | Show that [imath]16[/imath] is a perfect [imath]8[/imath]th power modulo [imath]p[/imath] for any prime number [imath]p[/imath]
Show that [imath]16[/imath] is a [imath]8th[/imath] power [imath]\mod{}[/imath] [imath]p[/imath] for any prime number [imath]p[/imath]. I have no idea how to approach this. I tried, [imath]a^8\equiv16\pmod{p}[/imath] [imath](a^4+4)(a^4-4)\equiv 0 \pmod{p}[/imath] [imath]a^4 \equiv \pm4\pmod{p}[/imath] [imath]a^4 \equiv 2,p-2\pmod{p}[/imath] But this doesn't seem to lead anywhere. | 6758 | Showing [imath]x^8\equiv 16 \pmod{p}[/imath] is solvable for all primes [imath]p[/imath]
I'm still making my way along in Niven's Intro to Number Theory, and the title problem is giving me a little trouble near the end, and I was hoping someone could help get me through it. Now [imath]x^8\equiv 16\pmod{2}[/imath] is solvable with [imath]x\equiv 0\pmod{2}[/imath], so I assume [imath]p[/imath] is an odd prime. From a theorem earlier in the text, If [imath]p[/imath] is a prime and [imath](a,p)=1[/imath], then the congruence [imath]x^n\equiv a\pmod{p}[/imath] has [imath](n,p-1)[/imath] solutions or no solution according as [imath]a^{(p-1)/(n,p-1)}\equiv 1\pmod{p}[/imath] or not. So since [imath](16,p)=1[/imath], the problem reduces to showing that [imath]16^{(p-1)/(8,p-1)}\equiv 1\pmod{p}[/imath] holds for all [imath]p[/imath]. I note that [imath](8,p-1)[/imath] can only take values [imath]2,4,8[/imath]. For [imath]2[/imath], the above equivalence is then [imath]4^{p-1}\equiv 1\pmod{p}[/imath], which is true by Fermat's little Theorem. For [imath]4[/imath], it is then [imath]2^{p-1}\equiv 1\pmod{p}[/imath], which again holds by FlT. However, the case where [imath](8,p-1)=8[/imath] is throwing me off. At best I see that [imath]16^{(p-1)/8}\equiv 2^{(p-1)/2}\pmod{p}[/imath], but I'm not sure how to show this is congruent to [imath]1[/imath] modulo [imath]p[/imath]. Maybe there's a more elegant way to do it without looking at cases. Thanks for any insight. |
729138 | Linearity of Expectation for Beer
Bud sells [imath]n[/imath] different brands of ales. When you place an order, Bud sends you one bottle of ale, chosen uniformly at random from the [imath]n[/imath] different brands, independently of previous orders. Jim wants to try all different brands of ale. He repeatedly places orders at Bud (one bottle per order) until he has received at least one bottle of each brand. Define the random variable [imath]X[/imath] to be the total number of orders that Jim places. Determine the expected value [imath]E(X)[/imath]. Use Linearity of Expectation. If Jim has received exactly [imath]i[/imath] different brands of ale, how many orders does he expect to place until he receives a new brand? I'm unsure of how to use Linearity of Expectation to solve this problem. I have the following so far. [imath]x[/imath] = the total number of orders that Jim places. Since we know with every new brand of ale Jim receives, then his chance to get a new ale is less. [imath]E(x) = \sum_{i=0}^n (1-\frac{i}{n})[/imath] First ale he receives, probability of a new type is and as follows: [imath]1, 1-\frac{1}{n}, 1-\frac{2}{n},...,1-\frac{n-1}{n},1-\frac{n}{n}[/imath] until he gets n different ales then there is no chance for him to receives new brands. Could someone assist me with this proof? | 287051 | Using Recursion to Solve Coupon Collector
I read a brilliant answer by Mike Spivey on one of the questions and I was wondering how I could use it to solve a coupon collectors problem. The problem is : There are coupons labelled 1,2,3...,10 how many coupons do I need to collect in order to have one of each labels. I know that the answer is [imath]\displaystyle \sum_{i=1}^{10} \dfrac{10}{i}[/imath] Here is my attempt : Let [imath]X_i[/imath] be the random variable corresponding to the number of coupons needed to be collected to have exactly [imath]i[/imath] unique labels. \begin{align} E(X_1)&=1\\ E(X_2|X_1)&=\dfrac{9}{10}.(X_1+1)+\dfrac{1}{10}.(E(X_2))\\ \implies E(X_2)&=\dfrac{9}{10}.(E(X_1)+1)+\dfrac{1}{10}.(E(X_2))\\ \text{Similarly,}\\ E(X_3)&=\dfrac{8}{10}.(E(X_2)+1)+\dfrac{2}{10}.(E(X_3))\\ \vdots \end{align} But this gives me the wrong answer. I know that there is a problem in my second equation but don't know why. My logic was as follows: Assuming I know how much it takes to get 1 coupon [imath](E(X_1))[/imath] with probability 9/10 I find 2 in [imath]E(X_1)+1[/imath] else, I just have [imath]E(X_2)[/imath]. Neither does the formula in the aforementioned question work. Can someone help me set up the recursion equation? (If possible, please retain my Random Variables. I am more interesting in knowing why my logic is failing in designing the recursion than answering the original question) |
729318 | Differentiability proof
Suppose [imath]f[/imath] is differentiable at [imath]0[/imath] and [imath]f(0)=0[/imath]. Prove that there exist some function [imath]g(x)[/imath] such that [imath]f(x)=xg(x)[/imath] and [imath]g[/imath] is continuous at [imath]0[/imath]. I think I am on the right track, here: If I write [imath]g(x)=\frac{f(x)}{x}[/imath] then [imath]g[/imath] will be discontinuous at [imath]0[/imath] because [imath]f(0)=0[/imath]. Now, as an example, let [imath]f(x)=x^2[/imath] and [imath]g(x)=x[/imath]. How can I proceed? | 338355 | Show that [imath]g:I\to\mathbb{R}[/imath] is continuous and [imath]f:(x-x_0)g(x)[/imath] is differentiable
(a) Let a function [imath]g:I\to\mathbb{R}[/imath] be continuous at [imath]x_0\in I[/imath]. Prove that the function [imath]f[/imath] given by [imath]f(x)=(x-x_0)g(x)[/imath] is differentiable at [imath]x_0[/imath]. (b) Let a function [imath]f:I\to\mathbb{R}[/imath] be differentiable at [imath]x_0\in I[/imath] and [imath]f(x_0)=0[/imath]. Prove that there exists a function [imath]g:I\to \mathbb{R}[/imath] that is continuous at [imath]x_0 \in I[/imath] such that [imath]f(x)=(x-x_0)g(x)[/imath]. Thought: (a) By [imath]Fermat's[/imath], since [imath]g[/imath] is continous, if [imath]g'(x_0)=0[/imath] then [imath]g(x)[/imath] is differentiable? but how does it connect back with [imath]f(x)[/imath]? (b) use the definition of every differentiable function is continuous? Thus [imath]f(x)[/imath] is continuous and also [imath](x-x_0)[/imath] so [imath]g(x)[/imath] too? Thank you!! |
729881 | weak convergence of a bounded linear operator
I need help with this problem Let [imath]X[/imath] be a reflexive Banach space and [imath]T: X \to X[/imath] a linear operator. Show that [imath]T[/imath] belongs to [imath]\mathcal{L}(X,X)[/imath] if and only if whenever [imath]\{x_n \}[/imath] converges weakly to [imath]x[/imath], [imath]\{T(x_n)\}[/imath] converges weakly to [imath]T(x)[/imath]. any hints or suggestions, thank you! | 5795 | weak sequential continuity of linear operators
Suppose I have a weakly sequentially continuous linear operator T between two normed linear spaces X and Y (i.e. [imath]x_n \stackrel {w}{\rightharpoonup} x[/imath] in [imath]X[/imath] [imath]\Rightarrow[/imath] [imath]T(x_n) \stackrel {w}{\rightharpoonup} T(x)[/imath] in [imath]Y[/imath]). Does this imply that my operator T must be bounded? |
718646 | Prove that [imath]\rm area(\triangle ABE) = area (ABCD)[/imath]
Here is a problem from a sample paper: I see that [imath]\rm area(ACE) = area(DCE)[/imath]. I also see that [imath]\rm AB[/imath] is the common base for [imath]\triangle \rm ABE[/imath] and [imath]\rm ABCD[/imath]. So this means that the height of [imath]\rm ABCD[/imath], i.e., [imath]\rm AD[/imath] is exactly the half of [imath]\rm BE[/imath]. I can't seem to get around the rest, though. It seems that [imath]\rm ACED[/imath] is a parallelogram, but that'd mean [imath]\rm BC = CE \Rightarrow AD = DC[/imath] which doesn't seem so looking at the figure. Could anyone please give me a complete proof using only the theorems: Two quadrilaterals lying on the same base and between the same parallels have the same area. Two triangles lying on the same base and between the same parallels have the same area. I also know that the area of 1 is twice the area of 2. | 666099 | When is [imath]Ar(APD)=Ar(ABCD)[/imath]?
This question arose while I was answering this question, (we need to show [imath]Ar(\Delta APD)=Ar(ABCD)[/imath]). First the original question: [imath]ABCD[/imath] is a quadrilateral. A line through [imath]D[/imath] parallel to [imath]AC[/imath] meets [imath]BC[/imath] produced at [imath]P[/imath] we need to show [imath]Area(\Delta APD)=Area(ABCD)[/imath] Its easy to see that the OP must have meant to prove, [imath]Area(\Delta ABP)=Area(ABCD)[/imath], the proof of which is given in my answer. However, it set me thinking when actually [imath]Area(\Delta ADP)=Area(ABCD)[/imath] is true? I have not been able to derive any good conclusion (I mean, in terms of the elements (diagonals, angles or sides) of [imath]ABCD[/imath]). Can anyone help? |
729894 | probability need help on correlation problem
A deck of 52 cards is shuffled you are dealt 13 cards. Let [imath]X[/imath] and [imath]Y[/imath] denote, respectively, the number of aces and the number of spades in your hand. Show that [imath]X[/imath] and [imath]Y[/imath] are uncorrelated. I try to write [imath]X=X_1+X_2+\cdots+X_{13}[/imath], such that [imath]X_1\ldots X_{13}[/imath] is the indicator of the [imath]i^{th}[/imath] card. But this doesn't work. please help. | 339454 | Let [imath]X[/imath] be the number of aces and [imath]Y[/imath] be the number of spades. Show that [imath]X[/imath], [imath]Y[/imath] are uncorrelated.
A deck of 52 cards is shuffled, and we deal a bridge of 13 cards. Let [imath]X[/imath] be the number of aces and [imath]Y[/imath] be the number of spades. Show that [imath]X[/imath], [imath]Y[/imath] are uncorrelated. Here is what I did: [imath]Cov(X,Y) = E[XY]-E[X]E[Y][/imath] uncorrelated means [imath]Cov(X,Y) = 0[/imath], hence [imath]E[XY]=E[X]E[Y][/imath] [imath]E[X] = \sum_{k=0}^{k=4} k \frac{\dbinom{4}{k} \dbinom{48}{13-k}}{\dbinom{52}{13}} [/imath] [imath]E[Y] = \sum_{k=0}^{k=13} k \frac{\dbinom{13}{k} \dbinom{39}{13-k}}{\dbinom{52}{13}} [/imath] Are the summations above correct? and how do I calculate [imath]E[XY][/imath]? |
730146 | Is the property of Euclidean domain inherited via surjective ring homomorphism?
Let [imath]f:R \to S[/imath] be surjective ring homomorphism and [imath]R,S[/imath] be integral domains. Could anyone advise me on how to prove/disprove this statement: If [imath]R[/imath] is Euclidean domain, then [imath]S[/imath] is Euclidean domain. Thank you. | 102513 | Is any quotient of a Euclidean domain by a prime ideal a Euclidean domain?
Let [imath]R[/imath] be a Euclidean domain, i.e., a ring with a norm [imath]N : R \rightarrow \mathbb N[/imath] such that for any [imath]a,b\in R[/imath] with [imath]b\not=0[/imath], we may write [imath]a = qb + r[/imath] for some [imath]q,r \in R[/imath] with [imath]N(r) < N(b)[/imath]. Let [imath]I[/imath] be a prime ideal of [imath]R[/imath], so [imath]R/I[/imath] is a domain. Must [imath]R/I[/imath] in fact be a Euclidean domain? I know that this is true if we replace "Euclidean domain" with "PID", and false if we replace it with "UFD", so there is no clear intuition to be gained from similar concepts. My first attempts at a proof was to borrow a construction from analysis and define a norm on [imath]R/I[/imath] by [imath]N(a + I) = \inf\{N(a+i) : i\in I\}[/imath] but this didn't really get me anywhere. So, this fact true? If so, how can I prove it? If not, whats a counterexample? |
722786 | What is the probability that [imath]x_1+x_2+...+x_n \le n[/imath]?
Given that [imath]X_1, X_2...[/imath] are mutually independent random variables. For each [imath]i[/imath] with [imath]1\le i \le n[/imath] the variable [imath]X_i[/imath] is equal to either [imath]0[/imath] or [imath]n+1[/imath] [imath]E(X_i)[/imath] = [imath]1[/imath] also.. if [imath]X_i[/imath] is equal to either [imath]0[/imath] or [imath]n+1[/imath], doesn't that mean that all [imath]X_i[/imath] need to be [imath]0[/imath]? Thanks | 736855 | Probability of mutually independent random variables
I have been doing an exam review and this is one of the problems I could not solve. Let [imath]X_1, X_2, ..., X_n[/imath] be a sequence of mutually independent random variables. For each [imath]i[/imath] with [imath]1 \leq i \leq n[/imath], the variable [imath]X[/imath], is equal to either [imath]0[/imath] or [imath]n+1[/imath] [imath]E(X_i) = 1[/imath] Determine [imath]Pr(X_1+ X_2 + ... + X_n \leq n)[/imath] |
605048 | When does Gâteaux imply Fréchet?
Speaking of the relation between Gâteaux and Fréchet, authors usually point out that [imath]\text{Fréchet} \implies \text{Gâteaux}[/imath] and then give a counterexample to illustrate that the converse doesn't hold. Say, we have a function which is known to be Gâteaux differentiable (all directional derivatives exist). I wonder what conditions need to be added in order to insure Fréchet differentiability? In case the function goes from [imath]\mathbb R^n[/imath], the continuity of partial derivatives does the job. But what if the function is between two Banach spaces? Would it suffice if all directional derivatives are continuous? Update: I have found the following theorem, unfortunately without any proof. Let [imath]X,Y[/imath] be Banach spaces and [imath]U\subset X[/imath] open. If the following conditions hold: [imath]f:U\to Y[/imath] has continuous Gâteaux derivative \begin{align} df: U \times X & \to Y \\ (x,h) & \mapsto df(x,h) \end{align} the linear operator [imath]df(x,\cdot)[/imath] is bounded for all [imath]x\in U[/imath] the function \begin{align}U & \to \mathcal{L}(X,Y) \\ x & \mapsto df(x,\cdot) \end{align} is continuous then [imath]f\in C^1(U,Y)[/imath]. Does it look reasonable? Where could I find the proof? | 145910 | Fréchet differentiability from Gâteaux differentiability
Let [imath]X[/imath] be a Banach space and [imath]\Omega \subset X[/imath] be open. The functional [imath]f[/imath] has a Gâteaux derivative [imath]g \in X'[/imath] at [imath]u \in \Omega[/imath] if, [imath]\forall h\in X,[/imath] [imath]\lim_{t \rightarrow 0}[f(u+th)-f(u)- \langle g,th \rangle]=0[/imath] How can I prove the following: If [imath]f[/imath] has a continuous Gâteaux derivative on [imath]\Omega[/imath], then [imath]f \in C^1(\Omega,\mathbb R)[/imath]. |
730411 | Behavior of [imath]f'(x)[/imath] as [imath]x \to \infty[/imath] given the behavior of [imath]f(x)[/imath] and [imath]f''(x)[/imath]
Suppose [imath]f[/imath] is twice differentiable on [imath](a, \infty)[/imath] and let [imath]\lim_{x \to \infty}f(x) = 0[/imath]. Also assume that [imath]f''(x)[/imath] is bounded for all [imath]x \in (a, \infty)[/imath]. Prove that [imath]f'(x) \to 0[/imath] as [imath]x \to \infty[/imath]. Note that if [imath]f'(x)[/imath] tends to a limit as [imath]x \to \infty[/imath] then this limit must be [imath]0[/imath] (because of the relation [imath]f(x) - f(x/2) = (x/2)f'(c)[/imath]). What we need to establish now is that [imath]f'(x)[/imath] does tend to a limit. To show that we need to somehow use the fact that [imath]f''(x)[/imath] is bounded. Let [imath]b > 0[/imath] be any arbitrary but fixed number then we have relation [imath]f(x + b) - f(x) = bf'(c)[/imath] and this again implies that for all sufficiently large [imath]x[/imath] the derivative [imath]f'[/imath] takes a small value in the interval [imath](x, x + b)[/imath]. What we need to show that eventually all the values of [imath]f'[/imath] are small in such intervals. Now we can take [imath]b[/imath] as small as we please, we can therefore find a value [imath]N > 0[/imath] such that [imath]f'[/imath] takes very small value for at least one point in [imath](x, x + b)[/imath] for all [imath]x > N[/imath]. If [imath]p, q[/imath] are two points of this interval [imath](x, x + b)[/imath] then [imath]|f'(p) - f'(q)| = |(p - q)f''(\xi)| \leq b|f''(\xi)|[/imath] and since [imath]f''[/imath] is bounded it follows that [imath]f'(p) - f'(q)[/imath] is small so that effectively all values of [imath]f'[/imath] in [imath](x, x + b)[/imath] are small. The above is a non-rigorous argument which I have not been able to make rigorous by using [imath]\epsilon, \delta[/imath] in proper manner. Maybe the approach used above can't be made rigorous and I am wrong path (or may be not!). Please help me in making above approach rigorous or suggest some alternative method. Further there is another generalization available: If [imath]f(x) [/imath] tends to a finite limit as [imath]x \to \infty[/imath] and [imath]f^{(n + 1)}(x)[/imath] is bounded then show that [imath]f^{(n)}(x) \to 0[/imath] as [imath]x \to \infty[/imath]. I have not tried to solve this general problem and any hint would be great. | 396707 | If [imath]f(x)\to 0[/imath] as [imath]x\to\infty[/imath] and [imath]f''[/imath] is bounded, show that [imath]f'(x)\to0[/imath] as [imath]x\to\infty[/imath]
Let [imath]f\colon\mathbb R\to\mathbb R[/imath] be twice differentiable with [imath]f(x)\to 0[/imath] as [imath]x\to\infty[/imath] and [imath]f''[/imath] bounded. Show that [imath]f'(x)\to0[/imath] as [imath]x\to\infty[/imath]. (This is inspired by a comment/answer to a different question) |
731535 | If Limit of function and derivative exist, then limit of derivative is 0
Any hints for this question , My attempt; Say $f(x):0[imath]\rightarrow[/imath]\mathbb{R}$ The by MVT, there exists a $c[imath]\in[/imath](0,\infty)$ , such that; $f'(c)=[imath]\frac{f(x)-f(0)}{x-0}$ but im not sure about this step.. $\lim_{c \to +\infty}[/imath]f'(c)$=$\lim_{x \to +\infty}[imath]\frac{f(x)-f(0)}{x-0}$ I am quite sure there is a mistake, any hints would be appreciated. [/imath] If $\lim_{x \to +\infty}f(x)$ exists and is finite and [imath]\lim_{x \to +\infty}[/imath][imath]f'(x)=b[/imath] then [imath]b=0[/imath]. | 42277 | Proving that [imath]\lim\limits_{x\to\infty}f'(x) = 0[/imath] when [imath]\lim\limits_{x\to\infty}f(x)[/imath] and [imath]\lim\limits_{x\to\infty}f'(x)[/imath] exist
I've been trying to solve the following problem: Suppose that [imath]f[/imath] and [imath]f'[/imath] are continuous functions on [imath]\mathbb{R}[/imath], and that [imath]\displaystyle\lim_{x\to\infty}f(x)[/imath] and [imath]\displaystyle\lim_{x\to\infty}f'(x)[/imath] exist. Show that [imath]\displaystyle\lim_{x\to\infty}f'(x) = 0[/imath]. I'm not entirely sure what to do. Since there's not a lot of information given, I guess there isn't very much one can do. I tried using the definition of the derivative and showing that it went to [imath]0[/imath] as [imath]x[/imath] went to [imath]\infty[/imath] but that didn't really work out. Now I'm thinking I should assume [imath]\displaystyle\lim_{x\to\infty}f'(x) = L \neq 0[/imath] and try to get a contradiction, but I'm not sure where the contradiction would come from. Could somebody point me in the right direction (e.g. a certain theorem or property I have to use?) Thanks |
731778 | How do I simplify [imath]\frac{\sqrt{4+h}-2}{h}[/imath]?
I know this looks like a dumb question, but how do I simplify this? Does it uses some square root property or factorization? The wolfram alpha has no step-by-step solution for this, so it may use some crazy technique, I guess. [imath]\frac{\sqrt{4+h}-2}{h} = \frac{1}{\sqrt{4+h}-2}[/imath] | 20583 | How to simplify [imath]\frac{\sqrt{4+h}-2}{h}[/imath]
The following expression: [imath]\frac{\sqrt{4+h}-2}{h}[/imath] should be simplified to: [imath]\frac{1}{\sqrt{4+h}+2}[/imath] (even if I don't agree that this second is more simple than the first). The problem is that I have no idea of the first step to simplify that.. any help? |
732351 | The solutions to [imath]x^2+y^2=5[/imath] in [imath]\mathbb{Q}[/imath].
Consider the following equation: [imath]x^2+y^2=5.\tag{1}[/imath] What are the solutions to this equation if [imath]x,y\in\mathbb{Q}[/imath], where [imath]\mathbb{Q}[/imath] is the set of all rational numbers? My attempt: Because [imath]x,y\in\mathbb{Q}[/imath], we can set [imath]x=a/b[/imath] and [imath]y=c/d[/imath], where [imath]a,b,c,d\in\mathbb{Z}[/imath], [imath]b\neq0[/imath], [imath]d\neq0[/imath], gcd[imath](a,b)=1[/imath], gcd[imath](c,d)=1[/imath]. [imath]\mathbb{Z}[/imath] is the set of all integers and gcd[imath](a,b)[/imath] denotes the greatest common divisor of [imath]a[/imath] and [imath]b[/imath]. Then the original equation (1) can be rewritten as [imath]\left(\frac{a}{b}\right)^2+\left(\frac{c}{d}\right)^2=5.\tag{2}[/imath] When [imath]a=0[/imath], we have [imath]c/d=\pm\sqrt{5}[/imath]. In this case, there is no solution for [imath]c[/imath] and [imath]d[/imath] in integers because [imath]\pm\sqrt{5}[/imath] are irrational numbers. Similarly, when [imath]c=0[/imath] we know that there is no solution for [imath]a[/imath] and [imath]b[/imath] in integers. When [imath]a\neq0[/imath] and [imath]c\neq0[/imath], we rewrite equation (2) as [imath]\begin{aligned} a&=\pm\sqrt{b^2[5-(c/d)^2]}\\ &=\pm|b|\sqrt{5-(c/d)^2}\\ &=\pm|b|\sqrt{\frac{5d^2-c^2}{d^2}}\\ &=\pm\left|\frac{b}{d}\right|\sqrt{5d^2-c^2}\tag{3}. \end{aligned}[/imath] In order to make [imath]a\in\mathbb{Q}[/imath], we must have [imath]5d^2-c^2=e^2\tag{4}[/imath] for some [imath]e\in\mathbb{Z}[/imath]. Then I don't know how to continue. Any comments and answers are welcome. Thank you very much! | 249735 | Generating Pythagorean triples for [imath]a^2+b^2=5c^2[/imath]?
Just trying to figure out a way to generate triples for [imath]a^2+b^2=5c^2[/imath]. The wiki article shows how it is done for [imath]a^2+b^2=c^2[/imath] but I am not sure how to extrapolate. |
733433 | Fibonacci Proof with Induction
[imath]f(n) = \left\{\begin{matrix} 0 & n=1\\ 1 & n=2\\ f_{n-1} + f_{n-2} & n\geqslant 2\end{matrix}\right.[/imath] How can I prove by induction that [imath]f_{n} \geq \left ( 1.5 \right )^{n-1}[/imath] for all[imath] n\geq l_{b}[/imath], I have to find the smallest value for [imath]l_{b}[/imath] | 733215 | Fibonacci proof by induction
I have fibonacci numbers defined as such: [imath] f(n) = f(n-1) + f(n-2)[/imath]with[imath] f(0) = 0 [/imath][imath]f(1) = 1[/imath] I have to prove that [imath] F(n) \geq 1.5^{n-1}, n \geq 6[/imath] Base Case: [imath] f(6) = 8 \geq (1.5)^5 = 7.6 [/imath] Base holds Inductive hypothesis [imath] k \geq 6 [/imath][imath] f(k) \geq (1.5)^{k-1} [/imath] Inductive Step Here is how I have started, any hints or pointers on how to continue are appreciated [imath] f(k-1) \geq (1.5)^{k+1-1}[/imath] [imath] f(k) + f(k-1) = (1.5)^{k-1} + (1.5)^{k-2}[/imath] now I think where I want to go from here is to expand the right side of this equation out, but I am unsure how to do this. |
665438 | Validity of Substitutions in Integrals
In integrals, we often make 'u' substitutions, which involve writing [imath]u=f(x)[/imath], and then also writing [imath]du=f'(x)dx[/imath]. How valid is this? I've heard that you aren't allowed to do things like 'break' up derivatives like they do to solve first order differential equations. So can you do this with integrals, or is it also a mnemonic? | 737928 | The formalism behind integration by substitution
When you are doing an integration by substitution you do the following working. [imath]\begin{align*} u&=f(x)\\ \Rightarrow\frac{du}{dx}&=f^{\prime}(x)\\ \Rightarrow du&=f^{\prime}(x)dx&(1)\\ \Rightarrow dx&=\frac{du}{f^{\prime}(x)}\\ \end{align*}[/imath] My question is: what on earth is going on at line [imath](1)[/imath]?!? This has been bugging me for, like, forever! You see, when I was taught this in my undergrad I was told something along the lines of the following: You just treat [imath]\frac{du}{dx}[/imath] like a fraction. Similarly, when you are doing the chain rule [imath]\frac{dy}{dx}=\frac{dy}{dv}\times\frac{dv}{dx}[/imath] you "cancel" the [imath]dv[/imath] terms. They are just like fractions. However, never, ever say this to a pure mathematician. Now, I am a pure mathematician. And quite frankly I don't care if people think of these as fractions or not. I know that they are not fractions (but rather is the limit of the difference fractions as the difference tends to zero). But I figure I should start caring now...So, more precisely, [imath]\frac{du}{dx}[/imath] has a meaning, but so far as I know [imath]du[/imath] and [imath]dx[/imath] do not have a meaning. Therefore, why can we treat [imath]\frac{du}{dx}[/imath] as a fraction when we are doing integration by substitution? What is actually going on at line [imath](1)[/imath]? |
725948 | Relationship for cosine of angle
If [imath]x[/imath] is the cosine of the angle between the vectors [imath]a[/imath] and [imath]b[/imath], [imath]y[/imath] is the cosine of the angle between the vectors [imath]a[/imath] and [imath]p[/imath], and [imath]z[/imath] is the cosine of the angle between the vectors [imath]b[/imath] and [imath]p[/imath], can we write [imath]z[/imath] in terms of [imath]x[/imath], [imath]y[/imath] and the magnitudes of [imath]a[/imath], [imath]b[/imath], and [imath]p[/imath]? That is, no dot products should appear in the relation. All vectors are in [imath]4[/imath]-dimensional Euclidean space. | 719622 | Relationship between cosines of angles in 4 dimensions
If the cosine of the angle between the vectors [imath]a[/imath] and [imath]b[/imath] is [imath]x[/imath] and the cosine of the angle between the vectors [imath]a[/imath] and [imath]p[/imath] is [imath]y[/imath], then, if we call [imath]z[/imath] the cosine of the angle between the vectors [imath]b[/imath] and [imath]p[/imath], can we write [imath]z[/imath] in terms of [imath]x[/imath] and [imath]y[/imath]? All vectors are in [imath]4[/imath]-dimensional Euclidean space. |
735216 | Classify this factor group
I'm having trouble classifying this factor group: ([imath]Z[/imath] x [imath]Z[/imath]) / [imath]\langle (2,2) \rangle[/imath] This is from section 15 of Fraleigh, so we don't have a general algorithm for classifying factor groups. In general, I figure out how many elements the factor group has, whether it's abelian (usually is), and consider [imath]Z_n[/imath] where n = order of factor group, amongst others. Ideas for how to think about this one? Thanks, Mariogs | 720900 | Classifying the factor group [imath](\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle[/imath]
We wish to classify the factor group [imath](\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle[/imath], that is, find a group to which it is isomorphic. (According to the fundamental theorem of finitely generated abelian groups. Initially, I thought the group had but two cosets, forcing an isomorphism to [imath]\mathbb{Z}_2[/imath]. Obviously, this is wrong, due to the existence of cosets such as [imath](1, 0) + \langle (2, 2) \rangle[/imath] However, I am unable to see how I am to find an isomorphism here. |
735097 | How can I prove this power set relation?
Please show me how can I prove this relation: [imath]P(A \cap B) = P(A) \cap P(B)[/imath] I have no idea how to prove these kind of relations that include power sets. | 490524 | The power set of the intersection of two sets equals the intersection of the power sets of each set
It would be great if someone could verify this proof. Theorem: [imath]\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)[/imath] Proof: First I prove that [imath]\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)[/imath]. Take any subset [imath]X \subseteq A \cap B[/imath]. Then [imath]X \in \mathcal{P}(A \cap B)[/imath]. Also, [imath]X \subseteq A \wedge X \subseteq B[/imath], meaning that [imath]X \in \mathcal{P}(A) \wedge X \in \mathcal{P}(B)[/imath]. This, again, means that [imath]X \in \mathcal{P}(A) \cap \mathcal{P}(B)[/imath], and this proves the statement. Then I prove that [imath]\mathcal{P}(A) \cap \mathcal{P}(B) \subseteq \mathcal{P}(A \cap B)[/imath]. Again, take any subset [imath]X \subseteq A \cap B[/imath]. Using just the same arguments as above, this is also a truth. The theorem follows. |
735377 | [imath]f:A \backslash U\to B \backslash f(U)[/imath] is also homeomorphic?
[imath]f:A \to B[/imath] a homeomorphic mapping from [imath]A[/imath] to [imath]B[/imath] [imath]=> f:A \backslash U\to B \backslash f(U)[/imath] is also homeomorphic, [imath]U \subseteq A[/imath]. Why does this implication hold? | 735121 | [imath]g[/imath] a restriction of a homeomorphic function [imath]f[/imath], [imath]g[/imath] also homeomorphic?
Let [imath]f:X \to Y[/imath] be a homeomorphism between topological spaces [imath]X[/imath] and [imath]Y[/imath]. Let [imath]A \subseteq X[/imath] and [imath]B = f(A) \subseteq Y[/imath] (given the subspace topology) and let [imath]g: A \to B[/imath] be the restriction of [imath]f[/imath] given by [imath]g(a) = f(a)[/imath] for [imath]a \in A[/imath]. Showing that [imath]g[/imath] is a homeomorphism. [imath]g[/imath] injective Let [imath]g(a_1) = g(a_2)[/imath], [imath]a_1, a_2 \in A[/imath]. [imath]\implies f(a_1) = f(a_2)[/imath] [imath]\implies a_1 = a_2[/imath] as [imath]f[/imath] homeomorphic. g surjective Let [imath]y \in B[/imath] [imath]\implies y = f(a)[/imath] for some [imath]a \in A[/imath]. [imath]\implies y = g(a)[/imath] for some [imath]a \in A[/imath]. [imath]g[/imath] continuous Let [imath]U \subseteq B[/imath] be an open set in [imath]B[/imath]. [imath]f(a) = g(a)[/imath] for all [imath]a \in A => f^{-1}(b) = g^{-1}(b)[/imath] for all [imath]b \in B[/imath]. So [imath]g^{-1}(U) = f^{-1}(U)[/imath] which is open in [imath]A[/imath] as [imath]f[/imath] homeomorphic. [imath]g^{-1}[/imath] continuous Let [imath]U \subseteq A[/imath] be an open set in [imath]A[/imath]. [imath]g(U) = f(U)[/imath] which is open in [imath]B[/imath] as [imath]f[/imath] homeomorphic. So [imath]g[/imath] is homeomorphic. Is my understanding correct, this seems almost too straightforward so I was wondering have I overlooked something? |
205661 | Open and Closed sets of [imath]\Omega[/imath]
Let be [imath]u[/imath] a numerical function defined over [imath]\Omega[/imath], with [imath]u[/imath] measurable, and let [imath](O_i)_{i\in I}[/imath] be a family of all open sub-sets [imath]O_i[/imath] of [imath]\Omega[/imath], such that [imath]u=0[/imath] almost always except on a set of measure [imath]0[/imath] in [imath]O_i[/imath]. Let [imath]O = \cup_{i\in I}O_i[/imath] (Then [imath]u=0[/imath] almost always except on a set of measure in [imath]O[/imath]). How can I define in explicit form the set [imath]\Omega\setminus O[/imath]? I trying this ... [imath]\Omega\setminus O = \{x\in \Omega; (u(x)=0 \text{ and } x\in K\subset \Omega, \text{ where } K \text{ is closed}) \text{ or } (u(x)\neq 0)\}[/imath]. This is correct? | 203603 | Open Set and Measure Theory
Let be [imath]u[/imath] a numerical function defined over [imath]\Omega[/imath], with [imath]u[/imath] measurable, and let be [imath](O_i)_{i\in I}[/imath] a family of all open sub-sets [imath]O_i[/imath] of [imath]\Omega[/imath], such that [imath]u=0[/imath] often in [imath]O_i[/imath]. Let be [imath]O = \cup_{i\in I}O_i[/imath]. Then [imath]u=0[/imath] often in [imath]O[/imath]. How I can be able to do this?. I am beginning make ... Let be [imath]u[/imath] defined than [imath]0[/imath] in [imath]O_i\setminus M_i[/imath] and [imath]\neq[/imath] [imath]0[/imath] in [imath]M_i[/imath], then [imath]O = \cup_{i\in I}O_i=\cup_{i\in I}[(O_i\setminus M_i)\cup M_i][/imath], ... but I don't know how find the subset of [imath]O[/imath] such that have measure zero. |
735935 | Boolean Function question
I need to know how I can prove this question. Prove that not every boolean function is equal to a boolean function constructed by only using And ([imath]\wedge[/imath]) and Or ([imath]\vee[/imath]) | 707337 | Not every boolean function is constructed from [imath]\wedge[/imath] (and) and [imath]\vee[/imath] (or)
Prove that not every boolean function is equal to a boolean function constructed by only using [imath]\wedge[/imath] and [imath]\vee[/imath]. Here is my solution, can I ask for a feed back on my solution please? [imath]p∧q[/imath] [imath]0 0 0 1 [/imath] [imath]p∨q[/imath] [imath]0 1 1 1[/imath] Not every boolean function is the same when using [imath] ∨ [/imath] and [imath]∧.[/imath] Edited part! [imath]p\vee q[/imath] [imath]0111[/imath] [imath](p\vee q)\wedge p = p[/imath] [imath]0 = 0[/imath] [imath]0 = 0[/imath] [imath]1 = 1[/imath] [imath]1 = 1[/imath] [imath](p\vee q)\vee p = p\vee q[/imath] [imath]0 = 0[/imath] [imath]1 = 1[/imath] [imath]1 = 1[/imath] [imath]1 = 1[/imath] here is my edited answer , can I ask for more feed back , thanks |
736238 | A group without proper subgroups is a cyclic group
If a group [imath]G[/imath] has no nontrival proper subgroups, prove that [imath]G[/imath] is a cyclic group with order [imath]p[/imath], where [imath]p[/imath] is a prime number. | 91850 | Determine group G as cyclic
If [imath]G[/imath] has no proper subgroup, prove that [imath]G[/imath] is cyclic of order [imath]p[/imath], where [imath]p[/imath] is a prime number. I know that since [imath]G [/imath]is a group with no proper subgroups, [imath]g \in G[/imath] is not just the identity. I don't know where to go from there. |
736334 | What about the index of this subgroup?
Let [imath]G[/imath] be a group, and let [imath]H[/imath] be a subgroup of finite index in [imath]G[/imath], and let [imath]N \colon = \cap_{x \in G} \ xHx^{-1}[/imath]. Then [imath]N[/imath] is clearly a subgroup of [imath]G[/imath] which is contained in [imath]H[/imath] and such that [imath]aNa^{-1} = N[/imath] for all [imath]a[/imath] in [imath]G[/imath]. Is [imath]N[/imath] of finite index in [imath]G[/imath]? And If so, what can we say about an upper bound for the index of [imath]N[/imath]? If not, then can we find a subgroup [imath]N^\prime[/imath] of finite index in [imath]G[/imath], contained in [imath]H[/imath], and such that [imath]aN^\prime a^{-1} = N^\prime[/imath] for all [imath]a[/imath] in [imath]G[/imath]? And what can we say about the index of [imath]N^\prime[/imath] in [imath]G[/imath]? | 680868 | Poincaré's theorem about groups
Let [imath]G[/imath] be a group and [imath]H<G[/imath] such that [imath][G:H]<\infty[/imath]. There exists a subgroup [imath]N\triangleleft G[/imath] such that [imath][G:N]<\infty[/imath]. I have to show this fact (that according to my book is due to Poincaré), but I think that the statement, written in this way, is trivial: for every group [imath]G[/imath], I can take [imath]N=G[/imath], in fact [imath]G\triangleleft G[/imath] and [imath][G:G]=1[/imath]. Where am I wrong? If I'm not wrong, do you know a similar statement? |
736725 | How find this limit [imath]\lim_{n\to\infty}\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n},x\in R[/imath]
Find this limit [imath]\lim_{n\to\infty}\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n},x\in R[/imath] My idea: let [imath]f(x)=\underbrace{\sin{\sin{\cdots\sin{x}}}}_{n}[/imath] then [imath]f(x+2\pi)=f(x)[/imath],so we only consider [imath]x\in[0,2\pi][/imath], so define sequence [imath]a_{n}[/imath] such [imath]a_{1}=\sin{x},a_{n+1}=\sin{a_{n}}[/imath] so I think we can Discussion of x.can you someone have methods? other idea: [imath]|a_{n+2}-a_{n+1}|=|\sin{a_{n+1}}-\sin{a_{n}}|\le |a_{n+1}-a_{n}|[/imath] | 732030 | Limit of infinite loops of sin x as n tends to infinity
Show that [imath]lim_{n\to\infty} \text {sin sin ... sin x} = 0 [/imath] for all x. Note that the n here refers to the number of sin in the expression above. |
736482 | Cardinalities of bases of a free [imath]R[/imath] module are same?
Let [imath]R[/imath] be a ring with no zero divisiors such that for all [imath]r,s\in R[/imath] there exist [imath]a,b\in R[/imath] not both zero with [imath]ar+bs=0[/imath]. If [imath]R=K\oplus L[/imath] then [imath]K=0[/imath] or [imath]L=0[/imath]. if [imath]R[/imath] has an idendity then any two bases of free [imath]R[/imath] module [imath]F[/imath] have the same cardinality. I am done with first statement, I guess that I will use first part to show the second one. I need help in second one. Thanks. | 540219 | Invariant dimension property on a ring without zero divisors
The following is exercise number 10 in Hungerford's Algebra, page 190: Let [imath]R[/imath] be a ring with no zero divisors such that for all [imath]r, s \in R[/imath] there exist [imath]a, b \in R[/imath], not both zero, such that [imath]ar + bs = 0[/imath]. a) If [imath]R = K \oplus L[/imath] (module direct sum) then [imath]K = 0[/imath] or [imath]L = 0[/imath] b) If [imath]R[/imath] has an identity, then [imath]R[/imath] has the invariant dimension property. I have proved part a), but I'm having difficulty with b). I realize that it suffices to show that [imath]R^m \cong R^n[/imath] implies [imath]m = n[/imath], but I can't seem to use part a) to conclude this. |
737064 | Let A and B be matrices with same dimension. Prove [imath]|\det({}^tA\times B)|^2\leq\det({}^tA\times A)\cdot \det({}^tB\times B)[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be matrices of the same dimension. Prove [imath]|\det({}^tA\times B)|^2\leq\det({}^tA\times A)\cdot \det({}^tB\times B)[/imath], where [imath]{}^tA[/imath] is the transpose of matrix [imath]A[/imath] and [imath]\det[/imath] is the determinant. Thank you so much, any help is appreciated :) | 721669 | A determinant inequality
Let [imath]A,B[/imath] be two [imath]m\times n[/imath] real matrices. Then [imath]|AA'|\cdot |BB'|\geq |AB'|^2.[/imath] For square matrices, it is the equality. How to prove this inequality then? |
737592 | Find the center of the group [imath]\operatorname{GL}(n,\mathbb R)[/imath] of invertible [imath]n \times n[/imath] matrices.
Find the center of the group [imath]\operatorname{GL}(n,\mathbb R)[/imath] of invertible [imath]n \times n[/imath] matrices. Please can someone please help me? I know that by definition the center [imath]Z[/imath] of a group [imath]G[/imath] is defined by [imath]Z(G) = \{g \in G\ |\ ag = ga ,\, \forall a \in G\}[/imath]. I know that the identity matrix commutes with any matrix. I also notice by computing several matrix products that if we have a matrix with a main diagonal and all other entries are zero, then the given matrix commutes. In addition, I know that the determinant cannot be zero since zero times another matrix will only be zero. Please I would really appreciate the help. Thank you. | 2726753 | Finding center of [imath]GL_n(\Bbb R)[/imath] using elementary matrices
I want center of [imath]GL_n(\Bbb R)[/imath]. Artin, in his book Algebra, hints that, 'You are asked to determine the invertible matrices [imath]A[/imath] that commute with every invertible matrix [imath]B[/imath]. Do not test with general matrix [imath]B[/imath], test with elementary matrices.' I know that Row-multiplying transformations and Row-addition transformations generate [imath]GL_n(\Bbb R).[/imath] I also see that Row-multiplying transformations (unless all diagonal entries are 1's) and Row-addition transformations (unless all off-diagonal entries are 0's) don not commute with each other, so only elenentary matrix in center is [imath]I[/imath]. What I don't get is: 1. How do we derive from this that [imath]cI[/imath] is in center for any [imath]c\in \Bbb R[/imath]? (I can see [imath]cI[/imath] in center, but, it did not click me, because I thought this way: since center is subgroup, [imath]\langle I\rangle[/imath] is in center, but [imath]\langle I\rangle=\{I\}[/imath]. Where did I not think through?) 2. No other matrix is in center? |
737684 | Union of preimages
Given [imath]f:X\rightarrow Y[/imath] as a function, the image of [imath]x[/imath] if [imath]f(x)[/imath]. The preimage of [imath]y[/imath] is [imath]f^{-1}(y)=\{x\ |\ f(x)=y\}[/imath], with the symbol PreIm[imath](Y)[/imath] Given the definition, could you prove the following statement? Thank you. Given that [imath]f:X\rightarrow Y[/imath] is a function, and that [imath]A, B \subseteq Z[/imath]. PreIm[imath](A\cup B)[/imath] = PreIm[imath](A) \cup[/imath] PreIm[imath](B)[/imath] | 174401 | Inverse image of a union equals the union of the inverse images
I just wonder if my following solution is true. Let [imath]X,Y[/imath] be sets, let [imath]f:X\to Y[/imath] be a function, let [imath]\{Y_i\}_{i\in I}[/imath] be a family of subsets of [imath]Y[/imath]. (Note: I use equalities instead of mutual containment) [imath]f^{-1}[\bigcup_{i\in I} Y_i] = \{x \in X: \mbox{there exists an}\quad i \in I\mbox{ such that } y \in Y_i,f(x)=y\} =\bigcup_{i \in I} f^{-1}[Y_i] [/imath] I initially do not know how to get from the left to right, but when I put both sets in set notation, they turn out to be the same, hence the one line proof. Something go ultimately wrong? |
737916 | Let [imath]H[/imath] be a subgroup of [imath]G[/imath] and let [imath]a[/imath] and [imath]b[/imath] belong to [imath]G[/imath]. Then [imath]aH=bH[/imath] or [imath]aH \bigcap bH =\emptyset[/imath]
Let [imath]H[/imath] be a subgroup of [imath]G[/imath] and let [imath]a[/imath] and [imath]b[/imath] belong to [imath]G[/imath]. Then : [imath]aH=bH[/imath] or [imath]aH \bigcap bH =\emptyset[/imath] ...... (1) Gallian gives the following proof which i have a little trouble understanding : By a previous result, we have : [imath]aH = bH [/imath] if and only if [imath]a \in bH[/imath] ....(2) It says [imath](1) [/imath] follows directly from [imath](2)[/imath] for if there is an element c in [imath]aH \bigcap bH[/imath], then [imath]cH = aH [/imath] and [imath]cH = bH[/imath] . I don't understand the above statement, so i have made the following attempt : Attempt: If [imath]aH \neq bH[/imath] , then [imath]a \notin bH[/imath] by [imath](2)[/imath] So , if [imath]a \notin bH[/imath], what can i say about [imath]aH \bigcap bH[/imath]? Thank you for the help | 99421 | Cosets of a subgroup do not overlap
This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post. This problem is from assignment [imath]5[/imath]. Prove directly that distinct cosets do not overlap. Let [imath]H[/imath] be a subgroup of [imath]G[/imath] and let [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] be elements of [imath]G[/imath] such that [imath]b\not\in aH[/imath] and [imath]c[/imath] is in both [imath]aH[/imath] and [imath]bH[/imath]. Then there are elements [imath]h[/imath] and [imath]h^\prime[/imath] in [imath]H[/imath] such that [imath]c=ah[/imath] and [imath]c=bh^\prime[/imath]. So [imath]ah=bh^\prime[/imath] and [imath]b=ahh^{\prime -1}[/imath]. But [imath]hh^{\prime -1}\in H[/imath] so [imath]b\in aH[/imath]. This contradicts our original assumption. Therefore, there can be no element in more than one distinct coset. Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem. Thanks. |
442459 | For the Fibonacci numbers, show for all [imath]n[/imath]: [imath]F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}[/imath]
The definition of a Fibonacci number is as follows: [imath]F_0=0\\F_1=1\\F_n=F_{n-1}+F_{n-2}\text{ for } n\geq 2[/imath] Prove the given property of the Fibonacci numbers for all n greater than or equal to 1. [imath]F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}[/imath] I am pretty sure I should use weak induction to solve this. My professor got me used to solving it in the following format, which I would like to use because it help me map everything out... This is what I have so far: Base Case: Solve for [imath]F_0[/imath] and [imath]F_1[/imath] for the following function: [imath]F_nF_{n+1}[/imath]. Inductive Hypothesis: What I need to show: I need to show [imath]F_{n+1}F_{n+1+1}[/imath] will satisfy the given property. Proof Proper: (didn't get to it yet) Any intro. tips and pointers? | 1133564 | Proving the Fibonacci identity [imath]\sum_{i=1}^n f_i^2=f_nf_{n+1}[/imath] by induction
I am having troubles with a proof question. Prove that for any [imath]n\ge1[/imath], [imath]\sum_{i=1}^n f_i^2=f_nf_{n+1}[/imath], where [imath]f_n[/imath] is the [imath]n[/imath]'th Fibonacci number. I have the base case and the induction hypothesis, and I know what I need to prove (substitute [imath]n+1[/imath] in for [imath]n[/imath]'s on both sides of the equation) If someone can just guide me in the right direction on where to go, using induction that would be helpful. Thank You |
738284 | What is the sum of [imath]\sum\limits_{n = 1}^\infty {\arctan \dfrac{2}{{{n^2}}}}[/imath] and why?
The series [imath]\sum\limits_{n = 1}^\infty {\arctan \dfrac{2}{{{n^2}}}}[/imath] converges because it is asymptotic to [imath]\dfrac{2}{n^2}[/imath] which is convergent. What is its sum and why? | 128357 | sum of arctangent
Here is an interesting topic. It comes from evaluating [imath]\sum_{k=1}^{\infty}tan^{-1}\left(\frac{1}{k^{2}}\right)[/imath] I managed to dig up an old paper I have on the sum of arctans by Boros and Moll. It is called the Method of Zeros. It is located here: http://www.mat.utfsm.cl/scientia/archivos/vol11/Art2.pdf on page 6-7 I will post, verbatim, what it says. Perhaps someone can figure out how they arrive at the solution. They skip over the details and I do not know how they arrived at the general form they establish in [2]. "Based on the factorization of the product [imath]p_{n}:=\prod_{k=1}^{n}(a_{k}+ib_{k}), \;\ a_{k},b_{k}\in R[/imath] The argument of [imath]p_{n}[/imath] is given by [imath]\text{arg}(p_{n})=\sum_{k=1}^{n}\tan^{-1}\left(\frac{b_{k}}{a_{k}}\right)[/imath] up to an integral multiple of [imath]\pi[/imath]. This can be applied to the case of a polynomial with real coefficients given by [imath]p_{n}(z)=\prod_{k=1}^{n}(z-z_{k})[/imath] Then, [imath]\text{arg}(p_{n}(z))=\sum_{k=1}^{n}\tan^{-1}\left(\frac{x-x_{k}}{y-y_{k}}\right)[/imath] up to an integral multiple of [imath]\pi[/imath]. The special case [imath]p_{n}(z)=z^{n}-1[/imath] has roots at [imath]z_{k}=\cos(\frac{2\pi k}{n})+i\sin\left(\frac{2\pi k}{n}\right)[/imath], so we get: [imath]\text{arg}(z^{n}-1)=\sum_{k=1}^{n}\tan^{-1}\left(\frac{x-\cos\left(\frac{2\pi k}{n}\right)}{y-\sin\left(\frac{2\pi k}{n}\right)}\right)[/imath] The classical factorization [imath]\sin(\pi z)=\pi z\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)....[1][/imath] shows that one can think of [imath]\sin(\pi z)[/imath] as a polynomial in z of infinite degree. Euler than derived [imath]\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}[/imath] by comparing the cubic terms of [1]. This now yields [imath]\sum_{k=1}^{\infty}\tan^{-1}\left(\frac{2xy}{k^{2}-x^{2}+y^{2}}\right)=\tan^{-1}(y/x)-\tan^{-1}\left(\frac{\tanh(\pi y)}{\tan(\pi x)}\right)....[2][/imath]" That is it. They mention Euler's famous sum of reciprocal of squares then just skip to "this then yields...." Does anyone see how?. Besides that, didn't Euler compare the square terms and not the cubic ones?. So, for the problem at hand, we would let [imath]x=y=\frac{1}{\sqrt{2}}[/imath] and arrive at: [imath]\sum_{k=1}^{\infty}\tan^{-1}\left(\frac{1}{k^{2}}\right)=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}{\tan\left(\frac{\pi}{\sqrt{2}}\right)}\right)\approx 1.42474......[/imath] |
738314 | The trace as an integral over a sphere
Let [imath]V[/imath] be a real vector space of dimension [imath]n[/imath] and let [imath]\langle \, \cdot\, , \,\cdot\, \rangle[/imath] be an inner product on [imath]V[/imath]. We can define a linear functional on the space of endomorphisms of [imath]V[/imath] by [imath] \alpha(A) := \int_{S^{n-1}} \langle Av, v \rangle d\mu, [/imath] where [imath]S^{n-1}[/imath] is the unit sphere defined by the inner product and [imath]d\mu[/imath] is the Lebesgue measure on [imath]S^{n-1}[/imath]. This functional is actually a multiple of the trace. Here's a nasty proof of this: Let [imath](e_1,\ldots,e_n)[/imath] be an orthonormal basis of [imath]V[/imath]. Given [imath]v \in V[/imath] we write [imath]v = v_1 e_1 + \cdots + v_n e_n[/imath], so [imath]v \in S^{n-1}[/imath] if and only if [imath]\sum |v_i|^2 = 1[/imath]. We now write [imath] Av = \sum_{j=1}^n v_j \, A e_j = \sum_{j,k=1}^n v_j \, a_{jk} e_k [/imath] so that [imath] \langle Av ,v \rangle = \sum_{j,k} a_{jk} \, v_jv_k. [/imath] It's now classically known that [imath]\int_{S^{n-1}} v_j v_k d\mu = c \, \delta_{jk}[/imath] for some constant [imath]c[/imath] that's not terribly important here (and depends on the normalization of [imath]\mu[/imath] anyway), so we get [imath] \alpha(A) = \int_{S^{n-1}} \langle Av, v \rangle d\mu = c \operatorname{tr}(A). [/imath] Question Is there not a better way of doing this? That is, is there not some way of seeing that [imath]\alpha = c \operatorname{tr}[/imath] for some nonzero [imath]c[/imath] without breaking out the local coordinates? This basically comes down to showing that [imath]\alpha(AB) = \alpha(BA)[/imath] for any [imath]A,B \in \operatorname{End} V[/imath], but I can't see how to show that (or equivalently that [imath]\alpha[/imath] vanishes on the commutator) without going through the same calculations as above. | 62358 | Integral around unit sphere of inner product
For arbitrary [imath]n\times n[/imath] matrices M, I am trying to solve the integral [imath]\int_{\|v\| = 1} v^T M v.[/imath] Solving this integral in a few low dimensions (by passing to spherical coordinates) suggests the answer in general to be [imath]\frac{A\,\mathrm{tr}(M)}{n}[/imath] where [imath]A[/imath] is the surface area of the [imath](n-1)[/imath]-dimensional sphere. Is there a nice, coordinate-free approach to proving this formula? |
512691 | Proving that if [imath]2a + 3b \ge 12m + 1[/imath], then [imath]a \ge 3m + 1[/imath] or [imath]b \ge 2m + 1[/imath]
Let [imath]a[/imath], [imath]b[/imath], [imath]m[/imath] be integers. Prove that if [imath]2a + 3b \ge 12m + 1[/imath], then [imath]a \ge 3m + 1[/imath] or [imath]b \ge 2m + 1[/imath]. I need help proving this. I am not sure what to do. Thank you for all of the edits. I am still unsure on how to word the problem. Right now. Proof - By contrapositive. Assume that a < 3m + 1 and b < 2m + 1 for the integers a, b, and m. Then, 2a + 3b < 12m + 1 = 2(3m + 1) + 3(2m+1) < 12 m + 1 = 6m + 2 + 6m + 3 < 12m + 1 = 12m + 5 < 12m + 1 Which is false because 5 < 1 So next, I would say... Assume that a < 3m + 1 and b < 2m + 1 for the integers a, b, and m. Then a < 3m < 3m + 1 and b < 2m < 2m + 1. Then plug in again to show 12m < 12m + 1 Then what? | 481401 | Basic Algebra Proof on Integers - Weak Inequalities Work but Strict Inequalities Don't?
Let [imath]a, b, \& \, m[/imath] be integers. Prove that if [imath]2a + 3b \geq 12m + 1[/imath], then [imath]a \geq 3m + 1[/imath] or [imath]b \geq 2m + 1[/imath]. My Attempt: I don't conceive apace how to contrive, from the one inequality in the antecedent, the two inequalities in the consequent. So a proof by contraposition may be more facile. The contrapositive is: [imath]\text{If } \color{Green}{a < 3m + 1} \; \& \; \color{#0073CF}{b < 2m + 1}, \text{ then } 2a + 3b < 12m + 1 \tag{*}.[/imath] From [imath](*)[/imath], [imath]\color{Green}{2a} + \color{#0073CF}{3b} < \color{Green}{6m + 2} \; + \;\color{#0073CF}{6m + 3} = 12m +5[/imath]. But this doesn't prove [imath](*)[/imath]. Given Solution: Assume that [imath]a < 3m+1[/imath] and [imath]b < 2m+1.[/imath] Since a and b are integers, [imath]a ≤ 3m[/imath] and [imath]b ≤ 2m.[/imath] Therefore, [imath]2a + 3b ≤ 2(3m) + 3(2m) = 12m < 12m+ 1,[/imath] as desired. [imath]\blacksquare[/imath] What went wrong in my attempt? I understand the given solution but I persisted with the more natural and direct green and blue, instead of deviating to weak inequalities. Shouldn't both solutions work? Source: Problem 4.19 on P102 (related to P90, Result 4.8) of Mathematical Proofs, 2nd ed by Chartrand et al. [imath]\large{\text{Supplement to Cameron Buie and pritam's Answers :}}[/imath] What overriding, catholic lessons can be extrapolated from this example to generalise about inequalities? Moreover, in order to circumvent the problem here (ie the loss of information from the weaker inequalities), should I always start with the strictest inequality? If so, should I always rework any weak inequality into the equivalent strict inequality? For example, define [imath]r \in \mathbb{R} [/imath] and [imath]n \in \mathbb{Z}[/imath] and [imath]g(...) \neq h(...)[/imath]. Then [imath] f(r) < h(r) \require{enclose} \enclose{updiagonalstrike}{\iff} f(r) \leq g(r).[/imath] But [imath] f(n) < h(n) \text{ MAY OR MAY NOT }\iff f(n) \leq g(n).[/imath] |
738893 | How to find this limit
Find the limit of: [imath]\lim_{x\to 0}\frac{x^2 \sin 1/x}{\sin x}[/imath] Answer: Since [imath]x^2 \sin\frac{1}{x}[/imath], so [imath]-x^2\leq x^2 \sin\frac{1}{x}\leq x^2[/imath], thus [imath]\lim_{x\to 0}x^2 \sin\frac{1}{x}=0[/imath]. I am now stuck with the [imath]\sin x[/imath] under it, how does this affect the limit? | 726976 | Solve [imath]\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}[/imath]
Find the limit: [imath]\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}[/imath] After treating it with l'Hopital rule, we get: [imath]\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}-\cos\frac{1}{x}}{\cos x}}[/imath] Now, the numerator of fraction doesn't have a limit, so I can't use l'Hopital rule again. What to do? I can split it into two fractions, but I'm not sure how would it help: [imath]\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}}{\cos x}} - \lim_{x\to0}{\frac{\cos\frac{1}{x}}{\cos x}}[/imath] |
738225 | Regular polygon with [imath]n[/imath] sides , the number of triangles
For a regular polygon with [imath]n[/imath] sides [imath](n>5)[/imath], the number of triangles whose vertices are joining non-adjacent vertices of the polygon is [imath]n(n-4)(n-5)[/imath]. When I take [imath]n=6[/imath], I get David's Star: So, only two triangles! How can I get [imath]n(n-4)(n-5)[/imath]? | 364151 | Number of triangles in a regular polygon
A regular polygon with [imath]n[/imath] sides. Where [imath](n > 5)[/imath]. The number of triangles whose vertices are joining non-adjacent vertices of the polygon is? |
701406 | Evaluating [imath]\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)[/imath]
[imath]\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)[/imath] Now, [imath]\log(n!) = \Theta (n\log(n))[/imath] so I think we could write, [imath]\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) [/imath] [imath]\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) [/imath] [imath]\log\left(2n\right) - \log\left(n\right) = \log(2)[/imath] Is this legitimate? I feel this might be wrong since, the [imath]\Theta[/imath] notation conceals a constant factor. | 1006140 | Evaluating [imath]\lim_{n\to \infty} \bigg(\frac{(2n!)}{n!^2}\bigg)^{\frac{1}{4n}}[/imath]
I am trying to evaluate [imath]\lim_{n\to \infty} \bigg(\frac{(2n!)}{n!^2}\bigg)^{\frac{1}{4n}},[/imath] which came from trying to find the radius of convergence of the complex power series [imath]\sum_{n\ge0} z^{2n}\frac{\sqrt{(2n)!}}{n!}[/imath] In the limit I we have some cancellation: [imath]\frac{(2n!)}{n!^2}=\frac{1\cdots n\cdot (n+1)\cdots(2n)}{1\cdots n\cdot 1\cdots n}=\frac{(n+1)\cdot(2n)}{1\cdots n},\,\,\,(*)[/imath] and the right hand side is less than [imath]2^n[/imath], so an upper bound on the limit is [imath](2^n)^{\frac{1}{4n}}=2^{1/4}[/imath] The right hand side of [imath](*)[/imath] is at least [imath]1[/imath], so I have the bounds [imath]1,2^{1/4}[/imath], but I don't know how to get a better estimate. How can I compute this limit? |
739104 | Proving an elementary inequality of real vectors related to the p-Laplacian
How would you prove the following inequality? [imath] \left|\left|a\right|^{p-2}a-\left|b\right|^{p-2}b\right|\leq C_p\left|a-b\right|^{p-1} [/imath] where [imath]a,b\in\mathbb{R}^{n}[/imath] and [imath]C_p>0[/imath] is some constant depending only on [imath]p[/imath], [imath]1<p<2[/imath]. The inequality is mentioned in http://www.math.ntnu.no/~lqvist/p-laplace.pdf on page 43 but proof is omitted. The same notes have a collection of inequalities at chapter 10, in particular the following inequality for euclidean inner product product is given [imath] (\left|a\right|^{p-2}a-\left|b\right|^{p-2}b)\cdot(a-b)\leq C_{p}\left|a-b\right|^{p} [/imath] but I'm not sure how to deduce the first inequality from this. | 328338 | Is this vector-valued map Hölder-continuous?
Pick [imath]0<q<1[/imath] and consider the map from [imath]\mathbb{R}^n[/imath] to [imath]\mathbb{R}^n[/imath] that sends [imath]x[/imath] to [imath]|x|^{q-1}x[/imath]. Is this map Hölder-continuous (I guess with exponent [imath]\leq q[/imath])? In dimension one, I can exploit the homogeneity of the inequality defined by Hölder-continuity; but how can I proceed in higher dimension? |
740173 | Dependent Product Functor
I am trying to finish the proof that in category C with Cartesian closed slices, the dependent product functor is the right adjoint of the pullback, so that C is locally Cartesian closed. The proof is here. I haven't been able to convince myself that the implication [imath]h\cdot u=f^{*}p\Rightarrow h^{f}\cdot v=s\cdot p[/imath] is correct. My diagram chase leads to [imath]f^{f}\cdot h^{f}\cdot v=f^{f}\cdot s\cdot p[/imath]. I had no trouble with the other direction. I may be missing something trivial here but we seem to require additionally that [imath]s\cdot f^{f}=id_{A^{f}}[/imath] for this to work. | 59552 | Constructing dependent product (right adjoint to pullback) in a locally cartesian closed category
I've been trying to find a proof that the pullback functors in a locally cartesian closed category have right adjoints (used to model the notion of indexed product inside a category (rather than indexed by a set), or, equivalently, dependent products in models of dependent type theories). I found a proof in Awodey's book, but I found it utterly incomprehensible (probably due to not having read the rest of the book and therefore missing something considred obvious by that point). Does anyone know of other references for this theorem (would it be worth the effort trying to understand Seely's original paper on models of dependent type theory in locally cartesian closed categories)? EDIT: I found a neat proof in Sheaves in Geometry and Logic where it is observed that one can add the assumption that the morphism [imath]f : I \to J[/imath] one takes pullbacks along is to a terminal object. First one notes that since a slice of a slice is isomorphic to a slice one can conclude that a slice of locally cartesian closed category is itself locally cartesian closed, with a terminal object given by the identity morphism of the object the slice is taken over. Now [imath]f[/imath] can be considered a morphism from "itself" [imath]I \; \xrightarrow {\; f} J[/imath] to [imath]J \; \xrightarrow {\textrm{Id}_J} J[/imath] in the slice category [imath]\mathcal{C}/J[/imath]. Given that one knows the right adjoint to exist in the case of a terminal object one can thus conclude, since [imath]\textrm{Id}_J[/imath] is terminal in [imath]\mathcal{C}/J[/imath], that pullback along [imath]f[/imath] as a functor [imath](\mathcal C/J)/\textrm{Id_J} \to (\mathcal C / J)/f[/imath] has a right adjoint. This functor can now be made a functor [imath]\mathcal C /J \to \mathcal C/I[/imath] by noticing that [imath](\mathcal C/J)/\textrm{Id_J} [/imath] and [imath](\mathcal C / J)/f[/imath] are isomorphic (essentially by identity) to [imath]\mathcal C/J[/imath] and [imath]\mathcal C / I[/imath], respectively. |
740834 | Prove [imath]\sum\limits_{n\mathop=0}^{\infty}\frac{1}{n^2+1}=\frac{1}{2}+\frac{\pi}{2}\left(\frac{e^{2\pi}+1}{e^{2\pi}-1}\right)[/imath]
How do I show that [imath]\sum\limits_{n\mathop=0}^{\infty}\frac{1}{n^2+1}=\frac{1}{2}+\frac{\pi}{2}\left(\frac{e^{2\pi}+1}{e^{2\pi}-1}\right)[/imath] | 736860 | Find the infinite sum of the series [imath]\sum_{n=1}^\infty \frac{1}{n^2 +1}[/imath]
This is a homework question whereby I am supposed to evaluate: [imath]\sum_{n=1}^\infty \frac{1}{n^2 +1}[/imath] Wolfram Alpha outputs the answer as [imath]\frac{1}{2}(\pi \coth(\pi) - 1)[/imath] But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked. |
656331 | Why is the empty set a subset of every set?
Take for example the set [imath]X=\{a, b\}[/imath]. I don't see [imath]\emptyset[/imath] anywhere in [imath]X[/imath], so how can it be a subset? | 2296328 | Is this proof correct? If not, where is the flaw?
It is to prove that the empty set is a subset of every set. Therefore the following proposition is to be proven: [imath] \forall z: ( \emptyset \subset z)~~.[/imath] Therefore, the converse [imath] \exists z: ( \emptyset \not \subset z)[/imath] must lead to a contradiction. Now, considering the set [imath] z = \{\{a\}\} \setminus \{a\} = \{\}~~, [/imath] if [imath] \emptyset \not \subset z~~, \text{and therefore} ~~ \emptyset \neq z ~~,[/imath] then [imath]z[/imath] wouldn't be a set, yet it is by definition a set, since the proposition [imath] a \in z [/imath] is still either true or false. Which is a contradiction. Assuming my reasoning is flawed, please don't give me the proof. If it is correcct though, I would appreciate an alternative proof. Thank you in advance. |
239875 | Element chasing proof that [imath](A\setminus C)\setminus(B\setminus C) = (A\setminus B)\setminus C[/imath]
I would like to construct a formal proof of the following: [imath](A\setminus C)\setminus(B\setminus C) = (A\setminus B)\setminus C[/imath] Let [imath]a∈A[/imath] be an arbitrary element, we will show that [imath]a\in A \cap \overline B \cap \overline C[/imath]. For LHS, since [imath]a\in A[/imath], we have that [imath]a\in (A \cap \overline C) \cap (\overline B \cap \overline C)[/imath]. This is equivalent to [imath]a\in (A \cap \overline B) \cap \overline C[/imath]. For RHS, since [imath]a\in A[/imath], we have that [imath]a\in (A \cap \overline B) \cap \overline C[/imath] [imath]\therefore lhs \equiv rhs[/imath] and this concludes the proof I would be grateful for any feed back on this element chasing proof. Is it flawed or where should improvements be made? Thanks | 322652 | Let [imath]F[/imath] be a field, show that [imath](A \setminus C) \setminus (B \setminus C) = (A \setminus B) \setminus C\; \forall A, B, C[/imath] in [imath]F[/imath]
Let F be a field, Show that: [imath](A \setminus C) \setminus (B \setminus C) = (A \setminus B) \setminus C,[/imath] [imath]\forall A, B, C \in F[/imath] where \ indicates set minus. I've started with left hand side and reached this point: [imath](A \cap C^c) \cap (B \cap C^c)^c[/imath] where ^c is the complement. I don't know where to go from here. |
741246 | Finite Series Inequality
For each [imath]n=1,2,{\dots}[/imath] and [imath]x\in(0,{\pi})[/imath], prove that the series [imath]S_n(x)=\sum_{k=1}^{n} \frac{\sin(kx)}{k}>0[/imath] | 376273 | Inequality [imath]\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0[/imath] (Fejer-Jackson)
Show the following inequality for any [imath]x\in [0, \pi][/imath] and [imath]n\in \mathbb{N}^*[/imath], [imath] \sum_{1\le k\le n}\frac{\sin kx}{k}\ge 0. [/imath] I have this question a very long time ago from a book or magazine but I cannot solve it by myself and did not know how to solve it until today. My try: for [imath]n=1, 2, 3[/imath], one can check this by hand. |
741750 | Discrete Math Induction: [imath]\sum^n_{i=1} \frac1{i(i+1)}[/imath]
For [imath]\sum^n_{i=1} \frac1{i(i+1)}[/imath] Find a formula and proofs that it holds for all n ≥ 1. How would I find the formula for this one that can hold for all n ≥ 1? | 741646 | Use Mathematical Induction to prove that [imath]\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}[/imath]
Use Mathematical induction to prove that for all integers, [imath]n[/imath] is greater than or equal to [imath]1[/imath]. I am confused on what to do after I do the the basis step that is using [imath]n[/imath] as [imath]1[/imath]. [imath]\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}[/imath] |
741727 | Discrete Math On Induction proof: [imath]\sum_{i=1}^n n2^n = (n-1)2^{n+1} + 2[/imath]
Show by induction that the following formulas hold. [imath]\sum_{i=1}^n n2ⁿ = (n-1)2^{n+1} + 2[/imath] What did a similar problem to this but this one is a little different. I think is because this one has a power n and I'm not sure how to start with this one? | 87030 | Proving [imath]\sum\limits_{i=0}^n i 2^{i-1} = (n+1) 2^n - 1[/imath] by induction
I'm trying to apply an induction proof to show that [imath]((n+1) 2^n - 1 [/imath] is the sum of [imath](i 2^{i-1})[/imath] from [imath]0[/imath] to [imath]n[/imath]. the base case: L.H.S = R.H.S we assume that [imath](k+1) 2^k - 1 [/imath] is true. we need to prove that [imath](k+2) 2^{k+1} - 1[/imath] My try to prove 3 is as follows: [imath](k+2) 2^{k+1} - 1[/imath] [imath](k+2) (2^k * 2) - 1[/imath] , from 2: [imath]2^k = 1/(k+1)[/imath] [imath](k+2) (2 / (k+1)) - 1[/imath] [imath](k+1) 2 - 1[/imath] My question is, how could I get to [imath]2^k[/imath] from the last line to prove this formula is right? |
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