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2067699
|
A question about fuzzy differential (Hukuhara and Generalized Hukuhara Diffrentiability):
I know this question has no relevance to the analysis, but it seems some friends about all the information they contain mathematical topics. So please forgive me I have a question about about fuzzy differential which i say it.My professor says that my solution is wrong,but i believe it is right.According to papers and books related to this problem,which i read,my solution is right. If you can, please send the correct answer for me. Let [imath]f:\mathbb R \to \mathbb R_f[/imath] with [imath]f(t)=(1,2,3)t[/imath] be a function. We're investigating whether or not the existing Hukuhara derivative and generalized Hukuhara derivative at [imath]t_0 =0[/imath] . According to definition: [imath]f'(x)=\lim_{h\to 0}\dfrac{f(x+h) \ominus_H f(x)}{h} =\lim_{h\to 0}\dfrac{[(1+\alpha)(x+h),(3-\alpha)(x+h)] \ominus_H [(1+\alpha)(x),(3-\alpha)(x)]}{h}=\lim_{h\to 0}\dfrac{[(1+\alpha)(x+h-x),(3-\alpha)(x+h-x)]}{h}=[(1+\alpha),(3-\alpha)]=(1,2,3) [/imath] And generalized Hukuhara derivative is equal with Hukuhara derivative.Can you tell me what is wrong? If writing answer was too hard take a picture and send me the photo.so Thanks. [email protected]
|
2068185
|
Hukuhara and Generalized Hukuhara Diffrentiability
I know this question has no relevance to the analysis, but it seems some friends about all the information they contain mathematical topics. So please forgive me I have a question about fuzzy differential which i say it.My professor says that my solution is wrong,but i believe it is right.According to papers and books related to this problem,which i read,my solution is right. If you can, please send the correct answer for me. Let [imath]f:\mathbb R \to \mathbb R_f[/imath] with [imath]f(t)=(1,2,3)t[/imath] be a function. We're investigating whether or not the existing Hukuhara derivative and generalized Hukuhara derivative at [imath]t_0 =0[/imath] . According to definition: \begin{align*}f'(x)&=\lim_{h\to 0}\dfrac{f(x+h) \ominus_H f(x)}{h}\\ &=\lim_{h\to 0}\dfrac{[(1+\alpha)(x+h),(3-\alpha)(x+h)] \ominus_H [(1+\alpha)(x),(3-\alpha)(x)]}{h}\\ &=\lim_{h\to 0}\dfrac{[(1+\alpha)(x+h-x),(3-\alpha)(x+h-x)]}{h}\\ &=[(1+\alpha),(3-\alpha)]\\ &=(1,2,3)\end{align*} And generalized Hukuhara derivative is equal with Hukuhara derivative.Can you tell me what is wrong? If writing answer was too hard take a picture and send me the photo.so Thanks. [email protected]
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2068213
|
Prove by induction that [imath]n^3-n[/imath] is divisible by [imath]24[/imath] for all odd positive integers
Q: Prove by induction that [imath]n^3-n[/imath] is divisible by 24 for all odd positive integers After proving the first part for n=1 Assume true for some positive integer [imath]n=k[/imath] ie [imath]k^3-k=24x[/imath] where x is an integer Prove true for [imath]n=k+2[/imath] ie [imath](k+2)^3-(k+2)=24y[/imath] where y is an integer [imath]=k^3+6k^2+11k+6[/imath] [imath]=24x+12k+6k^2+6[/imath] But how do I get this in the form [imath]24y[/imath]? Am i supposed to use [imath]2k+1[/imath] instead?
|
1427734
|
prove that [imath]24 \mid a(a^2-1)[/imath]
If [imath]a[/imath] is an odd integer, prove that [imath]24 \mid a(a^2-1)[/imath]. I proved in my class that [imath]a^2=8k+1[/imath] for some [imath]k[/imath], provided [imath]a[/imath] was odd. So [imath]a(a^2-1)=(2k-1)(8k+1-1)=8k(2k-1)[/imath] but I can't pull a [imath]24[/imath] out of it.
|
2067667
|
Prove that edge coloring produces a cycle with all edges having the same color
For a complete graph with vertices [imath]2n+1[/imath] and its edges colored with [imath]n[/imath] colors, how to prove that there exists a cycle with all edges of the same color. I have proceeded by calculating the number of edges in a complete graph of [imath]2n+1[/imath] vertices which gives [imath] \binom{2n+1}{2} = n(2n+1) [/imath] edges. Now using the pigeonhole principle there must be at least [imath]2n+1[/imath] edges of the same color. Now how do I show that this forms a cycle in the graph?
|
1110332
|
Prove that a graph with the same number of edges and vertices contains one cycle
We have a connected graph [imath]G(V,E)[/imath] that has [imath]|E|= n[/imath] edges and [imath]n[/imath] vertices. Prove that the graph has one cycle in it. I'm little bit confused here. I tried some ways but failed. Can you direct me?
|
2066620
|
How to find limit?
Consider the sequence [imath]\displaystyle a_{n} = \sum_{i = 1}^{n}i^{-1/3} [/imath] and [imath]b_{n} = n^{2/3}[/imath]. No we want to find [imath]\displaystyle \lim_{n\rightarrow \infty}a_{n}-\frac{3}{2}b_{n}[/imath]? Actually I find it lower bound , but I used Abel summation, have no idea how to find it easier.
|
1961275
|
Limit related to [imath]\zeta(x)[/imath]
I'm noticing some things: [imath]\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/3}-\frac{3}{2}n^{2/3} \right)=\zeta(1/3)[/imath] Note [imath]\int n^{-1/3} dn=\frac{3}{2}n^{2/3}+c[/imath] [imath]\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/2}-\frac{2}{1}n^{1/2} \right)=\zeta(1/2)[/imath] And [imath]\int n^{-1/2}dn=2n^{1/2}+c[/imath] It seems as though [imath]\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/s}-\frac{s}{s-1}n^{(s-1)/s} \right)=\zeta(1/s)[/imath] If [imath]s \neq 1[/imath], may someone please explain why.
|
2068243
|
Number of edges of any spanning tree of a graph G coming from a subgraph of G
Consider a simple undirected, unweigthed graph [imath]G=(V,E)[/imath] and its edge-induced subgraph [imath]G^{*}=(V^{*},E^{*})[/imath]. Suppose that [imath]G[/imath] and [imath]G^{*}[/imath] have [imath]c[/imath] and [imath]c^{*}[/imath] connected components, respectively. Is it true that exactly [imath]|V^{*}|-\min\{0,c-c^{*}\}[/imath] edges can be selected for any spanning tree of [imath]G[/imath] from [imath]E^{*}[/imath]? Could somebody give a proof/counterexample? Thank you for the help!
|
2067796
|
Number of edges of any spanning tree of a graph [imath]G[/imath] coming from a subgraph of [imath]G[/imath]
Consider a simple undirected, unweigthed graph [imath]G=(V,E)[/imath] and its edge-induced subgraph [imath]G^{*}=(V^{*},E^{*})[/imath]. Suppose that [imath]G[/imath] and [imath]G^{*}[/imath] have [imath]c[/imath] and [imath]c^{*}[/imath] connected components, respectively. Is it true that any spanning tree of [imath]G[/imath] has exactly [imath]|V^{*}|-\min\{c^{*},c\}[/imath] edges from [imath]E^{*}[/imath]? Could somebody give a proof/counterexample? Thank you for the help!
|
2068301
|
Which is greater [imath](1000)^{1000}[/imath] or [imath](1001)^{999}[/imath]
I know this has to be done using binomial expansion. I tried [imath](1001)^{999}=(1000+1)^{999}[/imath] but didn't reach anywhere.
|
794183
|
Which is greater: [imath]1000^{1000}[/imath] or [imath]1001^{999}[/imath]
Question: Find the greater number: [imath]1000^{1000}[/imath] or [imath]1001^{999}[/imath] My Attempt: I know that: [imath](a+b)^n \geq a^n + a^{n-1}bn[/imath]. Thus, [imath](1+999)^{1000} \geq 999001[/imath] And [imath](1+1000)^{999} \geq 999001[/imath] But that doesn't make much sense. I want some hints regarding how to solve this problem. Thanks.
|
2048779
|
How can I prove a function on product space lebesgue measurable?
Let [imath]X[/imath] and [imath]Y[/imath] be two measurable spaces. I want to construct a function [imath]f : X\times Y \to \mathbb{R}[/imath] such that [imath]f(x,\cdot) : Y \to \mathbb{R}[/imath] is measurable for all [imath]x \in X[/imath] and [imath]f(·,y) : X \to \mathbb{R}[/imath] is measurable for all [imath]y \in Y[/imath], yet [imath]f[/imath] is not measurable in the product space. How can I construct such a function?
|
647235
|
Counterexample to "Measurable in each variable separately implies measurable"
Some fellow classmates are preparing for a qualifying exam on real analysis, and asked me for help on the following question: Let [imath] \ f:[0,1]^2\longrightarrow\mathbb{R}[/imath] be such that: (i) [imath]\ f(x,\cdot)[/imath] is measurable for each fixed [imath]x\in[0,1][/imath]; (ii) [imath]\ f(\cdot,y)[/imath] is continuous for each fixed [imath]y\in[0,1][/imath]. Show that [imath]\ f[/imath] is measurable. If we only assume that [imath]\ f(\cdot,y)[/imath] is measurable for each [imath]y\in[0,1][/imath], rather than continuous, can we still conclude that [imath]f[/imath] is measurable? I was able to come up with the following simple proof (at least I hope it is a proof) of the first statement using some standard arguments: Proof. [imath] \ [/imath] Define a sequence of functions [imath]\{f_n:[0,1]^2\longrightarrow\mathbb{R}\}_{n\geq 1}[/imath] by: [imath] f_n(x,y)=f\bigg(\frac{i}{n},y\bigg), \qquad \text{for } x\in\bigg[\frac{i}{n},\frac{i+1}{n}\bigg], \ i=1,\cdots,n [/imath] Moreover, since [imath] \ f(\cdot,y)[/imath] is continuous at [imath]x=\tfrac{i}{n}[/imath] for each [imath]y[/imath], for each [imath]\epsilon>0[/imath] there is a [imath]\delta_y>0[/imath] so that [imath] \bigg| \ x-\frac{i}{n}\bigg|<\delta_y \qquad\Rightarrow\qquad \bigg| \ f(x,y)-f\bigg(\frac{i}{n},y\bigg)\bigg|<\epsilon; [/imath] therefore, for any [imath]n>\tfrac{1}{\delta_y}[/imath] and any fixed [imath](x,y)\in[0,1]^2[/imath] with [imath]x\in\big[\tfrac{i}{n},\tfrac{i+1}{n}\big][/imath], we have [imath] \bigg| \ x-\frac{i}{n}\bigg|\leq \frac{1}{n}<\delta_y [/imath] and so [imath] | \ f_n(x,y)-f(x,y)|=\bigg| \ f\bigg(\frac{i}{n},y\bigg)-f(x,y)\bigg|<\epsilon. \tag{$\ast$} [/imath] In other words, [imath] \ f_n\longrightarrow f[/imath] pointwise. Furthermore, we clearly have that [imath] A_n:=\{(x,y)\in[0,1]^2 \ | \ f_n(x,y)>\alpha\} \qquad\qquad\qquad [/imath] [imath] \qquad\qquad \ =\bigcup_{i=1}^{n-1}\bigg(\bigg[\frac{i}{n},\frac{i+1}{n}\bigg)\times \bigg\{y\in[0,1] \ \bigg| \ \ f\bigg(\frac{i}{n},y\bigg)>\alpha \bigg\}\bigg). [/imath] Since [imath] \ f\big(\tfrac{i}{n},\cdot\big)[/imath] is measurable for each [imath]i=1,\cdots,n[/imath], the latter expression is a union of measurable sets. This means that [imath] \ A_n[/imath] is measurable, and therefore [imath] \ f_n[/imath] is measurable for each [imath]n\geq 1[/imath]. Finally, as the limit of measurable functions is measurable, we conclude that [imath] \ f[/imath] is measurable. [imath]\hspace{6.25in}\square[/imath] I'm fairly certain that the above proof is correct (anyone care to confirm?); however, that still leaves the following: Question: If [imath] \ f[/imath] is merely measurable in each variable separately, is it measurable? The proof I gave above uses the continuity assumption in [imath](\ast)[/imath], so obviously the same proof will not work in this case, but I cannot come up with any counterexamples. Any suggestions?
|
2067220
|
How to show [imath]\frac{2x}{2+x}<\log(x+1)[/imath]
How to show [imath]\frac{2x}{2+x}<\log(x+1)[/imath] for [imath]x>0[/imath] Without differentiating, more elementary (it looks then more complicated but OK) [imath]\log(x+1)=\int\limits_1^{x+1}\frac 1udu[/imath] and [imath]\frac{2x}{2+x}=x\frac{1}{ 1+\frac x2}[/imath] hence [imath]\log(x+1)-\frac{2x}{2+x}=\int\limits_1^{x+1}\frac 1u-\frac{1}{ 1+\frac x2}du=\int\limits_1^{x+1}\frac{1+\frac x2-u}{u\left(1+\frac x2\right)}du=\int\limits_1^{\frac x2+1}\frac{1+\frac x2-u}{u\left(1+\frac x2\right)}du-\int\limits_{\frac x2+1}^{x+1}\frac{u-\left(1+\frac x2\right)}{u\left(1+\frac x2\right)}du[/imath] The numerator of the integrand in the first integral is positive and negative in the [imath]2[/imath]nd. But for the first one the denominatior is smaller, so the fraction is always bigger than its corresponding fraction in the second one, and so the difference is always positive is this correct, do you have an alternative proof
|
448162
|
Why does [imath] \frac{2x}{2+x}[/imath] provide a particularly tight lower bound for [imath]\ln(1+x)[/imath] for small positive values of [imath]x[/imath]?
EDIT: My question was poorly worded. I wasn't asking about showing [imath]\ln(1+x) > \frac{2x}{2+x}[/imath] for [imath]x>0[/imath]. What I wanted to know is why the lower bound provided by [imath] \frac{2x}{2+x}[/imath] was so tight for small positive values of [imath]x[/imath]. This is addressed in robjohn's answer.
|
2068074
|
find limit when x is infinity
How can we find the following limit. [imath]\lim_{x\to\infty}(x-\ln\cosh x)[/imath] where [imath]\cosh t=\frac{e^t+e^{-t}}{2}.[/imath] I thought about it alot but didn't get any start
|
1452299
|
Prove that [imath]\lim_{x\to \infty}\left(x-\ln\cosh x\right)=\ln 2[/imath]
Prove that [imath]\lim_{x\to \infty}\left(x-\ln\cosh x\right)=\ln 2[/imath] I used L Hospital Rule but it does not simplify.Then i expanded [imath]\cosh x[/imath] by using McLaurin series but due to [imath]x\to \infty[/imath],this is also not working.How should i evaluate this limit?
|
2066375
|
Every compact subset of [imath]\mathbb{R}[/imath] is the support of a Borel measure.
This is an exercise from Rudin's Real and Complex analysis pg 58. I have been stuck as to What the support of a measure even is. Originally Rudin defines it on a compact Hausdorff space, but since [imath]\mathbb{R}[/imath] is only locally compact how does this definition extend. How to define a positive linear functional [imath]\Lambda[/imath] that would help generate this measure? I have already seen the other post here about the same question and not only is the poster much farther than me the hint doesn't really give me any insights as to how to solve it.
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1507443
|
Is every compact subset of [imath]\Bbb{R}[/imath] the support of some Borel measure?
I have tried to prove the exercise 2.12 in Rudin's RCA: 12 Show that every compact subset of [imath]\Bbb{R}[/imath] is the support of a Borel measure. For perfect (i.e. no isolated point) compact [imath]K[/imath] with non-zero Lebesgue measure, we just take the Lebesgue measure. However I struck on the general case. My attempt is as follows: For every compact [imath]K[/imath], we can find a decreasing sequence of open set [imath]\langle W_n\rangle_n[/imath] s.t. [imath]\bigcap_n W_n =K[/imath] and [imath]m(W_n-K)<1/n[/imath]. Find a continuous function [imath]h_n[/imath] s.t. [imath]h_n(x) = 1[/imath] for [imath]x\in K[/imath] and [imath]h_n(x) = 0[/imath] outside of [imath]V_n[/imath]. (such function exists by Urysohn's lemma.) For continuous function [imath]f[/imath] with compact support define [imath]\Lambda f = \lim_{n\to\infty}\frac{\int_\Bbb{R} fh_ndm}{\int_\Bbb{R} h_ndm}.[/imath] I can not certain that the [imath]\Lambda[/imath] is well-defined. But if it is well-defined, then it gives an regular Borel measure (by Riesz representation theorem) [imath]\mu[/imath] s.t. [imath]\mu(K)=1[/imath]. I guess that the [imath]\mu[/imath] has a support [imath]K[/imath] but I don't know how to get it. Thanks for any help!
|
2068608
|
What is the minimal polynomial of [imath]\sqrt{3} + \sqrt[3] {2}[/imath] over [imath]\mathbb{Q}[/imath]?
What is the minimal polynomial of [imath]\sqrt{3} + \sqrt[3]{2}[/imath] over [imath]\mathbb{Q}[/imath]? I know the basic idea of what a minimal polynomial is--it is the lowest degree monic polynomial in [imath]\mathbb{Q}[x][/imath] that has the above as a root. But how do you go about calculating it? How do you know intuitively if it does or doesn't exist?
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1804435
|
Minimal polynomial of [imath]\sqrt[3]{2} + \sqrt{3}[/imath]
Suppose I want to find the minimal polynomial of the number [imath]\sqrt[3]{2} + \sqrt{3}[/imath]. Now that means I want to find a unique polynomial that is irreducible over [imath]\Bbb Q[/imath] such that [imath]f(x)=0[/imath]. Now I know that because [imath]\sqrt[3]{2} + \sqrt{3}[/imath] belongs to [imath]\Bbb Q( \sqrt[3]{2} , \sqrt{3})[/imath] it might be degree [imath]2[/imath], [imath]3[/imath] or [imath]6[/imath] and does not belong to [imath]\Bbb Q( \sqrt[3]{2})[/imath] so it cannot be of degree [imath]3[/imath] or [imath]\Bbb Q( \sqrt{3})[/imath] so it cannot be of degree [imath]2[/imath]. So it is of degree [imath]6[/imath]. I think my sayings are a bit intuitive and not formal and lack rigorous. Couldn't it belong to another extension of degree [imath]2[/imath]? or [imath]3[/imath]? I can't answer that. Why checking only those [imath]2[/imath] is enough? Or is it wrong at all to say that?
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2067276
|
Open sets in Q wrt euclidean topology.
What are the open sets in [imath]\mathbb{Q}[/imath] wrt euclidean topology? Basically I want to know whether an arbitrary open set in a metric space is countable union of closed sets or not, with or without depending on the locally compactness of the metric space.
|
498266
|
Open set in a metric space is union of closed sets
Show that every open set [imath]A[/imath] is in a metric space [imath](X,d)[/imath] is the union of closed sets. This is a question on my analysis homework. I understand that this can only be true if we consider the union of infinite closed sets. However, I am not sure what I can do. I understand to prove a set is open, then a ball centered at an arbitrary point with a radius will be completely contained in the set. But what is the radius? Is this the correct approach?
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2040306
|
Latus rectum of a parametric parabola
The latus rectum of the parabola defined parametrically by [imath]x=at^2+bt+c[/imath] and [imath]y=a't^2+b't+c'[/imath] is--- I tried to eliminate [imath]t[/imath] from both the equations and I got the following equation [imath](a'x-ay+c'a-ca')^2=(ab'-ba')(b'x-by+c'b-cb')[/imath] However I could not find out the length of latus rectum. I also tried to find out the equation of axis using this but it turned out a bit complex. Please help me in this regard. Thanks. :)
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2942682
|
What is the focal length of an arbitrary parabola?
Given a parabola of the form [imath]Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0[/imath], where [imath]B^2 - 4AC = 0[/imath], what is the formula for the focal length?
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2049377
|
[imath](\mathbb{Z}\times\mathbb{Z} \times\mathbb{Z} )/ \langle(3,3,3)\rangle[/imath] is isomorphic to what?
The solution told me it is isomorphic to [imath]\mathbb{Z}_3\times \mathbb{Z} \times \mathbb{Z}[/imath], but why couldn't we argue that [imath](1,0,0), (0,1,0)[/imath] and [imath](0,0,1)[/imath] all generates infinite groups, so it is isomorphic to [imath]\mathbb{Z}\times\mathbb{Z} \times\mathbb{Z}[/imath]?
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1003603
|
Classifying [imath]\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle(3,3,3)\rangle[/imath]
Again, my instructor's explanation in class is so confusing to me. Here we cannot take the original approach of finding the order and limiting our choices. So instead my instructor says that this is isomorphic to [imath]\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}_3[/imath]. To justify it he said that [imath]\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}[/imath] is 3 dimensional and the denominator is one dimensional but this doesn't quite explain where the [imath]\mathbb{Z}_3[/imath] comes from? Can anyone provide a better way to think about this? It is making no sense to me at all.
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49136
|
Is [imath]\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A[/imath] true?
Let [imath]A[/imath] be an integral domain of finite Krull dimension. Let [imath]\mathfrak{p}[/imath] be a prime ideal. Is it true that [imath]\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A[/imath] where [imath]\dim[/imath] refers to the Krull dimension of a ring? Hartshorne states it as Theorem 1.8A in Chapter I (for the case [imath]A[/imath] a finitely-generated [imath]k[/imath]-algebra which is an integral domain) and cites Matsumura and Atiyah–Macdonald, but I haven't been able to find anything which looks relevant in either. (Disclaimer: I know nothing about dimension theory, and very little commutative algebra.) If it is true (under additional assumptions, if need be), where can I find a complete proof? It is obvious that [imath]\operatorname{height} \mathfrak{p} + \dim A/\mathfrak{p} \le \dim A[/imath] by a lifting argument, but the reverse inequality is eluding me. Localisation doesn't seem to be the answer, since localisation can change the dimension...
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2847488
|
Krull dimension of [imath]\mathbb{R}[x_1, \ldots, x_n]/(F)[/imath]?
Let [imath]F \in \mathbb{R}[x_1, \ldots, x_n][/imath] be irreducible over [imath]\mathbb{R}[/imath]. I am trying to understand why the Krull dimension of the ring [imath]\mathbb{R}[x_1, \ldots, x_n]/(F)[/imath] is [imath]n-1[/imath]. Any comments are appreciated! Thank you! PS In fact is it always the case that if [imath]P[/imath] is a prime ideal of height [imath]k[/imath] then the Krull dimension of the ring [imath]\mathbb{R}[x_1, \ldots, x_n]/P[/imath] is [imath]n-k[/imath]?
|
806636
|
Question on primitive roots of unity
Let [imath]p[/imath] be an odd prime and [imath]\omega[/imath] be a primitive [imath]p[/imath]th root of unity. The question is to prove that: [imath](1-\omega)(1-\omega^2) \cdots (1-\omega^{p-1})=p[/imath] What I have done so far is: I can see that this is true for [imath]p=3[/imath] [imath](1-\omega)(1-\omega^2)=1-(\omega+\omega^2)+w^3=1-(-1)+1=3=p[/imath] I am not able to prove this in general.... [imath](1-\omega)(1-\omega^2) \cdots (1-\omega^{p-1})=1-(\omega+\omega^2+\cdots+\omega^{p-1})+????+\omega^{\frac{p(p-1)}{2}}[/imath] I do not have any idea what that [imath]????[/imath] could be but all I can say is: [imath](\omega+\omega^2+\cdots +\omega^{p-1})=-1[/imath] [imath]\omega^{\frac{p(p-1)}{2}}=1[/imath] So, [imath](1-\omega)(1-\omega^2)\cdots (1-\omega^{p-1})=3+????[/imath] I am not able to do more than this. There could be some (hard) way of doing it by hand multiplying all those things but I am looking for a more theoretical idea. Please help me to clear this up.
|
2680801
|
If [imath]1,a_1,a_2,...,a_{n-1}[/imath] are the [imath]n[/imath] roots of [imath]1[/imath],then [imath](1-a_1)(1-a_2)...(1-a_{n-1}) \space ?[/imath]
If [imath]1,a_1,a_2,...,a_{n-1}[/imath] are the [imath]n[/imath] roots of unity, then how can we find the value of [imath](1-a_1)(1-a_2)...(1-a_{n-1}) \space ?[/imath] My Approach: If [imath]n=2[/imath] , then [imath]1,-1[/imath] are the roots of unity [imath]\therefore (1-a_1)=(1-(-1))=2[/imath] for [imath]n=3 \space :[/imath] [imath]1,\omega,\omega^2[/imath] are the roots of unity [imath]\therefore (1-a_1)(1-a_2)=(1-\omega)(1-\omega^2)[/imath] [imath]\quad \quad \quad =1-\omega^2 -\omega +1=3[/imath] so we conclude for [imath]n[/imath] the value [imath](1-a_1)(1-a_2)...(1-a_{n-1}) \space = n[/imath] but i want a direct process to evaluate [imath](1-a_1)(1-a_2)...(1-a_{n-1}) [/imath] (without generalisation),so how can i do so?
|
688305
|
Why is an algebra not a [imath]\sigma[/imath]-algebra by induction?
I am studying probability theory by reading Sidney Resnick's "A Probability Path". On page 12 and 13, algebra and [imath]\sigma[/imath]-algebra are defined. The only difference between the two is the third requirement. For algebra, it is required that [imath]A, B\in \mathcal A \Rightarrow A\cup B\in \mathcal A.[/imath] That is, algebra is closed under finite union; For [imath]\sigma[/imath]-algebra, it is required that [imath]B_i\in \mathcal B, i\geq 1 \Rightarrow \bigcup_{i=1}^\infty B_i\in\mathcal B[/imath]. That is, [imath]\sigma[/imath]-algebra is closed under countable union. I find the definition for algebra confusing. If the union of [imath]A[/imath] and [imath]B[/imath] is in [imath]\mathcal A[/imath], then for any [imath]C\in\mathcal A[/imath] one has [imath](A\cup B)\cup C\in \mathcal A[/imath] and so on. Then by induction, this extends to countable union. Why does induction NOT work in this case, please? Thank you!
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2066805
|
Why is this 'Proof' by induction not valid?
I am trying to understand why induction is valid. For instance why would this 'proof' not be valid under the principle of proof by induction ? : [imath] \sum_{k=1}^{\infty} \frac{1}{k} \lt \infty[/imath] because using induction on the statement [imath]S(n) = \sum_{1}^{n} \frac{1}{k} \lt \infty[/imath] - "[imath]S(1) < \infty[/imath] is true and "[imath]S(n) < \infty[/imath]" implies "[imath]S(n+1) < \infty[/imath]" since [imath]S(n+1) \lt S(n) + \frac{1}{n}[/imath]
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2069426
|
Multiple irrational roots of irreducible polynomial over Q
Is it possible for irreducible polynomial [imath]f \in Q[x][/imath] to have irrational multiple root? I guess not, so I supposed to assume that my polynomial has factor [imath](x - a)^k[/imath] in some extention, where [imath]a[/imath] is irrational, [imath]k>1[/imath] and come to contradiction, but I don't know what could it be.
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1644079
|
An irreducible polynomial in a subfield of [imath]\mathbb{C}[/imath] has no multiple roots in [imath]\mathbb{C}[/imath]
Let [imath]K\subset \mathbb{C}[/imath] be a subfield and [imath]f\in K[t][/imath] an irreducible polynomial. Show that [imath]f[/imath] has no multiple roots in [imath]\mathbb{C}[/imath]. If I understand this question correctly, I must show that there is no [imath]a \in \mathbb{C}[/imath] such that [imath](t-a)^n|f[/imath] in [imath]F[t][/imath] with [imath]n>1[/imath]. So suppose [imath](t-a)^2|f[/imath] and [imath]f=(t-a)^2h[/imath]. Then we have [imath]f'=2(t-a)h+(t-a)^2h' \Rightarrow (t-a)|f'[/imath] so [imath]\gcd(f,f'[/imath]) is not constant. Therefore [imath]f[/imath] is divisible by some square of non-constant polynomial in [imath]F[t][/imath], which is a contradiction. Is my argument correct? Thank you.
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2042931
|
Evaluate the sum [imath]\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}[/imath]
Evaluate the sum [imath]\sum ^{\infty}_{n=1} \frac{1}{n^2+a^4}[/imath] I don't to know how to processed this problem. Can any one help with problem please. thanks
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1064217
|
How to sum [imath]\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}[/imath]?
Does anyone know the general strategy for summing a series of the form: [imath]\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2},[/imath] where [imath]a[/imath] is a positive integer? Any hints or ideas would be great!
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1576925
|
Evaluating limit by sandwich theorem: [imath]\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}[/imath]
[imath]\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}[/imath] For using Sandwich theorem I need two functions such that [imath]g(x)<f(x)<h(x)[/imath] [imath]\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{n+n^2} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2} [/imath] But I can't find function greater than the given which will help me evaluate the limit.
|
617407
|
Evaluate the limit [imath]\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}[/imath]
Evaluate the limit [imath]\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}[/imath] My approach : If I divide numerator and denominator by [imath]n^2[/imath] I get : [imath]\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0[/imath] but the answer is [imath]\dfrac{1}{2}[/imath] please suggest how to solve this.. thanks.
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2069244
|
Evaluation of limit [imath]\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right)[/imath]
Evaluation of the limit [imath]\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right).[/imath] I don't understand how can I start the above problem because Riemann sums and integrals are not applicable here. It seems that we can solve it using the Squeeze theorem, but I don't understand which inequality I have to use. Help required, thanks.
|
532404
|
Find the limit of [imath]\sum\limits_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)[/imath]
[imath]\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)[/imath] Note that [imath]\forall x\ge 0, \sqrt{x}-1\le\sqrt{1+x}-1\le x[/imath] Then [imath]\sum_{k=1}^n\left(\sqrt{\frac{k}{n^2}}-1\right)\le S_n\le \sum_{k=1}^n\frac{k}{n^2}=\frac{1}{2}-\frac{1}{2n}[/imath] How do I evaluate the sum on the left?
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2069819
|
Pascal triangle - how to derive row-specific formula
Is there any way I can intuitively demonstrate or remember the formula listed at Calculating a row or diagonal by itself for the Pascal triangle? I'm talking of [imath] {n\choose k}= {n\choose k-1}\times \frac{n+1-k}{k} [/imath] I tried with a piece of paper and it definitely works.. but I don't know why. There is no proof or demonstration.. so: why does it work?
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25068
|
How can I prove the formula for calculating successive entries in a given row of Pascal's triangle?
I've found in Wikipedia the formula for calculating an individual row in Pascal's Triangle: [imath]v_c = v_{c-1}\left(\frac{r-c}{c}\right).[/imath] where [imath]r = \mathrm{row}+1[/imath], [imath]c[/imath] is the column starting from [imath]0[/imath] on left and [imath]v_0 = 1[/imath]. Now, I've tried to do by hand and it works, but I don't understand how to find this magic formula when I don't have access to Wikipedia...:)
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2069778
|
Complex Numbers : Proof
Prove that if [imath]w,z[/imath] are complex numbers such that [imath]|w|=|z|=1[/imath] and [imath]wz\ne -1[/imath], then [imath]\frac{w+z}{1+wz}[/imath] is a real number. Do I substitute [imath]w[/imath] and [imath]z[/imath] for [imath]a+bi[/imath] and [imath]c+di?[/imath] How do I do this?
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1765255
|
Prove that a product of two complex numbers has zero imaginary part
This is my homework, which reads as follows: Let [imath]z_1, z_2[/imath] be complex numbers. Prove that when [imath]z_1z_2 \neq -1[/imath] and [imath]|z_1| = |z_2| = 1[/imath], then the imaginary part of [imath] \frac{z_1 + z_2}{1 + z_1z_2} [/imath] is zero. I've tried to approach this in several ways, but whichever one I try, at some point the expression gets too large and doesn't simplify (at least I cannot find a way to simplify it). The way I tried to approach this problem is to: Set [imath]z_1 = a+bi[/imath], [imath]z_2 = c+di[/imath], then [imath]z_1+z_2=(a+c)+(b+d)i[/imath], and in polar form, [imath]\sqrt{(a + c)^2 + (b + d)^2}\times (\cos\theta + i\sin\theta)[/imath], [imath]\tan \theta = \frac{b+d}{a+c}[/imath]. Find [imath]\tan \phi[/imath], where [imath]\phi[/imath] is the angle of [imath]\frac{1}{1 + z_1z_2}[/imath]. Substitute into [imath]\tan(\theta + \phi) = \frac{\tan(\theta)+\tan(\phi)}{1-\tan(\theta)\tan(\phi)} = 0[/imath]. The problem is [imath]\tan(\phi)[/imath] is a huge expression, and I cannot get it to simplify, so I cannot also show that the last formula is actually true. Since this is a homework, and we are expected to do this with pan and paper (not even a calculator...), I don't believe we are to do this many calculations by hand. Hence, there must be some "trick" to get this to simplify, or a completely different approach. Would you care to give me a hint?
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2069737
|
Linear Algebra - single solution
I have a question and I'm not sure about my solution. For which values of [imath]\lambda[/imath] the system has a single solution, an infinite number of solutions, and no solution? If there are solutions find them. [imath]\left\{ \begin{eqnarray} -2x_1 + x_2 + x_3 =& -\lambda \\ x_1 + 2\lambda x_2 + x_3 =& 1 \\ x_1 + x_2 + 2\lambda x_3 =& 1\end{eqnarray} \right.[/imath] my answer: infinite number of solutions: [imath]\lambda \in \{-1,2\}[/imath] no solution: [imath]\lambda = -1[/imath] single solution: did not make it
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567213
|
find all the values of a and b so that the system has a) no solution b) 1 solution c) exactly 3 solutions and 4) infinitely many solutions
[imath]\begin{cases} &x &- &y &+ &2z &= 4 \\ &3x &- &2y &+ &9z &= 14 \\ &2x &- &4y &+ &az &= b \end{cases} [/imath] I know that [imath]a[/imath] and [imath]b[/imath] has to either equal to something or not in order to satisfy the [imath]4[/imath] conditions stated above. my matrices looked like [imath] \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & a+12 \\ \end{bmatrix} \begin{bmatrix} x & \\ y &\\ z & \\ \end{bmatrix}= \begin{bmatrix} 6 & \\ 2 & \\ b+8 & \\ \end{bmatrix} [/imath]
|
1921395
|
Proof of this simple inequality: [imath]\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4[/imath]
Let [imath]a, b, c, d \in \mathbb{R}_{>0}[/imath], then prove that [imath]\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4[/imath] Can this be done without using AM-GM inequality, or without using any identity/theorem of inequality? I don't want it to be concise or elegant, I just want rigorous steps that show explicitly how we achieve that result. Thanks in advance
|
686791
|
Prove that, [imath]\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\ge 4[/imath] where we do not use AM-GM inequality on the given statement to prove it.
Prove that, [imath]\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4[/imath] where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for a little advanced and elegant solution. EDIT: [imath]a,b,c,d>0[/imath]
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2070390
|
Prove that a given function is bijective
Let [imath]p[/imath] be a prime number such that [imath]p \equiv 2 \pmod 3[/imath]. Prove that the function [imath]f : \mathbb{Z}_p \to \mathbb{Z}_p[/imath], with [imath]f(x) = x^3[/imath] is bijective. Besides the fact that [imath]p = 3k+2[/imath], for some integer [imath]k[/imath], I haven't figured out anything meaningful yet. Thank you in advance!
|
425683
|
If [imath]p[/imath] is congruent to [imath]2 \pmod 3[/imath], how can I prove that all [imath]a[/imath], [imath]1 \le a \le p-1[/imath] are cubic residues [imath]\mod p[/imath]?
If [imath]p[/imath] is congruent to [imath]2 \pmod 3[/imath], how can I prove that all [imath]a[/imath], [imath]1 \le a \le p-1[/imath] are cubic residues [imath]\mod p[/imath]? Here's what I've done: [imath]1^3[/imath] congruent to [imath]1 \pmod p[/imath] thus, 1 is a cubic residue, Also [imath](p-1)^3=(p-1) \pmod p[/imath] implies [imath](p-1)^3 - (p-1) = (p-1)[(p-1)^2 -1] = (p-1)(p^2 -2p) = p(p-1)(p-2) [/imath] implies [imath]p|(p-1)^3 -(p-1)[/imath] Thus [imath]p-1[/imath] is a cubic residue [imath]\mod p[/imath]. Now I don't know how to show for [imath]1\lt a \lt p-1[/imath] are all cubic residues when [imath]p[/imath] is congruent to [imath]2\pmod 3[/imath].
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2070492
|
Prove that a subgroup is dense in [imath]\mathbb{R}[/imath]
Let [imath]G[/imath] be a subgroup of [imath](\mathbb{R}, +)[/imath]. Prove that [imath]G[/imath] is dense in [imath]\mathbb{R}[/imath] or any element in [imath]G[/imath] has the form [imath]a\mathbb{Z}[/imath], with [imath]a \in \mathbb{R}[/imath]. (We denote that a group [imath]G[/imath] is dense in [imath]\mathbb{R}[/imath] if any element in [imath]\mathbb{R}[/imath] is the limit of a sequence of elements from [imath]G[/imath]) I haven't done anything notable yet. Thank you!
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153058
|
How can we find and categorize the subgroups of [imath]\mathbb{R}[/imath]?
[imath]\newcommand{\R}{\Bbb R}\newcommand{\Q}{\Bbb Q}\newcommand{\Z}{\Bbb Z}[/imath] What are all the subgroups of R = [imath](\R, +)[/imath] and how can we categorize them? I started thinking about this question last night after looking at the structure of the cosets of [imath]\R / \Q[/imath] What do the cosets of [imath]\mathbb{R} / \mathbb{Q}[/imath] look like?. I did some searching on SO and google but didn't find anything giving a full categorization (or even a partial one) of the subgroups of [imath]\R[/imath]. Here are the subgroups that I came up with so far: [imath]\Z[/imath] (there are no finite subgroups and [imath]\Z[/imath] is the universal smallest subgroup I think) n[imath]\Z[/imath] eg 2[imath]\Z[/imath] all even numbers a[imath]\Z[/imath] where a is any real number, including a in [imath]\Q[/imath] which "nest" nicely in each other [imath]\Z[/imath][a] - group generated by adding one real a to [imath]\Z[/imath] n[imath]\Z[/imath][a] which equals [imath]\Z[/imath][na] and so is just a case of the one above Dyadic rationals eg a numbers of the form a/2b or similar subgroups such as a/3b, a/2b7c etc [imath]\Q[/imath] [imath]\Q[/imath][a] [imath]\Q[/imath][a in A] where A is a subset of [imath]\R[/imath] - could be finite, countable or uncountable. Group generated by adding all elements of A to [imath]\Q[/imath] eg [imath]\Q[\sqrt2][/imath] It is clear that the "n[imath]\Z[/imath] subgroups" n[imath]\Z[/imath] and m[imath]\Z[/imath] are related according to the gcd(n,m) Also when H is a subgroup of R looking at the structure of the cosets of R / H. eg for H any of the Z subgroups we get R / H homomorphic to [0,1) or the circle. For H one of the Q subgroups it is more complex and I currently don't have ideas on the larger subgroup cosets I am not clear how "big" a subgroup H can get before it becomes the whole of R. I do know that if it contains any interval then it is the whole of R. But what about H with dimension less than 1? I am aware of one question on SO about the proper measurable subgroups of R having 0 measure Proper Measurable subgroups of [imath]\mathbb R[/imath], one on dense subgroups Subgroup of [imath]\mathbb{R}[/imath] either dense or has a least positive element? and one on the subgroups of Q How to find all subgroups of [imath](\mathbb{Q},+)[/imath] but that is all my searching found so far. Why is this question interesting? 1) there seem to be so many subgroups and they are related in many groupings 2) I think the subgroups related to the structure of the reals in some subtle ways 3) I know the result for the complete classification of all finite subgroups was a major result so wondering what has been done in this basic uncountable case. If anyone has any insight, intuition, info, papers or theorems on subgroups of R and how they are interrelated that would be interesting.
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1904154
|
Problem about solvable groups
Let G be a group of order 80 a) Show that G has a normal Sylow 5-subgroup or a normal Sylow 2-subgroup b) Show that G is solvable Approach: [imath]80=5*2^4[/imath] [imath]Syl_5(G) \equiv 1mod \text{ } 5[/imath] and [imath]Syl_5(G) | 16[/imath], so the only possibilities are 1 and 16 [imath]Syl_2(G) \equiv 1 mod \text{ }2[/imath] and [imath]Syl_2(G)|5[/imath], so the only possibilities are 1 and 5 How do we discard one of the possibilities to imply there exists a normal sylow p subgroup in G?
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2592083
|
Showing that 2-Sylow subgroups intersect trivially
I'm looking over a proof for something and I can't understand the last bit of a certain part (counting argument). [imath]G[/imath] is a group of order 80 and [imath]n_2 = 1[/imath] or [imath]5[/imath] (number of 2-Sylow subgroups). We assume that it's not one so it is five. Then let [imath]P_1 \neq P_2[/imath] be 2-Sylow subgroups with [imath]D[/imath] being the intersection of these. So, we know that [imath]P_1, P_2 \in C_G(D)[/imath] but if the centralizer has two 2-Sylow subgroups it must have at least 1+2=3 2-Sylow Subgroups. So, [imath]|C_G(D)| \ge 16(3) = 48 \implies C_G(D) = G[/imath]. I am fine with all of this, just the final bit... this is a contradiction. Am I missing some obvious reason that [imath]C_G(D) = G[/imath] is a contradiction? Then we can conclude that the subgroups intersect trivially, which is what I want. EDIT: The objective of the proof is simply to show that either the 2-Sylow or 5-Sylow subgroup must be normal. There is no mention of the group being abelian or non-abelian, so this is unknown. Before this part we assume that [imath]n_5 = 16[/imath] giving 64 elements of order 5. Thanks!
|
2070505
|
Expectation of [imath]W(t)e^{\lambda aW(t)}[/imath]
Let [imath]\{W_t\}_{t \in \mathbb{R}_+}[/imath] be a Wiener process. I wish to calculate [imath]E \left[ W(t)e^{\lambda W(t)} \right], \lambda \in \mathbb{R}[/imath]. Solution attempt First define a function given by [imath]f(t,x) = xe^{\lambda x}[/imath]. Then, \begin{align} \frac{\partial f(t,x)}{\partial x} &= e^{\lambda x} + \lambda x e^{\lambda x}, \\ \frac{ \partial^2 f(t,x)}{\partial x^2} &= 2\lambda e^{\lambda x} + \lambda^2 xe^{\lambda x}.\end{align} So by Ito's formula, the differential of [imath]f(t,W(t))[/imath] is given by \begin{align} df = \left[ \lambda e^{\lambda W} + \frac{\lambda^2}{2}We^{\lambda W} \right]dt + \left[ e^{\lambda W} + \lambda We^{\lambda W} \right] dW\end{align} Or, rather, [imath]f_s = f_0 + \int_0^s \left[ \lambda e^{\lambda W} + \frac{\lambda^2}{2}We^{\lambda W} \right]dt + \int_0^s \left[ e^{\lambda W} + \lambda We^{\lambda W} \right] dW.[/imath] Then we take expectations, so that the stochastic integral disappears, then we define [imath]m(t) = E f_s = E f(s,W(s))[/imath], and then we take the derivatives. We then get [imath]m'(t) = E \lambda e^{\lambda W} + \frac{\lambda^2}{2}m(t).[/imath] The first term on the left hand side equals [imath]\lambda e^{t\frac{\lambda^2}{2}}[/imath]. At this point, as I am not good with differential equations, I am kind of stuck. And, either way, I would like to have my solution so far checked out as well, as I am unsure whether the stochastic integral really does disappear when I take expectations. What condition needs to be satisfied for this to hold, and how do I check whether it does?
|
2068309
|
What is [imath]\mathbb{E}[W(s)\mathrm{e}^{W(S)}][/imath] where W(S) is a standard Brownian Motion?
I know that since it is a Brownian motion that [imath]W(s)\sim{N(0,s)}[/imath]. One approach I tried, was to see that [imath]W(s)-W(0)\sim N(0,1)[/imath] and to let [imath]X = W(s)[/imath] and then solve [imath]\mathbb{E}[X\mathrm{e}^X][/imath] but I end up with some awful integral. I was wondering whether there was an easier way? Or perhaps a simple way to solve this integral: [imath]\int_{-\infty}^{\infty} \frac{1}{\sqrt{2{\pi}s}}x\mathbb{e}^{x}\mathbb{e}^{-\frac{x^2}{2s}}[/imath]
|
2070618
|
Prove inequality [imath]e^{x}>1+x[/imath]
Prove that [imath]e^{x}>1+x[/imath] when [imath]x\neq0[/imath]. I can't make for all [imath]x[/imath]. But as [imath]e^{x}>0[/imath] for all [imath]x[/imath], it's obvious that [imath]e^{x}>1+x[/imath] when [imath]x\le-1[/imath]. It's where I got got stuck
|
694255
|
Showing that [imath]e^x > 1+x[/imath]
Problem. Show that [imath]e^x > 1 + x , \ \ x \neq 0[/imath] My attempt. Using Mean Value Theorem: [imath]f'(c) = \frac{f(b) - f(a)}{b-a}[/imath] [imath]\Rightarrow e^c = \frac{e^x - 1}{x}[/imath] [imath]\Rightarrow xe^c = e^x -1[/imath] [imath]\Rightarrow x+1 = \frac{e^x}{e^c}[/imath] I'm not feeling very comfortable with these calculations...Am I on the right track? Can I even choose [imath]e^o[/imath] in my initial calculation as it says that [imath]x \neq 0 [/imath] ?
|
2070248
|
Slope of an Angled Line Intersecting a Tangent Line
I know that to find the slope of a tangent line at a point [imath](x, y)[/imath] the general form is [imath]y-y_1 = f'(x_1)(x-x_1)[/imath], and the equation of a line perpendicular to that tangent line is [imath]y-y_1 = -\frac{1}{f'(x_1)} (x-x_1)[/imath]. So my question is, what is the general form for an equation that intersects the tangent line of a function at an angle of [imath]\theta[/imath]? Thanks in advance!
|
105770
|
Find the slope of a line given a point and an angle
I'm trying to figure out this problem and feel like it's something that must be so simple that I could've done in high school no problem, but for some reason my brain is frozen this morning. I would really appreciate any help, and want to say thanks in advance. I tried to draw a picture below; I want to find the slope of a line given a point [imath](x,y)[/imath] and [imath]\theta[/imath].
|
711018
|
Generating the symmetric group [imath]S_n[/imath]
I know that [imath]\sigma =(1 2 \ldots n)[/imath] and [imath]\tau =(1 2)[/imath] together should generate the symmetric group by virtue of conjugation, i.e. [imath](\sigma)^k \circ \tau \circ (\sigma^{-1})^k = (k+1, k+2)[/imath]; we know that the set of adjacent transpositions generates [imath]S_n[/imath], so we're done. However-- and I realize that this question is incredibly dumb-- when I try this for [imath]n=3[/imath], I should have that [imath]\sigma \circ \tau \circ \sigma^{-1} = (2 3)[/imath]. But I get [imath]\sigma \circ \tau = (1 2 3) \circ (1 2) = (2 3)[/imath] when I work it out by hand, writing out the digits and physically switching them; then [imath](2 3)\circ \sigma^{-1}= (1 3)[/imath]. What am I doing wrong? EDIT: I'd confused sigma and tau. I've edited the question to reflect this.
|
470557
|
Show that the group [imath] S_n [/imath] is generated by two sets
Show that the group [imath] S_n [/imath] is generated by [imath] \{(1,2), (1,2,3,...,n) \} [/imath] and also [imath] \{(1,2), (2,3,...,n)\}[/imath]. How I should start, maybe use induction?
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2071075
|
Stars and bars with at least 1 group of 2 adjacent bars
I am trying to figure out the number of [imath]n[/imath] permutations for [imath]N[/imath] stars and [imath]K \ge{2}[/imath] bars such that there is at least 1 group of 2 adjacent bars. For example, in this trivial example [imath]N = 1[/imath] and [imath]K = 3[/imath] then [imath]n = 4[/imath] [imath]|||* \\ ||*| \\ |*|| \\ *|||[/imath] This coincides with your typical counting argument [imath]\frac{4!}{3!1!}[/imath]. However, for a more non-trivial example where [imath]N = 2[/imath] and [imath]K = 3[/imath], we have [imath]n = 9[/imath]. My initial instinct was to group the 2 bars as a "group" and permute it with the rest of the stars and bars (so for [imath]N = 2[/imath] and [imath]K = 3[/imath], we have [imath]\frac{4!}{1!1!2!} = 12 \ne 9[/imath]). The only counting argument I can come up with is to find the total # of permutations and subtract the # of permutations where there are no adjacent bars ([imath]\frac{5!}{3!2!} - 1 = 9[/imath]), but this would involve tedious case work with groups of [imath]*|*[/imath] for the # of perms without adjacent bars. Clearly this does not work when [imath]K > 2N[/imath].
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1973512
|
Find the number of [imath](m+n)[/imath] digit binary sequences with [imath]m[/imath] [imath]0[/imath]'s and [imath]n[/imath] [imath]1[/imath]'s such that no two [imath]1[/imath]'s are adjacent
A [imath]k[/imath]-digit binary sequence is a word of length [imath]k[/imath] in the alphabet [imath]\{0, 1\}[/imath]. Let [imath]n\le m+1[/imath], find the number of [imath](m+n)[/imath] digit binary sequences with [imath]m[/imath] [imath]0[/imath]'s and [imath]n[/imath] [imath]1[/imath]'s such that no two [imath]1[/imath]'s are adjacent.
|
2070170
|
Solving [imath]2x^4+2x^2y^2+y^4 = z^2[/imath]
Find all integer pairs [imath](x,y,z)[/imath] that satisfy [imath]2x^4+2x^2y^2+y^4 = z^2.[/imath] We can rewrite the given equation as [imath](x^2+y^2)^2+(x^2)^2 = z^2[/imath]. Thus, [imath](x^2+y^2,x^2,z)[/imath] must be a Pythagorean triple. How do we continue? Also, for [imath]x = 0[/imath] we get [imath]y^4 = z^2[/imath] and so [imath]z = \pm y^2[/imath]. Can we show that [imath]x[/imath] must be [imath]0[/imath]?
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2070219
|
Prove that the equation [imath]x^4+(2k^2)x^2+2k^4[/imath] is not a perfect square
Prove that any quadratic of the form [imath]x^4+(2k^2)x^2+2k^4[/imath] for any nonzero integer [imath]k[/imath] is never a perfect square for all integers [imath]x[/imath]. I thought about proving this by induction. We first prove that [imath]x^4+2x^2+2[/imath] is not a perfect square for all integers [imath]x[/imath] (which is true since [imath]x^4+2x^2+2 = (x^2+1)^2+1[/imath]) and then proceed with the inductive hypothesis. Is there an easier way of solving this?
|
152516
|
Evaluating [imath]\int \sin^2 x \cos^3 x \operatorname dx[/imath]
[imath]\int \sin^2 x \cos^3 x dx[/imath] [imath]\int (1-\cos^2 x) \cos^3 x dx[/imath] I know there is a rule in my book (with little explanation) that tells me when I had an odd and an even degree on two trig functions I should split the odd and convert it to an identity but this way seems easier, and I can't get an answer either way. [imath]\int \cos^3 dx - \int \cos^ 5 x dx[/imath] I am not sure where to go from here, I don't know how to get the integral of [imath]\cos^3 x[/imath]
|
21589
|
Evaluate [imath]\int \cos^3 x\;\sin^2 xdx[/imath]
Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer. [imath]\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx = \int (\cos x)(1-\sin^2x)(\sin^2x)dx.[/imath] let [imath]u = \sin x[/imath], [imath]du = \cos xdx[/imath] [imath]\int(1-u^2)u^2du = \int(u^2-u^4)du = \frac{u^3}{3} - \frac{u^5}{5} +C[/imath] Plugging in back [imath]u[/imath], we get [imath]\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}[/imath] + C
|
1254395
|
If [imath]a,b[/imath] are positive rational numbers and [imath]\sqrt a+\sqrt b[/imath] is rational, then both of [imath]\sqrt a,\sqrt b[/imath] are rational numbers
I'm trying to show that If [imath]a,b[/imath] are positive rational numbers and [imath]\sqrt a+\sqrt b[/imath] is rational, then both of [imath]\sqrt a,\sqrt b[/imath] are rational numbers. I squared the number [imath]\sqrt a+\sqrt b[/imath] and found that [imath]\sqrt {ab}[/imath] is rational…
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136594
|
Show that [imath] a，b, \sqrt{a}+ \sqrt{b} \in\mathbb Q \implies \sqrt{a},\sqrt{b} \in\mathbb Q [/imath]
Inspired by this, I was wondering if there is a simple logical argument to Show that [imath] a,b, \sqrt{a}+ \sqrt{b} \in\mathbb Q \implies \sqrt{a},\sqrt{b} \in\mathbb Q [/imath] Note that the original link is using a computational method, where as I am looking for a simple logical argument. I tried (unjutifiably) to argue that if some of two square roots is rational then each one is rational, this is a different than the (incorrect) argument that if sum of two algebraic numbers is rational then each one is rational ( counter example [imath]a=1-\sqrt{2},b= 1+\sqrt{2} [/imath])
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2071875
|
Inflate a cosine wave
I have a simple cosine wave: [imath]y=\cos(\pi x/2)[/imath]. This equation passes through the points [imath](0,1), (1,0), (3,0)[/imath] and so on (the black line in the picture). I want to keep these points, but "inflate" the graph more to look like the green graph in the illustration. Illustration How would I go about doing this? EDIT: I have tried using roots of 2 through 3 (including square roots, cube roots, and even the 2.3rd root). However, it appears that none of them seem to line up with the green equation (made up of portions of circles. Is there any way to have a "perfectly circular" cosine equation (in terms of radius of the circle for general use) like the green graph? If so, what method or root should I use to achieve this? To clarify: I am looking for the equation of an "inflated" graph, and the equation of a graph that has been "inflated" to its fullest while still being a function (a "perfectly round" cosine equation).
|
1390198
|
What amplification can I apply to [imath]y=\sin x[/imath] for it to be a perfect oscillating arc?
A perfect arc is [imath]y=\sqrt{|1-(x-1)^2|}[/imath]. A sin wave is [imath]y=\sin({\pi x\over2})[/imath] I am curious how I can amplify the sin wave so that it's a perfect alternating arc. In the link below you can see graphed each of these two equations, as well as the delta between them. So in the sage code there's function [imath]i(x) = [/imath]SOMETHING [imath]\times f(x)[/imath], where [imath]f(x) = \sin({\pi x\over 2})[/imath], but I want to replace the 'SOMETHING' with an expression that transforms the wave to [imath]y=\sqrt{|1-(x-1)^2|}+\sqrt{|1-(x-3)^2|}+\sqrt{|1-(x-5)^2|}... + \sqrt{|1-(x-n)^2|}[/imath] In other words, I want [imath]g(x):g(x)\sin({\pi x\over 2}) = \sqrt{|(1-(x-1)^2|}[/imath] where [imath]0<x<2[/imath] I have no particular goal, I am just curious how I would go about solving that as currently I'm a bit stumped... sage math link to the various graphs thank you kind internet maths wizards!
|
2072072
|
Prove that for all [imath]a, b > 0, a^b + b^a > 1[/imath]
This seems an innocent inequality. The trivial case when [imath]a \ge 1[/imath] or [imath]b \ge 1[/imath] is easy to prove: if [imath]a \ge 1[/imath], then [imath]a^b \ge a^0 = 1[/imath]. So the hard part of the problem is when [imath]a, b \in (0, 1)[/imath] .
|
381090
|
[imath]x^y+y^x>1[/imath] for all [imath](x, y)\in \mathbb{R_+^2}[/imath]
Prove that [imath]x^y+y^x>1[/imath] for all [imath](x, y)\in \mathbb{R_+^2}[/imath].
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1009304
|
a two-variable cyclic power inequality [imath]x^y+y^x>1[/imath] intractable by standard calculus techniques
If [imath]x[/imath] is in the open interval (0,1) and so is [imath]y[/imath], prove that [imath]x^y+y^x>1[/imath] A direct two-variable application of maxima and minima seems difficult.
|
319578
|
Exponential teaser
Find the minimum value for [imath]x^x[/imath] for [imath]x[/imath] a positive real number If [imath]x[/imath] and [imath]y[/imath] are positive real numbers, show that [imath]y^x + x^y > 1[/imath]
|
1042080
|
An inequality concerning a particular function of two variables
How do you prove that [imath]{x^y+y^x>1}[/imath] if [imath]{x,y>0}[/imath] and both real? I saw a related problem on a Web puzzle page and it appears that the above statement is true when examined graphically in the unit square using a computer package like Mathematica. The result is obvious outside the unit square. I tried very hard to obtain an analytical proof (over several days) using my background in undergraduate mathematical analysis. Unfortunately I found the problem seems to be quite slippery - there are one or two bogus solutions on the Web that I found and discovered to be incorrect.
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482549
|
A classic exponential inequality: [imath]x^y+y^x>1[/imath]
This is a classic problem, but I haven't seen it on the site before. Suppose [imath]x[/imath] and [imath]y[/imath] are real numbers that both lie in the interval [imath](0,1)[/imath]. Prove [imath]x^y+y^x>1.[/imath] The following spoiler box contains a hint that indicates the solution I know. Does anyone know a more natural solution? Use Bernoulli's inequality to show [imath]x^y\ge \frac{x}{x+y}.[/imath] Edit: I recently learned the source is the 1996 French math olympiad.
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2070615
|
Formula for calculating the total number of solution pairs to an equation
I was solving a mathematical olympiad problem, which reads as follows: How many positive integer solution pair [imath](x, y)[/imath] are there for the equation [imath]y^2=\frac{x^5-1}{x-1}[/imath], where [imath]x\ne1[/imath]? My first act was to simplify the equation to this: [imath]y^2=x^4+x^3+x^2+x+1[/imath] Seeing that this did no good, I hit on the idea that this problem was actually very simple. A good, little trick would do it. I decided to work with the first five integers (excluding [imath]1[/imath], of course). By trial and error, I discovered that there was only one integer solution pair i.e. [imath](3, 11)[/imath] that satisfies the equation. This was as I suspected before. I also build a C program to confirm this; I was right. But... if I were to solve another such problem that doesn't involve tricks, then what? There should be a formula for calculating total number of solution pairs to equations of that kind. Is there? Could I solve this very problem using a formula? Please answer bearing in mind that I'm only 12 years and a 7th grader. How my question isn't a duplicate of any other question? At first look, my question might seem to be the exact duplicate of this: [imath]y^2 = \frac{x^5 - 1}{x-1}[/imath] & [imath]x,y \in \mathbb{Z}[/imath]. However, it isn't. First of all, it's obvious that I didn't notice that question before posting this question. The main difference lies in the question content and purpose. The other question was solely based on how to solve this particular problem. But, my question isn't that. I have solved the problem. My question is: is there a formula in general to find out the number of rational solutions of a diophantine equation? My question has a much more wider purpose. These reasons should be enough to convince you that this question isn't a duplicate of the question at all.
|
2066356
|
[imath]y^2 = \frac{x^5 - 1}{x-1}[/imath] & [imath]x,y \in \mathbb{Z}[/imath]
My question is this: [imath]y^2 = \frac{x^5 - 1}{x-1}[/imath] & [imath]x,y \in \mathbb{Z}[/imath] Source: BdMO 2016 Regionals Set 1 Question 3. An equation in [imath]\mathbb{Z}[/imath] I think this is the exact same question. Actually I didn't get the solution in the link. Please don't mark as duplicate. I tried solving this same problem by writing code; there seems to be [imath]6[/imath] solution pairs to this equation. How do I get the [imath]6[/imath] solutions mathematically? I indeed need a full solution. Please help.
|
2071756
|
Show if [imath]f(x+y)=f(x)+f(y)[/imath] for all [imath]x,y[/imath] and [imath]f[/imath] is Lebesgue measurable, then [imath]f[/imath] is continuous.
I'm trying to solve this problem [imath](5.8)[/imath] from Bass' Real Analysis for Graduate Students: Show that if [imath]f:\Bbb{R}\to\Bbb{R}[/imath] is Lebesgue measurable and [imath]f(x+y)=f(x)+f(y)[/imath] for all [imath]x,y\in\Bbb{R}[/imath], then [imath]f[/imath] is continuous. The first thing I note is that it suffices to show that [imath]f[/imath] is continuous at [imath]0[/imath], since if this is the case then for any [imath]\epsilon>0[/imath] and [imath]x_0\in\Bbb R[/imath], there exists [imath]\delta>0[/imath] such that [imath]|x|<\delta[/imath] implies [imath]|f(x)|<\epsilon[/imath]; in particular, replacing [imath]x[/imath] with [imath]x-x_0[/imath], we see [imath]|x-x_0|<\delta[/imath] implies [imath]|f(x)-f(x_0)|=|f(x-x_0)|<\epsilon[/imath]. Now, what I need to show then is that [imath]f^{-1}(-\epsilon,\epsilon)[/imath] contains some interval at the origin. I know that this set is Lebesgue measurable because [imath](-\epsilon,\epsilon)[/imath] is Borel measurable, and I know it contains [imath]0[/imath] since [imath]f(0)=0[/imath], but I'm not sure how to show it contains [imath](-\delta,\delta)[/imath] for some [imath]\delta[/imath]. I haven't used the fact that the set is Lebesgue measurable, so obviously theres some way I can use that but I'm not seeing it. Any suggestions? edit: because this has been tagged as a possible duplicate, I should point out that I'd prefer to figure out if the solution I am working on will work, rather than just copy a completely different solution outlined by some other user.
|
2067152
|
Let [imath]g:\mathbb{R}\to\mathbb{R}[/imath] be a measurable function such that [imath]g(x+y) =g(x)+g(y).[/imath] Then [imath]g(x) = g(1)x[/imath] .
Let [imath]g:\mathbb{R}\to\mathbb{R}[/imath] be a measurable function such that [imath]g(x+y) =g(x)+g(y).[/imath] How to prove that [imath]g(x) = cx[/imath] for some [imath]c\in \mathbb{R}?[/imath] The main thing to do here relies upon the fact that such function should be continuous and therefore by natural argument the answer will follow. Using this Additivity + Measurability [imath]\implies[/imath] Continuity Therefore I found out that there is nothing missing in this question.
|
2070439
|
What about the reciprocal of Heine theorem?
[imath]i.e.[/imath] What are the subsets [imath]H[/imath] [imath]\subset[/imath] [imath]\mathbb R[/imath] such that all continuous function [imath]f[/imath] [imath]:[/imath] [imath]H[/imath] [imath]\to[/imath] [imath]\mathbb R[/imath] is uniformly continuous ? Compact sets of course, but think about [imath]\mathbb Z[/imath] or more complicated subset.
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1426874
|
For which conditions on countable sets does continuity implies uniform continuity
This is an exercise from Thomson and Bruckner's "Elementary Real Analysis" (Exercise 5.6.5 page 239): Let [imath]X=\{x_1,x_2,\dots,x_n,\dots\}[/imath]. What property must [imath]X[/imath] have so that every function continuous on [imath]X[/imath] is uniformly continuous on [imath]X[/imath]? The set [imath]X[/imath] is a subset of [imath]\mathbb {R}[/imath]. I worked on the [imath]\varepsilon-\delta[/imath] definitions of limits and uniform continuity, but I coudln't link between them. Could you please help me? Edit: I removed the part My attempt because my result was wrong.
|
2070349
|
Prove that any base [imath]a>1[/imath] , each positive integer [imath]m[/imath] has a unique expression of the form [imath]a^nr_n+a^{n-1}r_{n-1}+...+ar_1+r_0[/imath]
I have the next problem. Prove that any base [imath]a>1[/imath], each positive integer [imath]m[/imath] has a unique expression as the form [imath]a^nr_n+a^{n-1}r_{n-1}+...+ar_1+r_0[/imath] where the integers [imath]r_k[/imath] satisfy [imath]0\le r_k \lt a, r_n\neq0.[/imath] I have tried this. Proof: By induction above n [imath]m_1=ar_1+r_0[/imath] We suppose that it is true for n=k [imath]m_k=a^kr_k+a^{k-1}r_{k-1}+...+ar_1+r_0[/imath] Prove for n=k+1. By hypothesis [imath]m_k=a^kr_k+a^{k-1}r_{k-1}+...+ar_1+r_0[/imath] We add [imath]a^{k+1}r_{k+1}[/imath]. [imath]m_k+a^{k+1}r_{k+1}=a^{k+1}r_{k+1}+a^kr_k+a^{k-1}r_{k-1}+...+ar_1+r_0[/imath] As [imath]a^{k+1}r_{k+1}[/imath] and [imath]m_k[/imath] are positive integers then be [imath]m_{k+1}=m_k+a^{k+1}r_{k+1}[/imath] a positive integer, hence [imath]m_{k+1}=a^{k+1}r_{k+1}+a^kr_k+a^{k-1}r_{k-1}+...+ar_1+r_0[/imath] but I am not sure and still I need to prove the uniqueness.Thank you for your advices.
|
780959
|
Base-b Representation of an Integer: Why can I make the assumption about number of terms in expansions in uniqueness part?
Base-b Representation of an Integer: Why can I make the assumption about number of terms in expansions in uniqueness part? Reading the book of Koshy I have come across the below theorem. It states that I can write any integer [imath]N[/imath] in a base [imath]b\ge2[/imath] representation [imath]N=a_kb^k+...+a_1b+a_0[/imath] such that [imath]0\le a_i<b[/imath], [imath]a_k \neq 0[/imath] for some [imath]k\ge0[/imath]. In the uniqueness part, why can I assume the expansions contains the same number of terms ? I must prove that all expansions of [imath]N[/imath] equal the one in the theorem, but if I add terms to the expansion by zero coefficients, well, then the expansion is not in the proper form anymore ? (the greatest term in the expansion is [imath]0[/imath] if I add zero coefficients) . Can someone help clear this out ? Why can I assume the expansions contain the same number of terms ? If I add terms to the expansion, well then I've violated the form of the expansion ?
|
2064255
|
Using matrix algebra prove that [imath] \left< A\textbf{x}, y \right> = \left< \textbf{x}, A^T \textbf{y} \right>[/imath]
If A is an [imath]m \times n[/imath] matrix using matrix algebra prove that [imath]\langle A\textbf{x}, \textbf{y} \rangle[/imath] = [imath]\langle \textbf{x}, A^T \textbf{y} \rangle[/imath] for [imath]x \in \mathbb{R}^n[/imath] and [imath]y \in \mathbb{R}^m[/imath]. My approach: let [imath]m = 2, n = 1[/imath] [imath]A = \begin{pmatrix} 1\\ 2 \end{pmatrix} [/imath], [imath] A^T = \begin{pmatrix} 1 & 2 \end{pmatrix} [/imath], [imath] \textbf{x} = (1)[/imath], [imath]\textbf{y} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}[/imath] Then [imath]A\textbf{x} = \begin{pmatrix} 1 \\2 \end{pmatrix}[/imath] [imath]\langle A\textbf{x}, y \rangle = 1*2 + 2*3 = 8[/imath] [imath]A^T\textbf{y} = 8 [/imath] [imath]\langle \textbf{x}, A^T \textbf{y} \rangle = 8 [/imath] [imath]\langle A\textbf{x}, \textbf{y} \rangle[/imath] = [imath]\langle \textbf{x}, A^T \textbf{y} \rangle[/imath] However I'm pretty sure this answer isn't satisfactory. What can I do to improve it?
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2063967
|
How to prove that [imath]\langle Av,w\rangle=\langle v,A^Tw\rangle[/imath]?
How to prove: [imath]\langle Av,w\rangle=\langle v,A^Tw\rangle[/imath] [imath]\langle,\rangle[/imath] represents inner-product, [imath]v,w[/imath] denote vectors and [imath]A[/imath] is a matrix.
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1215186
|
[imath]\mathbb{Q}(\sqrt{2+\sqrt{2}})[/imath] is Galois over [imath]\mathbb{Q}[/imath]?
I need to show that the extension [imath]\mathbb{Q}(\sqrt{2+\sqrt{2}})[/imath] is Galois over [imath]\mathbb{Q}[/imath], and compute its Galois group. I am learning Galois theory by myself and got stuck in this exercise. I know the fundamental theorem of Galois theory. Any help would be useful Thanks
|
575171
|
Galois Group of [imath]\sqrt{2+\sqrt{2}}[/imath] over [imath]\mathbb{Q}[/imath]
So I want to show that [imath]\mathbb{Q}(\sqrt{2+\sqrt{2}})[/imath] is Galois over [imath]\mathbb{Q}[/imath] and determine its Galois group. My thoughts are as follows: Define [imath]\alpha := \sqrt{2+\sqrt{2}}[/imath]. Then it is easily shown that [imath]\alpha[/imath] satisfies [imath]\alpha^4-4\alpha^2+2=0[/imath]. Define [imath]f(x) := x^4-4x^2+2[/imath]. Then [imath]f[/imath] is irreducible over [imath]\mathbb{Q}[/imath] by Eisenstein with [imath]p=2[/imath]. So we have that [imath]f[/imath] is the irreducible polynomial for [imath]\alpha[/imath] over [imath]\mathbb{Q}[/imath]. Further [imath]|\mathbb{Q}(\alpha):\mathbb{Q}|=4[/imath]. For [imath]\mathbb{Q}({\alpha})[/imath] to be Galois, it must contain all roots of [imath]f[/imath]. Define [imath]K=\mathbb{Q}(\alpha)[/imath] for convenience. Define [imath]\alpha := \alpha_1[/imath]. Since [imath]f[/imath] has only even powers, we know that [imath]-\alpha := \alpha_2[/imath] is a root, and therefore contained in [imath]K[/imath] since [imath]K[/imath] is a field. We note that the other two roots are [imath]\alpha_3=\sqrt{2-\sqrt{2}}[/imath] and [imath]\alpha_4=-\sqrt{2-\sqrt{2}}[/imath]. So in order to show [imath]K[/imath] is Galois, it must be shown that [imath]\alpha_3[/imath] and [imath]\alpha_4[/imath] lie in [imath]K[/imath]. Now [imath]\alpha_1^2=2+\sqrt2[/imath] and so [imath]\sqrt2 \in K[/imath]. Thus [imath]-\sqrt2 \in K[/imath] since [imath]K[/imath] is a field. Can somebody explain why [imath]\alpha_3[/imath] and [imath]\alpha_4[/imath] lie in [imath]K[/imath]? Next we are to determine the Galois group of [imath]K[/imath]. Assuming [imath]K[/imath] is Galois, since it has degree [imath]4[/imath] over [imath]\mathbb{Q}[/imath] (shown earlier), we know that its Galois group has size [imath]4[/imath]. There are only two groups of size [imath]4[/imath], namely [imath]V_4[/imath] and [imath]C_4[/imath], the Klein four group and the cyclic group of order [imath]4[/imath]. How do we determine which of these choice is in fact the Galois Group?
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1313272
|
Galois group of [imath](x^2-2)(x^3-2)(x^3-3)[/imath] over the field [imath]\mathbb{Q}(i\sqrt{3})[/imath]
I am trying to find the Galois group of [imath](x^2-2)(x^3-2)(x^3-3)[/imath] over the field [imath]\mathbb{Q}(i\sqrt{3})[/imath]. The roots of this polynomial are [imath]\pm \sqrt{2}[/imath], [imath]\zeta_3^k \sqrt[3]{2}[/imath], and [imath]\zeta_3^j \sqrt[3]{3}[/imath] for [imath]j,k=0,1,2[/imath], so that the splitting field over [imath]\mathbb{Q}(i\sqrt{3})[/imath] is [imath]\mathbb{Q}(i\sqrt{3}, \sqrt{2}, \sqrt[3]{2}, \sqrt[3]{3}, \zeta_3)[/imath]. Can someone help me find: The Galois group over [imath]\mathbb{Q}[/imath] The Galois group over [imath]\mathbb{Q}(i\sqrt{3})[/imath]
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1694952
|
Determine the Galois group of the polynomial [imath](x^3-2)(x^3-3)(x^2-2)[/imath] over [imath]\mathbb{Q}(\sqrt {-3})[/imath]
Determine the Galois group of the polynomial [imath](x^3-2)(x^3-3)(x^2-2)[/imath] over [imath]\mathbb{Q}(\sqrt {-3})[/imath]. I tried to decompose the splitting field into a tower, but then it is very complicated in each step and since the degree is large, I cannot determine the group by guessing. It seems to me that Kummer theory might work, but there are [imath]2[/imath]-degree and [imath]3[/imath]-degree factors.
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1291929
|
Showing any metric space is a Hausdorff space
This is the question i've been given along with the solution i have written, however, could someone explain why showing [imath]z \notin V[/imath] shows any metric space is a Hausdorff space
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1862598
|
Show that any metrizable space [imath]X[/imath] is Hausdorff
I wish to show that any metrizable space [imath](X,\mathcal{T})[/imath] is Hausdorff Proof attempt: Let [imath]d[/imath] be the metric that generates the topology on [imath]X[/imath]. Pick two points [imath]x,y \in X[/imath], we wish to produce two disjoint open sets that separates [imath]x,y[/imath]. Let [imath]B_\epsilon(x)[/imath] and [imath]B_\delta(y)[/imath] be two metric balls containing [imath]x,y[/imath] respectively. Suppose that [imath]B_\epsilon(x) \cap B_\delta(y) \neq \varnothing[/imath] Stuck Here: Hmm...How should I adjust [imath]\epsilon, \delta[/imath] so that these balls are separated? Idea: Reduce [imath]\epsilon[/imath] by half. If they are still intersecting...reduce [imath]\delta[/imath] by half. Continue ad infinitum Is there a more satisfying solution i.e. closed form expression for reduced [imath]\epsilon, \delta[/imath] so they are no longer intersecting. Thanks!
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778035
|
The Galois Group of [imath]x^4 - 5x^2 + 6[/imath]
As the title suggests, I'm asked to describe the Galois group of the polynomial [imath]x^4 - 5x^2 + 6 \in \mathbb{Q}[x][/imath] over [imath]\mathbb{Q}[/imath]. I am pretty certain I have 95% of the problem completed. I'm just having difficulty fully putting together the last step (I think it may boil down to a lack of knowledge in the "types" of groups). The roots of this polynomial are [imath]\pm \sqrt{2}[/imath] and [imath]\pm \sqrt{3}[/imath]. The automorphisms are thus: [imath]f_1(a + b\sqrt{2}) = a - b\sqrt{2} \\ f_2(a - b\sqrt{3}) = a - b\sqrt{3}[/imath] This resembles... something like... a group [imath]\{e, \sigma_1, \sigma_2\}[/imath] that has two coordinates? Something like [imath]\sigma_1 \times \sigma_2[/imath] where [imath]\sigma_1[/imath] cycles between itself and its negative and [imath]\sigma_2[/imath] cycles between itself and its negative. I feel like there's just one little thing I'm missing. (Edit: I think I've got it. This group is isomorphic to [imath]\mathbb{Z}_2 \times \mathbb{Z}_2[/imath], right?)
|
204709
|
Computing the Galois group of [imath]x^4+ax^2+b \in \mathbb{Q}[x] [/imath]
I want to compute the Galois group of some polynomials, but I want to see some examples first. For example this proposition could be helpful. I don't know how to prove it <.< Let's consider a polynomial [imath]f(x)= x^4+ax^2+b \in \mathbb{Z} [/imath]. Let [imath] \pm\alpha , \pm\beta[/imath] denote the roots of [imath]f(x)[/imath]. Note that [imath]f[/imath] is irreducible iff [imath]\alpha^2, \alpha\pm\beta \notin \mathbb{Q}[/imath]. So let's suppose that [imath]f(x)[/imath] is irreducible. Let's denote [imath]G[/imath] the galois group of [imath]f[/imath] (the Galois group of the splitting field [imath]\mathbb{Q}(\alpha,\beta)/\mathbb{Q}[/imath]) [imath]i)[/imath] [imath]G\cong V[/imath] the Klein 4-group iff [imath]\alpha\beta \in \mathbb{Q}[/imath] [imath]ii)[/imath] [imath]G\cong \mathbb{Z}_4[/imath] iff [imath]\mathbb{Q}(\alpha\beta)=\mathbb{Q}(\alpha^2)[/imath] [imath]iii)[/imath] [imath]G\cong D_8[/imath] the dihedral group of order 8, iff [imath]\alpha\beta \notin \mathbb{Q}(\alpha^2)[/imath] I need help with the proofs of this propositions... Well I'm very lost with the computation of Galois group, some care must be exercised, I don't have it. For example I know that Galois group acts transitively, in the sense that [imath]u\in K/F[/imath] a galois extension, the other roots of the minimal polynomial of [imath]m_u(x)\in F[x][/imath] are precisely the elements [imath] \sigma(u)[/imath] , where [imath]\sigma \in Gal(K/F)[/imath]. And clearly any automorphism, is determined by only the value of a basis. In this case we are working on [imath]\mathbb{Q}(\alpha,\beta)[/imath], so I have to know the values of both. But I have to respect the algebraic relations involving [imath]\alpha , \beta[/imath] I proved that in [imath]i),ii)[/imath] [imath]\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha) [/imath] So I have to know the values on [imath]\alpha[/imath]. My first question is: If in both cases [imath]i),ii)[/imath], I have an extension of degree 4 (so [imath]|G|=4[/imath]) I have to map [imath]\alpha[/imath] to determine the automorphism, but I have also to map [imath]\alpha[/imath] to all of it's four roots. Why the galois group in this case are different? My second question is about [imath]iii)[/imath] I have no idea How to attack it.
|
2072765
|
Show that [imath]\lim_{r\to 0}\int_{|z|=r}{f(z)\over z}dz=2\pi if(0)[/imath]
Let [imath]f:\Bbb{C}\to \Bbb{C}[/imath] be continuous at [imath]0[/imath]. Show that [imath]\lim_{r\to 0}\int_{|z|=r}{f(z)\over z}dz=2\pi if(0)[/imath]. I barely saw anything like that. In fact, I had a difficult time finding references for Complex Analysis topics regarding complex continuous functions that don't have to do with (n?)either analyticity or holomorphicity. My only problem, at this point, seems to be showing formally that [imath]f[/imath] is bounded on [imath]|z|=r[/imath] for sufficiently small [imath]r[/imath]. Regardless of that, under the assumption that for sufficiently small [imath]r[/imath], [imath]|f|[/imath] is bounded on [imath]\{z:|z|\le r\}[/imath] by M, I let [imath]\lim_{r\to 0}|\int_{|z|=r}{f(z)\over z}dz|\le\lim_{r\to 0}\max_{|z|=r}({f(z)\over r})2\pi r=2\pi f(0)[/imath] where [imath]M[/imath] is the maximum of [imath]f(z)[/imath] on the whole disc given by [imath]r[/imath] sufficiently small. Could you guid me through this problem? Though the question is not quite different from the one to which it was claimed to be identical in theory, I had not derived much from the answers, despite asking for explanations there myself, and I cannot make my misunderstanding a public thing, unless I either make another question or start a bounty. How is an absolute value of an integral an indication to the value of the integral itself? The value of the integral itself can be multiplicated by a branch of functions whose absolute value is 1.
|
1555195
|
Show that [imath]\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a} \, dz=f(a)[/imath]
Let [imath]a\in\Bbb C[/imath] and [imath]r>0[/imath] and denote by [imath]B(a,r)\subseteq \Bbb C[/imath] the open ball of center [imath]a[/imath] and radius [imath]r[/imath]. Assume that [imath]f:B(a,r)\to\Bbb C[/imath] is a continuous function and for each [imath]\epsilon>0[/imath] let [imath]\gamma_\epsilon:[0,2\pi]\to \Bbb C[/imath] be given by [imath]\gamma_\epsilon(t)=a+\epsilon e^{it}[/imath]. Show that [imath]\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a} \, dz = f(a).[/imath] I tried the following [imath]\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a}dz=\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(a+\epsilon e^{it})}{a+\epsilon e^{it}-a}i\epsilon e^{it} \, dt = \lim_{\epsilon\to0^{+}} \frac{1}{2\pi} \int_0^{2\pi}f(a+\epsilon e^{it}) \, dt[/imath] If I can change the limit and the integral, then it is obvious. I tried to use the Lebesgue bounded convergence theorem to argue that, since [imath]f:B(a,r)\to\Bbb C[/imath] thus on [imath]B(a,r)[/imath] we have [imath]|f(a+\epsilon e^{it})|\le M[/imath], where M is the maximum of [imath]|f(x)|[/imath] on [imath]B(a,r)[/imath]. Is that valid?
|
2072609
|
Find algebraic equation for epicycloid [imath]\theta \mapsto 4 e^{i\theta} - e^{4i\theta}[/imath]
I am reading about epicycloids and I wonder if they are algebraic. Here is in polar coordinates: \begin{eqnarray*} x(\theta) &=& (R+r) \cos \theta - r \cos \left( \frac{R+r}{r} \theta\right) \\ y(\theta) &=& (R+r) \sin \theta - r \sin \left( \frac{R+r}{r} \theta\right) \\ \end{eqnarray*} Then is this curve algebraic? If I add the real and imaginary parts we obtain: [imath] \theta \mapsto (R+r) e^{i\theta} - r e^{i\frac{R+r}{r} \theta} = (R+r) z - r z^{\frac{R+r}{r} } [/imath] Still is still not an equation of the type [imath]f(x,y) = 0[/imath]. The cardioid has an equation: [imath] (x^2 + y^2 +ax)^2 = a^2 (x^2 + y^2) [/imath]
|
1840563
|
Proof that Epicycloids are Algebraic Curves?
Epicycloids are most commonly described by the parametric equations, [imath]x(t) = (R + a)\cos(t) – a \cos \left(\frac{R + a}{a} t \right),[/imath] [imath]y(t) = (R + a)\sin(t) – a \sin \left(\frac{R + a}{a} t \right).[/imath] Where [imath]R[/imath] is the radius of the fixed circle and [imath]a[/imath] is the radius of the rolling circle. With [imath]R = ka[/imath] we also have, [imath]x(t) = a[(k + 1)\cos(t) - \cos((k + 1)t)],[/imath] [imath]y(t) = a[(k + 1)\sin(t) - \sin((k + 1)t)].[/imath] Several books discussing epicycloids mention that if the ratio of the radii of the circles [imath]\left( \frac{R}{a} = k \right)[/imath] is rational, then they are algebraic curves. However, I’ve only been able to find the Cartesian equations for the cardioid, nephroid and ranunculoid. With the cardioid being, [imath](ax + x^2 + y^2)^2 = a^2(x^2 + y^2).[/imath] The nephroid, [imath](-4a^2 + x^2 + y^2)^3 = 108a^4y^2.[/imath] And the ranunculoid, [imath]-52521875a^{12} – 1286250a^{10} (x^2 + y^2) – 32025a^8 (x^2 + y^2)^2 + 93312a^7 (x^5 – 10x^3y^2 + 5xy^4) – 812a^6 (x^2 + y^2)^3 – 21a^4 (x^2 + y^2)^4 – 42a^2(x^2 + y^2)^5 + (x^2 + y^2)^6 = 0.[/imath] Clearly this doesn’t cover all epicycloids where [imath]\frac{R}{a}[/imath] is rational. What is the proof that shows that epicycloids, where the ratio of the radii are rational, are algebraic curves?
|
2072746
|
Finding all [imath]\mathbb Z[/imath]-module homomorphisms [imath]\mathrm{Hom}(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)[/imath]
I am pretty sure that it has been already discussed many times, but I couldn't find. Well, the problem is to show that [imath]\mathrm{Hom}(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)\simeq\mathbb Z/(n,m)\mathbb Z[/imath]. I have proven that [imath]Hom(Z/nZ, Z/mZ)[/imath] is precisely [imath]Ann(nZ) = \{x \in Z/mZ: nx = 0\}. [/imath] Thus my problem reduces to proving [imath]Ann(nZ) = Z/(n,m)Z[/imath], but I haven't gotten success so far.
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381891
|
Computing [imath]\mathrm{Hom}(\mathbb Z_n,\mathbb Z_m)[/imath] as [imath]\mathbb Z[/imath]-module
My algebra is weak I need help computing [imath]\mathrm{Hom}(\mathbb Z_n,\mathbb Z)[/imath], [imath]\mathrm{Hom}(\mathbb Z_n,\mathbb Z_m)[/imath] and also [imath]\mathrm{Hom}(\mathbb Z,\mathbb Z)[/imath] as [imath]\mathbb Z[/imath]-modules. Also books suggestion to improve my basic. Thank you. Regards
|
562260
|
Rationalizing the denominator with 3 roots
Well, I can't find the example on how to solve this. If I multiply [imath] \dfrac{2}{\sqrt[3]{9}+\sqrt[3]{15}+\sqrt[3]{25}} [/imath] with [imath] \dfrac{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}}{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}} [/imath] or similar, it just gets even more complicated, and I get 4 terms instead of 3 in the denominator, and more can't be better... Can someone tell me some principle by which all kinds of expressions with 3 or more terms and with different roots could be rationalised? On a test, I haven't got much time to use appropriate formula, as I can for square difference and so on, I need a principle.
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1076199
|
Rationalizing a Denominator with Cube Roots
Rationalize [imath]\dfrac{1}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}.[/imath] How would I go about doing this without wading through lots of algebra? Is there a trick similar to how you can multiply by [imath]\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}[/imath] with square roots? Thanks in advance!
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2073295
|
Interection of conjugate elements forms a normal subgroup
Let [imath]G[/imath] be a group and [imath]H[/imath] be its subgroup. Then, is the set [imath]W=\cap_{g\in G} gHg^{-1}[/imath] a normal subgroup? I think the answer is yes, because the set is equal to the stabilizer of the action of [imath]H[/imath] on [imath]G[/imath] with action as conjugation, which is the normalizer of [imath]H[/imath]. Any ideas. Thanks beforehand.
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369629
|
Intersection of conjugate subgroups is normal
Is there a better (more direct or intuitive) proof for this proposition than I have come up with below? I am not sure whether it could be simiplified: Let [imath]G[/imath] be a group with [imath]H \leq G[/imath]. Then [imath]K = \bigcap_{g \in G} gHg^{-1}[/imath] is normal in [imath]G[/imath]. Let [imath]a \in K[/imath]. Then [imath]a \in gHg^{-1}[/imath] for all [imath]g \in G[/imath]. Therefore for all [imath]g_1,g \in G[/imath], [imath]g_1ag_1^{-1} \in g_1gHg^{-1}g_1^{-1} = (g_1g)H(g_1g)^{-1}[/imath] and so [imath]g_1ag_1^{-1} \in K[/imath] since [imath]g_1g \in G[/imath]. Then [imath]K[/imath] is normal in [imath]G[/imath].
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2072629
|
Can two continuous random variables not be jointly continuous?
In my book, lots of statements begin with "Let [imath]X[/imath] and [imath]Y[/imath] be jointly continuous random variables." and then some example or a derivation of a formula follows. But can two continuous random variables not be jointly continuous? What does it mean for two continuous random variables to not be jointly continuous? Does it have something to do with the sample spaces on which they are defined on?
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226648
|
Problem about jointly continuous and linearity of expectation.
When defining jointly continuous, we are saying: if there exists a function [imath]f(x,y)[/imath] satisfying that [imath] P\{X\in A, Y\in B\}=\int_B\int_a f(x,y)dxdy [/imath] then [imath]X,Y[/imath] are called jointly continuous. So not every pair of random variable [imath]X[/imath] and [imath]Y[/imath] are jointly continuous, because such [imath]f(x,y)[/imath] may not exists, right? Then in Ross's book Introduction to Probability Models, there is something I'm confused: He proved the formula [imath] E[X+Y]=E[X]+E[Y] [/imath] for jointly continuous random variables, but then use it for arbitrary pair of random variables. When [imath]X,Y[/imath] is jointly continuous, it is easy to prove, since [imath] E[g(X,Y)]=\int_{\mathbb{R}}\int_{\mathbb{R}}g(x,y)f(x,y)dxdy [/imath] Then let [imath]g(X,Y)=X+Y[/imath]. To be clear, my question is : (1) Isn't every pair of random variable [imath]X[/imath] and [imath]Y[/imath] are jointly continuous, right? (2) The formula [imath]E[X+Y]=E[X]+E[Y][/imath] is valid for any pair [imath]X, Y[/imath] even if they are not independent, right? (3) Ross in his book proved [imath]E[X+Y]=E[X]+E[Y][/imath] under the assumption [imath]X[/imath] and [imath]Y[/imath] is jointly continuous. It is quite easy. How to prove general case?
|
2073219
|
What are the parts of a logarithm called?
[imath]\log_x y = z[/imath] [imath]x[/imath] is the base. [imath]z[/imath] is the exponent or power. What's [imath]y[/imath] called?
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4214
|
What is the [imath]x[/imath] in [imath]\log_b x[/imath] called?
In [imath]b^a = x[/imath], [imath]b[/imath] is the base, a is the exponent and [imath]x[/imath] is the result of the operation. But in its logarithm counterpart, [imath]\log_{b}(x) = a[/imath], [imath]b[/imath] is still the base, and [imath]a[/imath] is now the result. What is [imath]x[/imath] called here? The exponent?
|
2073685
|
Raising differential to a power n in an integral
I was reading a paper and encountered an integral where the differential, [imath]dx[/imath], was raised to the power [imath]4[/imath]. What does this mean? Is this some kind of special integral? [imath]\int d^4x[/imath]
|
426263
|
Meaning of [imath]\int\mathop{}\!\mathrm{d}^4x[/imath]
What the following formula mean? [imath]\int\mathop{}\!\mathrm{d}^4x[/imath] I know that this [imath]\int f(x)\mathop{}\!\mathrm{d}x[/imath] is the integral of the function [imath]f[/imath] over the [imath]x[/imath] variable, but the following [imath]\int\mathop{}\!\mathrm{d}^4x[/imath] leave empty the argument and also have the [imath]\mathrm{d}[/imath] elevated to the fourth power. What that mean? Update. In the Einstein-Hilbert action we have (note that I have understand that the other parts of the integral are relevant only after your answers, thus sorry.): [imath]S=\frac{c^4}{16\pi G}\int\mathop{}\!\mathrm{d}^4x \, \ R \sqrt{-g}[/imath]
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2073663
|
How to proove the following?
For all non vanishing natural numbers prove: [imath]\sum_{k=1}^n\frac 1{\sqrt k} \leqslant 2 \sqrt n [/imath] I tried a proof by recurrence but I can't figure out the solution
|
768916
|
Prove that [imath]2\sqrt n \ge 1/\sqrt1+1/\sqrt2+\cdots+1/\sqrt n\ge2(\sqrt{n+1}-1) [/imath]
For each positive integer [imath]n[/imath], prove that [imath]2\sqrt n \geq 1/\sqrt1+1/\sqrt2+\cdots+1/\sqrt n\geq 2(\sqrt{n+1}-1) [/imath]
|
2073821
|
Proof that [imath]f(z) = cosec(\pi z)[/imath] has simple poles
In my book they say that [imath]f(z) = \csc(\pi z)=\frac{1}{\sin(\pi z)}[/imath] has simple poles, however doesn't explain why. I tried do to the following: I know it should have simple poles where [imath]g(z) = \sin(\pi z)[/imath] has zeroes, i.e. [imath]z = k\in\mathbb{Z}[/imath]. Then I can check if they are indeed simple poles with the following:[imath]\lim_{z\to k}(z-k)\csc(\pi z) = \lim_{z\to k}\frac{z-k}{\sin(\pi z)} = \lim_{z\to k}\frac{1}{\pi\cos(\pi z)}=\begin{cases} \frac{1}{\pi} & k \,\,\text{even} \\ -\frac{1}{\pi} & k \,\, \text{odd} \end{cases} \neq 0[/imath] hence a simple poles at [imath]z = k\in\mathbb{Z}[/imath]? I am not sure this is correct or if this is the shorter/best way to show this. Are there other options?
|
122825
|
Is there a fast technique to tell that the poles of [imath]\frac{1}{\sin z}[/imath] are simple?
Is there a systematic way to determine the order of a pole of a function? For instance, I want to calculate the residues of [imath]\frac{1}{\sin z}[/imath], for the poles at [imath]k\pi[/imath]. If the poles are simple, and apparently they are, it's easy to see the residues to be [imath](-1)^k[/imath]. However, how can one tell the poles are simple?
|
2073792
|
Proof of the Famous Relation
We all know that if we Highest Common Factor (or Greatest Common Divisor {As you prefer}) and the Least Common Multiple two numbers , then ALL WE KNOW is -> H.C.F [imath]/cdot[/imath] L.C.M [imath]=[/imath] Product of 2 numbers {Well,as pef my knowledge , this is only applicable for 2 numbers} Is there any kind of proof of this famous relation ??
|
349858
|
Easiest and most complex proof of [imath]\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.[/imath]
I'm looking for an understandable proof of this theorem, and also a complex one involving beautiful math techniques such as analytic number theory, or something else. I hope you can help me on that. Thank you very much
|
2073899
|
How can I prove this formula? [imath]\cdots=\sqrt{\frac{e\pi}2}[/imath]
I found in the book Escapades Arithmétiques written by Frédéric Laroche this formula: [imath]1+\frac 1{1\cdot 3}+\frac 1{1\cdot 3\cdot 5}+\cdots+\frac 1{1+\frac 1{1+\frac 2{1+\frac 3{1+\cdots}}}}=\sqrt{\frac{e\pi}2}.[/imath] Perhaps in a more explicit way, the first part of this formula is: [imath]\sum_{k=0}^\infty \left(\prod_{j=0}^k (2j+1)\right)^{-1}.[/imath] What I do not like is that this formula (beautiful in my opinion) is written without any proof nor reference. Do you have an idea on how to prove such a result? Do you know a book that gives the proof of this formula?
|
833920
|
An infinite series plus a continued fraction by Ramanujan
Here is a famous problem posed by Ramanujan Show that [imath]\left(1 + \frac{1}{1\cdot 3} + \frac{1}{1\cdot 3\cdot 5} + \cdots\right) + \left(\cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}\right) = \sqrt{\frac{\pi e}{2}}[/imath] The first series seems vaguely familiar if we consider the function [imath]f(x) = x + \frac{x^{3}}{1\cdot 3} + \frac{x^{5}}{1\cdot 3\cdot 5} + \cdots[/imath] and note that [imath]f'(x) = 1 + xf(x)[/imath] so that [imath]y = f(x)[/imath] satisfies the differential equation [imath]\frac{dy}{dx} - xy = 1, y(0) = 0[/imath] The integrating factor here comes to be [imath]e^{-x^{2}/2}[/imath] so that [imath]ye^{-x^{2}/2} = \int_{0}^{x}e^{-t^{2}/2}\,dt[/imath] and hence [imath]f(x) = e^{x^{2}/2}\int_{0}^{x}e^{-t^{2}/2}\,dt[/imath] Thus the sum of the first series is [imath]f(1) = \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt[/imath] But I have no idea about the continued fraction and still more I am not able to figure out how it would simplify to [imath]\sqrt{\pi e/2}[/imath] at the end. Please provide any hints or suggestions. Update: We have [imath]\begin{aligned}f(1) &= \sqrt{e}\int_{0}^{1}e^{-t^{2}/2}\,dt = \sqrt{e}\int_{0}^{\infty}e^{-t^{2}/2}\,dt - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\\ &= \sqrt{\frac{\pi e}{2}} - \sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt\end{aligned}[/imath] and hence we finally need to establish [imath]\sqrt{e}\int_{1}^{\infty}e^{-t^{2}/2}\,dt = \cfrac{1}{1+}\cfrac{1}{1+}\cfrac{2}{1+}\cfrac{3}{1+}\cfrac{4}{1+\cdots}[/imath] On further searching in Ramanujan's Collected Papers I found the following formula [imath]\int_{0}^{a}e^{-x^{2}}\,dx = \frac{\sqrt{\pi}}{2} - \cfrac{e^{-a^{2}}}{2a+}\cfrac{1}{a+}\cfrac{2}{2a+}\cfrac{3}{a+}\cfrac{4}{2a+\cdots}[/imath] and it seems helpful here. But unfortunately proving this formula seems to be another big challenge for me.
|
2074159
|
Prove that there exists a continuous function [imath]g[/imath] such that [imath]\int_a^b|f-g| \lt \epsilon[/imath]
Assume that [imath]f[/imath] is Riemann-integrable on the interval [imath][a,b][/imath] and [imath]\epsilon[/imath] is a positive number. Prove that there exists a continuous function [imath]g[/imath] such that [imath]\int_a^b|f-g| \lt \epsilon[/imath] . Note : The problem is the existence of the function. I don't know how to show it. Any hint or clue would be useful. An answer would be great. Thanks in advance.
|
43782
|
Approximation of Riemann integrable function with a continuous function
I have proved that if [imath]f \in R[a,b][/imath] and given [imath]\epsilon > 0[/imath] there exists a continuous function [imath]g[/imath] such that [imath]\int_a^b |f-g| < \epsilon[/imath] I was wondering if using this fact there is some way to show that there is also some continuous function [imath]h[/imath] such that [imath]\int_a^b |f-h|^2 < \epsilon[/imath] Any help will be appreciated, thanks :)
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1239046
|
Evaluate [imath]\int\sin(x^2)\mathrm{d}x[/imath]
Can you evaluate [imath]\int\sin(x^2)\mathrm{d}x \quad ?[/imath] I have tried substituting [imath]p=x^2[/imath] as well as integrating by parts, but then I came across an answer here which says that there is no way of working it out without using a computer.
|
613037
|
what is the[imath] \int \sin (x^2) \, dx[/imath]?
[imath]u[/imath] substitution doesnt work. I don't see any connection with the Weierstrass substitution either. integration by parts results in a infinite integral series.
|
2074797
|
Find area of triangle RTS when [imath]R=(-3,-4),[/imath] [imath]S=(3,-1),[/imath] [imath]T=(2,4)[/imath]
I can't from find this answer I tried whole day please help
|
516219
|
Finding out the area of a triangle if the coordinates of the three vertices are given
What is the simplest way to find out the area of a triangle if the coordinates of the three vertices are given in [imath]x[/imath]-[imath]y[/imath] plane? One approach is to find the length of each side from the coordinates given and then apply Heron's formula. Is this the best way possible? Is it possible to compare the area of triangles with their coordinates provided without actually calculating side lengths?
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2074965
|
Does the Fourier transform map [imath]L^1(\mathbb{R})[/imath] into itself?
Is the Fourier transform of an integrable function ( in [imath]L^1(-\infty, \infty)[/imath] ) also integrable? Put another way if: [imath]f \in L^1(\mathbb{R})[/imath] then is [imath] \hat{f}(t) = \int_{-\infty}^\infty e^{i xt} f(t) \, dt \stackrel{?}{\in} L^1(\mathbb{R})[/imath] It would seem to me perfectly reasonable the Fourier transform [imath]\mathcal{F}[/imath] takes [imath]L^1[/imath] to itself since it is an isometry on [imath]L^2[/imath].
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66173
|
Integrable function whose Fourier transform is not integrable
I am looking for an example of a function [imath]f: \mathbb R \rightarrow \mathbb R[/imath] such that [imath]f \in L^1[/imath] in the sense that [imath]\int_{\mathbb R} |f| < \infty[/imath] but its Fourier transform [imath]\hat f[/imath] is not in [imath]L^1[/imath]. Does anyone have one? Thanks.
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2075291
|
Is a product of derivatives still a derivative?
Given [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] and [imath]g : \mathbb{R} \rightarrow \mathbb{R}[/imath] differentiable functions, does [imath]f'g'[/imath] always have an antiderivative ? Origin of the question : [imath]\ [/imath] At first I considered two functions s.t. both [imath]f'[/imath] and [imath]g'[/imath] are locally bounded (thus locally Lebesgue integrable), and s.t. every point in [imath]\mathbb{R}[/imath] is a Lebesgue point for [imath]f'[/imath] and [imath]g'[/imath]. Then it is not hard to show that every [imath]x \in \mathbb{R}[/imath] is also a Lebesgue point for [imath]f'g'[/imath], and thus [imath]x \mapsto \displaystyle{\int_0^x} f'(s)g'(s)ds[/imath] is differentiable and its derivative is [imath]f'g'[/imath]. My problem is that in the general case, I do not manage to find a form for the antiderivative without, at some point, using integrals which are not always defined...
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1572255
|
Is the property "being a derivative" preserved under multiplication and composition?
Since differentiation is linear, we therefore have that if [imath]f, g: I\to \mathbb{R}[/imath] is a derivative (where [imath]I\subset \mathbb{R}[/imath] is an interval), then so does their linear combination. What if we consider their multiplication and composition? Due to the forms of the product rule of differentiation of product function and chain rule of differentiation of composition, I highly doubt their product or composition necessarily is still a derivative, but I cannot construct counterexamples.
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2073578
|
regular tempered distributions
All regular* distributions of [imath]\mathcal{D}'(\mathbb{R}^n)[/imath] can be characterised as [imath]L^1_{loc}(\mathbb{R}^n)[/imath]. Is there a similar characterisation of the regular distributions of [imath]\mathcal{S}'(\mathbb{R}^n)[/imath]? I often found, that all allgebraically bounded [imath]f \in L^1_{loc}(\mathbb{R}^n)[/imath], i.e. [imath]\exists C>0, N>0:\ |f(t)|\le C(1+|t|)^N[/imath], give regular tempered distributions, but for example this does not contain [imath]\chi_{[-1,1]} \frac{1}{\sqrt{|t|}}\in L^1(\mathbb{R}^n)[/imath], which should also be a regular tempered distribution? On the other hand, it should be "smaller" than [imath]L^1_{loc}(\mathbb{R}^n)[/imath] as [imath]e^t[/imath] won't give a regular tempered distribution. (*) A tempered distribution [imath]T\in \mathcal{S}'(\mathbb{R}^n)[/imath] is called regular in this context, if it can be represented by an (ordinary) function [imath]f\in L^1_{loc}(\mathbb{R}^n)[/imath], i.e. [imath]\exists f\in L^1_{loc}(\mathbb{R}^n):\quad T(\phi)=\int_{\mathbb{R}} f \phi\ dx \ \qquad\forall \phi \in \mathcal{S}(\mathbb{R}^n)[/imath]
|
34696
|
Which functions are tempered distributions?
Today's problem originates in this conversation with Willie Wong about the Fourier transform of a Gaussian function [imath]g_{\sigma}(x)=e^{-\sigma \lvert x \rvert^2},\quad x \in \mathbb{R}^n;[/imath] where [imath]\sigma[/imath] is a complex parameter. When [imath]\Re (\sigma) \ge 0[/imath], [imath]g_\sigma[/imath] is a tempered distribution[imath]^{[1]}[/imath] and so it is Fourier transformable. On the contrary, it appears obvious that if [imath]\Re(\sigma) <0[/imath] then [imath]g_\sigma[/imath] is not tempered. Question 1. What is the fastest way to prove this? My guess is that one should exploit the fact that the pairing [imath]\int_{\mathbb{R}^n} g_\sigma(x)\varphi(x)\, dx[/imath] makes no sense for some [imath]\varphi \in \mathcal{S}(\mathbb{R}^n)[/imath]. But is it enough? I am afraid that this argument is incomplete. Question 2. More generally, is there some characterization of tempered functions, that is, functions which belong to the space [imath]L^1_{\text{loc}}(\mathbb{R})\cap \mathcal{S}'(\mathbb{R})[/imath]? The only tempered functions that I know are polynomially growing functions. By this I mean the functions of the form [imath]Pu[/imath], where [imath]P[/imath] is a polynomial and [imath]u \in L^p(\mathbb{R}^n)[/imath] for some [imath]p\in[1, \infty][/imath]. Question 3. Is it true that all tempered functions are polynomially growing functions? [imath]^{[1]}[/imath] The definition of tempered distribution I refer to is the following. A distribution [imath]T \in \mathcal{D}'(\mathbb{R}^n)[/imath] is called tempered if for every sequence [imath]\varphi_n \in \mathcal{D}(\mathbb{R}^n)[/imath] such that [imath]\varphi_n \to 0[/imath] in the Schwartz class sense, it happens that [imath]\langle T, \varphi_n \rangle \to 0[/imath]. If this is the case then [imath]T[/imath] uniquely extends to a continuous linear functional on [imath]\mathcal{S}(\mathbb{R}^n)[/imath] and we write [imath]T \in \mathcal{S}'(\mathbb{R}^n)[/imath].
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2075217
|
How to find all analytic functions such that [imath]f(z)=-f''(z)[/imath]
I'm trying to find all analytic functions such that [imath]f(z)=-f''(z)[/imath] I keep on trying but the solution seems a bit elusive... I'm thinking of [imath]\{a\cdot e^{iz},a\cdot e^{-iz}|a\in \mathbb{C}\}[/imath] but cannot prove it... any advice? :)
|
194204
|
[imath]f'' + f =0[/imath]: finding [imath]f[/imath] using power series
How can one solve the following differential equation [imath]f'' + f =0[/imath] with the usage of power series? Writing: [imath]f(x) = \displaystyle\sum_{k=0}^{\infty} a_k x^k[/imath] [imath]f'(x) = \displaystyle\sum_{k=1}^{\infty} a_k k x^{k-1}[/imath] [imath]f'' (x) = \displaystyle\sum_{k=2}^{\infty} k (k-1) a_k x^{k-2}[/imath] How to continue?
|
2074768
|
How to express the condition in permutation formula.
I have a very basic and conceptual doubt regarding the function of the permutation formula. I am going to ask them through some examples in which I applied them and could not proceed. The first basic thing is that let us say I have four dice and I want two of my dice to show exactly [imath]3[/imath] and the rest can show any number except [imath]3[/imath]. Now my line of thought was that we fix two places where we fill in [imath]3[/imath] and the rest two places can be filled in [imath]5[/imath] ways each. the first die can show in [imath]5[/imath] ways, the second can also show [imath]5[/imath] ways, the third and fourth can show in [imath]1[/imath] ways each, so the total number of arrangement should be [imath]5 \cdot 5 \cdot1 \cdot 1[/imath]. My first doubt is - Does this thing need arrangement and rearrangement or is this automatically arranged? If not then how do I write it using the permutation formula so that the direct result which comes out is automatically arranged and I don't need to look for ways to arrange it.
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2074774
|
Solving permutation problem sequentially.
We have four dice and we want to know the number of ways in which at lease one dice shows 3. This is pretty simple if we judge it directly by saying that the number of ways in which all the dice can show any number is [imath]6^4[/imath] and the number of ways in which [imath]3[/imath] does not appear anywhere is [imath]5^4[/imath]. Subtracting the two gives us the right answer i.e [imath]679[/imath]. I am new to combinatorics and hence when I tried to solve it sequentially by writing out things like the number of ways through permutation formula in which we count the number of ways one by one, first assuming that we have only one [imath]3[/imath], then two [imath]3s[/imath] and so on, I get confused a lot. Can anybody help me with the procedure of solving this by writing permutation formula for each and every sequence (one three at a time, two threes at a time etc.) That would be of great help since I can find no other way through which I can remove my doubts.
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942263
|
Really advanced techniques of integration (definite or indefinite)
Okay, so everyone knows the usual methods of solving integrals, namely u-substitution, integration by parts, partial fractions, trig substitutions, and reduction formulas. But what else is there? Every time I search for "Advanced Techniques of Symbolic Integration" or "Super Advanced Integration Techniques", I get the same results which end up only talking about the methods mentioned above. Are there any super obscure and interesting techniques for solving integrals? As an example of something that might be obscure, the formula for "general integration by parts " for [imath]n[/imath] functions [imath]f_j, \ j = 1,\cdots,n[/imath] is given by [imath] \int{f_1'(x)\prod_{j=2}^n{f_j(x)}dx} = \prod_{i=1}^n{f_i(x)} - \sum_{i=2}^n{\int{f_i'(x)\prod_{\substack{j=1 \\ j \neq i}}^n{f_j(x)}dx}} [/imath] which is not necessarily useful nor difficult to derive, but is interesting nonetheless. So out of curiosity, are there any crazy unknown symbolic integration techniques?
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2409326
|
Favorite definite integral trick?
I'm compiling a list of interesting definite integrals for an upcoming blog post, and I thought that the math SE community might have a few interesting problems to offer. I am especially interested in integrals that use "tricks" that are hard to spot at first or are particularly elegant ways of solving seemingly difficult problems. An example of such an integral is the following: [imath]\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx[/imath] [imath]=\int_0^{\pi/2} 1-\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx[/imath] [imath]=\int_0^{\pi/2} 1-\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx[/imath] [imath]=\frac{\pi}{2}-\int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx[/imath] [imath]x\to \frac{\pi}{2}-x[/imath] [imath]=\frac{\pi}{2}-\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx[/imath] And so if one lets [imath]I[/imath] be the value of the integral, [imath]I=\frac{\pi}{2}-I[/imath] and [imath]I=\frac{\pi}{4}[/imath] This integral holds a very good example of what an "integral trick" is, because it uses the properties of integrals to evaluate a very difficult-looking definite integral without actually evaluating it. I look forward to seeing what everyone has to say!
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2075486
|
Why this inequality holds for any finite ring?
Let [imath]A[/imath] be a finite ring and [imath]N[/imath] the non-units of [imath]A[/imath]. If [imath]\vert N \vert \geq 2[/imath], why [imath]\vert N \vert \geq \sqrt{\vert A\vert }[/imath] ? I think that I must look for something to get a quotient that let's me the inequality, but I don't know which ideal must be the one I need. Can anyone help me? EDIT: The ring is commutative with unit.
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289794
|
Ring with finitely many zerodivisors
Show that a ring [imath] R [/imath] with exactly [imath] n [/imath] zero divisors (different from [imath]0[/imath]) has cardinality atmost [imath] (n+1)^2 [/imath]. I have shown that annihilator of any element among the zero divisor is a subset of the zero divisor which proves it is finite. now i think i have to show that it's coset space is finite.please help me.
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2076398
|
What is the remainder when [imath]45![/imath] is divided by [imath]47[/imath]?
What is the remainder when [imath]45![/imath] is divided by [imath]47[/imath] ? Is there any method to approach such questions ?
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1510718
|
Find the remainder when [imath]45![/imath] is divided by [imath]47[/imath]?
Find the remainder when [imath]45![/imath] is divided by [imath]47[/imath]? My approach I am using Wilson's theorem to solve the problem. I reduced the expression into ([imath]47[/imath]-[imath]1[/imath]-[imath]1[/imath])!/[imath]47![/imath]=[imath](47[/imath]-[imath]1[/imath])/[imath]47[/imath]!-[imath]1[/imath]/[imath]47![/imath]=-[imath]1[/imath]-[imath]1[/imath]=-[imath]2[/imath] Am I right in my approach.Please correct me if I am wrong? Please correct me how to approach towards the problem.
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2076386
|
How to study the convergence of [imath]\int^{\infty}_{0}\frac{2\cos(t)\sin(t)}{t}dt[/imath]?
By using Taylor series, I managed to see that the [imath]t \rightarrow 0[/imath], [imath]\frac{2\cos(t)\sin(t)}{t} = 1 -\frac{4t^2}{6}[/imath] whose integral converges. As for the case when [imath]t \rightarrow \infty[/imath], I don't know that to do. It equals to [imath]\int^{\infty}_{0}\frac{\sin(2t)}{t}dt[/imath]. What can I do?
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390809
|
Convergence of [imath]\int_0^\infty \sin(t)/t^\gamma \mathrm{d}t[/imath]
For what values of [imath]\gamma\geq 0[/imath] does the improper integral [imath]\int_0^\infty \frac{\sin(t)}{t^\gamma} \mathrm{d}t[/imath] converge? In order to avoid two "critical points" [imath]0[/imath] and [imath]+\infty[/imath] I've thought that it would be easier to test the convergence of the sum (is this coherent?): [imath] \int_0^1 \frac{\sin(t)}{t^\gamma}\mathrm{d}t + \int_1^\infty \frac{\sin(t)}{t^\gamma}\mathrm{d}t. [/imath] For the second integral, it converges if [imath]\gamma > 1[/imath] (comparision) and also converges if [imath]0 <\gamma \leq 1[/imath]. I'm stuck on proving the last part and the fact that the first integral converges for [imath]\gamma < 2[/imath]. Any help would be appreciated. Thanks in advance. PD: I've checked the answers for this question but I would not like to solve this integral using [imath](n\pi,(n+1)\pi)[/imath] intervals.
|
1996814
|
Taylor Series - How to evaluate and prove it?
Question: How would you prove the Taylor series for a function, and how would you find the exact value of it? I know that[imath]e^x=\sum_{k=0}^{\infty}\frac {x^k}{k!}\\\frac 1{1-x}=\sum_{k=0}^\infty x^k\\\sin x=\sum_{k=0}^\infty \frac {(-1)^k}{(2k+1)!}x^{2k+1}\\\vdots[/imath] My question is: How would you prove the taylor expansion for any polynomial [imath]f(x)[/imath], and how would you evaluate the convergence? For Example: With [imath]e^x[/imath], letting [imath]x[/imath] be a simple integer like [imath]2[/imath], we have\begin{align*}e^2=1+2+\frac 4{2!}+\frac 8{3!}+\ldots\tag1\end{align*} How would you solve for the exact value of [imath]e^2[/imath] when you have the infinite sequence? (Note that it does converge) And [imath]x[/imath] doesn't have to be an integer. It can also be a number such as [imath]\pi\sqrt{19}[/imath].\begin{align*}e^{\pi\sqrt{19}}=1+\pi\sqrt{19}+\frac {19\pi^2}{2!}+\frac {19\pi^3\sqrt{19}}{3!}+\frac {19^2\pi^4}{4!}+\ldots\tag2\end{align*} Which is intimating.
|
706282
|
How are the Taylor Series derived?
I know the Taylor Series are infinite sums that represent some functions like [imath]\sin(x)[/imath]. But it has always made me wonder how they were derived? How is something like [imath]\sin(x)=\sum\limits_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)!}\cdot(-1)^n = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}\pm\dots[/imath] derived, and how are they used? Thanks in advance for your answer.
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2077277
|
How do I prove that [imath]n^n > (n+1)^{n-1}[/imath]?
It seems simple enough, and by using brute force it's easy to see that [imath]n^n[/imath] will always be slightly larger for any [imath]n \ge 3[/imath]. I tried comparing ratios and also using induction, but nothing is conclusive. I would imagine that the epsilon proof would be the correct way to do this but I'm not very familiar with it and I would rather avoid such arduous proofs when I have to solve similar problems in a limited time frame.
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1146294
|
Prove that [imath](n+1)^{n-1}[/imath]
How would one prove that [imath](n+1)^{n-1}<n^n \ \forall n>1[/imath] I have tried several methods such as induction.
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1401868
|
Fibonacci proof question [imath]\sum_{i=1}^nF_i = F_{n+2} - 1[/imath]
The sequence of numbers [imath]F_n[/imath] for [imath]n \in N[/imath] defined below are called the Fibonnaci numbers. [imath]F_1 = F_2 = 1[/imath], and for [imath]n \geq 2[/imath], [imath]F_{n+1} = F_n + F_{n-1}[/imath]. Prove the following facts about the Fibonnaci numbers For each of the following, [imath]n \in N[/imath] a) [imath]\displaystyle \sum_{i=1}^nF_i = F_{n+2} - 1[/imath] I started this proof by using strong induction To prove something by strong induction, you have to prove that if all natural numbers strictly less than [imath]N[/imath] have the property, then [imath]N[/imath] has the property. [imath]n \geq 2[/imath], [imath]F_{n+1} = F_n + F_{n-1}[/imath] Check basis step [imath]n=2[/imath]: [imath]F_1 + F_2 = 1 + 1 = 2 = 3-1 = F_4 - 1[/imath], therefore TRUE I'm unsure how to go any further, when it comes to strong induction I'm lost on how to set up my IH, and proceed for the rest of the steps
|
680957
|
For the Fibonacci sequence prove that [imath]\sum_{i=1}^n F_i= F_{n+2} - 1[/imath]
For the Fibonacci sequence [imath]F_1=F_2=1[/imath], [imath]F_{n+2}=F_n+F_{n+1}[/imath], prove that [imath]\sum_{i=1}^n F_i= F_{n+2} - 1[/imath] for [imath]n\ge 1[/imath]. I don't quite know how to approach this problem. Can someone help and explain please?
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2077624
|
How to simplify an expression that does not have a common factor
I am trying to simplify this expression : [imath]9a^4 + 12a^2b^2 + 4b^4[/imath] So I ended up having this : [imath](3a^2)^2 + 2(3a^2)(2b^2) + (2b^2)^2[/imath] However, after that I don't know how to keep on simplifying the equation, it is explained that the answer is [imath](3a^2 + 2b^2)^2[/imath] because the expression is equivalent to [imath](x + y)^2[/imath] but I don't understand how they get to that ?
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1437215
|
How to factor[imath] x^2+2xy+y^2 [/imath]
I know the formula [imath]x^2 + 2xy + y^2 = (x + y)^2[/imath] by heart and in order to understand it better I am trying to solve the formula myself; however, I don't understand how step 4 is derived from step 3. [imath]x^2+2xy+y^2\qquad1[/imath] [imath]x^2+xy+xy+y^2\qquad2[/imath] [imath]x(x+y) + y(x+y) \qquad3[/imath] [imath](x+y)(x+y) \qquad 4[/imath] How does one get from step 3 to step 4? I don't know how to do that process.
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2077057
|
Calculate limit or prove it does not exist
[imath] \lim_{x\to \pi}\frac{\cos(\frac{x}{2})}{\pi^2-x^2} [/imath] I thought to change [imath]x[/imath] with [imath]t[/imath] like this: [imath] \lim_{t\to 0}\frac{\cos(\frac{t+\pi}{2})}{\pi^2-(t+\pi)^2} [/imath] Now I'm not sure how to continue.. Also I'm not allowed to use l'hopital rule or derivative at all. Thank you
|
1066330
|
Limits of trigonometric functions: [imath]\lim _{x\to \pi } \frac{\cos(x/2)}{\pi ^2-x^2}[/imath]
So I ran into this problem: [imath]\lim _{x\to \pi }(\frac{\cos\left(\frac{x}{2}\right)}{\pi ^2-x^2})[/imath] We haven't studied L'Hôpital's rule in my class yet, we basically need to transform the expression using trigonometric identities and use known limits like [imath]\lim_{x \to 0} \frac{sinx}{x} = 1[/imath] to calculate in case we get an indefinite form. I couldn't find anything I could do to calculate this limit :( Some help? Thanks.
|
2077288
|
solving the equation [imath]X^2-nY^2=d[/imath]
we know that equations of the form [imath]X^2-dY^2=Z^2[/imath] have parameter solutions in [imath]\mathbb{Z}[/imath] see here but what about equations of the form [imath]X^2-nY^2=d[/imath] where [imath]x,y,z,d \in \mathbb{Z}[/imath] ([imath]X[/imath] and [imath]Y[/imath] are variables) do they have solutions in general? what about special case below: [imath]X^2-17Y^2=14[/imath]
|
529480
|
Integers of the form [imath]x^2-ny^2[/imath]
Is there any algorithm to represent the given integers in the form [imath]N=x^2-ny^2[/imath], [imath]n>1[/imath], say for [imath]n=2[/imath]. As we have for any prime, [imath]p[/imath], a prime [imath]q[/imath] can be written as [imath]q=x^2+py^2[/imath],whenever [imath]q[/imath] is a quadratic residue of [imath]p[/imath] called cornacchia algorithm
|
2078040
|
How to prove that [imath](AB)^*=B^*A^*[/imath]
[imath]A,B\in M_n(F)[/imath] where [imath]F[/imath] is a field. Prove that [imath](AB)^*=B^*A^*[/imath]. I know it's correct if [imath]A,B[/imath] are both invertible. But how to prove the general case? I think it can be proved by Cauchy–Binet formula, but there should be better approaches. [imath]A^*[/imath] means the adjugate of [imath]A[/imath].
|
799284
|
[imath]\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A[/imath]
How can I prove that [imath]\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A[/imath], if [imath]A[/imath] and [imath]B[/imath] are any two [imath]n\times n[/imath]-matrices. Here, [imath]\operatorname{adj} A[/imath] means the adjugate of the matrix [imath]A[/imath]. I know how to prove it for non singular matrices, but I have no idea what to do in this case.
|
2078153
|
Limit of two sequences V(n) and U(n)
[imath]U(n)[/imath] and [imath]V(n)[/imath] are sequences. [imath]U(n)> 0[/imath] and [imath]V(n) >0[/imath] such that: [imath]{V(n+1)\over V(n)}≤ {U(n+1)\over U(n)}[/imath]. Prove that: If [imath]\,\,\lim\limits_{n\to\infty} V(n)= +\infty\,\,[/imath] then [imath]\,\,\lim\limits_{n\to\infty} U(n) = +\infty[/imath] If [imath]\,\,\lim\limits_{n\to\infty} U(n) = 0 \,\, [/imath] then [imath]\,\,\lim\limits_{n\to\infty} V(n) = 0[/imath] Can anyone help ?
|
2077589
|
Limit of sequences [imath]V(n)[/imath] and [imath]U(n)[/imath]
[imath]U(n)[/imath] and [imath]V(n)[/imath] are sequences. [imath]U(n)> 0[/imath] and [imath]V(n) >0[/imath] such that: [imath]{V(n+1)\over V(n)}≤ {U(n+1)\over U(n)}[/imath]. Prove that: If [imath]\,\,\lim\limits_{n\to\infty} V(n)= +\infty\,\,[/imath] then [imath]\,\,\lim\limits_{n\to\infty} U(n) = +\infty[/imath] If [imath]\,\,\lim\limits_{n\to\infty} U(n) = 0 \,\, [/imath] then [imath]\,\,\lim\limits_{n\to\infty} V(n) = 0[/imath] Can anyone help ?
|
2078129
|
Proof question of integers.
Let [imath]a[/imath] and [imath]b[/imath] be positive integers such that [imath]ab+1[/imath] divides [imath]a^2+b^2[/imath]. Show that [imath]\frac{a^2+b^2}{ab+1}[/imath] is the square of an integer.
|
1992951
|
Vieta Jumping: Related to IMO problem 6, 1988: If [imath]ab + 1[/imath] divides [imath]a^2 + b^2[/imath] then [imath]ab + 1[/imath] cannot be a perfect square.
The famous IMO problem 6 states that if [imath]a,b[/imath] are positive integers, such that [imath]ab + 1[/imath] divides [imath]a^2 + b^2[/imath], then [imath]\frac{ a^2 + b^2}{ab + 1 }[/imath] is a perfect square, namely, [imath]gcd(a,b)^2[/imath]. How about a modification of this problem: If [imath]a,b[/imath] are (strictly) positive integers, such that [imath]ab + 1[/imath] divides [imath]a^2 + b^2[/imath], then [imath]ab + 1[/imath] cannot be a perfect square. I am looking for a proof of the claim above, or a counter-example. One possible approach towards a proof is the following: Suppose there is such a pair [imath](a,b)[/imath] as above, then by the famous IMO problem 6 from 1988, [imath]\frac{ a^2 + b^2}{ab + 1 } = g^2[/imath] where [imath]g = gcd(a,b)[/imath]. Since [imath]ab + 1[/imath] is a perfect square, [imath]a^2 + b^2 = c^2[/imath] for some integer [imath]c[/imath]. So that [imath](a,b,c)[/imath] is a Pythagorean tripple, therefore there exists positive integers [imath]n,m,l[/imath] with [imath]n[/imath] coprime to [imath]m[/imath], such that [imath]a = l(n^2 - m^2)[/imath] and [imath]b = 2lnm[/imath] - by Euclids formula. Then by plugging in [imath]a[/imath] and [imath]b[/imath] in terms of [imath]n,m,l[/imath] into the original equation, and solving for [imath]l[/imath], it is possible to obtain the following: There exists positive coprime integers [imath]n,m[/imath] such that [imath]\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}[/imath] is a perfect square. If we put the quotient above into a program, as in this python code snipet: N = 1000 for n in range(1,N): for m in range(n+1, N): A = (n*n + m*m + 1)*(n*n + m*m - 1) B = 2*m*n*(n+m)*(m-n) if A % B == 0: print(A/B) The quotient always is 2, regardless of whether or not [imath]n[/imath] and [imath]m[/imath] are co prime! So if the very strong implication [imath]2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2[/imath] holds, then the original problem will be solved.
|
2005897
|
Every well-ordered set is isomorphic to a unique ordinal
I'm following a proof in Jech's book that every well ordered set is isomorphic to a unique ordinal and hitting a point where I'm not sure why a certain move is justified. He writes Proof. The uniqueness follows from Lemma 2.7. Given a well-ordered set [imath]W[/imath], we find an isomorphic ordinal as follows: Define [imath]F(x) = \alpha[/imath] if [imath]\alpha[/imath] is isomorphic to the initial segment of [imath]W[/imath] given by [imath]x[/imath]. If such an [imath]\alpha[/imath] exists, then it is unique. By the Replacement Axioms, [imath] F(W) [/imath] is a set. For each [imath]x \in W[/imath], such an [imath]\alpha[/imath] exists (otherwise consider the least [imath]x[/imath] for which such an [imath]\alpha[/imath] does not exist). If [imath]\gamma[/imath] is the least [imath]\gamma \not\in F(W)[/imath] then [imath]F(W) =\gamma[/imath] and we have an isomorphism of [imath]W[/imath] onto [imath]\gamma[/imath]. I've filled in all the details of this proof for myself up to the point where he assumes the class of ordinals not in the image of the function has a least element. Since this isn't a set, I'm not sure what can justify this. I could try to prove that every nonempty class of ordinals has a least element but that seems so substantial that it's hard to imagine it was a detail intentionally omitted. Am I missing something simple?
|
934852
|
There is a well ordering of the class of all finite sequences of ordinals
I am trying to solve this exercise from Jech's book on set theory: Ex. 3.6: There is a well ordering of the class of all finite sequences of ordinals such that for each [imath]\alpha[/imath], the set of all finite sequences in [imath]\omega_{\alpha}[/imath] is an initial segment and its order-type is [imath]\omega_{\alpha}[/imath]. Definition: If [imath]W[/imath] is a well ordered set and [imath]u \in W[/imath], then the set [imath]\{ x \in W \mid x < u \}[/imath] is an initial segment in W Any help?
|
2078144
|
Prove that [imath]\frac{1}{k+1} \le \ln(1+\frac{1}{k}) \le \frac{1}{k}[/imath]
I am trying to prove that [imath]\frac{1}{k+1} \le\ln(1+\frac{1}{k}) \le \frac{1}{k}[/imath]. This showed up as a known fact in a proof for [imath](1+\frac{1}{k})^k = e[/imath] as [imath]k[/imath] goes to infinity, so I would like to prove it without using that fact. I tried to prove the first part of the inequality by using another known inequality (one which I am comfortable proving): [imath]1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k} >\ln(1+k)[/imath] Let [imath]a[/imath] denote [imath]1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-2}[/imath] [imath]a+\frac{1}{k-1}+\frac{1}{k} >\ln(1+k)[/imath] Scaling this down 1 term: [imath]a + \frac{1}{k-1} >\ln(k)[/imath] Subtracting the second inequality from the first, we get: [imath]\frac{1}{k} >\ln(1+\frac{1}{k})[/imath] However, this seems to contradict the statement I'm trying to prove. Could anyone help?
|
1887583
|
Show that, for all [imath]n > 1: \frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.[/imath]
I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice: Show that, for all [imath]n > 1[/imath] [imath]\frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.[/imath] I tried to follow the below steps in order to prove the RHS inequality: Proving that [imath]f < g[/imath] on [imath]I[/imath] from [imath]a[/imath] to [imath]b[/imath]: Step [imath]1[/imath]. Prove that [imath]f' < g'[/imath] on [imath]\operatorname{Int}(I)[/imath]. Step [imath]2[/imath]. Show that [imath]f(a) \leq g(a)[/imath] or that [imath]f(a^+) \leq g(a^+)[/imath] Following the above steps, let [imath]f(x) = \log(1 + \frac1x)[/imath] and [imath]g(x) = \frac1x[/imath], for all [imath]x > 1[/imath]. One has [imath]f'(x) = -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.[/imath] We note that, for every [imath]x > 1[/imath], [imath]f'(x) > g'(x)[/imath]. Moreover, [imath]f(1^+) = \log(2) < 1 = g(1^+).[/imath] My problem is that I got the wrong inequality sign in Step [imath]1[/imath]. Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case.
|
2078552
|
Prove [imath]0 and \prod\limits_{k=1}^n a_k =1, then \prod\limits_{k=1}^n (1+a_k) \ge 2^n[/imath]
Prove:[imath]0<a_k\in \mathbb R\quad and\quad\prod_{k=1}^n a_k =1,\quad then\quad \prod_{k=1}^n (1+a_k) \ge 2^n[/imath] (*) I guess that the minimum of [imath]\prod\limits_{k=1}^n (1+a_k)[/imath] happens when all [imath]a_k[/imath]'s are [imath]1[/imath], but can't prove it. Thanks.
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547505
|
What is the minimum value of [imath](1 + a_1)(1 + a_2). . .(1 + a_n)[/imath]?
Suppose [imath]a_1, a_2,\dots , a_n[/imath] are [imath]n[/imath] positive real numbers with [imath]a_1a_2 \dots a_n = 1[/imath]. Then what is the minimum value of [imath](1 + a_1)(1 + a_2). . .(1 + a_n)[/imath] ? I think [imath](1 + a_1)(1 + a_2). . .(1 + a_n)[/imath] takes its minimum value when [imath]a_1=a_2=\dots=a_n=1[/imath] and thus the minimum value is [imath]2^n[/imath]. I don't know how to prove it. Please help.
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2078122
|
Let [imath]P(x)=4x^2+6x+4[/imath] and [imath]Q(x) =4y^2-12y+25[/imath]. If [imath]x,y[/imath] satisfy the equation [imath]P(x)Q(y)=28[/imath], then find the value of [imath]11y-26x[/imath].
Let [imath]P(x)=4x^2+6x+4[/imath] and [imath]Q(y) =4y^2-12y+25[/imath]. If [imath]x,y[/imath] satisfy the equation [imath]P(x)Q(y)=28[/imath], then find the value of [imath]11y-26x[/imath]. I know this question has been asked earlier here but the answer given states "Note that we have...". I want to know how we can arrive at that statement mathematically instead of trial and error method, because I wasn't able to get that while trying to solve it. Any other way of solving this question would also be helpful.
|
1595275
|
Find the unique pair of real numbers [imath](x,y)[/imath] that satisfy [imath]P(x)Q(y)=28[/imath]
Let [imath]P(x)=4x^2+6x+4[/imath] and [imath]Q(y)=4y^2-12y+25[/imath]. Find the unique pair of real numbers [imath](x,y)[/imath] that satisfy [imath]P(x)Q(y)=28[/imath] I can solve this question graphically. [imath]P(x)Q(y)=28\implies(4x^2+6x+4)(4y^2-12y+25)=28[/imath] [imath]4x^2+6x+4=\frac{28}{4y^2-12y+25}[/imath] I drew the graph of [imath]4x^2+6x+4[/imath] and found it is a upward parabola with minimum value at [imath](-\frac{3}{4},\frac{7}{4})[/imath] and I drew the graph of [imath]\frac{28}{4x^2-12x+25}[/imath]. I found that it is a bell shaped curve whose maximum value occurs at [imath](\frac{1}{4},\frac{7}{4})[/imath]. So I found that the common abscissa is [imath]y=\frac{7}{4}[/imath] which occurs at [imath]x=-\frac{3}{4}[/imath] for the first curve and at [imath]x=\frac{1}{4}[/imath] for the second curve, so solution should be [imath](-\tfrac{3}{4},\tfrac{1}{4})[/imath] But I do not know how to solve it algebraically.
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2078301
|
Combinatorics-set of divisors
It is given to me that r, s and t are prime numbers and that p, q are two positive integers such that the LCM of p and q is [imath]N=r^2t^4s^2 [/imath] and we want to find the numer of ordered pairs (p, q). Here since LCM of (p, q) is N then both the numbers must be divisors of N. If we calculate the total number of divisors of N we get 45 divisors which means that our (p, q) belongs to a set of these 45 divisors. Hence number of ways of choosing the two numbers are C(45,2) and then we arrange them in 2! ways thus getting [imath] \binom{45 } { 2 } \cdot 2![/imath] But my answer deviates completely from the actualy answet which truly implies that maybe my whole method is wrong. Can anybody help me in this problem?
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467569
|
If [imath]r,s,t[/imath] are prime numbers and [imath]p,q[/imath] positive integers s.t. [imath]\text{lcm}(p,q)[/imath] is [imath]r^2s^4t^2[/imath], then the number of ordered pairs [imath](p,q)[/imath] is?
If [imath]r,s,t[/imath] are prime numbers and [imath]p,q[/imath] are the positive integers such that LCM of [imath]p,q[/imath] is [imath]r^2s^4t^2[/imath], then the number of ordered pairs [imath](p,q)[/imath] is? My attempt to the solution: Let [imath]r^a s^b t^c[/imath] be the LCM where [imath]a=2,b=4,c=2[/imath] Then the cases that arise are I case 1) [imath]p[/imath] has [imath]r^as^bt^c[/imath] then there are a total of [imath](2+1)(4+1)(2+1)[/imath] options II case 1) [imath]p[/imath] has [imath]r^as^b[/imath] or [imath]s^bt^c[/imath],[imath]r^at^c[/imath] then there are total of [imath](2+1)(4+1)(2+1)(3)[/imath] options. The answers comes out to be [imath]45*4[/imath] but the answer is [imath]45*5[/imath]. Which case am I missing?
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2078853
|
Why does this process generate the factorial of the exponent?
Consider the process of taking a series of numbers and constructing a new series consisting of the difference between consecutive terms, and repeating this until a constant is reached: [imath]2,8,18,32,50\\6,10,14,18\\4,4,4[/imath] When this process is applied to sequences of the form [imath]f(n) = n^a[/imath], the constant reached seems to always be [imath]a![/imath]: [imath]1,2,3\\1,1[/imath] [imath]1,4,9,16\\3,5,7\\2,2\\[/imath] [imath]1,8,27,64,125\\7,19,37,61\\12,18,24\\6,6[/imath] [imath]1,16,81,256,625,1296\\15,65,175,369,671\\50,110,194,302\\60,84,108\\24,24[/imath] Can it be proven?
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1441049
|
Property of [imath]\{0^n, 1^n, \ldots\}[/imath]
The differences between adjacent squares are the odd numbers: 0 1 4 9 16 25 … 1 3 5 7 9 … The differences between adjacent odd numbers are 2 = 2! I found that this truth is more general: If the differences between adjacent powers of [imath]n[/imath] are written out, and the differences of those differences etc, the [imath](n+1)[/imath]th sequence is {n!, n!, …} Is this a well-known theorem? Does it have an easy proof?
|
2069264
|
If [imath]\int_{0}^{2} p(x) dx = p(\alpha) + p(\beta)[/imath] for all polynomials of degree at most [imath]3[/imath], what's [imath]3(\alpha - \beta)^2[/imath]?
If for some [imath]\alpha , \beta \in \mathbb{R}[/imath] , the integration formula [imath]\int_{0}^{2} p(x) dx = p(\alpha) + p(\beta)[/imath] , holds for all polynomials [imath]p(x)[/imath] of degree atmost [imath]3[/imath], then the valueof [imath]3(\alpha - \beta)^2 = ?[/imath]. How to approach this question? i thought of taking Polynomial as [imath]x^3[/imath] and i am getting the answer but is there any formal way of doing it without explicitly substituting any polynomial of degree atmost [imath]3[/imath] ?. Any suggestion , hints , solution ?
|
1129017
|
Finding the value of [imath]3(\alpha-\beta)^2[/imath] if [imath]\int_0^2 f(x)dx=f(\alpha) +f(\beta)[/imath] for all [imath]f[/imath]
Let [imath]f[/imath] be a polynomial of degree [imath]n[/imath] at most [imath]3[/imath] such that there exists some [imath]\alpha,\beta[/imath] satisfying [imath]\int_0^2 f(x)dx=f(\alpha) +f(\beta)[/imath] for all such [imath]f[/imath]. Find the value of [imath]3(\alpha-\beta)^2[/imath] Attempt: We need to find the value of [imath]3(\alpha-\beta)^2[/imath]. Now, this expression reminds of the function [imath]f(x) = (x-\beta)^3[/imath]. Differentiating this function at [imath]x=\alpha[/imath] will give us the desired result. Hence, we just need to know the derivative of [imath]k(x)=(x-\beta)^3[/imath] at every point. However, I am unable to move forward with this intuition. Could anyone please help me to move ahead in this problem. Thank you very much for your help in this regard.
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2079057
|
Discuss the continuity and differentiability of [imath]e^x[/imath]?
Discuss the continuity and differentiability of [imath]e^x[/imath] ? I know its continuous at all points, but how to check for differentiability ?
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803155
|
How to show that [imath]e^x[/imath] is differentiable?
I tried to search for a few minutes but I didn't find this question so I hope it's not a duplicate. So I want to show that [imath](e^x)' = e^x[/imath]. To do that, I must proof that the limit: [imath]\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to0}\frac{e^{x+h} - e^x}{h}[/imath] exists and equals to [imath]e^x[/imath]. So I have: [imath]\frac{e^{x+h} - e^x}{h} = \frac{e^x \cdot e^h - e^x}{h} = e^x \bigg(\frac{e^h - 1}{h}\bigg) \\ e^h - 1 = z \implies e^h = 1+z >0 \implies h = \ln(1+z)[/imath] Because of the continuity of the [imath]\ln[/imath] and [imath]e^x[/imath] functions, we have: [imath]h\to 0 \iff e^h \to 1 \implies z\to0 \\ z > 0 \implies \frac{1}{z} \to +\infty \\ z < 0 \implies \frac{1}{z} \to -\infty \\ \frac{e^h - 1}{h} = \frac{z}{\ln(1+z)} = \frac{1}{\frac{1}{z}\ln(1+z)} = \bigg[ y = \frac{1}{z} \bigg] = \frac{1}{y\ln(1+\frac{1}{y})} = \frac{1}{\ln(1+\frac{1}{y})^y} \\ h \to 0 \implies |y| \to +\infty \implies \bigg(1+\frac{1}{y}\bigg)^y \to e \implies \\ \lim_{h\to0}\frac{e^h - 1}{h} = \frac{1}{\ln e} = 1 \implies (e^x)' = e^x \cdot 1 = e^x[/imath] But how do I prove that [imath]\lim_{\pm\infty}(1 + \frac{1}{x})^x = e[/imath] ? We defined [imath]e[/imath] like this: [imath]\lim_{n\to\infty} \bigg(1 + \frac{1}{n}\bigg)^n = e \space \,, \space n\in\mathbb{N}[/imath] I thought of the sandwich theorem: [imath]\forall x \geq 1 \,, x\in\mathbb{R} \,, \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1[/imath] and [imath]\lfloor x \rfloor , \lfloor x \rfloor + 1 \in \mathbb{N} [/imath] and if I denote [imath]\mathbb{N} \ni n = \lfloor x \rfloor[/imath] I have: [imath] \bigg(1 + \frac{1}{n+1}\bigg)^n \leq \bigg(1 + \frac{1}{x}\bigg)^x \leq \bigg(1 + \frac{1}{n}\bigg)^{n+1} \\ \lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)^n = \frac{\lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)^{n+1}}{\lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)} = \frac{e}{1+0} = e \\ \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg)^{n+1} = \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg)^n \cdot \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg) = e \cdot (1 + 0) = e[/imath] and now I could apply the theorem but then again, how do I show that this: [imath]\bigg(1 + \frac{1}{n+1}\bigg)^n \leq \bigg(1 + \frac{1}{x}\bigg)^x \leq \bigg(1 + \frac{1}{n}\bigg)^{n+1}[/imath] is true? I mean, I can say that (left side): [imath] n+1 > x \implies \frac{1}{n+1} < \frac{1}{x} \implies 1 + \frac{1}{n+1} < 1 + \frac{1}{x} \\ \forall y\in\mathbb{R_+} \space n \leq x \implies y^n \leq y^x \\ \implies \bigg(1 + \frac{1}{n+1}\bigg)^n \leq \bigg(1 + \frac{1}{x}\bigg)^x[/imath] and by analogy the right side but is this a proof in terms of limits? and that's just [imath]+\infty[/imath] for [imath]x \geq 1[/imath], what do I do for [imath]-\infty[/imath] ? For [imath]x\in\langle 0, 1\rangle[/imath], a substitution [imath]y = \frac{1}{x}[/imath] gives me: [imath]\lim_{y\to 0}\bigg(1 + y\bigg)^\frac{1}{y}[/imath] I'll stop here to see if I'm on the right track, any hints, suggestions, edits, comments and answers are welcome!
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2079400
|
Is there a polynomial whose Coefficients in [imath]\mathbb Q[/imath] such that none of roots of this polynomial be in the root closure of [imath]\mathbb Q[/imath]?
For example, we know that [imath]x^5-x+1[/imath] is a polynomial of degree five over [imath]\mathbb Q[/imath] that is not solvable by radicals. Which means that one of the roots of this polynomial is not in root closure of [imath]\mathbb Q[/imath]. Is there a polynomial whose coefficients in [imath]\mathbb Q[/imath] such that none of the roots of this polynomial be in the root closure of [imath]\mathbb Q[/imath]? In the other hand, can we find a polynomial over [imath]\mathbb Q[/imath] such that none of the roots of this polynomial Obtained by the operations multiplying, suming, Subtracting, Divising and rooting from elements of [imath]\mathbb Q[/imath]
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2077766
|
Are there irreducible polynomials that are partially solvable by radicals?
In general, polynomials of degree 5 or higher, with rational coefficients, are not solvable by radicals. There are some exceptions and trivial cases, for example [imath]x^5-5x+12[/imath] and [imath]x^{100}+x^{25}+3[/imath], respectively. All irreducible polynomials with rational coefficients, that I am aware of, have either none or all roots that can be expressed by radicals. Do partially-solvable irreducible polynomials exist? If they do, are there any examples known? I mean, for example, an irreducible polynomial of [imath]n[/imath]th degree that can be split over [imath]Q(\sqrt{2})[/imath] into two polynomials of 3rd and [imath](n-3)[/imath]th degrees, where the polynomial of [imath](n-3)[/imath]th is not solvable by radicals. At https://math.stackexchange.com/q/662972 Alexander Gruber wrote: Polynomials which are not solvable by radicals have (at least one) root that cannot be written by any combination of the operations of addition, multiplication, and the taking of nth roots. I am not sure if he meant only reducible polynomials, and/or with non-rational coefficients, as polynomials that have "at least one" root that cannot be expressed by radicals.
|
2079664
|
What will be the remainder when [imath]24^{1202}[/imath] is divided by [imath]1446[/imath]?
What is the remainder when [imath]24^{1202}[/imath] is divided by [imath]1446[/imath]? My Try : [imath]\gcd(24,1446) = 6[/imath] So, By using Euler's theorem, I can write [imath]24^{240} = 1 \pmod {241}[/imath] Hence, [imath]24^{1202} = 24^2 \pmod {241}[/imath] How to solve this further to get [imath]1446[/imath] ?
|
808747
|
What is the remainder when [imath]24^{1202}[/imath] is divided by [imath]1446[/imath]?
I tried remainder theorem but that does not simplify it. I tried factorizing [imath]1446[/imath] as [imath]2\cdot3\cdot241[/imath] and got remainders when numerator is divided by [imath]2,3[/imath] and [imath]241[/imath] individually but then I did not know what to do next? Stuck!
|
2079414
|
Showing this set is an algebra
Suppose [imath]\Omega[/imath] [imath]=[/imath] [imath]\Bbb N[/imath]. Verify that the following set of subsets of [imath]\Omega[/imath] is an algebra: [imath]\mathcal A[/imath] [imath]=[/imath] [imath]\{[/imath] [imath]A[/imath] [imath]\subset[/imath] [imath]\Bbb N[/imath] [imath]:[/imath] [imath]A[/imath] is countable or [imath]A[/imath][imath]^c[/imath] is countable [imath]\}[/imath]
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1667469
|
Showing that a collection of sets is a [imath]\sigma[/imath]-algebra: either set or complement is countable
I wish to show that [imath]\mathcal{F} = \{A \subset \Omega: A \text{ countable or }A^c\text{ countable}\}[/imath] is a [imath]\sigma[/imath]-algebra. We have [imath]\Omega^c = \emptyset[/imath], so obviously [imath]\Omega \in \mathcal{F}[/imath]. Suppose [imath]A \in \mathcal{F}[/imath]. Assuming [imath]A[/imath] is countable, [imath](A^c)^c = A[/imath] is countable, so [imath]A^c \in \mathcal{F}[/imath]. If we assume [imath]A^c[/imath] is countable, [imath]A^c \in \mathcal{F}[/imath] follows trivially. Now let [imath]A, B \in \mathcal{F}[/imath]. I wish to show that either [imath]A \cap B[/imath] or [imath](A \cap B)^c = A^c \cup B^c[/imath] is countable. How would I start this part? I am only looking for hints.
|
2079801
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Evaluating the integral [imath]\int_0^\infty\frac{x^3}{e^{ax}-1}dx[/imath]
When describing black body radiation in Physics, one comes across an integral of the form [imath]I(a)=\int_0^\infty\frac{x^3}{e^{ax}-1}dx \tag1[/imath] for [imath]a>0[/imath]. In our lecture, we were given that this integral evaluates to [imath]I(a)=\frac{\pi^4}{15a^4}[/imath] without a derivation of this result. While reviewing my course notes over the Christmas break, I came across this integral and have tried to evaluate it myself, but I have not been successful so far. My first step, of course, was to substitute [imath]u:=ax[/imath], which gives us [imath]I(a) = \frac{1}{a^4}\int_0^\infty\frac{u^3}{e^u-1}du \tag2[/imath] but beyond that, I'm stuck. I've tried differentiation under the integral sign on the original integral in [imath](1)[/imath], which lead to the integral [imath]\int_0^\infty\frac{\log^4 u}{(u-1)^2}du[/imath] which I also no have no clue how to tackle. I've also thought about using contour integration, but I'm unsure as to what would be an appropriate contour to consider. I know that the integrand has singularities at [imath]z=2\pi ik[/imath] for [imath]k\in\mathbb{Z},k\neq 0[/imath] but I'm not even sure how one would go about calculating the residues at these singularities due to the peculiar form of the denominator, which is unlike any I have previously encountered. How would one go about deriving that [imath]\int_0^\infty\frac{x^3}{e^{ax}-1}dx = \frac{\pi^4}{15a^4}[/imath] is true?
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92793
|
Calculating [imath] \int _{0} ^{\infty} \frac{x^{3}}{e^{x}-1}\;dx[/imath]
how to calculate [imath]\int_0^\infty \frac{x^{3}}{e^{x}-1} \; dx[/imath] Be [imath]q:= e^{z}-1 , p:= z^{3}[/imath] , then [imath]e^{z} = 1 [/imath] if [imath]z= 2\pi n i [/imath], so the residue at 0 is : [imath]\frac{p(z_{0})}{q'(z_{0})} = 2\pi i n ^{3}[/imath] problem is that this is not symmetric, so how does one find the definite integral?
|
2079942
|
Why does [imath]2+2=5[/imath] in this example?
I stumbled across the following computation proving [imath]2+2=5[/imath] Clearly it doesn't, but where is the mistake? I expect that it's a simple one, but I'm even simpler and don't really understand the application of the binomial form to this...
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2078064
|
What is going wrong in this "proof" of [imath]0=1[/imath]?
[imath]\begin{align} -20 &= -20\\ 16-36 &= 25-45\\ 4^2-4\times 9&=5^2-5\times 9\\ 4^2-4\times 9+81/4&=5^2-5\times 9+81/4\\ 4^2-4\times 9+(9/2)^2&=5^2-5\times 9+(9/2)^2\\ \end{align}[/imath] Considering the formula [imath]a^2+2ab+b^2=(a-b)^2[/imath], one has [imath]\begin{align} (4-9/2)^2&=(5-9/2)^2\\ \sqrt{(4-9/2)^2}&=\sqrt{(5-9/2)^2}\\ 4-9/2&=5-9/2\\ 4&=5\\ 4-4&=5-4\\ 0&=1 \end{align}[/imath]
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2080584
|
Question about sum of primes
Let [imath]P(x)[/imath] be the xth prime, [imath]P(0)=2[/imath] Given a prime [imath]p[/imath], does it always exists a [imath]n[/imath] such [imath]\sum_{x=0}^{n}P(x) = 0[/imath] [imath](mod[/imath] [imath]p)[/imath] ? Example : [imath]p=7[/imath] [imath]2+3+5+7+11=28=4*7[/imath] Best regards
|
1882613
|
Every prime number divide some sum of the first [imath]k[/imath] primes.
Let [imath]S_n=\Sigma^n_{k=1}p_k[/imath], where [imath]p_k[/imath] is the [imath]k[/imath]-th prime number. Conjecture: [imath]\forall p\in\mathbb P\exists n\in\mathbb N: p|S_n[/imath] Verified for the [imath]1000[/imath] first primes. Is there a proof for this result in general? In the diagram primes are projected on the x-axis and [imath]n[/imath] (as in [imath]S_n[/imath]) on the y-axis. As H.H.Rugh commented there is a stronger conjecture for all positive integers m. Below a table of [imath]n[/imath]-records for different [imath]m\in\mathbb Z^+[/imath] (some of them primes): m n factorization of Sn 1 (1) 1 (2) 3 (3) 10 (3,43) 6 (2,3) 57 (2,3,5,229) 12 (2,2,3) 97 (2,2,3,1879) 18 (2,3,3) 113 (2,3,3,41,43) 35 (5,7) 180 (5,7,2531) 42 (2,3,7) 305 (2,2,2,3,7,7,239) 90 (2,3,3,5) 357 (2,2,2,3,3,3,5,367) 101 (101) 422 (5,101,1129) 137 (137) 861 (2,137,9739) 163 (163) 902 (5,7,11,47,163) 195 (3,5,13) 907 (2,3,3,5,13,2551) 202 (2,101) 1207 (2,2,19,101,719) 222 (2,3,37) 1359 (2,2,2,3,3,37,2671) 252 (2,2,3,3,7) 1683 (2,2,3,3,7,17,37,71) 326 (2,163) 1765 (2,2,3,23,163,277) 474 (2,3,79) 2077 (2,2,2,3,5,79,1861) 504 (2,2,2,3,3,7) 2133 (2,2,2,3,3,3,7,53,233) 522 (2,3,3,29) 2379 (2,3,3,3,3,3,7,29,239) 643 (643) 2529 (2,3,3,11,211,643) 647 (647) 3092 (11,647,5791) 658 (2,7,47) 3353 (2,3,7,43,47,577) 700 (2,2,5,5,7) 3593 (2,2,5,5,7,103,787) 817 (19,43) 4683 (2,3,11,19,43,1847) 995 (5,199) 5329 (2,3,5,17,199,1291) 1004 (2,2,251) 6415 (2,2,2,251,96643) 1204 (2,2,7,43) 6533 (2,2,2,2,7,7,31,43,193) 1459 (1459) 7241 (2,2,3,3,5,5,191,1459) 1488 (2,2,2,2,3,31) 7307 (2,2,2,2,2,3,31,85909) 1610 (2,5,7,23) 8079 (2,5,7,23,43,4567) 1677 (3,13,43) 10171 (2,2,2,3,3,3,13,43,4259) 1870 (2,5,11,17) 10331 (2,5,11,17,71,4003) 2035 (5,11,37) 11459 (2,2,3,3,5,11,37,9029) 2616 (2,2,2,3,109) 11753 (2,2,2,3,3,3,3,3,37,89,109) 2672 (2,2,2,2,167) 18137 (2,2,2,2,7,167,93047) 3420 (2,2,3,3,5,19) 21709 (2,2,3,3,3,5,13,19,137,139) 3830 (2,5,383) 27617 (2,5,53,383,20749) 4232 (2,2,2,23,23) 38861 (2,2,2,2,3,3,23,23,113189) 7394 (2,3697) 45381 (2,107,3697,15083) 7450 (2,5,5,149) 47323 (2,3,3,5,5,7,41,149,677)
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2080698
|
Square root multiplication
[imath]\sqrt {-3}[/imath] multiplied by [imath]\sqrt{-3}[/imath] is [imath]-3[/imath]. But this can also be written as [imath]\sqrt {-3} \cdot \sqrt{-3} = \sqrt {(-3).(-3)}= \sqrt{9} =3[/imath] So my question is why is this not possible?
|
534940
|
Why [imath]\sqrt {-1}\cdot \sqrt{-1}=-1[/imath] rather than [imath]\sqrt {-1}\cdot \sqrt{-1}=1[/imath]. Pre-definition reason!
It is for years that I teach complex numbers following a historical route. I start with the famous problem of Cardano: Find two numbers whose sum is equal to 10 and whose product is equal to [imath]40[/imath]. After "solving" we try to check the "things" we have found actually satisfies the conditions of the problem. To do so, we need to multiply [imath]\sqrt {-15}[/imath] by [imath]\sqrt {-15}[/imath]. "Relying on" and at the same time "extending" the rule of real numbers, we have two choices: take the product to be [imath]-15[/imath] or [imath]15[/imath]. We choose the first choice since it gives the solutions of the original problem. I wonder whether there is a better pre-definition explanation of this choice. Update Please consider that you all know the right answer! Go back in time for a moment and remember that the Great Euler took [imath]\sqrt {-1}[/imath] . [imath]\sqrt {-4} = 2[/imath]. See for example here. And remember that I follow a historical route. And I have no [imath]i [/imath], no nothing yet. So later on, I use something like [imath]\sqrt {-4} = \sqrt {4} \sqrt {-1} = 2\sqrt {-1}[/imath]. Now from a student's point of view, I guess everything seems like an ad hoc game: we use whatever we need whenever we want!
|
2080657
|
Hi i have a problem with a trigonomic excercise about tan
How can i solve this? the exercise is: [imath]\tan{20^{\circ}}\cdot\tan{40^{\circ}}\cdot\tan80^{\circ}[/imath] I have to define the exact answer. I have no idea how to start this exercise so please help. I think that would be so easy if I just multiply them. Thank for your help.
|
805821
|
How can I find the following product? [imath] \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.[/imath]
How can I find the following product using elementary trigonometry? [imath] \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.[/imath] I have tried using a substitution, but nothing has worked.
|
1620893
|
Converse Noetherian Relation
I have looked around and cannot find the answer so I ask here. It is well known that if [imath]R[/imath] is noetherian then [imath]R[X][/imath] is too, but what about the converse? If [imath]R[X][/imath] is noetherian can we say [imath]R[/imath] is? My gut feeling says no and the fact it seems so hard to find an answer makes me feel certain about it.
|
240555
|
Converse to Hilbert basis theorem
Prove the converse to Hilbert basis theoren: If the polynomial ring [imath]R[x][/imath] is Noetherian, then [imath]R[/imath] is noetherian.
|
2080800
|
How to find the given limits
I now that: [imath]\lim_{x\to 0}\frac{\sin x}{x}=1,[/imath] but I didnt now how to prove that:[imath]\lim_{x\to 0}\frac{1-\cos x}{x^2}[/imath] Please help me. Thanky very much for your help.
|
1334164
|
Finding the limit of [imath](1-\cos x)/x^2[/imath]
[imath]\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2[/imath] now [imath]\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}=\left({\sin\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)}\right)^2=1^2[/imath] So we have [imath]\frac{2}{x^2}\cdot \left(\frac{x}{2}\right)^2=\frac{2}{x^2}\cdot \left(\frac{x^2}{4}\right)=\frac{1}{2}[/imath] Are the moves right?
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