qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
370119 | If [imath]K/L[/imath] is normal and [imath]L/F[/imath] is purely inseparable, then [imath]K/F[/imath] is normal
This is a problem in Morandi's Field and Galois Theory, on page 49: Let [imath]F\subset L\subset K[/imath] be field extensions such that [imath]K/L[/imath] is normal and [imath]L/F[/imath] is purely inseparable. Show that [imath]K/F[/imath] is normal. | 367344 | If [imath]F/L[/imath] is normal and [imath]L/K[/imath] is purely inseparable, then [imath]F/K[/imath] is normal
Let field extensions [imath]K \subset L \subset F[/imath] such that [imath]F/L[/imath] is normal extension, and [imath]L/K[/imath] is purely inseparable extension. Show that [imath]F/K[/imath] is normal extension. My strategy: let [imath]f(x)\in K[x][/imath] is irreducible polynomial, and [imath]\alpha \in F[/imath] is solution of [imath]f(x)=0[/imath]. So if [imath]\alpha \in L[/imath] then [imath]f(x)=0[/imath] has only solution [imath]\alpha[/imath]. Thus [imath]f(x)[/imath] split on [imath]F[/imath],then [imath]F/K[/imath] is normal. PS: I don't what should I do when [imath]\alpha \notin L[/imath]. |
370599 | Adjugate invertible matrix
If A is an invertible [imath]nxn[/imath] matrix prove that:[imath] adj(adjA)=(A)(detA)^{n-2}[/imath] I have done this but it somewhere went wrong: [imath] adj(adjA)=adj(A^{-1} detA)=(A^{-1}detA)^{-1} det(A^{-1}detA)=AdetA det(A^{-1}detA)= Adet(AA^{-1}detA)=A (detA)^n [/imath] | 92837 | Proof: [imath]\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A[/imath] for [imath]A \in \mathbb{R}^{n\times n}[/imath]
I had my exam of linear algebra today and one of the questions was this one. Given [imath] A \in \mathbb{R}^{n \times n}[/imath], prove that: [imath]\mathrm{adj}(\mathrm{adj}(A)) = (\mathrm{det}(A))^{n-2} \cdot A.[/imath] Of course I was not able to prove this identity, otherwise I wouldn't post it here. But I'm still curious how one can prove this identity. Could someone point me in the right direction? |
370987 | Transformation Existence Proof: A Call for Critique
QUESTION Prove that there exists a [imath]T:V\rightarrow W[/imath] such that [imath]N(T)=V'\subset V[/imath] and [imath]R(T)=W'\subset W[/imath] ATTEMPTED ANSWER Let [imath]V[/imath] and [imath]W[/imath] be finite-dimensional vector spaces over [imath]F[/imath]. Let [imath]A=\{a_1,\dots,a_l\}[/imath] be a basis for [imath]V'\subset V[/imath], let [imath]B=\{a_1,\dots,a_l,b_1,\dots,b_m\}[/imath] be a basis for [imath]V[/imath], let [imath]C=\{c_1,\dots,c_n\}[/imath] be a basis for [imath]W'\subset W[/imath], and let [imath]D=\{c_1,\dots,c_n,d_1,\dots,d_p\}[/imath] be a basis for [imath]W[/imath]. Thus [imath]N(T)=V'[/imath] and [imath]R(T)=W'[/imath] means that \begin{eqnarray} T(a_1)&=&0+\cdots +0\\ T(a_2)&=&0+\cdots +0\\ \vdots\\ T(a_l)&=&0+\cdots +0\\ T(b_1)&=&k_{11}c_1+k_{21}c_2+\cdots+k_{n1}c_n\\ T(b_2)&=&k_{12}c_1+k_{22}c_2+\cdots+k_{n2}c_n\\ \vdots\\ T(b_n)&=&k_{n1}c_1+k_{n1}c_2+\cdots+k_{nm}c_n, \end{eqnarray} or simply \begin{eqnarray} \begin{pmatrix} c_1&c_2&\cdots&c_n \end{pmatrix} \begin{pmatrix} 0&\cdots&0&k_{11}&\cdots&k_{n1}\\ \vdots&&\vdots&\vdots&&\vdots\\ 0&\cdots&0&k_{n1}&\cdots&k_{nm} \end{pmatrix}, \end{eqnarray} which is an [imath](l+n)\times (l+m)[/imath] matrix. Thus, such a [imath]T[/imath] exists and has the above form. JUST CURIOUS As far as I understand, we're dealing with something that looks like this: | 370933 | [imath]T:V\rightarrow W[/imath] such that [imath]N(T)=V'\subset V[/imath] and [imath]R(T)=W'\subset W[/imath]
How should one prove that there exists a linear map [imath]T:V\rightarrow W[/imath] such that [imath]N(T)=V'\subset V[/imath] and [imath]R(T)=W'\subset W[/imath] if [imath]\dim(V')+\dim(W')=\dim(V)[/imath], where [imath]V[/imath] and [imath]W[/imath] are finite-dimensional vector spaces? My "answer" is just a guess really... It seems pocketed with holes. What do you think? |
371187 | A question on Cosets
Let [imath]G[/imath] be a group and [imath]H[/imath] , [imath]K[/imath] be subgroups of [imath]G[/imath] such that [imath][G:H][/imath] and [imath][G:K][/imath] are finite. Then is it true that [imath][G:H∩K][/imath] is also finite ? | 128538 | Does the intersection of two finite index subgroups have finite index?
Let [imath](G,*)[/imath] be a group and [imath]H,K[/imath] be two subgroups of [imath]G[/imath] of finite index (the number of left cosets of [imath]H[/imath] and [imath]K[/imath] in [imath]G[/imath]). Is the set [imath]H\cap K[/imath] also a subgroup of finite index? I feel like need that [imath][G\colon(H\cap K)][/imath] is a divisor of [imath][G\colon H]\cdot[G\colon K][/imath], but I dont't know when this holds. Can somebody help me out? |
340534 | Integrate [imath]\ln(x^2+1)/(x^2+1)[/imath]
How to evaluate [imath]\int_0^\infty \frac{\ln(x^2+1)}{x^2+1} \mathrm{d}x[/imath] using complex analysis? I've spent ages trying to think of some clever contour integral which will give it, but I can't seem to get anywhere. Any hints would be appreciated. | 358386 | Evaluating [imath]\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx[/imath]
How would I go about evaluating this integral? [imath]\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx.[/imath] What I've tried so far: I tried a semicircular integral in the positive imaginary part of the complex plane, excluding the negative real axis, but had trouble calculating the residue at [imath]z=i[/imath] (perhaps there is a way of doing this that I don't know of). After that didn't work, I tried a rectangular box integral from [imath]\epsilon[/imath] to [imath]R[/imath], from [imath]R[/imath] to [imath]R+i/2[/imath], from [imath]R+i/2[/imath] to [imath]-S+i/2[/imath], from [imath]-S+i/2[/imath] to [imath]-S+i\epsilon[/imath], from [imath]-S+i\epsilon[/imath] to [imath]-\epsilon+i\epsilon[/imath] and finally a semicircle around the origin, radius [imath]\epsilon[/imath]. Any help would be appreciated. |
371696 | Product of integrable functions also integrable?
I can't find in any book on measure and integral whether a product of two Lebesgue integrable functions is also integrable. Could someone clarify for me under what circumstances it is true? To be specific, I have an assumption [imath]fg\in L_{\mathrm{loc}}^1 (\Omega)[/imath] and I want to determine what requirements on [imath]f[/imath] a [imath]g[/imath] are sufficient to hold this assumption. | 80016 | Product of two Lebesgue integrable functions not Lebesgue integrable
I have a homework problem that says; Give Borel functions [imath]f,g: \mathbb{R} \to \mathbb{R}[/imath] that are Lebesgue integrable, but are such that [imath]fg[/imath] is not Lebesgue integrable. I saw this page too: Product of two Lebesgue integrable functions, but the question does not mention boundedness. I also am not sure what to do with the fact that the functions are Borel. (Any help on this would be especially appreciated) I know that if [imath]fg[/imath] were Lebesgue integrable then both [imath]\int (fg)^+\,d\mu[/imath] and [imath]\int (fg)^-\,d\mu[/imath] would be finite. This could lead to utilizing the finiteness of their difference (the function's integral) or their sum (the absolute value). I also know that [imath]f+g[/imath] are Lebesgue integrable if [imath]f[/imath] and [imath]g[/imath] are so I thought of using [imath]fg = \frac{1}{4}\,\big( (f+g)^2 - (f-g)^2 \big)\longrightarrow \int (fg)\,d\mu = \frac{1}{4}\,\int (f+g)^2\,d\mu - \frac{1}{4}\,\int (f-g)^2\,d\mu,[/imath] assuming linearity of the integral etc. I also thought of the Hölder inequality, [imath]\int \mid fg \mid d\mu \leq \bigg( \int \mid f \mid^p d\mu \bigg)^{(1/p)}\,\bigg( \int \mid g \mid^q d\mu \bigg)^{(1/q)},[/imath] but there was no mention in the question of what [imath]L^p[/imath]-space this was in. Maybe by the definition I gave it is such that [imath]p=1[/imath] and [imath]q=1[/imath]? Then [imath]\int \mid fg \mid d\mu \leq \bigg( \int \mid f \mid d\mu \bigg)\,\bigg( \int \mid g \mid d\mu \bigg).[/imath] However, I still can't seem to think of an approach to show that [imath]fg[/imath] is not Lebesgue integrable, while [imath]f[/imath] and [imath]g[/imath] are. Thanks for any guidance! |
157294 | Why is [imath]M_{mn}(R)\simeq M_m(M_n(R))[/imath]?
Intuitively, it's not hard to believe that for a ring [imath]R[/imath], the matrix ring [imath]M_{mn}(R)[/imath] is isomorphic to [imath]M_m(M_n(R))[/imath]. Taking a matrix in [imath]M_{mn}(R)[/imath] and turning the [imath]n\times n[/imath] blocks into single entries would give a matrix in [imath]M_m(M_n(R))[/imath], and taking a matrix in [imath]M_m(M_n(R))[/imath] and "erasing" the brackets of each entry would give something that looks like it belongs to [imath]M_{mn}(R)[/imath]. This is embarrassingly informal though, what's the proper way to show they are isomorphic? | 371697 | What is the isomorphism function in [imath]M_m(M_n(\mathbb R))\cong M_{mn}(\mathbb R)[/imath]?
What is the isomorphism function in [imath]M_m(M_n(\mathbb R))\cong M_{mn}(\mathbb R)[/imath]. I tried this [imath][[a_{ij}]_{kl}]\mapsto[a_{ijkl}][/imath] , but I couldn't prove all steps. |
371752 | Prove that [imath]f[/imath] is one-to-one an interval
Suppose that [imath]f:(a, b) \to \mathbb R[/imath], has a continuous derivative on [imath](a, b)[/imath]. If [imath]f'(x_0) \ne 0[/imath], prove that there is an interval [imath](c, d)[/imath] containing [imath]x_0[/imath] such that [imath]f[/imath] is one-to-one on [imath](c, d)[/imath]. | 367520 | If [imath]f'(x_0) \ne 0[/imath], prove that there is an interval [imath](c,d)[/imath] containing [imath]x_0[/imath] such that [imath]f[/imath] is one to one on [imath](c,d)[/imath].
Suppose that [imath]f:(a,b)\to\mathbb{R}[/imath] has a continuous derivative on [imath](a,b)[/imath]. If [imath]f'(x_0) \ne 0[/imath], prove that there is an interval [imath](c,d)[/imath] containing [imath]x_0[/imath] such that [imath]f[/imath] is ont to one on [imath](c,d)[/imath]. My solution: Suppose f is not one to one on [imath](c,d)[/imath]. If we define [imath](x,y)[/imath] to be an open interval in [imath](c,d)[/imath]. By Rolle's theorem, there exists [imath]z \in (a,b)[/imath] such that [imath]f'(z) \ne 0[/imath] so z cannot be in that interval. which is a contradiction. The ending feels weak, any suggestions? Thanks |
371992 | Combinatorial Proof of Binomial Coefficient Identity, summing over the upper indices
Consider the sum [imath]\displaystyle\sum_{j=r}^{n+r-k} \binom{j-1}{r-1}\binom{n-j}{k-r} = \binom{n}{k}[/imath] I am looking to show this identity combinatorially. Is the general idea perhaps to remove j from n and k from r then pick some 'middle' element, and form a final subset of size r-1? The bounds on the sum are a big roadblock for me. | 53307 | Combinatorial proof of binomial coefficient summation
While doing some Computer Science problems, I found one which I thought could be solvable using combinatorics instead of programming: Given two positive integers [imath]n[/imath] and [imath]k[/imath], in how many ways do [imath]k[/imath] numbers, all of which are between [imath]0[/imath] and [imath]n[/imath] (inclusive), add up to [imath]n[/imath]? In the problem, order does matter. For example, 0+20 is not equivalent to 20+0. This led me to the following solution: Assume you do want to create the sum without any of the terms being zero. Then, the problem can be looked at as if you have [imath]n[/imath] rocks, and you want know the number of ways you can place [imath]k-1[/imath] sticks between them, such that no stick may be placed between the same two rocks. This is obviously equal to [imath]\binom{n-1}{k-1}[/imath]. If you then want to count the number of ways you can create the sum with exactly one of the terms being zero. Then the solution is (number of ways to get to [imath]n[/imath] using [imath]k-1[/imath] non-zero terms)*(the number of ways we can choose 1 element among k). Then, when wanting to count the number of ways with exactly [imath]x[/imath] zeroes, the answer is (number of ways to get to n using [imath]k-x[/imath] non-zero terms)*(the number of ways we can choose [imath]x[/imath] elements among [imath]k[/imath]). The solution to the original problem will basically the sum of the above expression, with x ranging between [imath]0[/imath] and [imath]k-1[/imath] (since there needs to be at least 1 non-zero term). This leads to [imath]\displaystyle\sum_{i=0}^{k-1}\left[ \binom{n-1}{k-i-1}\binom{k}{i}\right][/imath] I later went on to search on Google if there existed any simpler solution, and found that [imath]\binom{n+k-1}{n}[/imath] worked as well, but without explanation and proof. Both solutions give the same answer for any positive [imath]n[/imath] and [imath]k[/imath], but I really have trouble understanding why. I've tried to simplify my summation, and proving it inductively, but I have not been successful. Could somebody help me prove the equality? [imath]\binom{n+k-1}{n} = \displaystyle\sum_{i=0}^{k-1}\left[ \binom{n-1}{k-i-1}\binom{k}{i}\right][/imath] |
372471 | All group homomorphism from [imath] \mathbb{Z} _m [/imath] to [imath]\mathbb{Z}_n [/imath]
All group homomorphism from [imath] \mathbb{Z} _m [/imath] to [imath] \mathbb{Z}_n [/imath] How could I find every group homomorphism? | 273169 | Number of Homomorphisms from [imath]\Bbb{Z}_m[/imath] to [imath]\Bbb{Z}_n[/imath]
This question coutesy of Allan Clark's "Elements of Abstract Algebra" (60[imath]\zeta[/imath]). Find the number of homomorphisms from [imath]\Bbb{Z}_m\to \Bbb{Z}_n[/imath] as a function of [imath]m[/imath] and [imath]n[/imath]. This is stumping me, can anyone help? |
372819 | Show that there is no rational number [imath]r=m/n[/imath] such that [imath]r^3=3[/imath]
How do I solve this by prime factorization? I came across a similar problem on MSE just recently, but I can't find it and I thoroughly searched for it. If anyone can find it, please post it in the comment so that I can delete this question. Update: I don't know if it makes a difference, but please note that this is cubed root | 131391 | Proving [imath]\sqrt 3[/imath] is irrational.
There is a very simple proof by means of divisibility that [imath]\sqrt 2[/imath] is irrational. I have to prove that [imath]\sqrt 3[/imath] is irrational too, as a homework. I have done it as follows, ad absurdum: Suppose [imath]\sqrt 3= \frac p q[/imath] with [imath]p/q[/imath] irreducible, then [imath]\begin{align} & 3q^2=p^2 \\ & 2q^2=p^2-q^2 \\ &2q^2=(p+q)(p-q) \\ \end{align}[/imath] Now I exploit the fact that [imath]p[/imath] and [imath]q[/imath] can't be both even, so it is the case that they are either both odd, or have different parity. Suppose then that [imath]p=2n+1[/imath] and [imath]q=2m+1[/imath] Then it is the case that [imath]\begin{align} &p-q=2(n-m) \\ &p+q=2(m+n+1) \\ \end{align}[/imath] Which means that [imath]\begin{align} &2q^2=4(m-n)(m+n+1) \\ &q^2=2(m-n)(m+n+1) \\ \end{align}[/imath] Then [imath]q^2[/imath] is even, and so is then [imath]q[/imath], which is absurd. Similarly, suppose [imath]q=2n[/imath] and [imath]p=2m+1[/imath]. Then [imath]p+q=2(m+n)+1[/imath] and [imath]p-q=2(m-n)+1[/imath]. So it is the case that [imath]\begin{align} &2q^2=(2(m-n)+1)(2(m+n)+1)\\ &2q^2=4(m^2+m-n^2)+1 \\ \end{align}[/imath] So [imath]2q^2[/imath] is odd, which is then absurd. Is this valid? |
373122 | Sum of square binomial coefficients
Please feel free to close this is necessary as I didn't see exactly this question (some variations that I tried but didn't seem to apply. Prove: [imath]\sum_{k=0}^{n}{\binom{n}{k}^2}=\binom{2n}{n}[/imath] I figured trying to muscle through it by expanding and then multiplying each term by variants of 1 such as [imath]\frac{n^2}{n^2}[/imath] and [imath]\frac{n^2(n-1)^2}{n^2(n-1)^2}[/imath] to get common denominators, but that was NOT the way it seemed as it got ugly pretty quickly. Then I tried thinking up formulas of the binomial theorem that I could differentiate like other sums of binomial coefficient problems and i couldn't think of one. Any help here would be great. | 320348 | Inductive proof that [imath]{2n\choose n}=\sum{n\choose i}^2.[/imath]
I would like to prove inductively that [imath]{2n\choose n}=\sum_{i=0}^n{n\choose i}^2.[/imath] I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively things like [imath]{2n+2\choose n+1}={2n+2\over n+1}{2n+1\choose n}=2\cdot {2n+1\over n+1}{2n\choose n},[/imath] But I don't think it can lead me anywhere. I would like the proof to be as simple as possible. |
370025 | If [imath]\phi(\alpha)[/imath] is prime in [imath]\mathbb{Z}[/imath], show that [imath]\alpha[/imath] is prime in [imath]\mathbb{Z}[i][/imath]
If [imath]\alpha=a+bi[/imath] is a Gaussian integer, let [imath]\phi(\alpha)=a^2+b^2[/imath]. If [imath]\phi(\alpha)[/imath] is prime in [imath]\mathbb{Z}[/imath], show that [imath]\alpha[/imath] is prime in [imath]\mathbb{Z}[i][/imath]. I use the idea that if [imath]a^2+b^2=p[/imath] where [imath]p[/imath] is prime number, then [imath]\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_p[/imath]. Hence, [imath](a+bi)[/imath] is a maximal ideal since [imath]\mathbb{Z}_p[/imath] is a field. So, [imath](a+bi)[/imath] is a prime ideal and hence [imath]a+bi[/imath] is a prime element. Is this proof works? EDIT: [imath]\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_p[/imath] is proved in my class so I think I can straight away use it here. By the way, in the proof of [imath]\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_p[/imath], my lecturer said that [imath]a^2+b^2=p[/imath] implies [imath]\gcd(a,b)=1[/imath]. I can't figure out the proof of this. I start with contradiction, i.e. [imath]d=\gcd(a,b)>1[/imath]. Then [imath]d(\frac{a^2}{d}+\frac{b^2}{d})=p \implies d |p \implies d=p \implies a^2+b^2=\gcd(a,b)[/imath], contradiction. Is this proof work ? | 367917 | Is it true that: [imath]a+bi[/imath] is prime in [imath]\mathbb{Z}[i][/imath] if and only if [imath]a^2+b^2[/imath] is prime in [imath]\mathbb{Z}[/imath]
Is it true that [imath]a+bi[/imath] is prime in [imath]\mathbb{Z}[i][/imath] if and only if [imath]a^2+b^2[/imath] is prime in [imath]\mathbb{Z}[/imath]? How can I prove this? Can anybody help me please? |
373192 | Reducing countable collections of countably infinite sets to pairwise disjoint sets
This might be obvious (or not); but I want to know if it is true that if [imath]\{X_i\}_{i\in\mathbb{N}}[/imath] is a collection of countably infinite sets, then there exists [imath]\{Y_i\}_{i\in\mathbb{N}}[/imath], pairwise disjoint, such that each [imath]Y_i[/imath] is an infinite subset of [imath]X_i[/imath]. | 368324 | Disjoint Refinement
Prove that for any countable family of infinite sets from [imath]\mathbb{N}[/imath], [imath]A = \{A_n \colon n \in \mathbb{N}\}[/imath], there is a disjoint refinement [imath]B = \{B_n \colon n \in \mathbb{N}\}[/imath] of infinite sets, that is for every [imath]n[/imath], [imath]B_n[/imath] is an infinite subset of [imath]A_n[/imath], and all the [imath]B_n[/imath]'s are pairwise disjoint. Also prove the version for [imath]\mathbb{R}[/imath]: for any family (with cardinality not surpassing [imath]\mathbb{R}[/imath]'s) of sets from [imath]\mathbb{R}[/imath], [imath]A = \{A_r \colon r \in \mathbb{R}\}[/imath] such that [imath]|A_r| = \mathbb{R}[/imath] for all [imath]r[/imath], there is a disjoint refinement [imath]B = \{B_r \colon r \in \mathbb{N}\}[/imath], that is for every [imath]r[/imath], [imath]B_r[/imath] is a subset of [imath]A_r[/imath], [imath]|B_r| = \mathbb{R}[/imath] for all [imath]r[/imath], and all the [imath]B_r[/imath]'s are pairwise disjoint. Sorry for my complete lack of knowledge in latex. |
371928 | What should be the intuition when working with compactness?
I have a question that may be regarded by many as duplicate since there's a similar one at MathOverflow. The point is that I think I'm not really getting the idea on compactness. I mean, in [imath]\mathbb{R}^n[/imath] the compact sets are those that are closed and bounded, however the guy who answered this question and had his answer accepted says that compactness is some analogue of finiteness. That's the first problem: In my intuitive view of finiteness, only boundeness would suffice to say that a certain subset of [imath]\mathbb{R}^n[/imath] is in some sense "finite". On the other hand there's the other definition of compactness (in terms of covers) which is the one I really need to work with and I cannot see how that definition implies this intuition on finiteness. Also, I feel it's pretty strange the covers people use when they want to deal with compact sets. To prove a set is compact I know they must show that for every open cover there's a finite subcover, the problem is that I can't see intuitively how one could show this for every cover. Also when trying to disprove compactness the books I've read start presenting strange covers that I would have never thought about. I think my real problem is that I didn't yet get the intuition on compactness. So, what intuition should we have about compact sets in general and how should we really put this definition to use? Can someone provide some reference that shows how to understand the process of proving (and disproving) compactness? | 1165263 | What does compactness actually mean
You're probably already thinking "silly person" but hear me out. Compactness: Every open cover (of a set [imath]A[/imath]) has a finite subcover (this means every cover [imath]\{U_\alpha\}_{\alpha\in I}[/imath] where [imath]A\subset\bigcup_{\alpha\in I}U_\alpha[/imath] or [imath]A=\bigcup_{\alpha\in I}U_\alpha[/imath] where the [imath]U_\alpha[/imath] are open with respect to the subset of topology) This means EVERY open cover, there's a finite bunch of open sets which cover it. So it's sort of like size of a set, it's "bounded" because you can always cover it finitely. I like that. But then consider the set [imath](0,1)[/imath] (or any open interval) on [imath]\mathbb{R}[/imath] this is bounded, it isn't a "big" set, but it isn't compact (yet [0,1] is) So my question is what does compactness actually try and do? "Oh great the set is compact" what does that mean? Can someone describe it qualitatively By the way I have read the wikipedia page, I've done some research. Compactness seems to only be useful for some proves but even then the proofs are not "and with this property we can do this" it's more like a "clever trick" |
373268 | Random variables and permutations
I'm trying find the number of ordered triples of non-negative integers [imath]a, b, c[/imath] whose sum [imath]a + b + c[/imath] is a given positive integer [imath]n[/imath]. I've related it to the concept of distinguishable balls in distinguishable boxes, but can't seem to see what's missing. | 372331 | Number of decompositions of 20 into four integer parts
How many solutions are there to the equation: [imath]a + b + c + d = 20[/imath], where [imath]a, b, c,[/imath] and [imath]d[/imath] are non-negative integers? I was thinking the answer would be [imath]\displaystyle{N+1-p \choose p}[/imath] something along those lines. We really have [imath]1[/imath] free variable I assume. This may be resemblant of the fact of derangements, combinations and/or factorials.. |
373504 | Field of fractions of [imath]\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle[/imath]
This problem goes as follows: Prove that [imath]\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle[/imath] is an integral domain and that its field of fractions is isomorphic to the ring of rational functions [imath]\mathbb{Q}(t)[/imath]. The first part is straightforward, [imath]x^2+y^2-1[/imath] is irreducible in [imath]\mathbb{Q}[x,y][/imath], and hence a prime ideal, as [imath]\mathbb[Q][x,y][/imath] is a euclidean and hence a UFD. The quotient by a prime ideal is an integral domain. How do I do the second part? I guess, I have to set a isomorphism by making a substitution such that only one variable remains. But I don't see how [imath]x^2+y^2-1[/imath] is going to help me make a choice. | 96400 | [imath]\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle[/imath] is an integral domain, and its field of fractions is isomorphic to [imath]\mathbb Q(t)[/imath]
How can I prove that [imath]\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle[/imath] is an integral domain? Also, I need to prove that its field of fractions is isomorphic to the field of rational functions [imath]\mathbb{Q}(t)[/imath]? (The question is taken from UC Berkeley Preliminary Exam, Fall 1995.) |
373618 | When does compactness imply sequential compactness?
It is well known that if [imath] X [/imath] is a metric space then sequential compactness and compactness are equivalent. Now we consider a normed vector space [imath] E [/imath] and its dual [imath] E^\ast [/imath]. From Banach Alauglou theorem we know that the closed unit ball in [imath] E^\ast [/imath] is compact in the weak [imath] \ast [/imath] topology. Moreover this topology is not metrizable in general. So it is natural to ask: Is this ball also sequentially compact in the weak [imath] \ast [/imath] topology? How can I prove it? In general, when does compactness imply sequential compactness? (I'm loooking for some results about it) Thanks | 27423 | Compactness in the weak* topology
Let [imath]X[/imath] be a Banach space, and let [imath]X^*[/imath] denote its continuous dual space. Under the weak* topology, do compactness and sequential compactness coincide? That is, is a subset of [imath]X^*[/imath] weakly* compact if and only if it is weakly* sequentially compact? Does one imply the other? Perhaps I should make this next one a separate question, but I'd prefer to keep all of this in one place. Is the weak* topology on [imath]X^*[/imath] Hausdorff? Is the weak topology on [imath]X[/imath] Hausdorff? Motivation: I would like to say that if a subset of [imath]X^*[/imath] is weakly* compact, then it is weakly* closed, and that if a subset of [imath]X[/imath] is weakly compact, then it is weakly closed. |
373678 | Prove That [imath]|a +b| = |a| +|b|[/imath] if [imath]a[/imath] and [imath]b[/imath] Have Same Signs, And [imath]|a +b| < |a| + |b|[/imath] if [imath]a[/imath] and [imath]b[/imath] Have Opposite Signs (Proved Differently)
My Proof: This problem has mainly four cases, they are as follows: 1) [imath]a, b > 0[/imath] 2) [imath]a, b < 0[/imath] 3) [imath]a > 0 > b[/imath] 4) [imath]a < 0 < b [/imath] Let suppose that the sum of the real numbers [imath]a +b[/imath] equals the third real number [imath]c[/imath] OR [imath]a +b = c[/imath]. Case 1 [imath]a, b > 0[/imath]: [imath]|a +b| = |a| + |b|[/imath] [imath]|c| = a +b[/imath] [imath]c = c[/imath] ..... Proved. Case 2 [imath]a, b <0[/imath]: [imath]|(-a) +(-b)| = |(-a)| + |(-b)|[/imath] [imath]|-a -b| = |-a| + |-b|[/imath] [imath]|-(a +b)| = -(-a) -(-b)[/imath] [imath]|-(c)| = a +b[/imath] [imath]|-c| = c[/imath] [imath]-(-c) = c[/imath] [imath]c = c[/imath] ..... Proved. Case 3 [imath]a>0>b[/imath]: [imath]|a +(-b)| < |a| + |(-b)|[/imath] [imath]|a -b| < a + |-b|[/imath] [imath]a -b < a -(-b)[/imath] [imath]a -b < a +b[/imath] Case 4: [imath]|(-a) +(b)| < [/imath]|(-a)| + |b|[imath][/imath] |-a +b| < |-a| +|b|[imath][/imath] |b -a| < -(-a) +b[imath][/imath] b -a < a +b[imath][/imath] As, b > a[imath] therefore their difference will be a positive real number but it will be less than the sum of [/imath]a[imath] and [/imath]b$, and Case 3 can also be reasoned in this way. Please check my proof. | 373524 | Prove That [imath]|a +b| = |a| +|b|[/imath] if [imath]a[/imath] and [imath]b[/imath] Have Same Signs, And [imath]|a +b| < |a| + |b|[/imath] if [imath]a[/imath] and [imath]b[/imath] Have Opposite Signs
My Proof: [imath]|a +b| = |a| +|b|[/imath] ..... [imath](i)[/imath] [imath]|a +b| < |a| + |b|[/imath] ..... [imath](i)[/imath] If [imath]'a'[/imath] and [imath]'b'[/imath] have same signs: Let [imath]a[/imath] and [imath]b[/imath] be equal to [imath]-x[/imath]. Replacing [imath]a[/imath] and [imath]b[/imath] with [imath]-x[/imath] in the equation [imath](i)[/imath], we get the following result: [imath]|-x +(-x)| = |-x| + |-x|[/imath] [imath]|-x -x| = -(-x) -(-x)[/imath] [imath]|-2x| = x +x[/imath] [imath]-(-2x) = 2x[/imath] [imath]2x = 2x[/imath] ..... Proved If [imath]'a'[/imath] and [imath]'b'[/imath] have opposite signs: Let [imath]a = x[/imath] and [imath]b = -x[/imath]. Then, by replacing [imath]a[/imath] and [imath]b[/imath] with [imath]x[/imath] and [imath]-x[/imath] respectively in the inequality [imath](i)[/imath], we get the following result: [imath]|x +(-x)| < |x| + |-x|[/imath] [imath]|x -x| < x -(-x)[/imath] [imath]|0| < x +x[/imath] [imath]0 < 2x[/imath] ..... Proved Is this way valid for proving what has been asked to prove in the problem? |
373725 | Let [imath]G[/imath] be a group in which [imath]a^2=e[/imath] for all elements of [imath]a[/imath] of [imath]G[/imath]. Show that [imath]G[/imath] is Abelian.
Let [imath]G[/imath] be a group in which [imath]a^2=e[/imath] for all elements of [imath]a[/imath] of [imath]G[/imath]. Show that [imath]G[/imath] is Abelian. I need help on this problem. Appreciated! | 2591602 | If a is self-inverse, then the group is abelian
Let [imath]\langle G; \star, \hat, e\rangle[/imath] be a group. Show that if for all [imath]a\in G[/imath], we have [imath]a\star a=e[/imath], then G is abelian. What I did: Assume G is abelian, then we have [imath]a\star b = b\star a \quad \forall a,b\in G[/imath] We also know that [imath] a = \hat a[/imath], so: [imath]a\star b = b \star a[/imath] [imath]\Leftrightarrow a\star a \star b = a \star b \star a[/imath] [imath]\Leftrightarrow b = a \star b \star a[/imath] [imath]\Leftrightarrow b \star b = (b \star a ) \star (b \star a)[/imath] [imath]\Leftrightarrow e = e[/imath] and since [imath]e\in G[/imath] we are finished. I'm very unsure about this proof. I know that there's a way better proof using [imath]\widehat{a\star b} = \hat b \star \hat a[/imath] Does what I did actually work? |
373724 | Analysis involving closed intervals and rationality
Can someone help me prove this problem? Prove that every closed interval [imath][a,b][/imath] is a subset of [imath]\mathbb{R}[/imath] contains at least one rational number. | 373169 | Analysis proof about irrational numbers
Prove that every closed interval [imath][a,b][/imath] is a subset of [imath]\mathbb{R}[/imath] contains at least one irrational number. |
370299 | Question on combinatorics, partitions.
Let [imath]p[/imath] ([imath]n|[/imath]distinct odd parts) be the number of partitions of [imath]n[/imath] into distinct odd parts. Prove that [imath]p(n)[/imath] is odd if and only if [imath]p[/imath]([imath]n|[/imath]distinct odd parts) is odd by using the theorem on self-conjugate partitions. Ok, so we count the hooks in the Ferrers diagram, but how? | 367645 | Proof on distinct odd parts, partitions
Let [imath]p (n|[/imath]distinct odd parts[imath])[/imath] be the number of partitions of [imath]n[/imath] into distinct odd parts. Prove that [imath]p(n)[/imath] is odd if and only if [imath]p(n|[/imath]distinct odd parts) is odd. I know we're suppose to use the theorem on self-conjugate partitions... |
373780 | Integral-Power series complex identity
Given the power series: [imath]f(z) = \sum_{n=0}^{\infty}a_{n} z^{n}[/imath], with radius of convergence R, prove that for r < R the following holds: [imath]\frac{1}{2\pi } \int_{0}^{2\pi} | f(r e^{i\theta }) |^{2} d \theta = \sum_{n=0}^{\infty }|a_{n}|^{2} r^{2n}[/imath] | 238734 | Integrating squared absolute value of a complex sequence
I was reading through my book in complex analysis and i encountered this problem. Given, [imath]F=\sum_{n=0}^{\infty} a_nX^n[/imath] is a convergent power series with radius of convergence R. We are asked to show that for every 0[imath]\leq[/imath]r[imath]<[/imath]R that [imath] \int_0^{2\pi}\mid F(re^{it})\mid^2\mathrm{d}t={2\pi}\sum_{n=0}^{\infty}\mid a_n\mid^2r^{2n}[/imath] This is not my homework but after some effort I have no clue how to solve this one. The assignment was in the chapter about convergent power series and there is nothing about integrating, thus I assume it should be solvable by real analysis calculus. I would appreciate if someone would help me with this one. Regards, Raxel. |
374278 | Complex Fourier series of a function
I need to find the complex Fourier series of this function, and I'm having problems calculating these integers: [imath]|a|<1[/imath] [imath]x\in [-\pi,\pi][/imath] [imath]f(x)=\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}[/imath] [imath]a_0=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}dx[/imath] [imath]b_n=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}\sin(nx)dx[/imath] Please help me on this integers, I can't solve them! I have no clue how to start... To get the complex serie.... Thanks | 374274 | Complex Fourier series
I need to find the complex Fourier series of this function, and I'm having problems calculating these integers: [imath]|a|<1[/imath] [imath]x\in [-\pi,\pi][/imath] [imath]f(x)=\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}[/imath] [imath]a_0=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}dx[/imath] [imath]b_n=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}\sin(nx)dx[/imath] |
374273 | Can we write [imath]\sqrt[w]{z}=z^\frac{1}{w}[/imath] when both [imath]w[/imath] and [imath]z[/imath] are complex numbers?
Let [imath]w[/imath] and [imath]z[/imath] be complex numbers defined in terms of real numbers [imath]a[/imath], [imath]b[/imath], [imath]c[/imath] and [imath]d[/imath] as follows: [imath] w = a+bi \\ z = c+di [/imath] Can we analogically write [imath] \sqrt[w]{z} = z^\frac{1}{w} \qquad \rightarrow \qquad \sqrt[a+bi]{c+di} = (c+di)^\frac{1}{a+bi} [/imath] from what we know about real numbers? | 201991 | For what values [imath]\alpha[/imath] for complex z [imath]\ln(z^{\alpha}) = \alpha \ln(z)[/imath]?
For example, when [imath]\alpha = 2[/imath], [imath]\ln(z^{2}) \neq 2\ln(z)[/imath], because argument z is determined up to constant [imath]2 \pi k[/imath]. So [imath] \ln(z^{2}) = \ln(z) + \ln(z) = \ln(z_{k_{1}}) + \ln(z_{k_{2}}) \neq 2\ln(z_{k_{3}}). [/imath] Of course, two of correct answers - [imath]1, -1[/imath]. But I know, that there are an infinite number of answers for [imath]\alpha[/imath]. Can you help me? |
374392 | Convergence of the series [imath]\sum a_n[/imath] implies the convergence of [imath]\sum \frac{\sqrt a_n}{n}[/imath], if [imath]a_n>0[/imath]
I need help to solve following problem from Rudin's Mathematical analysis book: Convergence of the series [imath]\sum a_n[/imath] implies the convergence of [imath]\sum \dfrac{\sqrt {a_n}}{n}[/imath], if [imath]a_n>0[/imath] I tried to construct a suitable convergence sequence [imath]b_n[/imath] such that [imath]\sum b_n[/imath] converges and [imath]a_n \leq b_n[/imath] but, I am not able to find such sequence [imath]b_n[/imath] . Thanks for the help and sugestions. | 163441 | If the series [imath]\sum_0^\infty a_n[/imath] converges, then so does [imath]\sum_1^\infty \frac{\sqrt{a_n}}{n} [/imath]
Problem: Suppose that for every [imath]n\in\mathbb{N}[/imath], [imath]a_n\in\mathbb{R}[/imath] and [imath]a_n\ge 0[/imath]. Given that [imath]\sum_0^\infty a_n[/imath] converges, show that [imath]\sum_1^\infty \frac{\sqrt{a_n}}{n} [/imath] converges. Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 7. |
374425 | Show that if [imath]\sum_{n=1}^{\infty}{a_n}[/imath] converges, then [imath]\lim_{n \rightarrow \infty}{na_n}=0[/imath]
Show that if [imath]\sum_{n=1}^{\infty}{a_n}[/imath] converges, then [imath]\lim_{n \rightarrow \infty}{na_n}=0[/imath] Suppose [imath]\lim_{n \rightarrow \infty}{na_n} \neq0[/imath]. Since [imath]\sum_{n=1}^{\infty}{a_n}[/imath] converges, by divergence test , we have [imath]\lim_{n \rightarrow \infty}{a_n}=0[/imath]. Then we have for large [imath]N[/imath], [imath]|a_n| < \epsilon[/imath] Since [imath]\lim_{n \rightarrow \infty}{na_n} \neq0[/imath], there exists a large [imath]N_1[/imath] such that [imath]|na_n-L|< \epsilon[/imath]. Then I get stuck at here. Can anyone guide me ? | 369669 | If [imath](a_n)[/imath] is a decreasing sequence of strictly positive numbers and if [imath]\sum{a_n}[/imath] is convergent, show that [imath]\lim{na_n}=0[/imath]
If [imath](a_n)[/imath] is a decreasing sequence of strictly positive numbers and if [imath]\sum{a_n}[/imath] is convergent, show that [imath]\lim{na_n}=0[/imath] Since [imath](a_n)[/imath] is decreasing and bounded below, by Monotonic Convergence Theorem, [imath](a_n)[/imath] converges. So, there exists [imath]N[/imath] such that [imath]|a_n-L|< \epsilon[/imath] Since [imath]\sum{a_n}[/imath] is convergent, by the Divergence test, we have [imath]\lim_n{a_n}=0[/imath], which means there exists [imath]N[/imath] such that [imath]|a_n-0|< \epsilon[/imath] Then I get stuck at here. I try to figure out the statement's meaning by inserting some examples, like [imath]a_n=\dfrac{1}{n^2}[/imath] Can anyone guide me on this question? |
374454 | Polynomial ring definition
Let [imath]F[/imath] be an arbitrary field (or just a ring). Usually [imath]F[t][/imath] is defined as a set of all finite sequences of 'numbers' which can be multiplied etc, [imath]F[t_1,t_2][/imath] is a ring of polynomials over [imath]F[t_1][/imath] i.e the set of finite sequences of polynomials over [imath]F[/imath]. Thus we can define [imath]F[t_1,t_2,...t_n][/imath]. The questions is how to define [imath]F[T][/imath] if T is an arbitrary infinite set? Is it possible to define [imath]F[T][/imath] a set of all finite formal linear combinations like [imath]a_1t_1t_2...t_k+....[/imath]? (where [imath]a_i\in F[/imath], [imath]t_i\in T[/imath] and it can be met more than once) We also consider the case of uncountable [imath]T[/imath]. | 374384 | Polynomial rings in an arbitrary set of indeterminates
Recently I was reading an article where the author described a strange polynomial ring that I had never seen before. Here it is (I changed some words): [imath]T[/imath] is a set (in the context of the article it is a partially ordered set but it is incidentally, I hope). Let [imath]\mathbb{Q}[/imath] be the field of rationals, and form the polynomial ring [imath]\mathbb{Q}[T][/imath] with the elements [imath]t\in T[/imath] as indeterminates. Also there is the following a phrase: If [imath]G\subset T[/imath], then [imath]G\mathbb{Q}[T][/imath] is a prime ideal of [imath]\mathbb{Q}[T][/imath]. Could you tell me what the object [imath]\mathbb{Q}[T][/imath] is? I have never met such constructions. Is it something like a ring of polynomials with infinitely many (if [imath]T[/imath] is infinite) variables, i.e. something like an [imath]m[/imath]-th tensor power of usual ring of polynomials [imath]\mathbb{Q}[t][/imath]? What is [imath]G\mathbb{Q}[T][/imath]? |
374570 | [imath]\sqrt9[/imath] has two solutions?
This question might seem to be a little ridiculous, but... [imath]\sqrt9=3[/imath] right? it's obvious, but [imath]\sqrt9=-3[/imath] what has happened here? We raise [imath]-3[/imath] to second power and we have 9, so nothing really wrong happened here? What kind of knowlege am I missing? Even wolfram says that :[imath]3[/imath] is principal root and [imath]-3[/imath] is real root. Thanks for any answer. | 268830 | Why do we choose [imath]3[/imath] to be positive after [imath]\sqrt{9 - x^2}[/imath] in the following substitution?
The integral [imath]\int \frac{\sqrt{9 - x^2}}{x^2}dx[/imath] is solved in my book by letting [imath]x = 3\sin\theta[/imath] where [imath]-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}[/imath]. Then, [imath]dx = 3\cos\theta\,d\theta[/imath] and, [imath]\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta[/imath] So, [imath]\int \frac{\sqrt{9 - x^2}}{x^2}dx = \int \cot^2 \theta \ d\theta = -\cot\theta - \theta + C[/imath] Returning to the original variable, [imath]\int \frac{\sqrt{9 - x^2}}{x^2}dx = -\frac {\sqrt{9 - x^2}}{x} - \sin^{-1}\left(\frac{x}{3}\right) + C[/imath] I don't understand why [imath]\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta \,[/imath] instead of [imath]\sqrt{9-x^2} = |3||\cos\theta| = |3|\cos\theta[/imath]. I feel like I have problems understanding this because I am not sure what is the purpose of the absolute value signs in this case, are they to indicate that, for example, [imath]|\cos\theta| = \pm\cos\theta[/imath]? If that's the case, why do we choose [imath]3[/imath] to be positive instead of negative? |
374928 | Definite integral with no closed form antiderivative
Now from wikipedia I know that [imath]\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi.[/imath] Also on wikipedia they have the following claim [imath]\int_{-\infty}^\infty x^{2n}e^{-x^2}dx=\frac{(2n-1)!!}{2^n}\sqrt\pi.[/imath] Where [imath]n\in \Bbb N[/imath]. I do not see how they got the second. | 369558 | General result of the following integral
Let [imath]I = \int\limits_0^\infty dx x^{2n} e^{-\alpha x^2}[/imath] I tried it as follows: with the substitution [imath]y = x^2 \implies \frac{dy}{dx} = 2x \implies dx = \frac{dy}{2x}[/imath] the integral transforms into [imath]\int\limits_0^\infty \frac{dy}{2x} y^n e^{-\alpha y}[/imath] with [imath]y = x^2 \implies x = \pm \sqrt{y} [/imath] we get: [imath]\frac{1}{2} \int\limits_0^\infty dy[/imath] [imath] y^{n \pm \frac{1}{2}} e^{-\alpha y} [/imath] now if I integrate by parts I get: [imath]I = \frac{1}{2}[- \frac{1}{\alpha} e^{-\alpha y} y^{n \pm \frac{1}{2}}]_0^\infty + \frac{1}{2} \int\limits_0^\infty -\frac{1}{\alpha} e^{-\alpha y} y^{n-\frac{3}{2}}(n \pm \frac{1}{2})[/imath] where the term on the left is zero. On the right side the exponent of the [imath]y[/imath] could also be [imath]y^{n-\frac{1}{2}}[/imath] in case we chose the plus. I would just go on partially integrating to get the solution in the end but I'm not really sure I did everything correct so far. Why is there an ambiguity with the [imath]\pm[/imath]? How do I get rid of it? Did I do everything correct so far? Cheers ! |
375107 | Finding the function
I would appreciate if somebody could help me with the following problem: Q: [imath]f(x):[/imath] conti-function and [imath]f(2x)-f(x)=x^3[/imath] find [imath]f(x)=?[/imath] | 373572 | Finding Value, Related To Functional Equation
[imath]f(x)[/imath] is continuous for [imath]\forall x \in R[/imath] and [imath]f(2x)-f(x)=x^{3}[/imath] (1) [imath]f(x)+f(-x)[/imath] is constant ? (2) [imath]f(0)=0[/imath] ? I don't know how to use the continuity. especially for [imath]f(0)=0[/imath] ? |
375058 | Factorization into irreducibles
Let [imath]R[/imath] be a ring with [imath]1[/imath]. Let [imath]r[/imath] be an element of [imath]R[/imath], [imath]r\neq 0[/imath], [imath]r[/imath] not a unit. Can I say that [imath]r[/imath] is a product of a finite number of irreducibles or not? | 37485 | Exhibit an integral domain [imath]R[/imath] and a non-zero non-unit element of [imath]R[/imath] that is not a product of irreducibles.
Exhibit an integral domain [imath]R[/imath] and a non-zero non-unit element of [imath]R[/imath] that is not a product of irreducibles. My thoughts so far: I don't really have a clue. Could anyone direct me on how to think about this? I'm struggling to get my head round irreducibles. Thanks. |
375527 | Proving that if [imath]a,b[/imath] are even, then [imath]\gcd(a,b) = 2 \gcd(a/2, b/2)[/imath]
Prove that if [imath]a, b[/imath] are both even then [imath]\gcd(a,b) = 2\cdot\gcd(a/2,b/2)[/imath]. Little confused here. I have tried the following but it's basically just repeating the proof unfortunately: [imath]a = 2 \cdot a_1 [/imath] and [imath] b = 2 \cdot b_1[/imath] (Take factor of 2 out) [imath]d = \gcd(a,b)[/imath] [imath]d = 2\gcd(a_1, b_1)[/imath] [imath]a/2 = a_1[/imath] and [imath]b/2 = b_1[/imath] I've even confused myself! Could someone help! | 202397 | How to prove that [imath]z\gcd(a,b)=\gcd(za,zb)[/imath]
I need to prove that [imath]z\gcd(a,b)=\gcd(za,zb)[/imath]. I tried a lot, for example, looking at set of common divisors of the two sides, but I can't conclude anything from that. Can you please give me some advice how I can handle this problem? And [imath]a,b,z \in \mathbb{Z}[/imath]. |
376130 | Showing that if [imath]\lim\limits_{x \to a} f'(x)=A[/imath], then [imath]f'(a)[/imath] exists and equals [imath]A[/imath]
Let [imath]f : [a; b] \to \mathbb{R}[/imath] be continuous on [imath][a, b][/imath] and differentiable in [imath](a, b)[/imath]. Show that if [imath]\lim\limits_{x \to a} f'(x)=A[/imath], then [imath]f'(a)[/imath] exists and equals [imath]A[/imath]. I am completely stuck on it. Can somebody help me please? Thanks for your time. | 257907 | Prove that [imath]f'(a)=\lim_{x\rightarrow a}f'(x)[/imath].
Let [imath]f[/imath] be a real-valued function continuous on [imath][a,b][/imath] and differentiable on [imath](a,b)[/imath]. Suppose that [imath]\lim_{x\rightarrow a}f'(x)[/imath] exists. Then, prove that [imath]f[/imath] is differentiable at [imath]a[/imath] and [imath]f'(a)=\lim_{x\rightarrow a}f'(x)[/imath]. It seems like an easy example, but a little bit tricky. I'm not sure which theorems should be used in here. ============================================================== Using @David Mitra's advice and @Pete L. Clark's notes I tried to solve this proof. I want to know my proof is correct or not. By MVT, for [imath]h>0[/imath] and [imath]c_h \in (a,a+h)[/imath] [imath]\frac{f(a+h)-f(a)}{h}=f'(c_h)[/imath] and [imath]\lim_{h \rightarrow 0^+}c_h=a[/imath]. Then [imath]\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)[/imath] But that's enough? I think I should show something more, but don't know what it is. |
376253 | show that [imath]f(z)=f(1)z. [/imath]
If [imath]f[/imath] is entire such that [imath]f(z_1 + z_2)=f(z_1)+f(z_2)[/imath] for all [imath]z_1,z_2 \in\Bbb C[/imath], then show that [imath]f(z)=f(1)z[/imath]. I can see if [imath]f=u+iv[/imath] then for all [imath](x_1,y_1),(x_2,y_2)\in\mathbb R^2[/imath] [imath]u(x_1+x_2,y_1+y_2)=u(x_1,y_1)+u(x_2,y_2)[/imath] and [imath]v(x_1+x_2,y_1+y_2)=v(x_1,y_1)+v(x_2,y_2)[/imath] | 325637 | [imath]f(z)= az[/imath] if [imath]f[/imath] is analytic and [imath]f(z_{1}+z_{2})=f(z_{1})+f(z_{2})[/imath]
If [imath]f[/imath] is an analytic function with [imath]f(z_{1}+z_{2})=f(z_{1})+f(z_{2})[/imath], how can we show that [imath]f(z)= az[/imath] where [imath]a[/imath] is a complex constant? |
376367 | Solve an equation of 3rd order
What is the simplest method to solve an equation of 3rd degree. For example: [imath]-x^{3} + x^{2} + x - 1 = 0[/imath] Please I don't want the resolution of this equation I just want the simplest method to use to solve it, then I'll try to solve it on my own. | 294331 | Factoring Cubic Equations
I’ve been trying to figure out how to factor cubic equations by studying a few worksheets online such as the one here and was wondering is there any generalized way of factoring these types of equations or do we just need to remember a bunch of different cases. For example, how would you factor the following: [imath]x^3 – 7x^2 +7x + 15[/imath] I'm have trouble factoring equations such as these without obvious factors. |
376448 | Holomorphic map from quarter plane to unit disk.
Let [imath]D[/imath] be the quarter plane where [imath]-\frac\pi4 < \arg(z) < \frac\pi4[/imath]. By Riemann mapping theorem, there exists a holomorphic bijection from the open unit disk to [imath] D[/imath] where [imath]f(0)=1[/imath] and [imath]f'(0)>0[/imath]. Calculate explicitly the inverse of this function. I am thinking maybe I can use the Cayley transform, but this is half the plane not a quarter. | 298075 | Find conformal mapping from sector to unit disc
Find a conformal mapping between the sector [imath]\{z\in\mathbb{C} : -\pi/4<\arg(z) <\pi/4\}[/imath] and the open unit disc [imath]D[/imath]. I know that it should be a Möbius transformation, but other than that I am very stuck, any help would be much appreciated. |
376622 | p-adic isomorphism [imath]\mathbb{Q}_p\not\cong \mathbb{Q}_q\iff p\ne q[/imath]
In class I learn that [imath]\mathbb{Q}_3\not\cong\mathbb{Q}_5[/imath] because one of them has [imath]\sqrt{2}[/imath] the other doesn't. Also professor asks us to find reference that [imath]\mathbb{Q}_p\not\cong \mathbb{Q}_q\iff p\ne q[/imath] but I can't find any. Could anyone provide me with a proof of this? I'm familiar with inverse limit definition of [imath]\mathbb{Z}_p[/imath] and its field of quotient [imath]\mathbb{Q}_p[/imath], plus Power series definition of [imath]\mathbb{Z}_p[/imath] and Laurant series defintion of [imath]\mathbb{Q}_p[/imath] and further completion of [imath]\mathbb{Q}[/imath] with respect to p-adic norm. Feel free to use whichever you want. Thanks a lot. | 93633 | Is [imath]\mathbb Q_r[/imath] algebraically isomorphic to [imath]\mathbb Q_s[/imath] while r and s denote different primes?
It is obvious that [imath]\mathbb{Q}_r[/imath] is topologically isomorphic to [imath]\mathbb Q_s[/imath] while [imath]r[/imath] and [imath]s[/imath] denote different primes. But I really don't know whether it is true in the aspect of algebra. As I failed to prove it, I think that it is false, but I can't give a counterexample. Last I'm quite sorry that I'm new to MathJax and I don't know how to use it properly.Thanks for reading and I would appreciate it if you could solve my problem. |
376754 | Division of a cubic equation by one of its factors
I'm trying to divide a cubic equation by a factor. This is the equation: [imath] -\lambda^3 -\lambda^2 + 10 \lambda - 8 = 0[/imath] and this is the factor : [imath](\lambda - 1)[/imath] I Googled about it and I found the Euclidean division, but I couldn't find some understandable way to do it. | 376726 | Solving cubic equations
I was trying to solve a cubic equation which is : [imath] -\lambda³ -\lambda² + 10 \lambda - 8 = 0[/imath] I googled about it and I found the Rational Root theorem which is takes time to do it, but I found that some people are solving cubic equation by factoring, like this equation : x3+4x2+x-6 = 0 it will be (x-1)(x+2)(x+3)=0 I tried to do the same with my equation but it's hard to find what is the common factorial. Is there any way to determine what is the common factorial as in the example I gave you (x-1)(x+2)(x+3)=0 |
376995 | If [imath]H,K \leq G[/imath] a finite group, then [imath]\left\lvert HK \right\rvert = \cdots[/imath]
If [imath]H,K \leq G[/imath] a finite group, then [imath]\left\lvert HK \right\rvert = \frac{\left\lvert H \right\rvert \cdot \left\lvert K \right\rvert}{\left\lvert H\cap K \right\rvert}.[/imath] The first part of the problem asked us to show this when [imath]\left\lvert H\cap K \right\rvert=1[/imath]. This was doable and I used a counting argument saying how if we had some [imath]h_ik_j=h_lk_m[/imath] then [imath]h_i=h_l[/imath] and [imath]k_j=k_m[/imath]. Then counting the size of the set [imath]HK[/imath] was straightforward. My attemps for this problem have been saying something like [imath]\frac{\left\lvert H \right\rvert \cdot \left\lvert K \right\rvert}{\left\lvert H\cap K \right\rvert}= \frac{\left\lvert H / (H \cap K) \right\rvert \cdot \left\lvert H \cap K \right\rvert^2 \cdot \left\lvert K / (H \cap K ) \right\rvert}{\left\lvert H \cap K \right\rvert}=\left\lvert H \cap K \right\rvert \cdot \left\lvert H / (H \cap K) \right\rvert \cdot \left\lvert K / (H \cap K) \right\rvert[/imath] [imath]= \left\lvert H \right\rvert \cdot \left\lvert K / (H \cap K) \right\rvert.[/imath] Any ideas for ways I could solve this? | 78395 | Given a finite Group G, with A, B subgroups prove the order of AB
How do you prove: Given a finite group [imath]G[/imath], with [imath]A,B[/imath] subgroups then [imath]|AB|=\frac{|A||B|}{|A \cap B|}.[/imath] |
377544 | Finding a primitive root modulo [imath]13[/imath]
Find a primitive root modulo each of the following integers. a) [imath]13[/imath] My TA said we are not going to go over this. We did not go over the topic. It seems like something good to know though. Is there anyone who can help me with this? | 377532 | Finding a primitive root modulo [imath]11^2[/imath]
Find a primitive root modulo each of the following moduli: a) [imath]11^2[/imath] My TA said he is not going to go over this so do not worry about it. He said you can try this if you want but he would not go over this. Can someone please show me how to solve this problem? It seems like something good to know. |
378031 | Does [imath]\lim\limits_{n\to\infty} \sum\limits_{r=0}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{n}f(\frac{r}{n})[/imath] imply [imath]\int\limits_{0}^{\frac{1}{2}}f(x)dx[/imath]?
We know that [imath]\lim\limits_{n\to\infty} \sum\limits_{r=0}^{n-1}\frac{1}{n}f(\frac{r}{n})[/imath] implies [imath]\int\limits_{0}^{1}f(x)dx[/imath]. Then does [imath]\lim\limits_{n\to\infty} \sum\limits_{r=0}^{\lfloor\frac{n}{2}\rfloor}\frac{1}{n}f(\frac{r}{n})[/imath] imply [imath]\int\limits_{0}^{\frac{1}{2}}f(x)dx[/imath]? | 375011 | integration as limit of a sum
If [imath]f[/imath] is continuous on [imath][0, 1][/imath] then [imath]\lim_ {n\to\infty}\sum_{j=0}^{\lfloor n/2\rfloor} \frac1{n}f\left(\frac {j}{n}\right) = ? [/imath] will the answer be that the limit exists and is equal to [imath] \int_0^1 f(x) dx[/imath] ? |
378303 | Show AB and BA have the same eigenvalues
If [imath]A[/imath] and [imath]B[/imath] are [imath]n[/imath] by [imath]n[/imath] matrices show that [imath]AB[/imath] and [imath]BA[/imath] have the same eigenvalues. I see why this is true if both are nonsingular. But does it still hold if they are not invertible? Thanks! | 232561 | Eigen Values Proof
a.) Let A and B be [imath]n[/imath] x [imath]n[/imath] matrices. Prove that the matrix products [imath]AB[/imath] and [imath]BA[/imath] have the same eigenvalues. b.) Prove that every eigenvalue of a matrix A is also an eigenvalue of its transpose [imath]A^T[/imath]. Also, prove that if v is an eigenvector of A with eigenvalue [imath]\lambda[/imath] and w is an eigenvector of [imath]A^T[/imath] with a different eigenvalue [imath]\mu \ne \lambda[/imath], then v and w are orthogonal vectors with respect to the dot product. For a, I know that if their eigenvalues are the same then their eigenvectors must relate too. For the first part of b, is it similar to proving that the [imath]det(A) =det (A^T)[/imath]? And, I do not know how to do the second part. |
378822 | Prove that [imath]( \frac{l}{(l,m)},\frac{m}{(l,m)}) = 1[/imath]
Prove that [imath]( \frac{l}{(l,m)},\frac{m}{(l,m)}) = 1[/imath], given that [imath]l, m \in \Bbb{N}[/imath]. I had this question on my number theory final that I took earlier today. This was the second part of a 2 part question, where the other piece is asked here. I would like to go through the solution that I put down and then ask whether or not my version is rigorous enough to be considered a good proof. Proof Let [imath]l = a_1 a_2 a_3...a_n x[/imath] and [imath]m = b_1 b_2 b_3...b_nx[/imath], where [imath](l,m) = x[/imath]. This means that the common factors Then if we look at [imath]\frac{l}{(l,m)}[/imath], we get [imath]\frac{l}{(l,m)} = \frac{a_1 a_2 a_3...a_n x}{x} = a_1 a_2 a_3...a_n[/imath]. Similarly, looking at [imath]\frac{m}{(l,m)}[/imath], we get [imath]\frac{m}{(l,m)} = \frac{b_1 b_2 b_3...b_nx}{x} = b_1 b_2 b_3...b_n[/imath]. Now if we look at [imath](\frac{l}{(l,m)},\frac{m}{(l,m)})[/imath], we get [imath](a_1 a_2 a_3...a_n, b_1 b_2 b_3...b_n)[/imath] and since the common factors of [imath]l[/imath] and [imath]m[/imath] have been divided away and thus the gcd of [imath](\frac{l}{(l,m)},\frac{m}{(l,m)}) = 1[/imath]. [imath]\blacksquare[/imath] | 350651 | How to prove [imath]\left(\frac{a}{(a,b)}, \frac{b}{(a,b)}\right) = 1[/imath]?
I have a number theory question that has me stumped. Let [imath]a, b, c \in \Bbb Z[/imath] with [imath]a[/imath] and [imath]b[/imath] both not zero. Prove: [imath](\frac{a}{(a,b)}, \frac{b}{(a,b)}) = 1[/imath]. Any help would be greatly appreciated! |
379074 | How do i prove that "every monic polynomial is the characteristic polynomial of some matrix?
Let [imath]F[/imath] be a field and [imath]f(X)\in F[X][/imath] be a monic polynomial such that [imath]deg(f(X))=n[/imath]. How do i prove that there exists [imath]A\in\mathscr{M}_n(F)[/imath] such that [imath]\det(XI_n - A)=f(X)[/imath]? I checked the solution and it says this can be proven using the fact that [imath]\det(XI-A)=X^n - tr(A)X^{n-1}+ ... + (-1)^n \det(A)[/imath], but i don't know how this could be a hint. I'm tried to prove it by induction, but it did't work well... How do i prove it? | 70763 | Do characteristic polynomials exhaust all monic polynomials?
Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix, then [imath]\mathrm{char}_A(x):=\det(xI-A)[/imath] is a monic polynomial of degree [imath]n[/imath]. It is called the characteristic polynomial of [imath]A[/imath]. My question is the converse: Let [imath]p(x)[/imath] be a monic polynomial of degree [imath]n[/imath]. Can we always find an [imath]n\times n[/imath] matrix such that [imath]p(x)=\mathrm{char}_A(x)[/imath]? |
73262 | Proving that [imath]m+n\sqrt{2}[/imath] is dense in [imath]\mathbb R[/imath]
I am having trouble proving the statement: Let [imath]S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}[/imath] Prove that for every [imath]\epsilon > 0[/imath], the intersection of [imath]S[/imath] and [imath](0, \epsilon)[/imath] is nonempty. | 852210 | Why does the additive subgroup of [imath]\mathbb{R}[/imath] generated by [imath]1[/imath] and [imath]\sqrt{2}[/imath] contain arbitrary small elements?
Let [imath]G\subset \mathbb{R}[/imath] be the additive subgroup of [imath](\mathbb{R},+)[/imath] defined by [imath]G=\mathbb{Z}+\sqrt{2}\mathbb{Z}[/imath]. I want to prove that for every [imath]\epsilon>0[/imath] there exists an element [imath]g_\epsilon\in G[/imath] with [imath]\epsilon>g_\epsilon>0[/imath]. Can anybody imagine a nice proof? (I coundn't think of an appropriate tag for this question. Please feel free to add or remove one) |
379330 | Repeating Decimals
I'm just wondering how do we simplify repeating decimals into a fraction in general? Like, for example, [imath]0.5656\dots[/imath] [imath]0.12424\dots[/imath] [imath]4.23777\dots[/imath] Thanks! | 295195 | H0w t0 prove that periodic decimal numbers are rational? [imath]a_1...a_k(b_1b_2..b_l)={m \over n}[/imath]
Given [imath]a_1...a_k(b_1b_2..b_l)={m \over n}[/imath] how can I prove that periodic decimal numbers are rational? Where do I even begin? |
380136 | If [imath]AB=BA[/imath] for all [imath]Bs[/imath], prove that [imath]A=kI[/imath]?
Given [imath]n\times n[/imath] matrix A, if for all [imath]n\times n[/imath] matrices [imath]B[/imath], [imath]AB=BA[/imath] is true, Prove that [imath]A=kI[/imath] Any hints or solutions would be appreciated. Thanks. | 142967 | A be a [imath]3\times 3[/imath] matrix over [imath]\mathbb {R}[/imath] such that [imath]AB =BA[/imath] for all matrices [imath]B[/imath]. what can we say about such matrix [imath]A[/imath]
Let [imath]A[/imath] be a [imath]3\times 3[/imath] matrix over [imath]\mathbb {R}[/imath] such that [imath]AB =BA[/imath] for all matrices [imath]B[/imath] over [imath]\mathbb {R}[/imath] then what can we say about such matrix [imath]A[/imath]. or such matrix [imath]A[/imath] must be orthogonal matrix? Can we say anything about its eigen values? I tried by taking random examples also tried to construct such [imath]3 \times 3[/imath] matrices. But i am not able to get any proper conclusion and proof. |
299626 | The Center of [imath]\operatorname{GL}(n,k)[/imath]
The given question: Let [imath]k[/imath] be a field and [imath]n \in \mathbb{N}[/imath]. Show that the centre of [imath]\operatorname{GL}(n, k)[/imath] is [imath]\lbrace\lambda I\mid λ ∈ k^∗\rbrace[/imath]. I have spent a while trying to prove this and have succeeded if [imath] k \subseteq \mathbb{R}[/imath]. So I imagine there is a nicer way to go about this. I have seen people saying take matrices where every element except one is zero in proving similar results but such matrices are not in [imath]\operatorname{GL}(n,k)[/imath] as they are not invertible. What I have done. I have said let [imath]B \in[/imath] Center then [imath]B[/imath] commutes with everything in [imath]\operatorname{GL}(n,k)[/imath]. So take a permutation matrix, [imath]P[/imath] which swaps rows [imath]i,j[/imath] [imath]P \neq I[/imath]. From this you can deduce that [imath]B^T = B[/imath] and that [imath]B_{ii} = B_{jj}[/imath]. We can use the fact that [imath]B[/imath] is symmetric to find an orthogonal diagonalisation of [imath]B[/imath] by the spectral theorem. This gives [imath]QBQ^T = D \rightarrow B=D[/imath]. And as [imath]B_{ii} = B_{jj}[/imath] we can say [imath]B = k * I[/imath]. But the spectral theorem requires [imath]B[/imath] to be a real matrix. How do I prove this for a general [imath]k[/imath]?. | 400315 | [imath]Z(GL_n(\mathbb R)) = \{aI : a\neq0\} [/imath]
[imath]Z(GL_n(\mathbb R)) = \{aI : a\neq0\} [/imath] This article is the general case for [imath]GL(n,k)[/imath] where [imath]k[/imath] is a field. Could I prove it only with a basic linear algebra? |
380609 | A rotation by an angle [imath]\theta [/imath] in two dimensions
Let [imath]x_i[/imath] be the coordinates of a two-dimensional vector and let [imath]x^\prime_i[/imath] be the coordinates of the vector rotated by an angle [imath]\theta[/imath] in the plane. The components of the two vectors are related by a transformation as [imath]x^\prime_j = R_{ij} x_i[/imath] where [imath]R_{ij}[/imath] is a rotation matrix. This is a representation of the rotation group. Specifically, a rotation by an angle [imath]\theta[/imath] in two dimensions matrix, [imath] R(\theta) = \begin{pmatrix} \cos\theta & \sin \theta\\ - \sin \theta & \cos\theta \\ \end{pmatrix} [/imath] we do we got this matrix, can someone show me the argument or visualization? | 363652 | Understanding rotation matrices
How does [imath] {\sqrt 2 \over 2} = \cos (45^\circ)[/imath]? Is my graph (the one underneath the original) accurate with how I've depicted the representation of the triangle that the trig function represent? What I mean is, the blue triangle is the pre-rotated block, the green is the post-rotated block, and the purple is the rotated change ([imath]45^\circ[/imath]) between them. How do these trig functions in this matrix represent a clockwise rotation? (Like, why does "[imath]-\sin \theta [/imath] " in the bottom left mean clockwise rotation... and "[imath]- \sin \theta [/imath] " in the upper right mean counter clockwise? Why not "[imath]-\cos \theta [/imath] "? [imath]\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta & \cos \theta \end{bmatrix}[/imath] Any help in understanding the trig representations of a rotation would be extremely helpful! Thanks |
380551 | Let G be a finite group and let H and K be subgroups of G. Suppose [G:K] and [G:H] are relatively prime. Prove G=HK
So I am rather confused on where to start this proof so all I've got is [imath][G:H \cap K]=[G:H][H:H \cap K][/imath] [imath][G:H \cap k]=[G:K][K:H \cap K][/imath] Thus that implies [imath][G:H][H:H \cap K]=[G:K][K:H \cap K][/imath] This is where I get stuck. | 372979 | If [imath][G:H][/imath] and [imath][G:K][/imath] are relatively prime, then [imath]G=HK[/imath]
I'm struggling to proof that if [imath]H[/imath] and [imath]K[/imath] are subgroups of finite index of a group [imath]G[/imath] such that [imath][G:H][/imath] and [imath][G:K][/imath] are relatively prime, then [imath]G=HK[/imath]. I don't know why I can't answer it, because this question seems easy. I'm stuck maybe because I've studied so far just Lagrange's theorem and some of its consequences. But I think we don't need much more, because this is the material covered so far by the Hungerford's book. I need help. Thanks. |
381041 | Number of distinct real roots
The equation [imath]x^6 − 5x^4 + 16x^2 − 72x+ 9 = 0[/imath] has (A) exactly two distinct real roots (B) exactly three distinct real roots (C) exactly four distinct real roots (D) six distinct real roots. [imath]f(0)>0[/imath] and [imath]f(1)<0[/imath] and [imath]f(3)>0[/imath], so there should be an odd no of roots between 0 and 1 and 1 and 3 but exactly how many? | 171881 | Multiple choice question - number of real roots of [imath]x^6 − 5x^4 + 16x^2 − 72x + 9[/imath]
The equation [imath]x^6 − 5x^4 + 16x^2 − 72x + 9 = 0[/imath] has (A) exactly two distinct real roots (B) exactly three distinct real roots (C) exactly four distinct real roots (D) six distinct real roots |
381123 | Number of real roots of an equation
What is the number of real roots of the equation [imath]2\cos(\frac{x^2 + x}{6}) = 2^x + 2^{-x}[/imath] How to solve this kind of problems. Any general methods ?? | 380896 | How to find the number of real roots of the given equation?
The number of real roots of the equation [imath]2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}[/imath] is (A) [imath]0[/imath], (B) [imath]1[/imath], (C) [imath]2[/imath], (D) infinitely many. Trial: [imath]\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}[/imath] Then I can't proceed. |
380918 | Prove that the complement of [imath]\mathbb{Q} \times \mathbb{Q}[/imath]. in the plane [imath]\mathbb{R}^2[/imath] is connected.
Prove that the complement of [imath]\mathbb{Q} \times \mathbb{Q}[/imath]. in the plane [imath]\mathbb{R}^2[/imath] is connected. I have no idea how can I do that. If [imath]\mathbb{R} \setminus\mathbb{Q}[/imath] is connected then the proof is easy which is not true. somebody help me please.thanks for your time. | 145912 | Formal proof that [imath]\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}[/imath] is connected.
Cam anyone provide me the proof of: that [imath]\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}[/imath] is connected. |
381243 | Integral of [imath]\sin x \cdot \cos x[/imath]
I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different? [imath] \int \sin x \cos x dx [/imath] 1) via subsitution [imath] u = \sin x [/imath] [imath] u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x [/imath] 2) via subsitution [imath] u = \cos x [/imath] [imath] u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x [/imath] 3) using [imath] \sin 2x = 2 \sin x \cos x [/imath] [imath] \int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x [/imath] So, we have: [imath] \frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x [/imath] | 233971 | A contradictory integral: $\int \sin x \cos x \, \mathrm dx$
I've been thinking about integration lately, and I've come up with a question that I'm not sure how to address. Consider [imath]$$ \int \sin x\cos x \, \mathrm dx = - \int -\sin x \cos x \, \mathrm dx $$[/imath] I started with the integral on the left hand side, which suggests a typical [imath]u[/imath]-substitution. Let [imath]u=\sin x[/imath] then [imath]$\, \mathrm du=\cos x \, \mathrm dx$[/imath]. So the integral evaluates to [imath]$$ \int \sin x\cos x \, \mathrm dx = \frac{\sin^2(x)}{2} $$[/imath] But the original integral also suggests an alternate substitution. Let [imath]u=\cos x[/imath] and then [imath]du=-\sin x \, dx[/imath]. So now [imath]$$ \int \sin x\cos x \, \mathrm dx =- \int -\sin x \cos x \, \mathrm dx= -\frac{\cos^2(x)}{2} $$[/imath] So now I have that the integral evaluates to two different functions. I've tried playing with some different trigonometric identities, but I haven't been able to show that this is true and I'm fairly certain I haven't had any success because the statement itself isn't true. What am I doing wrong? How do you evaluate [imath]$\int \sin x\cos x \, \mathrm dx$[/imath]? |
380967 | Example of a finite, non-abelian group in which left invariant metric is also right invariant
I need an example of a finite, non-abelian group [imath](G, \cdot)[/imath] which satisfies the following condition: If [imath]d[/imath] is a metric on [imath]G[/imath] such that [imath]d(ax, ay)=d(x,y), \ \ \ \ \forall a,x,y \in G[/imath], then [imath]d(xa, ya)=d(x,y), \ \ \ \ \forall a,x,y \in G[/imath]. Could you help me find it or maybe tell me where to look for it? Thank you. | 370552 | Metric on a group
Is there a non-abelian finite group [imath]G[/imath] with the property: If metric [imath]d[/imath] on [imath]G[/imath] is left invariant then is also right invariant? |
381936 | Euler's totient function maximum value for a range
For the euler's totient function, we have a number [imath]n<10^{18}[/imath] we have to find the value of [imath]i[/imath] between [imath]2[/imath] and [imath]n[/imath] (both inclusive) such that the value of [imath]\phi(i)/i[/imath] is maximum. I have have observed that this value will be equal to the largest prime number less than or equal to n. Now since [imath]n[/imath] is upto [imath]10^{18}[/imath], what will be the most efficient way to do this? | 381053 | Maximum of [imath]\frac{\phi(i)}i[/imath]
For a given [imath]N[/imath], is there an approach to find the maximum [imath]\frac{\phi(i)}i[/imath] ([imath]2\le i\le N[/imath])? Like for [imath]n=2[/imath], [imath]\frac{\phi(2)}2=\frac12[/imath] is maximum for [imath]n=3[/imath], [imath]\frac{\phi(3)}3=\frac23[/imath] is maximum for [imath]n=4[/imath], [imath]\frac{\phi(3)}3=\frac23[/imath] is maximum From wikipedia's definition : [imath]\frac{\phi(n)}n = \prod_{p|n}\left (1 - \frac1p\right)[/imath]. So, I guess [imath]p[/imath] should be very big for this approach. Still, I can't find a generalised approach. Please help. |
381987 | A problem on calculating a limit as [imath]n \to \infty[/imath]
What is the value of [imath]\lim _{n\to\infty} (1-\frac{1}{\sqrt2}) \dots (1-\frac{1}{\sqrt{n+1}})[/imath] | 381293 | How can I find [imath]\lim_{n\to \infty} a_n[/imath]
Let [imath]a_n=\left(1-\dfrac{1}{\sqrt2}\right)\dots \left(1-\dfrac{1}{\sqrt{n+1}}\right),n\ge1[/imath] Then find [imath]\lim_{n\to \infty} a_n[/imath]. How can I proceed? I am stuck at the first step. Please help. |
382125 | Is my technique valid?
I have serious doubts about this, but I thought you guys might at least fix this and suggest something useful which would make this approach work. The question is to prove that [imath]2=2\cos(x)+x\sin(x)[/imath] has no solution on [imath](0, 2\pi)[/imath]. So [imath] 2(1-\cos(x))=x\sin(x).[/imath] If for some value of [imath]x[/imath] [imath]LHS<RHS[/imath], and for some other value of [imath]x[/imath] [imath]LHS>RHS[/imath], then there exists a value in between where [imath]LHS=RHS[/imath] (this follows from the fact that both [imath]LHS[/imath] and [imath]RHS[/imath] are continuous functions). So the min of the [imath]LHS[/imath] is at x=0, and it so happens that 0=0 so this is a solution, but is not in the given domain. The max of the [imath]LHS[/imath] is given by [imath]pi/2[/imath], and [imath]2>pi/2[/imath]. Therefore, there is no solution on [imath](0, 2\pi)[/imath]. | 381962 | Show that equation has no solution in [imath](0,2\pi)[/imath]
Hi I want to show that the equation [imath]2=2 \cos(x)+x \sin(x) [/imath] has no solution in [imath](0,2 \pi)[/imath]. Since it is algebraically impossible to solve this equation for [imath]x[/imath] I wanted to ask you whether one of you has an idea how to show this |
382233 | Problem # 25, page 95, from Stein and Rami
Let [imath](X,M,\mu)[/imath] be a measure space with [imath]\mu(X) < 1[/imath]. Show that for any [imath]1\le p<q[/imath], we have [imath]L^q (X,\mu)\subset L^p(X,\mu).[/imath] Let [imath]\ell^p(Z)[/imath] denote the [imath]L^p[/imath] space of the integers equipped with the counting measure. Show that [imath]\ell^p(Z)\subset \ell^q(Z)[/imath] for any [imath]1\le p < q[/imath]. Okay, first Part of the problem already has been discussed and been answered by Davide Giraudo as following:[imath] [/imath] let [imath]1\leq p<q<\infty[/imath] and [imath]f\in L^q[/imath]. We put [imath]E_n:= \left\{x\in X: \frac 1{n+1}\leq |f(x)|\leq\frac 1n\right\}[/imath] for [imath]n\in\mathbb N^*[/imath]. The sets [imath]\{E_n\}[/imath] are pairwise disjoint and by [imath]2[/imath] we get [imath]\displaystyle\sum_{n=1}^{\infty} m(E_n)<\infty[/imath]. We have \begin{align*} \int_X |f|^pdm &=\int_{\{|f|\geq 1\}}|f|^pdm+\sum_{n=1}^{+\infty}\int _{E_n}|f|^pdm\\\ &\leq\int_X |f|^qdm+\sum_{n=1}^{+\infty}\frac 1{n^p}m(E_n)\\\ &\leq \int_X |f|^qdm+\sum_{n=1}^{+\infty}m(E_n)<\infty. \end{align*} Now we look at the case [imath]q=+\infty[/imath]. If [imath]m(E)<\infty[/imath], since for each [imath]f\in L^q[/imath] we can find [imath]C_f[/imath] such that [imath]|f|\leq C_f[/imath] almost everywhere, we can see [imath]f\in L^p[/imath] for all [imath]p[/imath]. Controversely, if [imath]L^{\infty}\subset L^p[/imath] for a [imath]p<\infty[/imath], then the function [imath]f=1[/imath] is in [imath]L^p[/imath] and we should have [imath]m(E)<\infty[/imath]. | 66029 | [imath]L^p[/imath] and [imath]L^q[/imath] space inclusion
Let [imath](X, \mathcal B, m)[/imath] be a measure space. For [imath]1 \leq p < q \leq \infty[/imath], under what condition is it true that [imath]L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)[/imath] and what is a counterexample in the case the condition is not satisfied? |
382575 | closed form for this sequence?
[imath]a_1 = 1; a_2 = 9; a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}[/imath] I need to find non-recurring formula for [imath]a_n[/imath]. Is there any good way to do this? The only one comes to mind is to guess the formula and then prove it using mathematical induction. Thanks in advance! I've got the result and it looks like this: [imath]a_n = \frac{-3*2^{n-1} + 2^{2n - 1} + 1}{3}[/imath] but I really don't like this way and would love to know how to solve this properly. | 375215 | Closed form for [imath](a_n)[/imath] such that [imath]a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}[/imath] with [imath]a_1=1[/imath], [imath]a_2=9[/imath]
[imath]a_1 = 1; a_2 = 9; a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}[/imath] I need to find non-recurring formula for [imath]a_n[/imath]. Is there any good way to do this? The only one comes to mind is to guess the formula and then prove it using mathematical induction. Thanks in advance! I've got the result and it looks like this: [imath]a_n = \frac{-3*2^{n-1} + 2^{2n - 1} + 1}{3}[/imath] but I really don't like this way and would love to know how to solve this properly. |
382085 | Is it possible to evaluate this integral?
Is it possible to evaluate this integral: [imath]\int_{0}^{\frac{\pi }{2}}\ln(\sin 2x){\rm d}x[/imath] | 354795 | Evaluate: [imath]\int_0^{\pi} \ln \left( \sin \theta \right) d\theta[/imath]
Evaluate: [imath] \displaystyle \int_0^{\pi} \ln \left( \sin \theta \right) d\theta[/imath] using Gauss Mean Value theorem. Given hint: consider [imath]f(z) = \ln ( 1 +z)[/imath]. EDIT:: I know how to evaluate it, but I am looking if I can evaluate it using Gauss MVT. ADDED:: Here is what I have got so far!! [imath]\ln 2 = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{i \theta}) d\theta = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{-i \theta}) d\theta[/imath] Hence, [imath] \displaystyle 2 \ln 2 = \frac{1}{2 \pi } \int_{0}^{2 \pi} \log(5 + 4 \cos \theta )d \theta = \frac{1}{\pi} \int_0^{\pi} \log(1 + 8 \cos^2 \theta) d \theta[/imath], now to problem is how to reduce it to the above form? |
377588 | How do you evaluate this line integral, where C is the given curve (Please help)?
Evaluate the line integral [imath]\int_C (x+2y)dx + x^2dy,[/imath] where [imath]C[/imath] consists of line segments from [imath](0,0)[/imath] to [imath](2,1)[/imath] and from [imath](2,1)[/imath] to [imath](3,0)[/imath]. How do you solve this by using the following parametrics? I split them up but got a negative answer of -1/3. What's wrong? For [imath]C_1[/imath] got, [imath]\langle t, t/2\rangle[/imath], [imath]0 \leq t \leq 2[/imath]. For [imath]C_2[/imath] got, [imath]\langle t, 3-y\rangle[/imath], [imath]2 \leq t \leq 3[/imath]. | 375956 | How do you evaluate this line integral, where C is the given curve?
Evaluate the line integral [imath]\int_C (x+2y)dx + x^2dy,[/imath] where [imath]C[/imath] consists of line segments from [imath](0,0)[/imath] to [imath](2,1)[/imath] and from [imath](2,1)[/imath] to [imath](3,0)[/imath]. How do you solve this? I split them up but got a negative answer. For [imath]C_1[/imath] got, [imath]\langle t, 1/2t\rangle[/imath], [imath]0 \leq t \leq 2[/imath]. For [imath]C_2[/imath] got, [imath]\langle t, t-3\rangle[/imath], [imath]2 \leq t \leq 3[/imath]. |
251310 | If [imath]n[/imath] is a natural number [imath]\ge 2[/imath] how do I prove that any graph with [imath]n[/imath] vertices has at least two vertices of the same degree?
Any help would be appreciated. If [imath]n[/imath] is a natural number [imath]\ge 2[/imath] how do I prove that any graph with [imath]n[/imath] vertices has at least two vertices of the same degree? | 1325896 | Why do graph degree sequences always have at least one number repeated?
Why do graph degree sequences always have at least one number repeated? [imath](1, 2, 2, 3)[/imath] = Valid, as you can see, because the [imath]2[/imath] is repeated. [imath](1, 2, 3)[/imath] = Not possible to construct a graph with because all numbers are unique. I know that the graph degree sum must be even, but in this case, both are even. Our professor said it had to do with Combinatorics, but we never touched that topic. EDIT: Loops are not allowed. |
383318 | What is the smallest natural number divisible by the first [imath]n[/imath] natural numbers?
For example, for the numbers 1 to 10, one can just find the necessary factors and multiply them: [imath]5 \times 7 \times 8 \times 9 = 2520[/imath], and all the other numbers in that range follow. But with larger ranges, larger numbers logically result. Is there a simpler way to find this smallest number other than straight multiplication filtering? | 80031 | Smallest integer divisible by all up to [imath]n[/imath]
What's the smallest positive integer which is divisible by all integers [imath]1, 2, \dots, n[/imath]? Is there a simple way to represent the answer? Call it [imath]f(n)[/imath] here. Clearly factorial ([imath]n![/imath]) satisfies the condition of divisibility. But once you get to [imath]f(4)[/imath] (which is [imath]12[/imath]), factorial ([imath]24[/imath]) is too large, because it includes as factors both [imath]4[/imath] and [imath]2[/imath], which are redundant. Multiplying all the prime numbers up to [imath]n[/imath] gives you another estimate, which is this time too low, because you need to include primes multiple times when they're repeated in the factorization of the inputs. So the answer is somewhere between factorial, and the product of primes. Is there a simple answer? |
383725 | How to determine the spectrum on Banach Space
On Banach Space [imath]C[0,1][/imath], T is a bounded linear operator and is defined by [imath]Tf(x)=\int_0^xf(y)dy[/imath], then how can I determine the spectrum of T? I was hinted to first show [imath]T^nf(X)=\frac1{(n-1)!} \int_0^x(x-y)^{n-1}f(y)dy[/imath] | 320336 | Spectral radius of the Volterra operator
The Volterra operator acting on [imath]L^2[0,1][/imath] is defined by [imath]A(f)(x)=\int_0^x f(t) dt[/imath] How can I calculate the spectral radius of [imath]A[/imath] using the spectral radius formula for bounded linear operators: [imath]\rho(A)=\lim_{n\rightarrow \infty} \|A^n\|^{1/n} \text{?}[/imath] This was given as an exercise in a book right after introducing the spectral radius formula, so it should be simple exercise, but I don't see immediately how to do the calculations. Any hint is appreciated. |
383503 | Show there exists a sequence of postive real numbers s.t.
Let [imath](f_n)_{n\in\mathbb N}[/imath] be a sequence of measurable functions on [imath][0,1][/imath] with [imath]\forall n\in\mathbb N, \lvert f_n(x) \rvert < +\infty[/imath] a.e. Show there exists a sequence [imath](c_n)_{n\in\mathbb N}[/imath] of positive real numbers s.t. [imath]f_n(x)/c_n \rightarrow 0[/imath] for almost every [imath]x[/imath] in [imath][0,1][/imath]. Hint: use the borel-cantelli lemma- pick the sequence [imath]c_n[/imath] so that [imath]m[x \in [0,1]:\frac{|f_n(x)|}{c_n}>1/n]<1/2^n[/imath] where m is the measure. | 381780 | Show there exists a sequence of positive real numbers s.t. ...
Let [imath]f_n[/imath] be a sequence of measurable functions on [imath][0,1][/imath] with [imath]|f_n(x)|\lt\infty[/imath] a.e. Show there exists a sequence [imath]c_n[/imath] of positive real numbers s.t. [imath]f_n(x)/c_n\to0[/imath] for almost every [imath]x[/imath] in [imath][0,1][/imath]. Hint: use the borel-cantelli lemma- pick the sequence [imath]c_n[/imath] so that [imath]m[x{\rm\ in\ }[0,1]:|f_n(x)|/c_n\gt1/n]\lt2^{-n}[/imath] where [imath]m[/imath] is the measure. |
382444 | Bound of partials sums [imath]\sum_{k=0}^n f(k)[/imath] for a generic, periodic function
There was a question recently about if [imath]\left|\sum_{k=0}^n \sin k\right| \leq M[/imath] for some [imath]M[/imath]. I wonder, how far could it be generalized? My idea is: Let [imath]f[/imath] be continuous(*), bounded, periodic function with an irrational period [imath]s[/imath] such that [imath]\int_0^s f(x)\,\mathrm{d}x = 0[/imath]. Then there is [imath]M[/imath] such that [imath]\left|\sum_{k=0}^n f(k)\right| \leq M.[/imath] (*) Or rather uniformly or absolutely continuous? The informal idea is that if the period is irrational, the terms of the sum will be "randomly" placed in the interval and since [imath]f[/imath] is continuous and its integral is 0, they should eventually cancel out, just as if we integrated [imath]f[/imath] using Monte-Carlo integration. If it's true, how to prove such a statement? If not, why, and how do we need to strenghten the premisses? | 369690 | Possible generalizations of [imath]\sum_{k=1}^n \cos k[/imath] being bounded
I just read in this month's MAA Math Horizons the problem: Show that [imath]\sum_{k=1}^n \cos k[/imath] is bounded. After a bit of floundering, I realized that the key is to write this as the real part of [imath]\sum_{k=1}^n e^{ik}[/imath] and sum the geometric series. I then realized that this also shows that [imath]\sum_{k=1}^n \cos(ak)[/imath] and [imath]\sum_{k=1}^n \sin(ak)[/imath] are also bounded for any real [imath]a[/imath] not a multiple of [imath]2\pi[/imath], since the denominator of the expression that results is [imath]e^{i a}-1[/imath], and this is nonzero for any such [imath]a[/imath]. So I wondered if there is a generalization that doesn't depend on the properties of [imath]e^{ix}[/imath], and came up with this: If [imath]f[/imath] is continuous and periodic with shortest period [imath]p[/imath], [imath]\int_0^p f(x)\, dx = 0[/imath], and [imath]q[/imath] is not a multiple of [imath]p[/imath], is [imath]\sum_{k=1}^n f(qk)[/imath] bounded? If this is false, are there any additional conditions on [imath]f[/imath] or [imath]q[/imath] (such as [imath]f[/imath] having a bounded derivative or [imath]q/p[/imath] being irrational) that would make it true? (added later, after the comments and the very nice work by Micah) Here is another possible condition: [imath]q/p[/imath] is not an integer and, in every period of [imath]f[/imath], for every [imath]f(x)[/imath], there is at most one other [imath]z[/imath] such that [imath]f(z) = f(x)[/imath]. (This would rule out the possibility suggested by Thomas Andrews.) |
239783 | Maps - question about [imath]f(A \cup B)=f(A) \cup f(B)[/imath] and [imath] f(A \cap B)=f(A) \cap f(B)[/imath]
I am struggling to prove this map statement on sets. The statement is: Let [imath]f:X \rightarrow Y[/imath] be a map. i) [imath]\forall_{A,B \subset X}: f(A \cup B)=f(A) \cup f(B)[/imath] ii) [imath]\forall_{A,B \subset X}: f(A \cap B) \subset f(A) \cap f(B)[/imath] iii) [imath]f[/imath] is injective [imath] \Longleftrightarrow[/imath] [imath]\forall_{A,B \subset X}: f(A \cap B)=f(A) \cap f(B)[/imath] My problem is: I know how to operate on sets, I know how to operate on sets, but I don't know how and where to start the proof, the biggest problem in mathematics, I think. Thanks for help! | 425326 | Prove [imath]f(A \cup B)=f(A) \cup f(B)[/imath] and [imath]f(A \cap B) \subseteq f(A) \cap f(B)[/imath]
Where [imath]S[/imath] and [imath]T[/imath] are metric spaces with metrics [imath]d_S[/imath] and [imath]d_T[/imath]. Given a function [imath]f: S \to T[/imath] and 2 arbitrary subsets [imath]A \subseteq S[/imath] and [imath]B \subseteq S[/imath]. Prove [imath]f(A \cup B)=f(A) \cup f(B)[/imath] and [imath]f(A \cap B) \subseteq f(A) \cap f(B)[/imath]. |
384002 | A counterexample for [imath]\operatorname{Ass}(M_1+M_2)[/imath]
[imath]\newcommand{\Ass}{\operatorname{Ass}}[/imath] Let [imath]A[/imath] be a Noetherian ring and let [imath]M[/imath] be an [imath]A[/imath]-module. Suppose [imath]M=M_{1}+M_{2}[/imath], then we have [imath]\Ass(M)\supset \Ass(M_{1})\cup \Ass(M_{2})[/imath]. What is an example when [imath]\Ass(M)\supsetneqq \Ass(M_{1})\cup \Ass(M_{2})[/imath]? | 73217 | Associated primes of a sum of modules
Let [imath]M[/imath] be a module with [imath]M_1[/imath] and [imath]M_2[/imath] submodules such that their sum (not necessarily a direct sum) is [imath]M[/imath]. Is it true in full generality that [imath]\text{Ass}(M) = \text{Ass}(M_1) \cup \text{Ass}(M_2)[/imath]? If so prove, if not, provide a counterexample. I believe the statement is false. The fact that [imath]\text{Ass}(M_i) \subset M[/imath] is obvious, and I know it holds for a direct sum, however I was hoping someone could clarify with a counter example. Thanks is advance. |
384034 | Show that [imath]\liminf a_n \le \liminf s_n[/imath]
Let [imath]a_1, a_2, a_3,\dotsc[/imath] be a bounded sequence of real numbers. Define [imath]s_n =\frac{(a_1 + a_2 + ...+ a_n)}{n}[/imath], [imath]n \in \mathbb{N}[/imath] Show that [imath]\liminf a_n \le \liminf s_n[/imath] Can I get some help? I am totally stuck on it. | 193157 | If [imath]\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}[/imath] then [imath]\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n[/imath]
This is a question from the book Methods of Real Analysis by R. R. Goldberg. If [imath](s_n)[/imath] is a sequence of real numbers and if [imath]\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}[/imath] then prove that: [imath]\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n[/imath]. I don't have any idea how to start working on this problem. Please help. Thanks. |
266364 | How to evaluate the series [imath]1+\frac34+\frac{3\cdot5}{4\cdot8}+\frac{3\cdot5\cdot7}{4\cdot8\cdot12}+\cdots[/imath]
How can I evaluate the following series: [imath]1+\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12}+\frac{3\cdot 5\cdot 7\cdot 9}{4\cdot 8\cdot 12\cdot 16}+\cdots[/imath] In one book I saw this sum is equal to [imath]\sqrt{8}[/imath], but I don't know how to evaluate it. | 789098 | [imath]1 + \frac34 + \frac34.\frac58 + \frac34.\frac58.\frac7{12} + ....... = ?[/imath]
What is the value of the following infinite sum: [imath]1 + \frac34 + \frac34.\frac58 + \frac34.\frac58.\frac7{12} + .......[/imath] I need a definitive hint, please help. |
384586 | If [imath]{x_n}[/imath] and [imath]{y_n}[/imath] are sequences of real numbers then are the followings true?
If [imath]{x_n}[/imath] and [imath]{y_n}[/imath] are sequences of real numbers then are the followings true? a) [imath]\limsup(x_n+y_n)\le\limsup x_n+\limsup y_n[/imath] b) [imath]\liminf x_n+\liminf y_n\le\liminf(x_n+y_n)[/imath] | 324981 | lim sup inequality proof - is this the right way to think?
I have tried to read many proofs of this but I'm not sure I get it, so please bare with me. Show that [imath]\lim_{n \rightarrow \infty} \sup (a_n+b_n) \leq \lim_{n \rightarrow \infty} \sup (a_n)+lim_{n \rightarrow \infty} \sup (b_n).[/imath] I care about the case when [imath](a_n + b_n)[/imath] is bounded. Take an arbitrary subsequence [imath]({a_n}_k + {b_n}_k)[/imath], it must converge to a limit [imath]l[/imath] since [imath](a_n + b_n)[/imath] is bounded. Take the subsequences [imath]({{a_n}_k}_j)[/imath] and [imath]({{b_n}_k}_j)[/imath], since [imath]({a_n}_k + {b_n}_k)[/imath] converges to [imath]l[/imath], [imath]({{a_n}_k}_j)[/imath] + [imath]({{b_n}_k}_j)[/imath] must also converge to [imath]l[/imath]. So we have [imath]\lim_{n \rightarrow \infty}({a_n}_k + {b_n}_k)=lim_{n \rightarrow \infty}({{a_n}_k}_j)+\lim_{n \rightarrow \infty}({{b_n}_k}_j) \leq \lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n)[/imath] which is true by the definition of supremum. (Is there some intuition here?) Well, any subsequence [imath](a_n+b_n)[/imath] that converges is bounded above by [imath]\lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n) [/imath], so certainly [imath]\lim_{n \rightarrow \infty} \sup (a_n+b_n)[/imath] is bounded above, or in other words [imath]\lim_{n \rightarrow \infty} \sup (a_n+b_n) \leq \lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n).[/imath] Questions: Is this a correct proof? The inequality seems obvious by itself, because the suprema of two sequences will always be an upper bound of those two sequences added together. The only other thing that can happen is that those suprema are an upper bound for [imath](a_n+b_n)[/imath], but not the least upper bound - if this is correct, is this a case of 'so easy it's hard' or am I missing an essential piece of the puzzle? None of the proofs I've seen online have made any sense to me, so I'm guessing it's the latter. |
384619 | Can we write every uncountable set [imath]U[/imath] as [imath]V∪W[/imath], where [imath]V[/imath] and [imath]W[/imath] are disjoint uncountable subsets of [imath]U[/imath]?
Is it true that for every uncountable set [imath]U[/imath], we can write [imath]U=V∪W[/imath], where [imath]V[/imath] and [imath]W[/imath] are disjoint uncountable subsets of [imath]U[/imath] ? | 17432 | Uncountable subset with uncountable complement, without the Axiom of Choice
Let [imath]X[/imath] be a set and consider the collection [imath]\mathcal{A}(X)[/imath] of countable or cocountable subsets of [imath]X[/imath], that is, [imath]E \in \mathcal{A}(X)[/imath] if [imath]E[/imath] is countable or [imath]X-E[/imath] is countable. If [imath]X[/imath] is countable, then [imath]\mathcal{A}(X)[/imath] coincides with the power set [imath]\mathcal{P}(X)[/imath] of [imath]X[/imath]. Now suppose that [imath]X[/imath] is uncountable. Assuming the axiom of choice, we can conclude that [imath]\mathcal{A}(X) \ne \mathcal{P}(X)[/imath], since [imath]|X| = |X| + |X|[/imath]. So the question is: Can we prove in ZF that [imath]\mathcal{A}(X) \ne \mathcal{P}(X)[/imath] for every uncountable set [imath]X[/imath]? I'm assuming that a set [imath]X[/imath] is uncountable if there is no injective function [imath]f : X \rightarrow \mathbb{N}[/imath]. |
384150 | prove [imath] e^{bt} \ \int_0^t \ f(s) ds=\int_0^t \ ( e^{-bs} \ f(s)-be^{-bs}\int_0^s\ f(u)\ du) \ ds [/imath]
please How can I prove that [imath] e^{-bt} \ \int_0^t \ f(s) ds=\int_0^t \ ( e^{-bs} \ f(s)-be^{-bs}\int_0^s\ f(u)\ du) \ ds [/imath] f non-negative measurable function I would appreciate it enormously if anyone could help best,educ | 383892 | prove Complicated Integral.
please How can I prove that [imath] e^{bt} \ \int_0^t \ f(s) ds=\int_0^t \ ( e^{-bs} \ f(s)-be^{-bs}\int_0^s\ f(u)\ du) \ ds [/imath] f non-negative measurable function best, Educ |
385796 | Density function of the sum of uniformly distributed random variables
Suppose we choose independently two numbers [imath]X[/imath] and [imath]Y[/imath] , at random, from the interval [imath][0,1][/imath] with uniform probability density. What is the density of their sum [imath]Z = X + Y[/imath] ? I know that the answer is: [imath] f_Z(z) = \left\{ \begin{array}{lr} z & : 0 \leq z \leq 1\\ 2-z & : 1 < z \leq 2\\ 0 & : \text{otherwise} \end{array} \right.[/imath] But I'm having a hard time seeing the translation from the P.D.F. of just a single random variable to the sum of two. I think an explanation for three random variables (i.e. [imath]F_Z'(z')[/imath] for [imath]Z' = A + B + C[/imath] where [imath]A[/imath],[imath]B[/imath], and [imath]C[/imath] are uniformly distributed on [imath][0,1][/imath]) would help illustrate this. | 357672 | density of sum of two uniform random variables [imath][0,1][/imath]
I am trying to understand an example from my textbook. Let's say [imath]Z = X + Y[/imath], where [imath]X[/imath] and [imath]Y[/imath] are uniform random variables with range [imath][0,1][/imath]. Then the PDF is [imath]f(z) = \begin{cases} z & \text{for $0 < z < 1$} \\ 2-z & \text{for $1 \le z < 2$} \\ 0 & \text{otherwise.} \end{cases}[/imath] How was this PDF obtained? Thanks |
386210 | Let [imath]{P_n}[/imath] be the sequence of all consecutive prime numbers. Is [imath]\sum_{n\geq 1} \frac{1}{p_n}[/imath] convergent?
Let [imath]{P_n}[/imath] be the sequence of all consecutive prime numbers. Is [imath]\sum_{n\geq 1}\frac{1}{p_n}[/imath] convergent? | 15946 | Does the sum of reciprocals of primes converge?
Is this series known to converge, and if so, what does it converge to (if known)? Where [imath]p_n[/imath] is prime number [imath]n[/imath], and [imath]p_1 = 2[/imath], [imath]\sum\limits_{n=1}^\infty \frac{1}{p_n}[/imath] |
94012 | Sum of reciprocal prime numbers
How can the following equation be proven? [imath] \forall n > 2 : \sum_{p \le n}{\frac1{p}} = C + \ln\ln n + O\left(\frac1{\ln n}\right), [/imath] where [imath]p[/imath] is a prime number. It's not homework; I just don't understand from where should I start. | 2315122 | Divergent series of [imath]\frac{1}{P_n}[/imath]
the series of terms [imath]\frac{1}{P_n}[/imath], where [imath]P_n[/imath] is the [imath]n^{th}[/imath] prime number, is divergent, I know only one proof using logarithm, is there another proof? |
385998 | computing a limit of integrals, maybe using the Convergence dominated theorem
Let's consider a measure space [imath](X,M,\mu)[/imath] and [imath]f[/imath] a measurable function [imath]f:X\to \mathbb R[/imath] such that [imath]\int f d\mu <\infty[/imath] , for every [imath]\alpha>0[/imath] compute the following limit: [imath] \mathop {\lim }\limits_{n \to \infty } \int {n\log \left( {1 + \left( {\frac{f} {n}} \right)^\alpha } \right)d\mu } [/imath] Maybe I have to use the Convergence dominated theorem, but I don't know how. Maybe [imath]f[/imath] is nonnegative. | 383199 | Calculate the following: [imath]\lim\limits_{n\to \infty}\int_X n \log(1+(\frac{f}{n})^{\alpha})d\mu[/imath]
Let[imath](X,\mathcal{M},\mu)[/imath] be measure space, [imath]f[/imath] is non-negative integrable function on [imath]X[/imath], and [imath]\int_X fd\mu=c,0<c<\infty[/imath].Then calculate the following: [imath]\lim\limits_{n\to \infty}\int_X n \log\left(1+\left(\frac{f}{n} \right)^{\alpha} \right)d\mu[/imath] depending on positive [imath]\alpha[/imath]. (1). When [imath]f>0[/imath], calculate [imath]\lim\limits_{n\to \infty}n \log\left(1+\left(\frac{f}{n} \right)^{\alpha} \right)[/imath]. (2). When [imath]0<\alpha<1[/imath], use Fatou's Lemma. (3). When [imath]\alpha=1[/imath], Using [imath]x\ge 0[/imath], then [imath]\log(1+x)\le x[/imath], then use Lebesgue convergence theorem. (4). When [imath]\alpha>1[/imath], Using [imath]x\ge 0[/imath], then [imath]1+x^{\alpha} \le (1+x)^{\alpha}[/imath]. My Solution: (1). [imath]n\log \left(1+\left(\frac{f}{n}\right)^{\alpha} \right) = \log(1+\frac{f^\alpha}{n^\alpha})^n=\log\sqrt[\alpha]{\left(1+\dfrac{f^\alpha}{n^\alpha}\right)^{n^\alpha}} \to \log\sqrt[\alpha]{e^{f^\alpha}}=f[/imath] as [imath]n\to \infty[/imath] (2). I don't know the answer. (3). When [imath]\alpha=1[/imath], [imath]|n\log(1+f/n)|\le n(f/n)=f[/imath], using domainted convergence theorem. [imath]\lim\limits_{n\to \infty}\int_X n\log(1+(\frac{f}{n})^{\alpha})d\mu=\int_X fd\mu=c[/imath] (4). As (3), [imath]|n\log(1+(f/n)^\alpha)|<n\log(1+f/n)^{\alpha}=\log(1+f/n)^{n\alpha}\le f\alpha[/imath], as [imath]n\to \infty[/imath]. Then [imath]\lim\limits_{n\to \infty}\int_X n\log(1+(\frac{f}{n})^{\alpha})d\mu=\int_X \alpha fd\mu=\alpha c[/imath]. |
38902 | The set of differences for a set of positive Lebesgue measure
Quite a while ago, I heard about a statement in measure theory, that goes as follows: Let [imath]A \subset \mathbb R^n[/imath] be a Lebesgue-measurable set of positive measure. Then we follow that [imath]A-A = \{ x-y \mid x,y\in A\}[/imath] is a neighborhood of zero, i.e. contains an open ball around zero. I now got reminded of that statement as I have the homework problem (Kolmogorov, Introductory Real Analysis, p. 268, Problem 5): Prove that every set of positive measure in the interval [imath][0,1][/imath] contains a pair of points whose distance apart is a rational number. The above statement would obviously prove the homework problem and I would like to prove the more general statement. I think that assuming the opposite and taking a sequence [imath]\{x_n\}[/imath] converging to zero such that none of the elements are contained in [imath]A[/imath], we might be able to define an ascending/descending chain [imath]A_n[/imath] such that the union/intersection is [imath]A[/imath] but the limit of its measures zero. I am in lack of ideas for the definition on those [imath]A_n[/imath]. I am asking specifically not for an answer but a hint on the problem. Especially if my idea turns out to be fruitful for somebody, a notice would be great. Or if another well-known theorem is needed, I surely would want to know. Thank you for your help. | 497904 | Let E be a measureable subset of [0,1] with positive measure. Let [imath]E-E= \{ x-y | x,y \in E\}[/imath]Show that this set contains a neighborhood of 0.
Let E be a measureable subset of [0,1] with positive measure. Let [imath]E-E= \{ x-y | x,y \in E\}[/imath]Show that this set contains a neighborhood of 0. (hint use inner approximations) I think : this set is measurable as it is a union of translates of E, 0 is contained in the set as [imath]x-x \in E-E [/imath] ( [imath] \forall x \in E[/imath]),and I suspect that E-E is contained in E but am not sure about this. Thanks for any help or hints |
386470 | Find the following integral:
Find [imath]\int \sqrt{\tan x}dx[/imath] My attempt: [imath]\text{Let}\ I=\int \sqrt{\tan(x)}dx[/imath] [imath]\text{Let}\ u=\tan(x), du=(1+\tan^{2}(x))dx[/imath] [imath]I=\int \frac{\sqrt{u}}{u^{2}+1}[/imath] [imath]\text{Let}\ v=\sqrt{u}, dv=\frac{du}{2\sqrt{u}}[/imath] [imath]I=2\int \frac{v^{2}}{v^{4}+1}[/imath] [imath]\int_0^\infty\frac{x^2}{1+x^4}dx[/imath] [imath]\text{Let}\ t=\frac{1}{v} \therefore dt=\frac{-dv}{v^2}[/imath] [imath]\therefore I=\int \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}[/imath] [imath]I=-\int \frac{dt}{1+t^4}[/imath] Where do I go from here? | 43457 | How can I compute the integral [imath]\int_{0}^{\infty} \frac{dt}{1+t^4}[/imath]?
I have to compute this integral [imath]\int_{0}^{\infty} \frac{dt}{1+t^4}[/imath] to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken this course yet. Thanks for any help. |
386373 | [imath]\operatorname{rank}(A + B) ≤ \operatorname{rank}(A) + \operatorname{rank}(B)[/imath]
Let [imath]A, B ∈ M_{m×n}(F)[/imath]. Could someone give a hint as to how to prove that [imath]\operatorname{rank}(A + B) ≤ \operatorname{rank}(A) + \operatorname{rank}(B).[/imath] | 246941 | Prove that [imath]\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A + B)[/imath]
Let [imath]A,B[/imath] be matrices [imath]m\times n[/imath] [imath](A, B \in M_{m\times n}(R))[/imath]. How can we prove that [imath]\operatorname{rank}(A+ B) \le \operatorname{rank}(A) + \operatorname{rank}(B)\ ?[/imath] |
375982 | Rank of the difference of matrices
Let [imath]A[/imath] and [imath]B[/imath] be to [imath]n \times n[/imath] matrices. My question is: Is [imath]\operatorname{rank}(A-B) \geq \operatorname{rank}(A) - \operatorname{rank}(B)[/imath] true in general? Or maybe under certain assumptions? | 853279 | Show that the rank of [imath]A+B[/imath] is no more than the sum of th ranks of [imath]A[/imath] and [imath]B[/imath]
Let [imath]k[/imath] and [imath]n[/imath] be positive integers and let [imath]F[/imath] be a field. For matrices [imath]A,B \in M_{k\times n} (F)[/imath], Show that the rank of [imath]A+B[/imath] is no more than the sum of th ranks of [imath]A[/imath] and [imath]B[/imath] I believe this question is addressed here but to be quite honest, I dont quite understand the explanations given. Can someone possibly help with a more detailed explanation? |
386917 | Show that [imath]h \equiv 1 \pmod p[/imath], where [imath]h[/imath] is the number of subgroups of order [imath]p[/imath] and [imath]p[/imath] divides the group order.
Let [imath]G[/imath] be a finite group and [imath]p[/imath] a prime number that divides the order of [imath]G[/imath]. Let [imath]h[/imath] be the number of subgroups of [imath]G[/imath] of order [imath]p[/imath]. Prove that there are [imath]h(p-1)[/imath] elements of order [imath]p[/imath] in [imath]G[/imath]. Furthermore, prove [imath]h \equiv 1 \pmod p[/imath]. How should I solve the second part? This is what I did to prove the first: Let [imath]H[/imath] be a subgroup of [imath]G[/imath] of order [imath]p[/imath]. Because the order of an element in [imath]H[/imath] must divide the order of the group, its order must be [imath]1[/imath] or [imath]p[/imath]. Since only the identity element has order [imath]1[/imath], the remaining [imath]p-1[/imath] elements have order [imath]p[/imath]. Assume that an element [imath]x[/imath] of order [imath]p[/imath] is contained in two subgroups [imath]H_1[/imath] and [imath]H_2[/imath]. Then [imath]x[/imath] is a generator of both [imath]H_1[/imath] and [imath]H_2[/imath], so [imath]\langle x \rangle = H_1 = H_2[/imath], so an element of order [imath]p[/imath] is in only one such group. So there are at least [imath]h(p-1)[/imath] elements of order [imath]p[/imath] in [imath]G[/imath]. Assume there is an element [imath]z[/imath] of order [imath]p[/imath] which is not contained in any of the [imath]h[/imath] subgroups of order [imath]p[/imath]. But we know that [imath]\langle x \rangle[/imath] is a subgroup of [imath]G[/imath] of order [imath]p[/imath]. So all elements of order [imath]p[/imath] must be in one of those [imath]h[/imath] subgroups. Therefor, there are exactly [imath]h(p-1)[/imath] elements of order [imath]p[/imath] in [imath]G[/imath]. Now, how do I prove that [imath]h \equiv 1 \pmod p[/imath]? This question is in a chapter about group actions, so that should maybe help. I'm assuming that I have to use the previously proven fact. And something that has to do with group actions. However, I'm not quite used to that concept and when it is useful to apply (more importantly, how to apply it). Could someone also explain to me its importance (especially in the case of this question)? Thanks in advance. | 97460 | Number of subgroups of prime order
I've been doing some exercises from my introductory algebra text and came across a problem which I reduced to proving that: The number of distinct subgroups of prime order [imath]p[/imath] of a finite group [imath]G[/imath] is either [imath]0[/imath] or congruent to [imath]1\pmod{p} [/imath]. With my little experience I was unable to overcome this (all I was able to conclude is that these groups are disjoint short of the identity), and also did not find any solution with a search on google (except for stronger theorems which I am not interested in because of my novice level). I remember that a similar result is widely known as one of Sylow Theorems. This result was proven by the use of group actions. But can my problem be proved without using the concept of group actions? Can this be proven WITH the use of that concept? EDIT: With help from comments I came up with this: The action Derek proposed is well-defined largely because in a group if [imath]ab = e[/imath] (the identity), then certainly [imath]ba = e[/imath]. By Orbit-Stabilizer Theorem we can see that all orbits are either of size 1 or [imath]p[/imath] (here I had most problems, and found out cyclic group of order [imath]p[/imath] acts on the set of solutions in the same way). The orbits of size 1 contain precisely the elements [imath](x,x,x....,x)[/imath] for some element x in G. In addition, orders of all orbits add up to [imath]|G|^{(p-1)}[/imath] because the orbits are equivalence classes of an equivalence relation. But certainly [imath](e,e,e....,e)[/imath] is in an orbit of size 1, and that means there has to be more orbits of exactly one element, actually [imath]p-1 + np[/imath] more for some integer [imath]n[/imath]. These elements form the disjoint groups I am looking for. if [imath]p-1[/imath] divides [imath](p-1 + np)[/imath], it's easy to check the result is 1 mod p. Could someone check if I understood this correctly? |
387070 | Prove that [imath](a+1)(a+2)...(a+b)[/imath] is divisible by [imath]b![/imath]
The problem is following, prove that: [imath](a+1)(a+2)...(a+b)\text{ is divisible by } b!\text{ for every positive integer a,b}[/imath] I've tried solving this problem using mathematical induction, but I don't think that i did it correctly. Here's what i've done [imath]1.\ b=1\ (Basis)[/imath] [imath]b!| (a+1)(a+2)...(a+b)[/imath] [imath]1 | a+1 \text{, which is true}[/imath] [imath]2.\ b=k\ (Induction\ Hypothesis)[/imath] [imath]k! | (a+1)(a+2)...(a+k)\text{, we assume it's true}[/imath] [imath]k!*n = (a+1)(a+2)...(a+k)\text{, n is some positive integer}[/imath] [imath]3.\ b=k+1\ (Inductive\ Step)[/imath] In order to prove I should get: [imath](k+1)!*m = (a+1)(a+2)...(a+k)(a+k+1)\text{, where m is some positive integer}[/imath] [imath](a+1)(a+2)...(a+k)(a+k+1) = k!*n*(a+k+1)[/imath] [imath](a+1)(a+2)...(a+k)(a+k+1) = (k+1)!*n + k!*n*a[/imath] [imath](a+1)(a+2)...(a+k)(a+k+1) - a(a+1)(a+2)...(a+k) = (k+1)!*n[/imath] [imath](a+1)(a+2)...(a+k)(k+1) = (k+1)!*n[/imath] And as you can see I'm returning at the beginning. Also I've tried to use the fact that in b consecutive numbers, there must be at least one that divides b, but since i eliminate that number (because I couldn't be sure that the quotient is divisible with (b-1) or (b-2) or ... or 2. And after this i can't continue with (b-1) using this method, because it's not necessary for the other (b-1) to be consecutive. | 12065 | The product of [imath]n[/imath] consecutive integers is divisible by [imath]n[/imath] factorial
How can we prove that the product of [imath]n[/imath] consecutive integers is divisible by [imath]n[/imath] factorial? Note: In this subsequent question and the comments here the OP has clarified that he seeks a proof that "does not use the properties of binomial coefficients". Please post answers in said newer thread so that this incorrectly-posed question may be closed as a duplicate. |
387135 | Any finite subgroup of the abelian group [imath](F-\{0\},\cdot)[/imath] is cyclic? ([imath]F[/imath] a field)
I found this problem: Suppose that [imath]F[/imath] is a field, and that [imath](F-\{0\},\cdot)[/imath] is an abelian group. Show that if [imath]H[/imath] is a finite subgroup of [imath]F-\{0\}[/imath], then [imath]H[/imath] is cyclic. What I have done is: Since [imath](F-\{0\},\cdot)[/imath] is an abelian group and [imath]H[/imath] is finite, [imath]H[/imath] is a finitely generated abelian group, and by the fundamental theorem of finitely generated abelian groups, we know that [imath]H[/imath] is isomorphic to something like [imath]\Bbb{Z}_{m_1} \times \cdots \times \Bbb{Z}_{m_k}[/imath] with [imath]m_{i}\mid m_{i+1}[/imath]. It means that [imath]\left|H\right|= m_1 \cdots m_k[/imath]. And I know that if I prove that [imath]m_i[/imath] is prime, it means that [imath]\left|H\right|=m^k[/imath] with [imath]m[/imath] prime, so it would be obvious that [imath]H[/imath] is cyclic. But I have no idea how to show this, so any ideas will help a lot. Thanks. P.S. If the problem is wrong, explain why. | 59903 | Finite subgroups of the multiplicative group of a field are cyclic
In Grove's book Algebra, Proposition 3.7 at page 94 is the following If [imath]G[/imath] is a finite subgroup of the multiplicative group [imath]F^*[/imath] of a field [imath]F[/imath], then [imath]G[/imath] is cyclic. He starts the proof by saying "Since [imath]G[/imath] is the direct product of its Sylow subgroups ...". But this is only true if the Sylow subgroups of [imath]G[/imath] are all normal. How do we know this? |
387146 | Why is this True? I find it False
Mostly I believe in math. However I have trouble in my economic textbook (which really should be right). I have the following equation: [imath] u(c,d)=\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}[/imath] where [imath]\theta=(1-\gamma)/(1-1/\psi)[/imath], [imath]a\in (0,1)[/imath],[imath]b\in (0,1)[/imath], [imath]c>0[/imath], [imath]d>0[/imath], [imath]\gamma>0[/imath], and [imath]\psi>0)[/imath]. So now my True or False question follows: Is it true or false that the derivative wrt. [imath]c[/imath] is: [imath] u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{1/\psi}[/imath] Can some tell me why this is true ? I would write it as [imath] u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{-1}[/imath] Hope to hear from someone, Thanks in advance | 387144 | Taking a complicated partial derivative
Mostly I believe in math. However I have trouble in my economic textbook (which really should be right). I have the following equation: [imath] u(c,d)=\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}[/imath] where [imath]\theta=(1-\gamma)/(1-1/\psi)[/imath], [imath]a\in (0,1),b\in (0,1), c>0, d>0, \gamma>0[/imath], and [imath]\psi\in(0,\infty)[/imath]. Is the derivative of [imath]u[/imath] wrt. [imath]c[/imath] really [imath] u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{1/\psi}[/imath] If so, can someone tell me why? I would write it as [imath] u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{-1}[/imath] Hope to hear from someone, thanks in advance. |
386907 | Isometry, compact space, there exists only one isometry other than identity
I'm solving a problem concerning isometry in a compact metric space [imath]([0,1],d)[/imath], the metric [imath]d[/imath] is consistent with natural topology of [imath][0,1][/imath]. Here is the problem: Prove that in the space defined above there exists only one isometry other than identity. Is it true that if [imath]f[/imath] is the isometry, then the sequence [imath](f^n(x)), \ n \in \mathbb{N}[/imath] is strictly monotone and if so, how to prove it? | 369454 | Metric space has at most one isometry other than identity
Could you help me with this problem? Let [imath]d[/imath] be a metric on [imath][0,1][/imath] consistent with the standard topology. Prove that the metric space: [imath]([0,1], d)[/imath] has at most one isometry (except for identity). I would really appreciate your help. Thank you. |
387345 | Isomorphism of polynomial rings implying isomorphism of the coefficient rings
Let [imath]R[/imath] and [imath]S[/imath] be commutative rings. Let [imath]x, y[/imath] be indeterminates, and assume that one has an isomorphism [imath]R[x] \rightarrow S[y][/imath] (not necessarily mapping [imath]x[/imath] to [imath]y[/imath] of course). Does this imply [imath]R \cong S[/imath]? If not, what is a counterexample? This may seem like a homework problem, but is not. | 13504 | Does [imath]R[x] \cong S[x][/imath] imply [imath]R \cong S[/imath]?
This is a very simple question but I believe it's nontrivial. I would like to know if the following is true: If [imath]R[/imath] and [imath]S[/imath] are rings and [imath]R[x][/imath] and [imath]S[x][/imath] are isomorphic as rings, then [imath]R[/imath] and [imath]S[/imath] are isomorphic. Thanks! If there isn't a proof (or disproof) of the general result, I would be interested to know if there are particular cases when this claim is true. |
387187 | Sum of squares of the degrees of irreducible representations equals order of group (positive characteristic case)
Suppose [imath]K[/imath] is a splitting field for a finite group [imath]G[/imath] such that [imath]p = \mathrm{char} K >0[/imath] and [imath]p \nmid |G|[/imath]. Let [imath]\{\rho_1, \ldots, \rho_s\}[/imath] be the set of all irreducible representations (up to equivalence) of [imath]G[/imath] over [imath]K[/imath] of degrees [imath]d_1, \ldots, d_s[/imath] respectively. Show that [imath]\sum_{i=1}^{s}{d_i^2}=|G|.[/imath] I'm aware of the general approach to proving this result for the case where [imath]K[/imath] is of characteristic [imath]0[/imath] but I'm struggling to see how the result may be proved for the positive characteristic case. Any help/guidance in this would be greatly appreciated. EDIT: This has been marked as a duplicate of: Sum of squares of dimensions of irreducible characters. This is not quite the same question as in the link, as the other question deals only with the characteristic [imath]0[/imath] case. | 234557 | Sum of squares of dimensions of irreducible characters.
For anyone familiar with Artin's Algebra book, I just worked through the proof of the following theorem, which can be seen here: (5.9) Theorem Let [imath]G[/imath] be a group of order [imath]N[/imath], let [imath]\rho_1,\rho_2,\dots[/imath] represent the distinct isomorphism classes of irreducible representations of [imath]G[/imath] and let [imath]\chi_i[/imath] be the character of [imath]\rho_i[/imath]. (a) Orthogonality relations: The characters [imath]\chi_i[/imath] are orthonormal. In other words [imath]\langle\chi_i,\chi_j\rangle=0[/imath] if [imath]i\ne j[/imath], and [imath]\langle\chi_i,\chi_i\rangle=1[/imath] for each [imath]i[/imath]. (b) There are finitely many isomorphism classes of irreducible representations, the same number as the number of conjugacy classes in the group. (c) Let [imath]d_i[/imath] be the dimension of the irreducible representation [imath]\rho_i[/imath], let [imath]r[/imath] be the number of irreducible representations. Then [imath]d_i[/imath] divides [imath]N[/imath] and [imath]N=d_1^2+\dots+d_r^2.[/imath] This theorem will be proved in Section 9, with the exception of the assertion that [imath]d_i[/imath] divides [imath]N[/imath], which we will not prove. The theorem was contained in the last section, but the proof for part (c) was missing completely. It is mentioned that the divisibility property would not be proved, but the sum of squares formula for [imath]N[/imath] is not verified at all. In applications on homework this property is used extensively to fill in missing characters for character tables of finite groups, so I would like to understand why it is true. Can anyone suggest a reference or sketch an argument? Thanks in abundance! |
387553 | Definite trig integral
How do I evaluate: [imath]\int_{0}^{\pi} \sin (\sin x) \ dx[/imath] I have seen a similar question here but can't find it. | 287395 | Evaluate [imath]\int\sin(\sin x)~dx[/imath]
I was skimming the virtual pages here and noticed a limit that made me wonder the following question: is there any nice way to evaluate the indefinite integral below? [imath]\int\sin(\sin x)~dx[/imath] Perhaps one way might use Taylor expansion. Thanks for any hint, suggestion. |
387647 | Finding number of functions in set A
Let [imath]A = \{1, 2, 3, 4, \ldots, n\}[/imath] (it follows that: [imath]|A| = n[/imath]). My objective is to count the number of functions: [imath]f: A \rightarrow A[/imath], that are monotonically increasing, i.e. [imath]f(x) \leq f(x + 1)[/imath], and the following equation holds: [imath]f(x) = f(f(x+1))[/imath]. It is trivial to see that any constant function of the form [imath]f(x) = k; x, k \in A[/imath] satisfies all conditions. There are exactly [imath]n[/imath] such functions, yet I want to disregard these and only count the non-constant functions. However, I have not had any luck in finding a function that satisfies the second condition. Could anyone assist in finding such functions and the process of counting all functions that satisfy the conditions? (Having studied some combinatorics, I know that the total number of functions (no conditions) from A to A, is [imath]n^n[/imath]. Further, I may subtract the number of all constant functions, for which we established that it is equal to [imath]n[/imath]. So far I have [imath]n^n - n[/imath]. Since the function should be monotonically increasing, the number of all decreasing functions should be subtracted from the result, and furthermore, all monotonically increasing functions that do not satisfy the second condition should be subtracted. I suppose that there is a different approach though.) | 379298 | Function mapping challange
For a given set [imath]A=\{1, 2, 3, 4, \ldots, n\}[/imath], find the number of non-constant mappings (associations ) [imath]f[/imath] from [imath]A[/imath] to [imath]A[/imath] such that [imath]f(k) \leq f(k + 1)[/imath] and [imath]f(k) = f(f(k + 1))[/imath]. This is the text of the problem. But I don't understand how can we have [imath]f(k)[/imath], when we are not given a function? Constant mapping is a mapping [imath]f[/imath] such that for some C, [imath]f(k)[/imath]=C for all k. Thanks to @ferson2020 for explanation |
260686 | Eigenvalues of complex [imath]2×2[/imath] matrices
Let [imath]A[/imath] be complex [imath]2×2[/imath] matrices s.t. [imath]A^2=0[/imath]. Which of the following statements are true? [imath]PAP^{-1}[/imath] is diagonal for some [imath]2×2[/imath] real matrix [imath]P[/imath]. [imath]A[/imath] has [imath]2[/imath] distinct eigenvalues in [imath]\Bbb C[/imath]. [imath]A[/imath] has [imath]1[/imath] eigenvalue in [imath]\Bbb C[/imath] with multiplicity [imath]2[/imath]. [imath]Av=v[/imath] for [imath]v\in \Bbb C^2 ,v≠0[/imath]. | 144656 | Eigenvalues of a matrix [imath]A[/imath] such that [imath] A^2=0[/imath].
Suppose the matrix [imath]A[/imath] is a [imath]2 \times 2[/imath] non-zero matrix with entries in [imath]\Bbb C[/imath]. Which of the following statements must be true? [imath]PAP^{-1}[/imath] is a diagonal matrix for some invertible matrix [imath]P[/imath] with entries in [imath]\Bbb R[/imath]. [imath]A[/imath] has only one distinct eigenvalue in [imath]\Bbb C[/imath] with multiplicity [imath]2[/imath]. [imath]A[/imath] has two distinct eigenvalues in [imath]\Bbb C[/imath]. [imath]Av = v[/imath] for a non-zero [imath]v[/imath]. Please suggest which of the possibilities hold. It seems to me that the characteristic poly is [imath] f(t) = t^2[/imath], which means option (2) holds that is only one eigenvalue zero with multiplicity [imath]2[/imath]. |
388202 | Let [imath]f(x)[/imath] be continuous on [[imath]0,1[/imath]] to [imath]f(x)=0[/imath] for each [imath]x\in[0,1][/imath]
Let [imath]f(x)[/imath] be continuous on [[imath]0,1[/imath]] such that [imath]f(x) \ge 0[/imath] for each [imath]x\in[0,1][/imath].If [imath]\int^{1}_{0}f(x)dx = 0[/imath] , Show that [imath]f(x)=0[/imath] for each [imath]x\in[0,1][/imath] | 388200 | A formal proof required using real analysis
If [imath]\int_0^1f^{2n}(x)dx=0[/imath], prove that [imath]f(x)=0[/imath], where [imath]f[/imath] is a real valued continuous function on [0,1]? It is obvious, since [imath]f^{2n}(x) \geq 0[/imath], the only way this is possible is when [imath]f(x)=0[/imath]. I am looking for any other formal way of writing this proof i.e. using concepts from real analysis. Any hints will be appreciated. |
261587 | How do I find the number of group homomorphisms from [imath]S_3[/imath] to [imath]\mathbb{Z}/6\mathbb{Z}[/imath]?
How do I find the number of group homomorphisms from the symmetric group [imath]S_3[/imath] to [imath]\mathbb{Z}/6\mathbb{Z}[/imath]? | 152853 | Count the number group homomorphisms from [imath]S_3[/imath] to [imath]\mathbb{Z}/6\mathbb{Z}[/imath] ?
I have to count the number of group homomorphisms from [imath]S_3[/imath] to [imath]\mathbb{Z}/6\mathbb{Z}[/imath] ? 1 2 3 6 I aware of the formula for calculating group homomorphisms defined on cyclic groups..here [imath]S_3[/imath] is not cyclic. Please suggest how to proceed. Thank you so much Egbert for the much detailed and patient reply. |
386423 | prove or disprove invertible matrix with given equations
Given a non-scalar matrix [imath]A[/imath] in size [imath]n\times n[/imath] over [imath]\mathbb{R}[/imath] that maintains the following equation [imath]A^2 + 2A = 3I[/imath] given matrix [imath]B[/imath] in size [imath]n\times n[/imath] too [imath]B = A^2 + A- 6I[/imath] Is [imath]B[/imath] an invertible matrix? | 379163 | Is [imath]B = A^2 + A - 6I[/imath] invertible when [imath]A^2 + 2A = 3I[/imath]?
Given: [imath]A \in M_{nxn} (\mathbb C), \; A \neq \lambda I, \; A^2 + 2A = 3I[/imath] Now we define: [imath]B = A^2 + A - 6I[/imath] The question: Is [imath]B[/imath] inversable? Now, what I did is this: [imath]A^2 + 2A = 3I \rightarrow \lambda^2v + 2\lambda v = 3v \rightarrow \lambda_1 = 1, \lambda_2 = -3[/imath] Is what I suggested correct? I know that if so, I just do the same to B and calculate the determinant. |
261886 | limit of Holder norms: [imath]\sup\limits_{x\in [a,b]} f(x) = \lim\limits_{n\rightarrow\infty} \left(\int_a^b (f(x))^n \;dx\right)^{\frac{1}{n}}[/imath]
Show that [imath]\sup_{x\in [a,b]} f(x) = \lim_{n\rightarrow\infty} \left(\int_a^b (f(x))^n \;dx\right)^{\frac{1}{n}}[/imath] for f continuous and positive on [a,b]. I can show that LHS is greater than or equal to RHS but I can't show the other direction. | 821836 | Limit Integral equal to a supremun: [imath]\lim_{n\to \infty} \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).[/imath]
I'd like a hint to solve the following problem: Given a [imath]f[/imath] continuos and positive on [imath][a,b][/imath] prove that [imath]\lim_{n\to \infty} \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).[/imath] Thanks a lot! |
388400 | Prove that for every simple graph [imath]G=(V,E):\, \frac{1}{2} \chi(G) (\chi(G)-1) ≤ |E|[/imath]
I am a first year math student, and I'm studying some exercises for my graph theory exam. This exercise I didn't understand very well: Prove that for every simple graph [imath]G=(V,E)[/imath] [imath]\frac{1}{2} \chi(G) (\chi(G)-1) ≤ |E|[/imath] A hint would be appreciated. | 167258 | Inequality between chromatic number and number of edges of a graph
I have not been able to find a proof to the statement that if a graph [imath]G[/imath] has [imath]\chi(G)=k[/imath], then it must have at least [imath]\binom{k}{2}[/imath] edges. Would you be able to show me a simple proof? |
388722 | proving that a certain function is not differentiable at [imath](0,0)[/imath]
Let's consider the function given by: [imath] F\left( {x,y} \right) = \begin{cases} \frac{{xy}}{{\sqrt {x^2 + y^2 } }}\text{ when }\,\,\,\,(x,y) \ne (0,0) \\ \\ \\ \\ \,\,\,\,\,\,0\hspace{10ex}\text{ if } \,\,\,\,\,\,\,\,\,(x,y)=(0,0) \end{cases} [/imath] I proved a lot of things of this function, the partial derivated exist everywhere, in particular at [imath](0,0)[/imath] , But I want to prove that [imath]F[/imath] is not differentiable at [imath](0,0)[/imath], I I tried by definition but it does not work. Please help me !! | 377935 | Show that a function [imath]f:\mathbb{R}^2 \to \mathbb{R}[/imath] is not differentiable at [imath](0,0)[/imath]
Let [imath]f:\mathbb{R}^2 \to \mathbb{R}[/imath] be a function defined as [imath] f(x,y) = \cases{\frac{xy}{\sqrt{x^2+y^2}}, (x,y)\neq (0,0) \\ 0, (x,y)=(0,0) }. [/imath] Then how can I show that this function is not differentiable at [imath](0,0)[/imath]? This lecture note says that this is because the "possible derivative is not a linear function." But I just don't understand what it really means by that. I'll appreciate any help. Thank you. |
389726 | Laurent Expansions about [imath]z_0=0[/imath]
I'm really struggling to understand the concept of Laurent Expanaions.. I have gone throu lectures notes and a couple of examples, but don't seem to be able to compute fully understand.. Any help with the following example would be greatly appreciated... [imath]f(z)= \dfrac{1}{z(z-1)(z-2)} [/imath] On (i) [imath]0< |z| < 1[/imath] (ii) [imath] 1< |z|< 2[/imath] (iii) [imath] 2<|z|[/imath] Workings I have worked out the partial fractions for this equation as follows.... [imath]\dfrac{1}{z(z-1)(z-2)} = \dfrac{A}{z} + \dfrac{B}{z-1} + \dfrac{C}{z-2}[/imath] [imath]\dfrac{1}{z(z-1)(z-2)}= \dfrac{1}{z} -\dfrac{1}{z-1}+ \dfrac{1}{2(z-2)}[/imath] But now I'm lost as to what to do.. | 387448 | How to find a Laurent Series for this function
How do I give a Laurent Series on various ranges of [imath]|z|[/imath]? I need to find the Laurent series expansion for [imath]f(z)=\frac{1}{z(z-1)(z-2)}[/imath] for the following ranges of [imath]|z|[/imath]: [imath]0<|z|<1[/imath] [imath]1<|z|<2[/imath] [imath]2<|z|[/imath] I've calculated the partial fractions expasion of [imath]f(z)[/imath] to be [imath]f(z)=\frac{1}{2z}+\frac{1}{z-1}+\frac{1}{2(z-2)}[/imath] |