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2598814 | If [imath]A[/imath] has eigenvalues [imath]\{\lambda_1,\lambda_2,...,\lambda_k\}[/imath] then does [imath]A^n[/imath] have only [imath]\{\lambda^n_1,\lambda^n_2,...,\lambda^n_k\}[/imath] as eigenvalues?
If [imath]A[/imath] has [imath]k[/imath] distinct eigenvalues [imath]\{\lambda_1,\lambda_2,...,\lambda_k\}[/imath] then does [imath]A^n[/imath] have only [imath]\{\lambda^n_1,\lambda^n_2,...,\lambda^n_k\}[/imath] as eigenvalues? It is well know that if [imath]\lambda[/imath] is an eigenvalues of [imath]A[/imath] then [imath]\lambda^n[/imath] is an eigenvalues of [imath]A^n[/imath] for [imath]n\in \mathbb{N}[/imath]. But I am wondering whether [imath]A^n[/imath] has only [imath]\lambda^n[/imath] as eigenvalues, or are there more? | 241764 | Eigenvalues and power of a matrix
Let [imath]A[/imath] be an n×n matrix with eigenvalues [imath]\lambda_i, i=1,2,\dots,n[/imath]. Then [imath]\lambda_1^k,\dots,\lambda_n^k[/imath] are eigenvalues of [imath]A^k[/imath]. I was wondering if [imath]\lambda_1^k,\dots,\lambda_n^k[/imath] are all the eigenvalues of [imath]A^k[/imath]? Are the algebraic and geometric multiplicities of [imath]\lambda_i^k[/imath] for [imath]A^k[/imath] same as those of [imath]\lambda_i[/imath] for [imath]A[/imath] respectively? Thanks! |
2598887 | Tate's thesis: why is the image of [imath]k[/imath] dense in its Pontryagin dual?
I'm trying to understand a lemma from the beginning of John Tate's PhD thesis. Here [imath]k[/imath] is a local field of characteristic zero. The character group [imath]\hat{k}[/imath] of [imath]k^+[/imath] is the group of continuous homomorphisms [imath]k^+ \rightarrow S^1[/imath], with the compact open topology, which has subbasis [imath]V(K,U) = \{ \chi \in \hat{k} : \chi(K) \subseteq U \}[/imath] for [imath]K[/imath] compact and [imath]U[/imath] open. Since [imath]S^1[/imath] is metrizable and [imath]k[/imath] is a union of increasing compact subsets, each contained in the interior of the next, the group [imath]\hat{k}[/imath] is metrizable, and [imath]f_n \to f[/imath] in this metric if and only if [imath]f_n \to f[/imath] uniformly on compact sets. For property (4), it seems all that has been proven is that the map [imath]k \rightarrow \hat{k}[/imath] is injective. What does this have to do with the image of [imath]k[/imath] being dense? | 715038 | Why is the image of [imath]k^+[/imath] dense in the character group?
Let [imath]k[/imath] be a local field, and consider the map [imath]\phi: k^+ \hookrightarrow \widehat {k^+}[/imath] given by [imath]\eta \mapsto \eta X(\cdot)=X(\eta \cdot)[/imath] where [imath]X[/imath] is a non-trivial character. Tate argues in his thesis that [imath]\phi[/imath] has dense image simply because we have the implication [imath]X(ηξ)=1 ∀η⟹ξ=0[/imath]. However, I am not sure how this shows that we can multiply [imath]X[/imath] by some [imath]\eta[/imath] to get arbitrarily close of another given [imath]X'[/imath]. For that we would have to show that for any finite collection of compact subsets [imath]B_{1,...n}[/imath] in [imath]k^+[/imath] and open sets [imath]U_{1...n}\subset S^1[/imath] such that [imath]X'(B_i)\subset U_i[/imath] we can find [imath]\eta[/imath] such that [imath]X(\eta B_i) \subset U_i[/imath] as well. |
2599419 | The value of the limit.
The value of the limit [imath] \lim _{n\rightarrow \infty} \sum _{j=n} ^{4n} {4n \choose j} \left(\frac{1}{4}\right)^j \left(\frac{3}{4}\right)^{4n-j} [/imath] | 1098004 | What is the value of [imath]\lim_{n\to \infty} \sum_{j=n}^{4n} {4n \choose j}\bigl(\frac{1}{4}\bigr)^{j} \bigl(\frac{3}{4}\bigr)^{4n-j}[/imath]?
Find the value of the limit [imath]\lim_{n\to \infty} \sum_{j=n}^{4n} {4n \choose j}\left(\dfrac{1}{4}\right)^{j} \left(\dfrac{3}{4}\right)^{4n-j}[/imath] First we take the limit of the summation from [imath]0[/imath] to [imath]\infty[/imath] then subtract the sum from [imath]0[/imath] to [imath]n-1[/imath]. The value of the first sum is [imath]1[/imath]. I can't determine the second sum with limit. How we find it? |
2598789 | Sigma algebra generator set of product space
Let [imath](X_1, \mathcal{F}_1)[/imath], [imath](X_2, \mathcal{F}_2)[/imath] be two measure spaces. Suppose [imath]D_i[/imath] is the generator of the [imath]\sigma-[/imath]algebra [imath]\mathcal{F}_i[/imath], [imath]i=1, 2[/imath]. Question:(1) Is [imath]D_1\times D_2=\{A\times B: A\in D_1, B\in D_2\}[/imath] a generator of [imath]\mathcal{F}_1\otimes \mathcal{F_2}[/imath]? (2) What if in addition [imath]D_i[/imath] is closed under intersection? | 195841 | Is the product of generators equal to the generator of the product?
Let [imath](X, \mathcal{J})[/imath] and [imath](Y, \mathcal{F})[/imath] be two measure spaces. Let us assume that [imath]J[/imath] is a collection of subsets of [imath]X[/imath] which generates [imath]\mathcal{J}[/imath], i.e. [imath]\sigma(J)=\mathcal{J}[/imath]. Similarly, assume [imath]\sigma(F)=\mathcal{F}[/imath]. Is it always true that [imath] \sigma(J)\times\sigma(F)=\sigma(J\times F)? [/imath] Here, [imath]J\times F[/imath] is the set of all cartesian products of sets in [imath]J[/imath] with sets in [imath]F[/imath]. Please note: in the above expression the two product signs mean different things. On the left I am considering the product sigma algebra [imath]\mathcal{J}\times \mathcal{F}[/imath], while on the right I am considering the [imath]\sigma[/imath]-algebra generated by the cartesian products of my ``elementary sets.'' |
2599926 | Prove that [imath](AB-BA)^n=0[/imath]
Let [imath]A,B[/imath] be two [imath]n \times n [/imath] matrices with real entries and [imath]\alpha \in \mathbb{R}, \alpha \neq 0[/imath]. It is given that [imath]A^2-B^2=\alpha(AB-BA)[/imath] Prove that [imath](AB-BA)^n=0[/imath]. At first, I tried to prove that [imath]A(AB-BA)=(AB-BA)A[/imath], as it would imply that [imath](AB-BA)[/imath] is nilpotent and the problem would be solved. But this didn't work. Then, after writing [imath]\det(A-B)(A+B)=\det(A+B)(A-B)[/imath] and using the given relation I obtained [imath]\det (AB-BA)=0[/imath]. But I got stuck and I tried to make a substitution: \begin{align*} X=\frac{1}{2}(A+B) \\ Y=\frac{1}{2}(A-B) \end{align*} The given relation then becomes [imath](\alpha+1)XY=(\alpha-1)YX \quad (1)[/imath] so [imath]XY[/imath] and [imath]YX[/imath] commute. Since [imath](AB-BA)^n=2^n(YX-XY)^n[/imath], we can expand the latter using the binomial theorem. Combined with [imath](1)[/imath], I obtained [imath](AB-BA)^n=\frac{4^n}{(\alpha-1)^n}(XY)^n=\frac{2^n(-2^n)}{(\alpha+1)^n}(YX)^n[/imath] and I don't see how to finish this. | 2590752 | Show matrix is nilpotent
I have matrices [imath]A,B[/imath] of dimension [imath]n[/imath] with real coefficients which satisfy the following: [imath]A^2-B^2=c(AB-BA)[/imath] where [imath]c[/imath] is a real number. If [imath]c\neq0[/imath] , prove that [imath](AB-BA)^n = 0[/imath]. So far, I've been able to show that [imath]AB-BA[/imath] is singular. Can someone help? |
2599439 | On Lagrange's Theorem
I have a group of order [imath]p^\alpha[/imath], [imath]p[/imath] is prime. I argued that by the Lagrange's theorem, the subgroups of this group are of prime power order. Just want to clarify if this also implies the existence of these subgroups? Can I say this group has subgroups of order [imath]p[/imath], [imath]p^2[/imath], ..., [imath]p^{\alpha-1}[/imath]? Or do I need to refer to other results/theorems? Thank you. | 567017 | If [imath]|G| = p^n[/imath] then [imath]G[/imath] has a subgroup of order [imath]p^m[/imath] for all [imath]0\le m [/imath]
Prove that if [imath]|G| = p^n[/imath] then [imath]G[/imath] has a subgroup of order [imath]p^m[/imath] for all [imath]0\le m <n.[/imath] Since [imath]G[/imath] is of prime-power order I know [imath]|Z(G)| \ne e[/imath] so there is an [imath]a\in Z(G)[/imath] with order [imath]p[/imath] such that [imath]p \mid |Z(G)|[/imath]. Now, the subgroup generated by is normal since it's a subgroup of the center. How can I get this normal subgroup to be less than [imath]n[/imath]? |
2594612 | Sum, series and Harmonic numbers
I found the solution of series on Wolfram Alpha http://www.wolframalpha.com/input/?i=sum+1%2F(2k%2B1)%2F(2k%2B2)+from+1+to+n%2F2 for [imath]\alpha = 1[/imath] [imath]\sum\limits_{k=1}^{n/2} \left(\frac{1}{\left(2k-1+2^{1/\alpha}\right)^\alpha} - \frac{1}{\left(2k+2^{1/\alpha}\right)^\alpha}\right)=\frac{1}{2} \left(-H_{\frac{n}{2}+1} + H_\frac{n+1}{2} -1 + \text{ln}(4)\right)[/imath] Can someone tell how to prove this in the form of Harmonic numbers? | 2579156 | Sums and harmonic series
I found the solution of series on Wolfram Alpha http://www.wolframalpha.com/input/?i=sum+1%2F(2k%2B1)%2F(2k%2B2)+from+1+to+n [imath] \sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = \sum\limits_{k=1}^{n} \frac{1}{(2k+1)(2k+2)} = \frac{1}{2} \left(H_{n+\frac{1}{2}} - H_{n+1} -1 + \text{ln}(4)\right)[/imath] Can someone tell how to prove this in the form of Harmonic numbers? |
2599544 | How does two dimensional induction work for my special case?
In one dimensional induction we show something holds for [imath]n=0[/imath] or [imath]n=1[/imath] and assume it is true at [imath]n=n[/imath] and prove for [imath]n=n+1[/imath]. In two dimensional induction is it something along show it is true for [imath](n,k)=(0,k)[/imath] or [imath](n,k)=(1,k)[/imath] at any [imath]k[/imath] and assume it is true at [imath](n,k)=(n,k)[/imath] and prove for [imath](n,k)=(n+1,k)[/imath]? In my case [imath]n\leq k[/imath] always. | 1821045 | Double induction - another method?
I am going through some good old Fibonacci proof by induction problems that require two counters [imath]m, n[/imath] instead of one. In order to prove [imath]P(m, n)[/imath] for all [imath]m,n \in \mathbb{N}[/imath], I am thinking of first proving [imath]P(0, 0)[/imath] and then proving that [imath]P(m,n) \implies P(m + 1, n)[/imath] [imath]P(m, n) \implies P(m, n + 1)[/imath] [imath]P(m, n) \implies P(m + 1, n + 1)[/imath] (EDIT: As noted by Arthur, this is unnecessary). but am unsure if this is a correct axiom. Isomorphically speaking, consider an integer grid. If we can show that the bottom left most tile is black, and that if any black tile implies that the tile to the top, the top-right diagonally and the right is also black, then we can show that the entire grid is black. This is in contrast with traditional double induction which first shows using induction that the bottom row is black, and then using induction again to show that all rows are black. |
2600224 | Find the number of sequences of letters: ’AAABBBCCC’ such that:
a)Three identical letters are not next to each other. So I came up with a solution but I have no possibility to check whether it's correct so I've decided to post it here. So |X| - number of all possible sequences, |X| = [imath]\frac{9!}{3!\cdot 3!\cdot 3!}[/imath] |A|, |B| and |C| are those sequences in which either 3 A's, B's or C's are next to each other |A| = |B| = |C| = [imath]7\cdot \binom{6}{3}\cdot \binom{3}{3}= 140[/imath] [imath]|A\cap B|, |A\cap C| ,|B\cap C|[/imath] are for use of Inclusion-exclusion [imath]|A\cap B|=|A\cap C|=|B\cap C|=7\cdot 4\cdot \binom{3}{3}= 28[/imath] [imath]|A\cap B\cap C|=1[/imath] [imath]|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|[/imath] And the number we're looking for [imath]=|X|-|A\cup B\cup C|[/imath] Is that by any chance a correct solution? | 1944807 | Arrangements of a,a,a,b,b,b,c,c,c in which no three consecutive letters are the same
Q: How many arrangements of a,a,a,b,b,b,c,c,c are there such that [imath]\hspace{5mm}[/imath] (i). no three consecutive letters are the same? [imath]\hspace{5mm}[/imath] (ii). no two consecutive letters are the same? A:(i). 1314. [imath]{\hspace{5mm}}[/imath] (ii). 174. I thought of using the General Principle of Inclusion and Exclusion along with letting [imath]p_i[/imath] denote a property that.. , and doing so, I will evaluate [imath]E(0)[/imath], which gives us the number of arrangements without any of the properties. How do I go about doing that? I am trying to use the method stated above in solving this question, but I am unable to generalize the properties [imath]p_i[/imath]. |
2470862 | [imath]c_n\to\infty[/imath] so that [imath]X_n/c_n\to 0[/imath] a.s.
Question: If [imath]X_n[/imath] is any sequence of random variables, there are constants [imath]c_n\to\infty[/imath] so that [imath]X_n/c_n\to 0[/imath] a.s. I know "[imath]X_n\to 0[/imath] iff for all [imath]\epsilon >0,\space P(|X_n|>\epsilon\space \text{i.o.})=0[/imath]", and for that I have to use Borel-Cantelli Lemma. By Chebychev's inequality, for any [imath]\epsilon>0[/imath], [imath]P(|X_n|> |c_n|\epsilon)\leq \dfrac{EX^2_n}{c^2_n\epsilon^2}[/imath]. Let, [imath]EX^2_n=O(n^k)[/imath], then I choose [imath]|c_n|=O(n^{\frac{k}{2}+1})[/imath]. In that case, [imath]P(|X_n|> |c_n|\epsilon)\leq\dfrac{\delta}{n^2}[/imath], for some positive constant [imath]\delta[/imath]. Since, [imath]\sum_{n=1}^{\infty}P\Big(\Big|\dfrac{X_n}{c_n}\Big|>\epsilon\Big)<\infty[/imath] then, by Borel-Cantelli Lemma, [imath]P\Big(\Big|\dfrac{X_n}{c_n}\Big|>\epsilon\space\text{i.o.}\Big)=0[/imath] Hence, [imath]X_n/c_n\to 0[/imath] a.s. Is my argument ok? Thanks for your help. Note: i.o. means infinitely often. | 2440688 | Can we find [imath]c_n[/imath] to make [imath]X_nc_n \rightarrow 0[/imath] almost surely, even if [imath]EX_n > \infty[/imath]?
Let [imath]X_n[/imath] be a sequence of random variables. Can we find [imath]c_n[/imath] (positive constants) so that [imath]X_nc_n \rightarrow 0[/imath]? It is easy with Markov's equality + Borel Cantelli, with assumptions on expectations, but what if we don't know anything about expectations? |
2601495 | Is there a distance on [imath]\mathbb{R}[/imath] so that a non-empty subset is open iff its complement is finite?
Is there a distance [imath]d[/imath] on [imath]\mathbb{R}[/imath] so that a non-empty subset [imath]\mathcal{U}[/imath] of [imath]\mathbb{R}[/imath] is open wrt to [imath]d[/imath] if and only if its complement [imath]\mathbb{R}\setminus\mathcal{U}[/imath] is finite? My observations so far: Let [imath]\emptyset\neq\mathcal{U}\subset \mathbb{R}[/imath]. Denote the metric topology by [imath]T(d)[/imath]. Then [imath]\mathcal{U} \in T(d) \iff \mathbb{R}\setminus\mathcal{U} \text{ finite} \implies\mathbb{R}\setminus{\mathcal{U}} \text{ compact}.[/imath] Since every finite set in a metric space is compact. So the question becomes, is there a [imath]d[/imath] on [imath]\mathbb{R}[/imath] such that every closed set is compact? | 145516 | Metrization of topological space
Can you help me please with this question? Let [imath]X[/imath] be a non-empty set with the cofinite topology. Is [imath]\left ( X,\tau_{\operatorname{cofinite}} \right ) [/imath] a metrizable space? Thanks a lot! |
2599547 | The exact value of [imath]\lim_{n \to \infty} \frac{a_n}{\sqrt[4]{n^3}}[/imath]
Let [imath]a_1=1[/imath] and [imath]a_{n+1}=a_n+\frac{\sqrt{n}}{a_n}[/imath]. If it exists, find [imath]\lim_{n \to \infty} \frac{a_n}{\sqrt[4]{n^3}}[/imath] Obviously, [imath]a_n \to \infty[/imath] and [imath]a_n[/imath] is strictly increasing. I tried to get rid of that [imath]n[/imath] in the limit and get only [imath]a_n[/imath] terms in the numerator so that I could substitute [imath]a_n[/imath] to [imath]x \to \infty[/imath] and maybe use L'Hospital to finish. But [imath]a_{n+1}, a_n[/imath] and [imath]\sqrt{n}[/imath] are so tightly connected that I didn't manage to do that. Another approach was Cesaro-Stolz, writing the limit as [imath]\lim_{n \to \infty}\left(\sqrt[4]{\frac{\sqrt[3]{a_n^4}}{n}}\right)^3=\left(\lim_{n \to \infty}\left(\sqrt[3]{a_{n+1}^4}-\sqrt[3]{a_n^4}\right) \right)^\frac{3}{4}[/imath] but I encountered the same issue. I'm even wondering if there is a nice closed form for the limit. Edit: As pointed below, if the limit exists, then its value is [imath]\frac{2}{\sqrt{3}}[/imath]. How can we prove that it exists? | 2591342 | The convergence of a sequence
This problem is form the Romanian GM. Let [imath](x_{n})_{n\geq1}[/imath] be a sequence of real numbers defined by [imath]x_{1}=1[/imath] and [imath]x_{n+1}=x_{n}+ \frac{\sqrt{n}}{x_{n}}[/imath] for all [imath]n \geq 1[/imath]. Study the convergence of the sequence [imath](\frac{x_{n}}{n^{\frac{3}{4}}})_{n\geq1}[/imath] I only managed to show from the given relation that [imath](x_{n})_{n\geq1}[/imath] is strictly increasing and tends to [imath]+\infty[/imath]. |
2600679 | why that normalized cross correlation(NCC) lies in [imath][-1,1][/imath]?
Provided two real number sequences: [imath]a_1,a_2,...,a_n[/imath];[imath]b_1,b_2,...,b_n[/imath], define their means respectively: [imath]\bar a=\frac{1}{n}\sum_{i=1}^n a_i,\bar b=\frac{1}{n}\sum_{i=1}^n b_i[/imath] and define their variances and covariance respectively: [imath]var(a)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)^2,var(b)=\frac{1}{n}\sum_{i=1}^n (b_i-\bar b)^2,cov(a,b)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)(b_i-\bar b)[/imath] naturally leads to the definition of normalized cross correlation: [imath]NCC=\frac{cov(a,b)}{\sqrt{var(a)var(b)}}=\frac{\sum_{i=1}^n(a_i-\bar a)(b_i-\bar b)}{\sqrt{\sum_{i=1}^n (b_i-\bar b)^2 \sum_{i=1}^n (a_i-\bar a)^2}}[/imath] Now how to show that [imath]NCC[/imath] lies in [imath][-1,1][/imath]? | 2600655 | How to show normalized cross correlation(NCC) lies in [imath][-1,1][/imath]?
Provided two real number sequences: [imath]a_1,a_2,...,a_n[/imath];[imath]b_1,b_2,...,b_n[/imath], define their means respectively: [imath]\bar a=\frac{1}{n}\sum_{i=1}^n a_i,\bar b=\frac{1}{n}\sum_{i=1}^n b_i[/imath] and define their variances and covariance respectively: [imath]var(a)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)^2,var(b)=\frac{1}{n}\sum_{i=1}^n (b_i-\bar b)^2,cov(a,b)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)(b_i-\bar b)[/imath] naturally leads to the definition of normalized cross correlation: [imath]NCC=\frac{cov(a,b)}{\sqrt{var(a)var(b)}}=\frac{\sum_{i=1}^n(a_i-\bar a)(b_i-\bar b)}{\sqrt{\sum_{i=1}^n (b_i-\bar b)^2 \sum_{i=1}^n (a_i-\bar a)^2}}[/imath] Now how to show that [imath]NCC[/imath] lies in [imath][-1,1][/imath]? |
645519 | When does equality in Markov's inequality occur?
Markov's inequality states that given any nonnegative random variable and [imath]a>0[/imath] then we have: [imath]P(X \geq a) \leq \frac{E(X)}{a}[/imath] At which [imath]a[/imath] is equality supposed to hold? | 349839 | Case of equality in Markov's inequality
Find an example where Markov's inequality is tight in the following sense: For each positive integer [imath]a[/imath], find a non-negative random variable [imath]X[/imath] such that [imath]P(X\ge a)=E(X)/a[/imath]. How to do this problem, I am really confused. Also, what is the definition of Markov's inequality? |
2602820 | How were the values of this trigonmetric ratio determined?
I'm reading a book that is pretty spartan about definitions. How did the book come up with the length of the sides of this triangle? I understand the trig ratios once we have the lengths... but how were the lengths of [imath]\sqrt{3}[/imath], 1, and 2 determined? I think the book is assuming that the radius is 2 and that the terminal angle (unsure if this is the right word... but the angle created by the terminal side and the x-axis) is [imath]\frac{\pi}{3}[/imath]. But how did we get the other two sides that are not the assumed radius? | 881520 | Why is [imath]\sin 30^\circ=\frac{1}{2}[/imath]
Take half a square with side length [imath]1[/imath]. The resulting right-angled triangle ABC has two angles of [imath]45^\circ[/imath]. By Pythagoras’ theorem, the hypotenuse AC has length [imath]\sqrt{2}[/imath]. Applying the definitions on the previous page gives the values in the table below. that [imath]\sin 30^\circ= \frac{1}{2}[/imath] Sorry I cannot provide diagram, but from my understanding [imath]\sin =[/imath] opposite / hypotenuse. How is the value [imath]0.5[/imath] derived then? No possible combination. What point of reference should I be looking from? |
1087676 | Coordinate relation under the change of basis
I want to find out the rules of coordinate under the change of basis. Suppose let [imath]V[/imath] be n-dimensional vector space and that [imath]B=\{\alpha_1,...,\alpha_n\},[/imath] [imath]B'=\{\alpha'_1,...,\alpha'_n\},[/imath] are two ordered basis for V. Then we can say there exists scalar [imath]P_{ij}[/imath] such that [imath]\alpha'_j =\sum_i P_{ij}\alpha_i \quad (1\leq j \leq n) [/imath] Let [imath]\alpha = \sum_j x'_j \alpha'_j[/imath] then [imath]\alpha = \sum_j x'_j \alpha'_j = \sum_{i,j} (P_{ij}x_j')\alpha_i[/imath] About the coordinates [imath](x_1 ,... x_n)[/imath] of [imath]\alpha[/imath] in the ordered basis B, [imath]x_i = \sum_j P_{ij}x'_j[/imath] We can reformulate it as [imath]X = PX'[/imath] In linear algebra text(Hoffman), [imath]X=0 \Leftrightarrow X'=0[/imath] so [imath]P[/imath] is invertible. I can't understand why [imath]X=0 \Leftrightarrow X'=0[/imath] is valid. | 2017995 | Help to understand a comment in Hoffman and Kunze's linear algebra book
I'm reading Hoffman and Kunze's Linear Algebra and on page 52, the authors said: Let [imath]P[/imath] be the [imath]n \times n[/imath] matrix whose [imath]i,j[/imath] entry is the scalar [imath]P_{ij}[/imath], and let [imath]X[/imath] and [imath]X'[/imath] be the coordinate matrices of the vector [imath]\alpha[/imath] in the ordered bases [imath]\mathscr{B}[/imath] and [imath]\mathscr{B}'[/imath]. Then we may reformulate (2-15) as [imath] X = PX'. \tag{2-16} [/imath] Since [imath]\mathscr{B}[/imath] and [imath]\mathscr{B}'[/imath] are linearly independent sets, [imath]X=0[/imath] if and only if [imath]X'=0[/imath]. Thus from (2-16) and Theorem 7 of Chapter 1, it follows that [imath]P[/imath] is invertible. Hence [imath] X' = P^{-1}X. \tag{2-17} [/imath] I didn't find any mention of this result (highlighted in bold) in the book. How can I prove this fact? |
2603012 | Prove that [imath]2^{3^n}+1[/imath] is divisible by [imath]3^{n+1}[/imath] for any non-negative integer [imath]n[/imath].
[imath]2^{3^n}[/imath] must be 2 mod 3, which it must be since it is 0 mod 2. However, for certain cases, this does not hold, and I have trouble showing why. | 2570643 | Proving that [imath]2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}[/imath] for every natural [imath]n[/imath]
Prove that [imath]2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}[/imath] for every natural [imath]n[/imath] Of course, should be done by induction. Base case ([imath]n=1[/imath]) is easy. I got stuck with the step: Let's assume that it's right for [imath]n[/imath]. Then, [imath]2^{2\cdot 3^{n-1}} = 1 + 3^n \pmod{3^{n+1}}[/imath] We want to evaluate [imath]2^{2 \cdot 3^n} \pmod {3^{n+2}}[/imath] I think we can somehow utilize the fact that [imath]2,3[/imath] are coprime and reduce [imath]3^{n+2}[/imath] to [imath]3^{n+1}[/imath]. I'd be glad for help on that. Thanks! |
2603433 | Calculating the radius of convergence for [imath]\sum _{n=1}^{\infty}\frac{\left(\sqrt{ n^2+n}-\sqrt{n^2+1}\right)^n}{n^2}z^n[/imath]
Calculate the radius of convergence for [imath]\sum _{n=1}^{\infty}\frac{\left(\sqrt{ n^2+n}-\sqrt{n^2+1}\right)^n}{n^2}z^n.[/imath] Do I have to do a index shift so the sum starts at [imath]0[/imath]? I tried it like here (without index shift) Finding the_radius of convergence, and my solution for the denominator is [imath]+\infty[/imath] so in the end my radius of convergence should be [imath]0[/imath]. Is that correct? Thanks! | 2603341 | Finding radius of convergence [imath]\sum _{n=0}^{}(2+(-1)^n)^nz^n[/imath]
1) [imath]\sum _{n=0}^{}(2+(-1)^n)^nz^n[/imath] 2) [imath]\sum _{n=0}^{}n!z^n[/imath] 3) [imath]\sum _{n=1}^{}\frac{1}{n^2}(\sqrt{ n^2+n}-\sqrt{n^2+1})^nz^n[/imath] Hello, I struggle to show the radius of convergence for the above functions. I began with the second one and came to the conclusion that the radius of convergence is [imath]0[/imath] but I'm not sure if that's correct. I would appreciate it if you could help me with the other ones! |
2603481 | Confirmation of Proof: [imath]\forall n \in \mathbb{N}, \ \pi (n) \geqslant \frac{\log n}{2\log 2}[/imath]
The question is as follows: How can I prove the following theorem? [imath]\forall n \in \mathbb{N}, \ \pi(n) \geqslant \frac{\log n}{2\log 2} \tag1 \label1[/imath] I have no idea where to begin. I came across this equation on a paper that apparently proved Legendre's Conjecture. You can review the paper here, which introduces the open conjecture, but it only reveals two proof even though the title states there are three proofs. In the paper, it mentioned that a demonstrated proof of [imath]\eqref1[/imath] was on pp. [imath]21[/imath] but for some reason, the paper that I observed, and that of which is accessible in the link, holds only [imath]7[/imath] pages. May somebody please provide a proof of [imath]\eqref1[/imath]? I am just curious. Thank you in advance. Edit: When I was about to post this question, I was unaware that such a similar question has already been posted and answered. Sorry about that. | 1502852 | [imath]\pi(x)\geqslant\frac{\log x}{2\log2}[/imath] for all [imath]x\geqslant2.[/imath]
Let [imath]\pi[/imath] be the prime counting function. Then [imath]\pi(x)\geqslant\log x/(2\log2)[/imath] for all [imath]x\geqslant2.[/imath] Maybe I am missing something pretty evident, but, so far, I have proved that [imath]\pi(x)\geqslant\log{\lfloor x\rfloor}/(2\log2)[/imath] using a method Paul Erdős used to prove that there are infinitely many primes, however, I don't know how to prove the original inequality. Any help is really appreciated!. |
2603934 | Limit of a sequence that looks like [imath]e[/imath]
How does one go about calculating the limit of the sequence [imath]a_n = \left(1 + \dfrac{1}{n^2} \right)^n.[/imath] I understand the cases of [imath] b_n = \left(\alpha + \dfrac{\beta}{n}\right)^n, [/imath] in terms of [imath]e[/imath], but I am missing the "trick" with regards to [imath]a_n[/imath]. | 2572209 | Test [imath]f_n = (1+\frac{1}{n^2})^n[/imath] for convergence and give its limit if it exists.
Now this exercise looks like I need to use [imath]\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e[/imath] somehow, but I'm unsure of how. Bernoulli would give me [imath]\left(1+\frac{1}{n^2}\right)^n > 1+n\cdot\frac{1}{n^2}=1+\frac{1}{n}[/imath] for which the limit is [imath]1[/imath], but I'm not sure how that helps. Can anybody give me a tip on how to solve this elegantly? |
2603788 | Surface by revolution
Prove that the equation [imath]x^3+y^3+z^3 -3xyz -1=0[/imath] defines a surface by revolution and determine the equation for the axis of revolution. I know I have to end up with something that looks like this: [imath](x-a)^2+(y-b)^2+(z-c)^2=r^2[/imath] [imath]dx+ex+fx = g[/imath] but I don't know where to start... (I am aware that there is a similar question on this site but I need a simpler answer than the one given there.) | 2075063 | Prove that [imath]x^3+y^3+z^3-3xyz=1[/imath] defines a surface of revolution
Prove that the equation [imath]x^3+y^3+z^3-3xyz=1[/imath] defines a surface of revolution and find the analytical equation of its axis of revolution. I think that I need to apply Euler's formula, so that I get rid of the third-grade polynomial there: [imath]x^3+y^3+z^3-3xyz=1 \Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-xz-yz)=1[/imath] but then I stuck on how to prove it defines a surface of revolution. The textbook notes that: [imath]f(x,y,z)=0[/imath] defines a surface of revolution around the axis with equation [imath]\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}[/imath] if and only if it can be written as a polynomial of [imath](x-x_0)^2+(y-y_0)^2+(z-z_0)^2[/imath] and [imath]ax+by+cz[/imath]. Any hint for that one? |
2603931 | How to prove that [imath]\pi^{e^{\pi^e}}[/imath]
I have a question: I Can to proof that [imath]\pi^e<e^{\pi}[/imath] (it is easy), and also I proof that [imath]\pi^{e^{\pi}}>e^{{\pi^e}}[/imath] (by contradiction I assume that [imath]\pi^{e^{\pi}}\leq e^{^{\pi^e}}[/imath] and via natural logaritms obtains that [imath]e^{\pi}<e^{\pi}\cdot \ln \pi <\pi^e[/imath], an absurd whit the above inequality!) But, the question is that I can not prove that [imath]\pi^{e^{\pi^e}}<e^{\pi^{e^{\pi}}}[/imath]. Any help! Thanks! | 1857969 | [imath]e^{\left(\pi^{(e^\pi)}\right)}\;[/imath] or [imath]\;\pi^{\left(e^{(\pi^e)}\right)}[/imath]. Which one is greater than the other?
[imath]\newcommand{\bigxl}[1]{\mathopen{\displaystyle#1}} \newcommand{\bigxr}[1]{\mathclose{\displaystyle#1}} [/imath] [imath]\large e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\quad\text{or}\quad\pi^{\bigxl(e^{(\pi^e)}\bigxr)}[/imath] Which one is greater? Effort. I know that [imath]e^\pi\ge \pi^e[/imath] Then [imath]\pi^{(e^\pi)}\ge e^{(\pi^e)}[/imath] But I can't say [imath]e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\le \pi^{\bigxl(e^{(\pi^e)}\bigxr)}[/imath] or [imath]e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\ge \pi^{\bigxl(e^{(\pi^e)}\bigxr)}[/imath] |
2603999 | Using trigonometric identities to simply the following expression $\tan\frac{\pi}{5} + 2\tan\frac{2\pi}{5}+ 4\cot\frac{4\pi}{5}=\cot\frac{\pi}{5}$
I know this is simple, but I just can't seem to get it... The question is to simplify: [imath]\tan\frac{\pi}{5} + 2\tan\frac{2\pi}{5}+ 4\cot\frac{4\pi}{5}[/imath] To get: [imath]\cot\frac{\pi}{5}[/imath] How does one go about this? | 2125682 | To prove [imath] \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) [/imath]
To prove [imath] \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) [/imath]. I tried to convert [imath]\tan(4A)[/imath] and [imath] \tan(8A)[/imath] to [imath]\tan(A[/imath]) and tried putting in L.H.S but it becomes a mess. Is there a easy and also intuitive way to do this? Thanks |
2604132 | I can't undestand why [imath] \{x \in X : f(x) > g(x) \} = \bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}} [/imath]
I'm trying to understand this equality: [imath] \{x \in X : f(x) > g(x) \} = \bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}} [/imath] [imath]X[/imath] is a measurable space and [imath]f,[/imath] [imath]g[/imath] are measurable functions. As, [imath]f[/imath] is measurable, then [imath]\{x\in X : f(x) > r\}[/imath] is measurable. The same goes for [imath]g.[/imath] The intersection between these sets would be measurable, and the countable union, [imath]\bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}}[/imath] would then be measurable. I have been trying to visualize what this union means, employing some examples (not numeric, but general shapes of graphs). For example, if [imath]f(x) = c, c\in \mathbb{R}[/imath] and, [imath]g(x) = d, d\in \mathbb{R}[/imath], and [imath]c \neq d,[/imath] [imath]\bigcup_{r \in \mathbb{Q}}{\{x\in X : f(x) > r\}\cap\{x\in X:g(x) < r\}} = \mathbb{R}.[/imath] Clearly, the domain where the functions are defined, should be equal, or have at least some intersecting region, because in other case, the union is empty. It is possible to create graphs, for which the union is empty, even though the functions are defined in a same interval. And in lots of cases, there are one ore several intervals of intersection. But I still don't have a clue as to why the equality holds. This would show that [imath] \{x \in X : f(x) > g(x) \}[/imath] is measurable. | 305107 | How to prove that [imath]\{f+g?[/imath]
Let [imath](X,{\mathcal A})[/imath] be a measurable space, and [imath]f,g:X\to{\Bbb R}[/imath] be measurable functions. It is a standard trick to use the following set equality to show that [imath]f+g[/imath] is also measurable. [imath]\{f+g<a\}=\bigcup_{r\in{\Bbb Q}}(\{f<r\}\cap\{g<a-r\})\quad\text{for any}\quad a\in{\Bbb R}[/imath] where [imath]\{f<a\}:=\{x\in X:f(x)<a\}[/imath]. One direction is easy: [imath]\{f+g<a\}\supset \bigcup_{r\in{\Bbb Q}}(\{f<r\}\cap\{g<a-r\})\quad\text{for any}\quad a\in{\Bbb R}.[/imath] Here are my questions: How can one prove that [imath]\{f+g<a\}\subset \bigcup_{r\in{\Bbb Q}}(\{f<r\}\cap\{g<a-r\})\quad\text{for any}\quad a\in{\Bbb R}?[/imath] Is there a counterexample of the following claim? [imath]\{fg<a\}= \bigcup_{r\in{\Bbb Q}\setminus\{0\}}(\{f<r\}\cap\{g<\frac{a}{r}\})\quad\text{for any}\quad a\in{\Bbb R}?[/imath] |
2599846 | Lagrange Interpolating Polynomial , Error Estimation
Suppose that we got [imath]f\in\mathcal{C}^{n+1}[-1,1][/imath] and [imath]x_{0},...,x_{n}\in [-1,1][/imath] , uniformly separated.With [imath]x_{i}=-1+ih, \ \ i=0,..,n[/imath] , [imath]h=2/n[/imath] and [imath]p_{n}\in \mathcal{P}_{n}[/imath](polynomial). We also know that the approximation of error can be expressed as [imath]f(x)-p_{n}(x)=\frac{1}{(n+1)!}\Phi_{n+1}(x)f^{(n+1)}({\xi})[/imath], [imath]\xi \in [-1,1] , x\in[-1,1][/imath] where [imath]\Phi_{n+1}(x)=\prod_{i=0}^{n}(x-x_{i})[/imath] My problem is that in my notes I have that [imath]\left \| \Phi_{n+1} \right \|_{\infty}\leq \frac{n!}{4}h^{n+1}[/imath], but there is no proof and I really don't know how to prove it. | 641841 | Runge's phenomenon-question
I am looking at the Runge's phenomenon and I have a question. We have the interval [imath][a,b]=[-1,1][/imath] and [imath]f\in C^{n+1}[-1,1][/imath]. We know that [imath]\forall x \in [-1,1] [/imath] [imath]\exists[/imath] [imath]\xi\in[-1,1][/imath] so that: [imath]f(x)-p_{n}(x)=\frac{1}{(n+1)!}f^{(n+1)}(\xi)\varPhi_{n+1}(x)[/imath] where [imath]\varPhi_{n+1}(x)=\prod_{i=0}^{n}(x-x_{i})[/imath] So,[imath]||f-p_{n}||_{\infty}=max|f(x)-p_{n}(x)|\leq\frac{1}{(n+1)!}||\varPhi_{n+1}||_{\infty}||f^{n+1}||_{\infty} [/imath] Then,we suppose that [imath]x_{i}=-1+ih , h=\frac{b-a}{n}[/imath] Then we say that [imath]||\varPhi_{n+1}||_{\infty}\leq \frac{n!}{4}h^{n+1}[/imath] ,but I haven't understood why..Could you explain it to me? |
1791396 | Recurrent sequence limit
Let [imath]a_n[/imath] be a sequence defined: [imath]a_1=3; a_{n+1}=a_n^2-2[/imath] We must find the limit: [imath]\lim_{n\to\infty}\frac{a_n}{a_1a_2...a_{n-1}}[/imath] My attempt The sequence is increasing and does not have an upper bound. Let [imath]b_n=\frac{a_n}{a_1a_2...a_{n-1}},n\geq2[/imath]. This sequence is decreasing(we have [imath]b_{n+1}-b_{n}<0[/imath]. How can I find this limit? | 121586 | If [imath]x_1=5[/imath], [imath]x_{n+1}=x_n^2-2[/imath], find [imath]\lim x_{n+1}/(x_1\cdots x_n)[/imath]
If [imath]\left\{x_{n}\right\}\mid x_{1}=5,x_{n+1}=x_{n}^{2}-2,\forall n\geq 1[/imath] find [imath]\lim_{n\to\infty}\frac{x_{n+1}}{x_{1}x_{2}\cdots x_{n}}.[/imath] If someone could help me out with tags, it'd be lovely. I think this is calculus and real-analysis, but I'm not sure--I had the problem scribbled down on a post-it, and I forget where it's from. |
2604956 | Sine of the sum of two solutions of [imath]a\cos\theta + b \sin\theta = c[/imath]
This is kind of a new type of question for me: If [imath]\alpha[/imath] and [imath]\beta[/imath] are two different roots of the equation [imath]a\cos\theta + b\sin\theta = c[/imath], then show that [imath]\sin(\alpha+\beta) = \frac{2ab}{a^2 + b^2}[/imath] Could someone please explain the procedure to me? | 927530 | Cosine of the sum of two solutions of trigonometric equation [imath]a\cos \theta + b\sin \theta = c[/imath]
Question: If [imath]\alpha[/imath] and [imath]\beta[/imath] are the solutions of [imath]a\cos \theta + b\sin \theta = c[/imath], then show that: [imath]\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}[/imath] No idea how to even approach the problem. I tried taking two equations, by substituting [imath]\alpha[/imath] and [imath]\beta[/imath] in place of [imath]\theta[/imath] in the equation and manipulating them, but that didn't get me anywhere. Please help! |
2598560 | Weak convergence in separable Hilbert spaces
Let [imath]H[/imath] be a Hilbert space with ONB [imath]\{b_1, b_2, \cdots \}[/imath]. I want to prove that the sequence [imath]x_n = \frac{1}{n}\sum_{i=1}^{n^2} b_i[/imath] converges weakly to [imath]0[/imath]. Notice also that [imath]\|x_n\|=1[/imath]. The claim is easily verified on the basis elements, i.e. for large [imath]n[/imath] [imath]\langle x_n, b_i \rangle= \frac{1}{n}\to 0[/imath] but I fail to see why it is true for an arbitrary element [imath]y = \sum_i \langle y, b_i \rangle b_i[/imath]. Why does [imath]\langle x_n, y\rangle = \frac{1}{n} \sum_{i=1}^{n^2} \langle y, b_i \rangle[/imath] converge to [imath]0[/imath]? | 2613300 | Infinite dimensional separable Hilbert space with orthonormal basis (weak convergence of [imath]\frac{1}{N} \sum^{N^2}_{n=1} e_n[/imath])
Let [imath]H[/imath] be an infinite dimensional separable Hilbert space with orthonormal basis [imath](e_n)_n≥1[/imath]. Let [imath]f_N = \frac{1}{N} \sum^{N^2}_{n=1} e_n[/imath], [imath]\forall N ≥ 1[/imath]. I am trying to show that [imath]e_n → 0[/imath] weakly, as [imath]n →∞[/imath] and [imath]f_N → 0[/imath] weakly, as [imath]N →∞[/imath], while [imath] \|f_N\| = 1[/imath], [imath]\forall N ≥ 1[/imath]. |
2604900 | How to approach ordinary differential equations?
[imath]y'' -4y' + 4y=0[/imath] My attempt: We solve [imath]r^2e^{rx}-4re^{rx}+e^{rx}=0[/imath] for [imath]r[/imath], this gives [imath]r=2[/imath] Then the solution is: [imath]y=c_1e^{2x}[/imath] But the solution given is [imath]y=c_1e^{2x}+c_2xe^{2x}[/imath]. I wonder where the [imath]x[/imath] in [imath]c_2xe^{2x}[/imath] comes from, why is it different? | 306622 | Differential Equation [imath]y'' - 4y' + 4y = 0[/imath]
[1] [imath]y'' - 4y' + 4y = 0[/imath] Usually problem like these will have the answer in the form [imath]C_1e^a + C_2e^b ... [/imath] where [imath]a [/imath] and [imath]b[/imath] are the roots of the characteristic equation [imath]e^{rt}[/imath] [imath] y = e^{rt} [/imath] [imath] y' = re^{rt}.. y'' = r^2 e^{rt} [/imath] [imath] r^2e^{rt} - 4e^{rt} + 4e^{rt} = 0[/imath] [imath] e^{rt}(r-2)(r-2) = 0[/imath] [imath] y= C_1e^{2t} + C_2e^{2t}[/imath] However, this is not correct as the answer is [imath] y = C_1e^{2t} + C_2te^{2t}[/imath] ! I just don't know why. [2] For a similar problem , [imath]y'' + 3y' - 4y = 0 [/imath] I did the exact same thing and the answer is [imath]y = c_1e^t + c_2e^{-4t} [/imath] How are [1] and [2] different? They look the same, why does [2]'s solution have an extra factor of t. |
2605197 | Total degree of a product of polynomials
Note: This is not a duplicate of this question. The accepted answer there gives some answer to my question here, but the question is not a duplicate... I'm currently brushing up on some algebra with "Abstract Algebra" by Grillet, Second Edition. We consider, for a unital Ring [imath]R[/imath] and a family of sets [imath](X_i)_{i\in I}[/imath] the polynomial ring [imath] R[(X_i)_{i\in I}] [/imath] I don't think the details of this construction are too important here; suffice it to say that we are dealing with the free [imath]R[/imath]-algebra over the free commutative monoid of all monomials [imath] X^k=\Pi_{i\in I}\, X^{k_i}\,,\; k \in \mathbb{N}^I\,, k_i = 0\, \text{ for almost all } i [/imath] with multiplication [imath]X^k X^l = X^{k+l}[/imath]. The degree of a polynomial [imath]0 \neq A \in R[(X_i)_{i\in I}][/imath] with coefficients [imath]a_k[/imath] is defined as [imath] \max\left\{\sum_{i \in I} k_i;\, k\in I^\mathbb{N}\,,\, k_i = 0\, \text{ for almost all } i\, \land \, a_k \neq 0\right\} [/imath] I generally put quite a lot of trust into textbooks as carefully written as this one, but the following statement on page 127 seems wrong to me and I am stuck trying to prove it: Proposition 6.4. For all [imath]A, B \neq 0[/imath] in [imath]R[(X_i)_{i\in I}][/imath]: ... (4) if [imath]R[/imath] has no zero divisors, then [imath]\deg(\,AB\,) = \deg A + \deg B[/imath] What I have proven is [imath]\deg(\,AB\,) \leq \deg A + \deg B[/imath], which was part (3) of the proposition. However, part (4) baffles me. I started out expecting this to be easy but got stuck; and now I'm really a little too exhausted because I couldn't let go and kept on thinking about how to prove this. I'm actually wondering if this is actually wrong and was just a copy-paste error from the corresponding page 121 for the univariate case, where we have the exact same statement, safe for there only being one variable. A counterexample is probably not that hard, but I'm really too tired right now, and then I actually already got sidetracked just brushing up on basics required for my bachelor's thesis with this already, in the first place. Why I feel like this isn't right: For [imath]k \in I^\mathbb{N}[/imath] with finite support, the [imath]k[/imath]-th coefficient of [imath]AB[/imath] is [imath] (AB)_k = \sum_{(m,\,n) \in (I^\mathbb{N})^2;\,m + n = k} a_m b_n [/imath] where the [imath]a_i, b_i[/imath] are the coefficients of [imath]A[/imath], respectively [imath]B[/imath]. In the univariate case, the degree formula holds if [imath]R[/imath] has no zero divisors, because the leading coefficient of the product is the product of the leading coefficients. However, in the multivariate case, each coefficient can be a sum of products: Picking indices [imath]\nu\,,\, \mu[/imath] with [imath]\sum_i \nu_i = \deg A[/imath] and [imath]\sum_i \mu_i = \deg B[/imath], I don't see how [imath]R[/imath] having no zero divisors would lead to the coefficient [imath] (AB)_{\nu + \mu} = \sum_{(m,\,n) \in (I^\mathbb{N})^2;\,m + n = \nu + \mu} a_m b_n [/imath] being nonzero, although it seems obvious that [imath]AB \neq 0[/imath] if we look at the [imath]i[/imath]-degree [imath] \max\left\{k_i;\, k\in I^\mathbb{N}\,,\, k_j = 0\, \text{ for almost all } j\, \land \, a_k \neq 0\right\} [/imath] for some [imath]i \in I[/imath] (should I be wrong with this reasoning, please tell me), so that [imath]R[(X_i)_{i\in I}][/imath] is a domain, if [imath]R[/imath] is. I also failed to find anything about this on good old Google or here. So, am I right in assuming this part of the proposition quoted above is wrong? I'm all out of creative mojo, so if someone could just plug me an easy counterexample, I'd be very, very happy... if the statement is actually true, I'd appreciate some help in proving it. | 1212103 | Example of [imath]\deg(fg)<\deg(f)+\deg(g)[/imath]
Let [imath]R[/imath] be an integral domain and [imath]f,g\in R[X_1,...,X_n][/imath] where [imath]n>1[/imath]. What is an example of a pair [imath]f,g[/imath] such that [imath]\deg(fg)<\deg(f)+\deg(g)[/imath]? Moreover, I have proven that the units of [imath]R[X_1,...,X_n][/imath] are just the units in [imath]R[/imath]. (Proof is done inductively) Is it true? EDIT Even though the each term in the underlined summation is nonzero, isn't it possible that the summation is 0? Why is this impossible ? |
2605840 | Prove that a ring of fractions is a local ring
I have no idea how to prove this: Let be [imath]B[/imath] a prime ideal of a commutative ring [imath]A[/imath]. Show that [imath]A_B[/imath], the ring of fractions respect [imath]B[/imath] as multiplicative set is a local ring. Any suggestions? | 300446 | Why is the localization at a prime ideal a local ring?
I would like to know, why [imath] \mathfrak{p} A_{\mathfrak{p}} [/imath] is the maximal ideal of the local ring [imath] A_{\mathfrak{p}} [/imath], where [imath] \mathfrak{p} [/imath] is a prime ideal of [imath] A [/imath] and [imath] A_{\mathfrak{p}} [/imath] is the localization of the ring [imath] A [/imath] with respect to the multiplicative set [imath] S = A -\mathfrak{p} [/imath] ? Thanks a lot. N.B. : I have to tell you that I'm not very good at Algebra, so please, be more kind and generous in your explanation, and give me a lot of details about this subject please. Thank you. |
2605759 | Proving addition is associative in [imath]\mathbb{R}[/imath]
I have to prove that [imath]\mathbb{R}[/imath] is a field. One of the properties for [imath]\mathbb{R}[/imath] to be a field is that addition in [imath]\mathbb{R}[/imath] is associative. I wanted to use this property as an example for the other ones. My question is, how can I show something so intuitive? Can I say that it is simply assumed that addition in [imath]\mathbb{R}[/imath] is associative? Or should I use [imath]a+(b+c)=(a+b)+c[/imath] for [imath]a,b,c \in \mathbb{R}[/imath] in some way? I am sorry for the silly question, it has been a while since I proved something. Thank you. | 2134588 | How does one prove that addition on [imath]\mathbb{R}[/imath] is associative?
How does one prove that addition on [imath]\mathbb{R}[/imath] is associative? We can define addition to be a binary function [imath]+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}[/imath], by that definition [imath]\mathbb{R}[/imath] is closed under addtition as for any [imath](a, b) \in \mathbb{R}^2[/imath] we have [imath]+(a, b) = a + b = c[/imath] for some [imath]c \in \mathbb{R}[/imath]. But if we take [imath]a, b, c \in \mathbb{R}[/imath], how can we prove rigorously that [imath](a + b) + c = a + (b + c)[/imath]? We know also that [imath](\mathbb{R}, +)[/imath] is a group, but most proofs of this fact assume that addition in [imath]\mathbb{R}[/imath] is associative, the proofs do not actually prove that addition is associative. |
2605996 | Prove [imath]\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}[/imath] using binomial and induction
Can anyone help? I've got to prove [imath]\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}[/imath] using binomial first and then induction. | 894159 | Find the value of [imath]\sum_{k=1}^{n} k \binom {n} {k}[/imath]
I was assigned the following problem: find the value of [imath]\sum_{k=1}^{n} k \binom {n} {k}[/imath] by using the derivative of [imath](1+x)^n[/imath], but I'm basically clueless. Can anyone give me a hint? |
2606306 | Possibly uncountable subset of the power set of the naturals
Let [imath]S \subset \mathcal P \Bbb N[/imath] with the condition that if [imath]A, B \in S[/imath] then [imath]A \subset B[/imath] or [imath]B \subset A[/imath]. Can [imath]S[/imath] be uncountable? I've been thinking about this problem for a while and I think it's probably the case that [imath]S[/imath] can be uncountable, but I'm not sure how to prove it. If that's not the case, please correct me. I would prefer hints to a whole solution please, because I'm really struggling to understand how one approaches this problem in the first place, and any suggestions about these thought-processes would be much appreciated. I've come across the useful heuristic of thinking of countable sets as those whose elements can be given specific finite descriptions, but otherwise I don't have much intuition here. Thank you very much in advance. | 2039204 | Geometric solution to find a Set of totally ordered Subsets of [imath] \mathbb{N} [/imath] is uncountable
My question is, does there exist a set of subsets of the naturals, [imath]S[/imath], such that [imath]S[/imath] is uncountable, where for any two subsets [imath]A,B \in S[/imath] we have either [imath]A \subset B [/imath] or [imath]B \subset A [/imath]? Initially my thoughts were that we could construct a chain [imath]A_{1} \subset A_{2} \subset \cdots [/imath] and find a bijection by picking out one element which is 'new' to each set by axiom of choice and so we have bijected to a subset of naturals, however I'm not convinced this fully works. Furthermore, I was thinking there may be a nice elegant solution by looking for such a set in either [imath] \mathbb{Q}^2 [/imath] or [imath] \mathbb{Z}^2 [/imath] for example, but any attempt seems to fail due to the 'ordering' part of the question. |
2606366 | Value of integration
Given that [imath]\int_{-\infty}^\infty e^{{-x}^2} dx = \sqrt{\pi}[/imath] The value of: [imath]\iint_{-\infty}^\infty e^{{{-(x}^2}+xy+{{y}^2)}}dxdy[/imath] is A) [imath]\sqrt{\frac{\pi}{3}}[/imath] B)[imath]\frac{\pi}{\sqrt3}[/imath] C)[imath]\sqrt{\frac{2\pi}{3}}[/imath] D)[imath]\frac{2\pi}{\sqrt3}[/imath] *Both the limits of double integral is from [imath]-\infty[/imath] to [imath]\infty[/imath] * I tried with integration by parts and it is leading nowhere.Please help. | 384732 | Find the value of [imath]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+xy+y^2)}dx\,dy[/imath]
Given that [imath]\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}[/imath]. Find the value of [imath]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+xy+y^2)}dx\,dy[/imath] I don't understand how I find this double integral by using the given data. Please help. |
2606284 | i dont know how to solve sum of series [imath]n \cdot \frac{1}{2^n}[/imath]
i need some ideas to solve [imath]\sum_{n=1}^\infty n\cdot\left(\frac12\right)^n[/imath] I prove that the series converges to using ratio method, but i dont know how to find the sum. | 1170897 | Find the sum of [imath]\sum_{n=1}^{\infty}\frac{n}{x^n}[/imath]
The problem states: Wherever it converges, find the exact sum of [imath]\displaystyle\sum_{n=1}^{\infty}\frac{n}{x^n}[/imath] Using ratio test I know the series converges iff [imath]|x|>1[/imath], but have not idea how to calculate the sum. Any help is highly appreciated. Thanks and regards. |
2596820 | Find the minimum value of the expression [imath]|z|^2 +|z-3|^2 +|z-6i|^2[/imath]
I have this question as homework: Find the minimum value of the expression: [imath]|z|^2 +|z-3|^2 +|z-6i|^2.[/imath] Here's what I did: I plotted the points (0,0), (0,6), and (3,0) on the argand plane and joined them to make a triangle. Now here is where I doubt myself. I found it centroid (1,2) and this should be the centre of mass if unit masses are kept at every vertex. Yes, centre of mass. Now the Moment of Inertia of a planar object is minimum about the axis passing perpendicular through its centre of mass and also from the parallel axis theorem [imath] I = Icom + md^2[/imath] where [imath] Icom = (1)|z|^2 + (1)|z-3|^2 + (1)|z-6i|^2[/imath] I is the moment of Inertia about any axis parallel to Icom at a distance d from it. So the minimum value of expression must be minimum about this point only. My answer is also correct. If I am correct how do to explain it mathematically or using mathematic theorems or axioms. Otherwise how do I do it. Thanks. I can shift this to Physics Stack Exchange if this doesn't fit here. Sorry for no LaTeX as I have just started using Stack Exchange from mobile. Please edit whatever required. | 1104165 | Complex numbers - minimum value proof
Question: For:[imath]|z - z_1|^2+|z - z_2|^2+|z - z_3|^2+\cdots+|z - z_n|^2 = S[/imath] Prove that the minimum value of [imath]S[/imath] is when:[imath]z = \frac{z_1+z_2+z_3+\cdots+z_n}{n}[/imath] I have no idea how to even start this question. I tried to do it graphically but that didn't get me anywhere. However, I feel that the easiest proof for this would be graphically. |
2606737 | What does this "\" mathematics symbol mean?
I am reading a paper which has a mathematical argument: [imath]X \subset V \setminus \{0\}[/imath]. Can anyone explain this statement to me? Especially the "[imath]\setminus \{0\}[/imath]". Many thanks in advance! | 669767 | What does "\" mean in math
In a Linear Algebra textbook I am reading, the following is stated: [imath]b\notin \operatorname{span}(A \cup \{a\})\setminus \operatorname{span}(A)[/imath]. It does so without explaining what "[imath]\setminus[/imath]" means. I apologize if this question does not belong here but I just want to understand what it means. I can close the question if someone just comments on its meaning. |
649971 | Consider the smallest number in each of the [imath]n\choose r[/imath] subsets (of size [imath]r[/imath]) of [imath]S=\{1,2,\ldots,n\}[/imath]...
Consider the smallest number in each of the [imath]n\choose r[/imath] subsets (of size [imath]r[/imath]) of [imath]S=\{1,2,\ldots,n\}[/imath].Show that the arithmetic mean of the numbers so obtained is [imath]{n+1}\over{r+1}[/imath]. I have no idea, how to solve it.It would be better if the solution has bijection. Please help! | 2266133 | Show that arithmetic mean of numbers obtained is [imath]\frac{n+1}{r+1}[/imath]
Consider the smallest number in each of the [imath]n \choose r[/imath] subsets (of size r) of [imath]S = \{1, 2, \ldots, n\}[/imath]. Show that the arithmetic mean of the numbers so obtained is [imath]\frac{n+1}{r+1}[/imath] I've tried counting the number of times every integer appears, but it seems too complicated (for me). Is there any way this can be done without induction? |
2606107 | Differentiability of the function:[imath]g(x,y)=f(\sqrt{x^2+y^2})[/imath]
Let [imath]g(x,y)=f(\sqrt{x^2+y^2})[/imath] with [imath]f[/imath] continuously differentiable([imath]\mathbb{R}\to\mathbb{R}[/imath]) and [imath]f'(0)=0[/imath]. Then, is [imath]g[/imath] differentiable on [imath]\mathbb{R^2}[/imath]? Specifically, does [imath]g(0,0)[/imath] and [imath]f(0)[/imath] have a role to play? | 603028 | Differentiability of [imath]g:=f(\sqrt{x^2+y^2})[/imath] for a [imath]C^1[/imath] function with [imath]f'(0)=0[/imath] NBHM [imath]2008[/imath]
Let [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] be a continuously differentiable function such that [imath]f'(0)=0[/imath]. Define for all [imath]x,y\in \mathbb{R}[/imath], [imath]g(x,y)=f(\sqrt{x^2+y^2})[/imath] Pick out the true statements : [imath]g[/imath] is differentiable on [imath]\mathbb{R}^2[/imath] [imath]g[/imath] is differentiable on [imath]\mathbb{R}^2[/imath] if and only if [imath]f(0)=0[/imath] [imath]g[/imath] is differentiable on [imath]\mathbb{R}^2- \{ (0,0)\}[/imath] I do not have much idea of differentiation of multivariable calculus. I just tried out some examples to get some idea.. I have to take some function whose derivative at [imath]0[/imath] has to be zero. for simplicity I would take [imath]f(x)=x^2[/imath] Then I would get : [imath]g(x,y)=x^2+y^2[/imath] which is differentiable on [imath]\mathbb{R}^2[/imath] This would suggest me that second option has more chances as [imath]f(0)=0[/imath] in this case. I would now take [imath]f(x)=x^2+1[/imath] then : [imath]g(x,y)=x^2+y^2+1[/imath] which is differentiable on [imath]\mathbb{R}^2[/imath] though [imath]f(0)\neq 0[/imath] So, I would now eliminate second option and third option also... So, It is now got confirmed that i have to work out to "Prove" that first option is correct. Now, I have to prove that [imath]g(x,y)=f(\sqrt{x^2+y^2})[/imath] is differentiable. I have to look for partial derivatives first : [imath]\frac{\partial g}{\partial x}= f'(\sqrt{x^2+y^2}).\frac{2x}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}[/imath] [imath]\frac{\partial g}{\partial y}= f'(\sqrt{x^2+y^2}).\frac{2y}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}[/imath] I have used here that [imath]f[/imath] is differentiable on [imath]\mathbb{R}[/imath] As [imath]f[/imath] is continuously differentiable, we have [imath]f'[/imath] is continuous and I know that : [imath]\frac{x}{\sqrt{x^2+y^2}}[/imath] and [imath]\frac{y}{\sqrt{x^2+y^2}}[/imath] are also continuous. As [imath]f'[/imath] and [imath]\frac{x}{\sqrt{x^2+y^2}}[/imath] are continuous So is the product [imath]\frac{\partial g}{\partial x}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}[/imath] As [imath]f'[/imath] and [imath]\frac{y}{\sqrt{x^2+y^2}}[/imath] are continuous So is the product [imath]\frac{\partial g}{\partial y}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}[/imath] So we have continuous partial derivatives for [imath]g(x,y)[/imath].. So, I am willing to conclude that: [imath]g(x,y)[/imath] is differentiable on [imath]\mathbb{R}^2[/imath] I would be thankful if some one can confirm if the answer correct. I would be much more thankful if some one can let me know if there are any gaps in my argument. This is just a proof verification Question.. Thank you. |
2606406 | proving bound on edges using Euler formula
I want to prove the following result Every [imath]n[/imath]-vertex planar graph with [imath]n\geq3[/imath] and with girth [imath]g[/imath] has at most [imath]\frac{g}{g-2}(n-2)[/imath] edges. Any hints? I have no idea. I don't see any useful link between the girth and the number of edges. | 2000888 | Prove that a graph on [imath]n[/imath] vertices with girth [imath]g[/imath] has at most [imath]\frac{g}{g-2}(n-2)[/imath] edges
I'm not sure how to go about this. One of my thoughts was to take the number of edges in a complete graph, and subtract all of the edges that would be needed to make a smaller cycle. If you have [imath]C_5[/imath], there is only one way to make a [imath]5[/imath]-cycle on [imath]5[/imath] vertices, but there's [imath]10[/imath] ways to make a [imath]3[/imath]-cycle, and... I have no idea how to do this without counting things multiple times. |
2607455 | Find nth term for below sequence
I need nth term for the below sequence [imath]1,2,(3/2),(5/3),(8/5),(13/8),(21/13),(34/21)........nth term[/imath] I know [(present number)= sum of previous(numerator + denominator ) / previous numerator] Thanks before | 460147 | Find [imath]n^{th}[/imath] term of [imath]\frac23,\frac35,\frac58,\frac8{13}\dots[/imath]
[imath]\frac23,\frac35,\frac58,\frac8{13},\frac{13}{21},\frac{21}{34},\frac{34}{55},\ldots[/imath] You would be able to find here a pattern that the denominator of previous term gets new term numerator and new term denominator gets the sum of numerator and denominator of previous term |
2606644 | Can you apply CRT to the congruence [imath]84x ≡ 68[/imath] [imath](mod[/imath] [imath]400)[/imath]?
this might be a (very) stupid question, however I need to ask it (why?). I have the congruence [imath]84x ≡ 68[/imath] [imath](mod[/imath] [imath]400)[/imath] from: Congruence equation ..., where the solution is given. However, I tried to solve this using the Chinese Remainder Theorem and got the wrong answer. So my question is, can I use CRT to solve this and if so please provide an answer, if I can't use CRT why not, what condition is not fullfilled? Ok, I might not have been clear. When applying CRT I get: Primefactorization of [imath]400 = 2^4*5^2[/imath] then: [imath]84x ≡ 68[/imath] [imath](mod[/imath] [imath]2^4)[/imath] ↔ [imath]4x ≡ 4[/imath] [imath](mod[/imath] [imath]2^4)[/imath] ↔ [imath]x ≡ 1[/imath] [imath](mod[/imath] [imath]2^4)[/imath] [imath](1)[/imath] [imath]84x ≡ 68[/imath] [imath](mod[/imath] [imath]5^2)[/imath] ↔ [imath]9x ≡ 14[/imath] [imath](mod[/imath] [imath]5^2)[/imath] ↔ [imath]x ≡ 21[/imath] [imath](mod[/imath] [imath]5^2)[/imath] [imath](2)[/imath] From (1) and (2) I get: [imath]x = 1 + 16s ≡ 21[/imath] [imath](mod[/imath] [imath]25)[/imath] → [imath]s = 20 + 25*t[/imath] → [imath]x = 321 + 400t[/imath] which is wrong :( Where am I going wrong? Regards | 2605392 | Congruence equation ...
I reduced the equation: [imath]84x \equiv 68\mod 400[/imath] [imath]21x \equiv 17\mod 100[/imath] [imath]100 = 4 \times 21 + 16[/imath] [imath]21 = 1 \times 16 + 5[/imath] [imath]16 = 3 \times 5 + 1[/imath] so we have... [imath]16 = 100 - 4 \times 21[/imath] [imath]5 = 21 - 1 \times 16[/imath] [imath]1 = 16 - 3 \times 5[/imath] Now, I need the equation in the form of [imath]21v + 100w = 1[/imath] and I have to find v and w [imath]1=16-3 \times 5[/imath] [imath]=16-3(21-1 \times 16)[/imath] [imath]=16 - 3 \times 21 +3 \times 16[/imath] [imath]=100 - 4 \times 21 -3 \times 21 +3 \times 16[/imath] At this point I don't know what to do. Please help. |
2607630 | how many solutions in natural numbers are there to the inequality [imath]x_1 + x_2 +....+ x_{10} \leq 70[/imath]?
how many solutions in natural numbers are there to the inequality [imath]x_1 + x_2 +....+ x_{10} \leq 70[/imath] ? I know it has to be solved with combinatorics and specifically with the inclusion exclusion principle but I have no idea where to start. Help ? | 2606420 | Number of total possibilities for an equation
I need to find the number of possibilities for which the following equation exists: [imath]x_1 + x_2 + x_3 + \cdots + x_{10} \leq 70[/imath] Each variable is a non-negative integer. I tried simplifying the question to the point of finding the number of possible solutions to each equation seperatly: [imath]x_1 + x_2 + x_3 + \cdots + x_{10} = 70,[/imath] which is [imath]{79}\choose{10}[/imath]. [imath]x_1 + x_2 + x_3 \cdots + x_{10} = 69,[/imath] which is [imath]{78}\choose{10}[/imath]. [imath]x_1 + x_2 + x_3 + \cdots + x_{10} = 68,[/imath] which is [imath]{77}\choose{10}[/imath], and so forth. Then I thought about adding all of these together. This seems a bit too much work considering that the textbook solution is: [imath]{80}\choose{10}[/imath], and I can't seem to figure out the idiomatic way of approaching these kind of questions. Any help is greatly appreciated. |
2607725 | [imath][Solved][/imath] Find all units in the ring [imath]M_2(\mathbb{Z})[/imath]
my problem here is: Find all units in the ring [imath]M_2(\mathbb{Z})[/imath] I've done it before but now it's a matrix and I'm lost. First I think on the determinant because I saw some other wrote about it, but I'm not sure. Any help? Thank you | 3655 | Units of [imath]M_2(Z)[/imath]
In one of my classes we discussed the ring of 2x2 matrices [imath]M_{2}(\mathbb{Z})[/imath]. We said that its group of units was [imath]SL_{2}(\mathbb{Z})[/imath] which means that it is the set of 2x2 with determinant equal to [imath]\pm[/imath]1. Why can't we have a 2x2 matrix with entries a,b,c, and d such that [imath]\frac{a}{ad-bc}[/imath],[imath]\frac{-b}{ad-bc}[/imath],[imath]\frac{-c}{ad-bc}[/imath], and [imath]\frac{d}{ad-bc}[/imath] are all integers? I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated |
2607370 | The subgroup of upper triangular matrices of [imath]GL_n(F)[/imath] is equal to its normalizer
Let [imath]U[/imath] be the subgroup of [imath]G = GL_n(F)[/imath] of upper triangular matrices. Prove that the normalizer of [imath]U[/imath] in [imath]G[/imath] is [imath]U[/imath], i.e [imath]N_G(U) = U[/imath]. Two observations: Let [imath]L[/imath] be the subgroup of [imath]G[/imath] of lower triangular matrices. This subgroup is conjugate to [imath]U[/imath] by the matrix [imath] P = \begin{bmatrix} 0 & 0 & \dots & 1 \\ 0 & & 1 & 0 \\ \vdots & && \vdots \\ 1 & 0 & \dots & 0 \end{bmatrix}. [/imath] Considering the conjugation action of [imath]G[/imath] acting on [imath]G/U[/imath] we have [imath]L \in O_U[/imath], which is restating what's above. I feel like there is something very natural I should do at this point, but don't see or know. More directly, we have [imath]U \subset N_G(U)[/imath]. We need to show [imath]N_G(U) \subset U[/imath], so looking for a contradiction suppose there exists a matrix [imath]B[/imath] such that [imath]B \in N_G(U)[/imath] and [imath]B \notin U[/imath]. That is [imath]BUB^{-1} = U[/imath] and [imath]B[/imath] is not upper triangular, thus in [imath]B[/imath] there exists some [imath]b_{ij} \neq 0[/imath] where [imath]1 \le j < i \le n[/imath]. Now maybe break up the matrix [imath]B = B_u + B_l[/imath] where [imath]B_u[/imath] is the upper diagonal values of [imath]B[/imath] and [imath]B_l[/imath] is the lower diagonal values below the diagonal. So [imath](B_u + B_l)U = U(B_u + B_l)[/imath], thus for any [imath]T_1 \in U[/imath], there exists [imath]T_2 \in U[/imath] such that [imath]B_uT_1 + B_l T_1 = T_2 B_u + T_2 B_l[/imath]. Then [imath]B_lT_1 - T_2 B_l = T_2 B_u - B_u T_1 \in U[/imath]. Now this seems to be getting a bit muddled. Can someone provide some guidance for either proving the statement with using that fact that upper and lower matrices are conjugate or showing [imath]N_G(U) \subset U[/imath]? Thanks. | 2087062 | Normalizer of upper triangular group in [imath]{\rm GL}(n,F)[/imath]
The following question has already appeared on mathstack: If [imath]B[/imath] is the subgroup of [imath]{\rm GL}(n,F)[/imath] consisting of upper triangular matrices then normalizer of [imath]B[/imath] in [imath]{\rm GL}(n,F)[/imath] is [imath]B[/imath] itself. I know a proof of this using Bruhat decomposition of [imath]{\rm GL}(n,F)[/imath]. Question: Can we prove above theorem without using Bruhat decomposition? Why came to this question: Consider the general linear Lie algebra [imath]L=\mathfrak{gl}(n,F)[/imath]; in it, let [imath]T=\mathfrak{t}(n,F)[/imath] be the upper triangular sub-algebra. Then normalizer of [imath]T[/imath] in [imath]L[/imath] is [imath]T[/imath] itslef, and this can be proved just by considering a very simple decomposition of [imath]{\mathfrak gl}(n,F)[/imath]: write any element as sum of upper triangular matrix and lower triangular matrix whose diagonal is [imath]0[/imath]. But then for problem above, is it necessary to use Bruhat decomposition? |
2607475 | Is it possible to construct an entire function that grows faster than any given continuous function.
Assume that [imath]f[/imath] is a given real-valued continuous function over [imath]\mathbb{R}[/imath]. How to construct an entire function [imath]F[/imath] such that [imath]F(z)[/imath] is real-valued when [imath]z\in \mathbb{R}[/imath] and there exists a real number [imath]a[/imath] such that [imath]f(z)<F(z)[/imath] for all [imath]z>a,z\in \mathbb{R}[/imath]. | 141946 | Can a function "grow too fast" to be real analytic?
Does there exist a continuous function [imath]\: f : \mathbf{R} \to \mathbf{R} \:[/imath] such that for all real analytic functions [imath]\: g : \mathbf{R} \to \mathbf{R} \:[/imath], for all real numbers [imath]x[/imath], there exists a real number [imath]y[/imath] such that [imath]\: x < y \:[/imath] and [imath]\: g(y) < f(y) \:[/imath]? |
2607176 | Let [imath]f:[0,a]\times[0,b]\rightarrow\mathbb{R}[/imath] be a continuous function. Prove a version of Fubini's Theorem:
[imath]\int_0^b\left(\int_0^af(x,y)dx\right)dy=\int_0^a\left(\int_0^bf(x,y)dy\right)dx.[/imath] I am given the following hint: Investigate [imath]p:[0,b]\rightarrow\mathbb{R}[/imath] defined by [imath]p(t)=\int_0^t\left(\int_0^af(x,y)dx\right)dy-\int_0^a\left(\int_0^tf(x,y)dy\right)dx.[/imath] I have read the classical proof of Fubini's Theorem that uses [imath]\sigma[/imath]-finite spaces, but this is beyond the scope of which this question is asked in. I am looking for advice on how to utilize the provided hint to establish a proof of this. | 1578462 | Prove that [imath]\int_a^b \left( \int_c^d f(x,y)dy\right) dx=\int_c^d \left( \int_a^b f(x,y)dx\right) dy[/imath]
Let [imath]f[/imath] be continuous function on [imath][a,b]\times [c,d][/imath]. Prove that Prove that [imath]\int_a^b \left( \int_c^d f(x,y)dy\right) dx=\int_c^d \left( \int_a^b f(x,y)dx\right) dy[/imath] First of all, note that [imath] \int_c^d f(x,y)dy[/imath] is continuous in [imath]x[/imath] and [imath] \int_a^b f(x,y)dx[/imath] is continuous in [imath]y[/imath] so both integrals exist. We have [imath]\frac{d\left[\int_a^b \left( \int_c^t f(x,y)dy\right) dx \right]}{dt}=\int_a^b f(x,t) dx[/imath] (We used differentiation under the integral sign and the fundamental theorem of calculus) Also [imath]\frac{d\left[\int_c^t \left( \int_a^b f(x,y)dx\right) dy \right]}{dt}=\int_a^b f(x,t) dx[/imath] (Here we used the fundamental theorem of calculus) Thus [imath]\int_a^b \left( \int_c^t f(x,y)dy\right) dx-\int_c^t \left( \int_a^b f(x,y)dx\right) dy[/imath] as a function of [imath]t[/imath] is constant on [imath](c,d)[/imath]. It remains to show that this constant is [imath]0[/imath]. How can I do this? |
2607665 | [imath]\mathrm{tr}(AB)=\mathrm{tr}(BA)[/imath] proof
I only want an intuitive proof or idea that underlines the essence of this equality. I proved this already using the summation but it doesn't help me to actually see why they are equal. I hope you could help me. Thanks in advance! | 2594277 | intuition behind cyclicality of the trace?
The trace of a matrix product is "commutative" (i.e. cyclical): [imath]\operatorname{trace}(AB)=\operatorname{trace}(BA)[/imath] I know how to prove this. I was wondering, are there any intuitive ways to understand why this must be true? One reason why I don't see it intuitively is that I don't really understand intuitively what the trace represents (e.g. geometrically). |
1646685 | How to prove this curiosity that has to do with cubes of certain numbers?
I saw on facebook some image on which these identities that I am going to write below are labeled as "amazing math fact" and on the image there are these identities: [imath]1^3+5^3+3^3=153[/imath] [imath]16^3+50^3+33^3=165033[/imath] [imath]166^3+500^3+333^3=166500333[/imath] [imath]1666^3+5000^3+3333^3=166650003333[/imath] and then it is written under these identities "and so on and on and on and on!" which suggests that for every [imath]k \in \mathbb N[/imath] we shuld have [imath](1 \cdot 10^k + \sum_{i=0}^{k-1} 6 \cdot 10^i)^3 + (5 \cdot 10^k)^3 + (\sum_{i=0}^{k} 3 \cdot 10^i)^3=16...650...03...3[/imath] (on the right hand side of the above stated identity the number of times that number [imath]6[/imath] is shown up is [imath]k-1[/imath], the number of times that number [imath]0[/imath] is shown up is [imath]k-1[/imath] and the number of times that number [imath]3[/imath] is shown up is [imath]k[/imath]) This problem seems attackable with mathematical induction but I would like to see how it could be proved without using mathematical induction in any step(s) of the proof. | 2714122 | a specific sum of cube property(narcissistic number)
So I am currently doing a problem about narcissistic number and I need to prove that [imath]a^3+b^3+c^3=a*10^{2n}+b*10^n+c[/imath], where [imath]a=1666..666, b=5000..00, c=33...333 [/imath] and [imath]a,b,c[/imath] has [imath]n[/imath] digits. I am trying to do an induction and have observed that [imath]3a+2=b[/imath], [imath]3c+1=2b[/imath] and [imath]a+b+c=10^n[/imath]; the difficult part when doing induction is I just cannot write the equation into [imath]b[/imath] and [imath]n[/imath] which would allow me to solve as I know that [imath]b=(10^n)/2[/imath]. Is there any idea that is better than induction? Thanks in advance. |
2515600 | How do I prove that [imath]1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n[/imath]?
How do I prove that [imath]1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n[/imath]? I've spent quite some time on this problem. So far, I've simplified the right-hand side to [imath]\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)][/imath]. But then, the algebra becomes very complicated when I add [imath](n+1)^4[/imath] to both sides of the inductive hypothesis. | 2434311 | A formula for [imath]1^4+2^4+...+n^4[/imath]
I know that [imath]\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)[/imath] and [imath]\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.[/imath] Here is the question: is there a formula for [imath]\sum^n_{i=1}i^4.[/imath] |
2341778 | Question about baby rudin theorem 5.12 corollary
The corollary says if [imath] f [/imath] is differentiable on [imath] [a,b] [/imath] then [imath] f' [/imath] cannot have any simple discontinuities on [imath][a,b] [/imath]. I just don't how to prove it. I think it should be proved on both two cases of simple discontinuities(first type and second type of simple discontinuities). Thanks in advance. | 622738 | Discontinuities of the derivative of a differentiable function on closed interval
I have a question about the corollary to theorem 5.12 in Rudin's Principles of Mathematical Analysis (page 108): Suppose [imath]f[/imath] is a real differentiable function on [imath][a,b][/imath] and suppose [imath]f'(a)< \lambda < f'(b)[/imath] then there is a point [imath]x \in (a,b)[/imath] such that [imath]f'(x) = \lambda[/imath] Corollary: If [imath]f[/imath] is differentiable on [imath][a,b][/imath] then [imath]f'[/imath] cannot have any simple discontinuities on [imath][a,b][/imath]. Can someone help me to show how he uses the result in the "main theorem" in the corollary? (There are two cases of simple discontinuities [imath]f(x+) = f(x-) \neq f(x)[/imath] and [imath]f(x +) \neq f(x-)[/imath] |
1888337 | If [imath]f,g[/imath] are continuous and [imath]g[/imath] is [imath]1[/imath]-periodic, [imath]\int_0^1 f(x)g(nx)dx \xrightarrow[n\to\infty]{} \int_0^1 f \int_0^1 g[/imath]
Let [imath]f,g\in C(\mathbb{R},\mathbb{R})[/imath] where [imath]g(x+1)=g(x) \; \forall x\in \mathbb{R}[/imath]. Prove that \begin{equation} \lim_{n\rightarrow\infty} \int_0^1 f(x)g(nx)dx=\int_0^1f(x)dx \int_0^1g(x)dx . \end{equation} | 105258 | integral involving a periodic function
Let [imath]f:[0,1] \longrightarrow \mathbb R[/imath] be a continuous function, and let [imath]g:\mathbb R \longrightarrow \mathbb R[/imath] be a continuous and periodic function with period [imath]1[/imath]. Prove that [imath]\displaystyle\lim_{n\to \infty}\int_0^1f(x)g(nx)dx=\left(\int_0^1f(x)dx\right)\left(\int_0^1g(x)dx\right)[/imath]. any Ideas? |
2608455 | How do I calculate the sum of this infinite series? [imath]\sum\limits_{n=1}^{\infty}\frac{1}{4n^{2}-1}[/imath]
Could someone please help me with how do I calculate the sum of the [imath]\sum_{n=1}^{\infty}\frac{1}{4n^{2}-1}[/imath] infinite series? I see that [imath]\lim_{n\rightarrow\infty}\frac{1}{4n^{2}-1}=0[/imath] so the series is convergent based on the Cauchy's convergence test. But how do I calculate the sum? Thank you. | 265277 | sum of this series: [imath]\sum_{n=1}^{\infty}\frac{1}{4n^2-1}[/imath]
I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: [imath]\sum_{n=1}^{\infty}\frac{1}{4n^2-1}[/imath] my steps: [imath]\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\sum_{n=1}^{\infty}\frac{2}{4n^2-1}-\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=..help..=sum[/imath] I am lacking some important properties, I feel I am coming to the right step and cannot spit that out.. |
2608175 | Prove that [imath]\lim_{n\to \infty} (a_1a_2\ldots a_n)^{\frac 1n} = L[/imath] given that [imath]\lim_{n\to \infty} (a_n) = L[/imath]
Let [imath]\{a_n\}[/imath] be a sequence of positive numbers such that [imath]\lim_{n\to \infty} a_n = L[/imath]. Prove that [imath]\lim_{n\to \infty} (a_1a_2\ldots a_n)^{\frac 1n} = L[/imath] (Also given HINT: Let [imath]a>0[/imath]. Then [imath]\lim_{n\to\infty} a^{\frac 1n}=1[/imath]) What I have so far: [imath]\text{Let}\:\epsilon>0.\:\text{Then}\;\exists\;N\in\mathrm{I}\!\mathrm{N}\;\text{such that}\;n\ge N.[/imath][imath]\text{Then}\;L-\epsilon\lt a_n \lt L + \epsilon[/imath] [imath]\text{Let}\; b_n =(a_1a_2\ldots a_n)^{\frac 1n} \text{for}\; n\ge N[/imath] [imath]\text{Now}\;b_n = (a_1a_2\ldots a_N)^{\frac 1n}(a_{N+1}\ldots a_n)^{\frac 1n}[/imath] I am struggling with how to use the given hint and maybe limit superior/inferior facts to prove the limit is L. Any tips would be greatly appreciated. Also this is my first time using MathJax so please let me know if I have made any mistakes. Thank you. EDIT: I'm looking specifically on how to prove it using [imath]\limsup[/imath] or [imath]\liminf[/imath] properties! | 1220020 | limit of a geometric mean
Suppose that a sequence [imath]a_n[/imath] of positive numbers converges to [imath]a[/imath]. Show that [imath]\lim_{n\rightarrow \infty}\left(\prod_{i=1}^{n}a_i\right)^{1/n}=a[/imath] This seems to be simple using that [imath]x=e^{\log x}[/imath], but I can't go any further after using that. PS. I'm almost sure that this post is a duplicate, but I wasn't able to find it here. I'm sorry if that's the case. |
1158322 | If [imath]a_n>0[/imath] converges to [imath]a>0[/imath], then [imath](a_0 a_1\cdots a_n)^{\frac{1}{n}}[/imath] converges to [imath]a.[/imath]
I would appreciate your help! How can we show that if a sequence of positive real numbers [imath]a_n[/imath] converges to [imath]a\in\mathbb{R}[/imath] with [imath]a>0[/imath], then [imath](a_0 a_1\cdots a_n)^{\frac{1}{n}}[/imath] converges to [imath]a[/imath]. | 2880000 | Geometric Mean of real, convergent sequence preserves it's limit
For a given sequence [imath]\left\{x_n\right\}[/imath] we define its sequence of geometric means to be [imath]\left\{G_n\right\}[/imath], where [imath]G_n=\sqrt[n]{x_1x_2 \cdots x_n}[/imath]. We have to show that if [imath]x_n \rightarrow L[/imath] then [imath]G_n \rightarrow L[/imath]. One method that I have implemented to prove this theorem is that I have used the property of continuous function (i.e if [imath]f : A \rightarrow \mathbb{R}[/imath] is a continuous function, then a convergent sequence [imath]\left\{x_n\right\}[/imath] in [imath]A[/imath] implies [imath]\left\{f(x_n)\right\}[/imath] is convergent, and [imath]f(x_n) \rightarrow f(x)[/imath] if [imath]x_n \rightarrow x[/imath] ) and the fact that the sequence of arithmetic mean of a convergent sequence preserves its limit (same statement as in the paragraph above but it's replaced with arithmetic mean in place of geometric mean, which I've proven). But suppose you don't know anything about continuous function. Can anyone give me a hint as to how could I solve my original problem, without using any properties of continuous function ? |
2607644 | Functions on [imath]\mathbb{R}^n[/imath] commuting with orthogonal transformations
I am interested in finding the functions [imath]f:\mathbb{R}^n \to \mathbb{R}^n[/imath] for which [imath]f \circ U = U \circ f[/imath] for all orthogonal transformations [imath]U:\mathbb{R}^n \to \mathbb{R}^n[/imath]. Note that [imath]f[/imath] need not be linear. Any ideas on the conditions on [imath]f[/imath] ? | 2591815 | Let [imath]V[/imath] be spherically symmetric and [imath]W[/imath] i.i.d. Gaussian. Then [imath]E[V|V+W=t]=g(\|t\|)[/imath]
A random vector [imath]V \in \mathbb{R}^n[/imath] is said to be spherically symmetric. If for every orthogonal matrix [imath]A[/imath] the distribution of [imath]AV[/imath] is the same as the distribution of [imath]V[/imath]. My question: Let [imath]W[/imath] be i.i.d. zero mean Gaussian random variable. Can we show that for a spherically symmetric [imath]V[/imath] the conditional expectation \begin{align} E[V|V+W=t]= g(\|t\|). \end{align} That is the conditional expectation is only a function of the magnitude. My attempt: \begin{align} E[V|V+W=t]= \int V \frac{f_W(t-v)}{f_{V+W}(t)} dP_V(v)= \frac{\int v f_W(t-v) dP_V(v)}{f_{V+W}(t)}= \frac{E[ V f_W(t-V) ]}{f_{V+W}(t)} \end{align} Therefore, we can show that the two functions in the above ratio only depend on [imath]\|t\|[/imath]. Observe that we can write [imath]t= \|t\| u_t[/imath] where [imath]u_t[/imath] is a unit vector. Then, for any orthogonal matrix [imath]A[/imath] \begin{align} f_{V+W}(t)=f_{V+W}(\|t\|, u_t)&= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t-V \|^2}{2}} ]\\ &= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t-AV \|^2}{2}} ]\\ &= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|tA^{-1}-V \|^2}{2}} ]\\ &= \frac{1}{(2\pi)^{n/2}} E[ e^{-\frac{\|t A^T-V \|^2}{2}} ]\\ &= f_{V+W}(tA^T)\\ &= f_{V+W}(\|t\|, u_t A^T) \end{align} Since, [imath]A[/imath] was arbitrary we have that [imath]f_{V+W}(t)=h(\|t\|)[/imath](i.e., only depends on the magnitute). Now I am not sure how to show that [imath]E[ V f_W(t-V) ][/imath] only depends on [imath]\|t\|[/imath]? |
2608764 | [imath]\displaystyle\mathbb{R}^n=\bigcup_{k=1}^\infty W_k[/imath]
Show that if [imath]\mathbb{R}^n=W_1\bigcup W_2\bigcup\ldots\bigcup W_k\bigcup\ldots[/imath], where each [imath]W_k[/imath] is a subspace , then [imath]\mathbb{R}^n=W_i[/imath] for some [imath]i[/imath]. | 2562815 | How to prove if [imath]\Bbb R^n=W_1\cup W_2\cup \cdots\cup [/imath] where each [imath]W_k[/imath] is a subspace, then [imath]\Bbb R^n=W_i[/imath] holds for some [imath]i[/imath]
I know if [imath]W_1,\ldots,W_k[/imath] are subspaces of a vector space, their sum is the span of their union. ie. [imath]W_1+W_2+\cdots+W_k=W_1\cup W_2\cup \cdots \cup W_k[/imath]. But how to prove [imath]\Bbb R^n=W_i[/imath] holds for some [imath]i[/imath]? Is there anything to do with direct sum? Because if [imath]\Bbb R^n[/imath] is the direct sum of [imath]W_1+W_2+\cdots+W_k\iff[/imath] the sum of the dimensions of the [imath]W_i[/imath] is [imath]n[/imath]. |
2608832 | Proof for f must be a constant polynomial
Suppose [imath]f[/imath] is a complex polynomial in [imath]z[/imath] and [imath]\bar z[/imath]. Assume that [imath]\frac{\partial f}{\partial z} = 0[/imath] and [imath]\frac{\partial f}{\partial \bar z} = 0[/imath]. How do you prove that [imath]f[/imath] here MUST be a constant polynomial? Thanks in advance!!! | 104607 | A complex polynomial with partial derivatives equal to zero is constant.
There is an exercise in Function Theory of One Complex Variable by Greene & Krantz that is very similar to a Proposition in the book, but I am having trouble getting to the conclusion. Let [imath]f : \mathbb{C} \to \mathbb{C}[/imath] be a polynomial. Suppose further that [imath]\frac{\partial f}{\partial z} = 0[/imath] and [imath]\frac{\partial f}{\partial \bar z} = 0[/imath] for all [imath]z \in \mathbb{C}[/imath] Prove that [imath] f \equiv[/imath] constant. Now, part of the proposition (1.3.2) proves that if [imath]p[/imath] is a polynomial, [imath]p(z, \bar z) = \sum a_{lm}z^l \bar z^m[/imath] with [imath]\frac{\partial f}{\partial \bar z} = 0[/imath], then [imath]\frac{\partial ^{l + m}}{\partial z^l \partial \bar z^m}p[/imath] evaluated at zero is [imath]l!m!a_{lm}[/imath]. Is this the constant that [imath]f[/imath] equals? It doesn't seem like I've used the hypothesis, or reached the conclusion! Maybe I should use the definition of the partial... [imath]\frac{\partial f}{\partial z} := 1/2 (\frac{\partial }{\partial z} - i \cdot \frac{\partial }{\partial z})f[/imath] Any hints would be greatly appreciated. |
2608874 | Need of Localisation? and Intuition.
I was going through some material related to localisation of rings. I am having a trouble with understanding why it is needed. I just see it as a way to create more rings but I have no idea what is special about the [imath]S^{-1}A[/imath] where [imath]S[/imath] is a multiplicative closed set and [imath]R[/imath] is the ring. My knowledge of tensors if very limited. | 27660 | Why do the Localization of a Ring
This question may be a bit vague, but neverthless, i would like to see an answer. Wikipedia tells me that: In abstract algebra, localization is a systematic method of adding multiplicative inverses to a ring. It is clear that for integral domains, we have the Field of Fractions, and we work on it. It's obvious that a Field has multiplicative inverses. Now if we consider any arbitrary ring [imath]R[/imath], by the definition of localization, it means that we are adding multiplicative inverses to [imath]R[/imath] thereby wanting [imath]R[/imath] to be a division ring or a field. My question, why is it so important to look at this concept of Localization. Or how would the theory look like if we never had the concept of Localization. |
2608902 | Show that [imath](x^2 + y^2) \sin\left(\frac{1}{x^2 + y^2}\right)[/imath] is differentiable.
In this exercise one must show that [imath]f(x,y) = \begin{cases} 0 \text{ if } (x,y) = (0,0)\\ (x^2 + y^2) \sin\left(\frac{1}{x^2 + y^2}\right) \text{ else} \end{cases}[/imath] is differentiable for all [imath](x,y) \in \mathbb{R^2}[/imath] In the lecture, we have proven that if all partial derivatives of [imath]f[/imath] exist and are continuous, then the derivative of [imath]f(x,y)[/imath] exists everywhere. I start by calculating the partial derivatives of [imath]f[/imath] at [imath](0,0)[/imath]: [imath]\partial_xf(0,0) = \lim_{x\to0} \frac{f(x,0)-f(0,0)}{x} = \lim_{x\to0} \frac{x^2\sin({1 \over x^2})}{x} = 0[/imath] Similar for [imath]\partial_y[/imath]: [imath]\partial_yf(0,0) = \lim_{y\to0} \frac{f(0,y)-f(0,0)}{x} = \lim_{y\to0} \frac{y^2\sin({1 \over y^2})}{y} = 0[/imath] We know that for all other values of [imath]x[/imath], [imath]\partial_x f = 2x\sin\left(\frac{1}{x^2+y^2}\right)-\frac{2x\cos \left(\frac{1}{x^2+y^2}\right)}{x^2+y^2}[/imath] I have tried to show that [imath]\lim _ {x,y \to (0,0)} \partial_x f = 0[/imath]. (to show f is continuous) this however is problematic, since this limit doesn't exist. This implies that the partial derivates are not continuous at all points. Which makes this approach of proving that [imath]f[/imath] is differentiable useless. How do I prove that [imath]f[/imath] is differentiable on all of [imath]\mathbb{R}^2[/imath]. I am open for all kind of suggestions. Thanks for your time. | 1818523 | Is [imath](x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}}[/imath] Differentiable in [imath](0,0,0)[/imath]?
[imath]f(x,y,z)=\begin{cases} (x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}} \qquad (x,y,z) \ne (0,0,0) \\ \\ 0 \qquad (x,y,z)=(0,0,0) \end{cases} [/imath] At first, I study the continuity in the origin. I apply the concept of sequential continuity: [imath]x_n \rightarrow x_0 \Rightarrow f(x_n) \rightarrow f(x_0)[/imath] So, [imath]x_n=\frac{1}{n} [/imath] [imath]x_n \rightarrow 0 \Rightarrow f(x_n,x_n,x_n) \rightarrow f(0,0,0)=0[/imath] [imath]\lim_{n\rightarrow +\infty} \frac{3}{n^2} \ \sin \frac{1}{\frac{\sqrt{3}}{n}}=0 [/imath] How can I continue the study of differentiation? Thanks! |
2584515 | Interesting short Inequality
This problem is from Romanian G.M. and although it is very short, it is also (I think) very hard. Let [imath]x, y, z[/imath] real non-negative numbers such that [imath]x+y+z=3[/imath]. Prove that [imath]27 \leq (x^2+2)(y^2+2)(z^2+2) \leq 44 [/imath] I didn't succeed in any of the inequalities, I only managed to find the equality cases ([imath]x=y=z=1[/imath] for the first one and one of them [imath]3[/imath] and the others [imath]0[/imath]). Any hint/idea/solution is welcome. | 2613849 | How can I prove that [imath]27\leq \left( x^{2}+2\right) \left( y^{2}+2\right) \left( z^{2}+2\right) \leq 44,\ x+y+z=3[/imath]
If [imath]x,y,z\in [0,\infty )[/imath] such that [imath]x+y+z=3[/imath] prove that [imath]27\leq \left( x^{2}+2\right) \left( y^{2}+2\right) \left( z^{2}+2\right) \leq 44[/imath]. I tried to use the relation [imath]x^{2}+2\leq x^{2}+3x+2,x\geq 0[/imath], but without success. |
2605285 | Normalizing a wave function
I need to normalize the wave function [imath]Axe^{-ax^2}[/imath]. But this asks me the integral of [imath]A^2x^2e^{-2ax^2}[/imath] over entire space. But how am I supposed t evaluate that? I know about improper integrals but I think I need to use by parts here and from there it is all getting messed up. Any help is appreciated. Thanks. | 670221 | Integral of [imath]x^2e^{-ax^2}[/imath]
Hey guys I need to find the following integral using integration by parts and not the gamma function. Also there is an a constant a in the exponential function. So it is actually [imath]x^2e^{-ax^2}[/imath]. Thanks for the help! |
2609487 | Integral of a Polynomial in Square Root
I need to solve the indefinite integral: [imath]\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} dx,[/imath] but I can not find any technique which could solve it. Can you help me please? | 2605013 | How to find [imath]\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x[/imath]?
The integral to be solved is [imath]\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x[/imath]. I tried substitution [imath]t=x^8[/imath] which got me to [imath]\frac{1}{8} \int \sqrt{t + 2 + \frac{1}{t}} t^{-7/8} \,\mathrm{d}t[/imath] and I'm stuck. Can you help? |
2609514 | Simple proof of [imath]x^n−y^n=(x−y)(x^{n−1}+x^{n−2}y+…+xy^{n−2}+y^{n−1})[/imath]
I'd like to see a simple proof of [imath]x^n−y^n=(x−y)(x^{n−1}+x^{n−2}y+…+xy^{n−2}+y^{n−1})[/imath] I don't know what proof by induction is... is there a simpler way? The previous post on this doesn't really solve my problem.... I don't understand the induction proofs. The text that I am reading with this is this: | 2601887 | Power rule step. Where does equation of the difference of two numbers raised to the same power come from?
I am looking at a proof of the power rule and I see this: [imath]x^n - a^n = (x-a)(x^{n-1} + x^{n-2}*a + {...} + xa^{n-2} + a^{n-1})[/imath] Where does this come from? It's used in this proof which I'm trying to understand: What is it called? |
2610222 | Help me simplify this expression
I've this expression to simplify: [imath]\frac {1}{2^{3N}}\frac{(2m\pi)^{3N/2}}{h^{3N}\sqrt{\frac{2\pi3N}{2}}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2}}U^{3N/2}\frac{V^N}{\pi^{3N}}[/imath] and I'm supposed to end with: [imath]V^N\left(\frac{U}{N}\right)^{3N/2}\left(\frac{me}{3h^2\pi}\right)^{3N/2}[/imath] But whenever I try I couldn't reach it | 2610259 | Help me simplify this expression (II)
I resume here (giving the original text) what I've written here. I've this expression to simplify: [imath]\frac {1}{2^{3N}}\frac{(2m\pi/h^2)^{3N/2}}{\Gamma(3N/2+1)}U^{3N/2}\frac{V^N}{\pi^{3N}}[/imath] To place it in context, it models the entropy of an ideal gas of [imath]N[/imath] monoatomic molecules. Using the Stirling approximation for the [imath]\Gamma[/imath], I've rewritten it as: [imath]\frac {1}{2^{3N}}\frac{(2m\pi)^{3N/2}}{h^{3N}\sqrt{\frac{2\pi3N}{2}}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2}}U^{3N/2}\frac{V^N}{\pi^{3N}}[/imath] and I'm supposed to end with: [imath]V^N\left(\frac{U}{N}\right)^{3N/2}\left(\frac{me}{3h^2\pi}\right)^{3N/2}[/imath] |
2608705 | [imath]\lim_{n\to \infty}\sum_{j=0}^{[n/2]} \frac{1}{n} f\left( \frac{j}{n}\right)[/imath]
If f is continuous in [0,1] then [imath]\lim_{n\to \infty}\sum_{j=0}^{[n/2]} \frac{1}{n} f\left( \frac{j}{n}\right)[/imath](where [y] is the largest integer less than or equal to y) (A) does not exist (B) exists and is equal to [imath]\frac{1}{2} \int_0^1 f(x)dx[/imath] (C) exists and is equal to [imath]\int_0^1 f(x)dx[/imath] (D) exists and is equal to [imath]\int_0^{\frac{1}{2}} f(x)dx[/imath] My approach: Let I = [imath]\lim_{n\to \infty}\sum_{j=0}^{n} \frac{1}{n} f( \frac{j}{n})[/imath] Then I= [imath]\int_0^1 f(x)dx[/imath] (by integration as the limit of a sum) The required integral is therefore ans (B) exists and is equal to [imath]\frac{1}{2} \int_0^1 f(x)dx[/imath] This is area under the function [imath]f(x)[/imath] upto [imath]x=\frac{1}{2}[/imath] which is also = [imath]\int_0^{\frac{1}{2}} f(x)dx[/imath] I am confused between the answers (B) and (D) Please help.Thank you. | 1817164 | Value of [imath]\lim_{n\to\infty}\sum_{j=0}^{[\frac{n}{2}]}\frac{1}{n}f(\frac{j}{n})[/imath]
If [imath]f[/imath] is continuous in [0,1] then what will be the value of [imath]\lim_{n\to\infty}\sum_{j=0}^{\left[\frac{n}{2}\right]}\frac{1}{n}f\left(\frac{j}{n}\right)?[/imath] We know that,[imath]\lim_{n\to\infty}\sum_{j=0}^{{n}}\frac{1}{n}f\left(\frac{j}{n}\right)=\int_0^1f(x)dx[/imath]But what will be the value if we have [imath]\left[\frac{n}{2}\right][/imath] instead of [imath]n?[/imath] |
2610116 | Value of [imath] (a+d)(b+c)[/imath]
[imath] \begin{cases} \begin{align} &\text{1) }a + 7b + 3c + 5d = 16 \\ &\text{2) }8a + 4b + 6c + 2d = -16 \\ &\text{3) }2a + 6b + 4c + 8d = 16 \\ &\text{4) }5a + 3b + 7c + d = -16 \\ \end{align} \end{cases} [/imath] Then [imath](a+d)(b+c)[/imath] equals (A) -4 (B) 0 (C) 16 (D) -16 Please note that this is a 1 mark MCQ sum.Solving for the linear equations will give you the answer but it will be a lengthy process.There must be some quick succinct step to reach the answer. Please help me out. | 725665 | Why [imath](a+d)(b+c)[/imath] equals [imath]-16[/imath]?
If [imath]a,b,c[/imath] and [imath]d[/imath] satisfy the equations:[imath]a+7b+3c+5d=0[/imath][imath]8a+4b+6c+2d=-16[/imath][imath]2a+6b+4c+8d=16[/imath][imath]5a+3b+7c+d=-16[/imath] then [imath](a+d)(b+c)[/imath] equals [imath]-16[/imath]. I can't understand why [imath](a+d)(b+c)[/imath] equals [imath]-16[/imath]? |
2610442 | Prove that [imath]5 \mid 3^{3n+1}+2^{n+1}[/imath]
I am working right now through a book: Number Theory: A Lively Introduction with Proofs, Applications, and Stories And I cant connect 5 to [imath]3^{3n+1}+2^{n+1}[/imath] | 1526409 | Prove that [imath]5[/imath] divides [imath]3^{3n+1}+2^{n+1}[/imath]
Prove that [imath]5[/imath] divides [imath]3^{3n+1}+2^{n+1}[/imath] I tried to prove the result by induction but I couldn't. The result is true for [imath]n=1[/imath]. Suppose that the result is true for [imath]n[/imath] i.e [imath]3^{3n+1}+2^{n+1}=5k[/imath] for some [imath]k\in \mathbb{N}[/imath]. We study the term [imath]3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2[/imath] I tried to prove that that the difference is a multiple of [imath]5[/imath]. [imath]3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)[/imath] Therefore it's enough to prove that [imath]3^{3n+1}\cdot 13+2^n[/imath] is a multiple of [imath]5[/imath]. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction. |
2610283 | Prove with the mean value theorem that [imath]x-\frac{x^2}{2} < \ln(1+x)[/imath]
Prove with the mean value theorem that [imath]x-\frac{x^2}{2}<\ln(1+x)<x[/imath] in [imath](0,\infty)[/imath] Approach [imath]f(x) := \ln(1+x) [/imath] with the mean value theorem in [imath][0,x][/imath] [imath]\frac{1}{1+\xi}= \frac{\ln(1+x)-0}{x-0}[/imath] [imath]\frac{1}{1+\xi}[/imath] takes the biggest value when [imath]\xi[/imath] is [imath]0[/imath] and so [imath]1 <\frac{\ln(1+x)}{x}[/imath] multiply with x and you get [imath]x<\ln(1+x)[/imath] I can't prove the second part. | 1110508 | How can I prove that [imath]x-{x^2\over2}<\ln(1+x)[/imath]
How can I prove that [imath]\displaystyle x-\frac {x^2} 2 < \ln(1+x)[/imath] for any [imath]x>0[/imath] I think it's somehow related to Taylor expansion of natural logarithm, when: [imath]\displaystyle \ln(1+x)=\color{red}{x-\frac {x^2}2}+\frac {x^3}3 -\cdots[/imath] Can you please show me how? Thanks. |
2610558 | Prove that the given series of functions is continuously differentiable.
[imath]f(x)= \Sigma_{n=1}^{\infty} \frac {\sin nx^2}{1+n^3}[/imath], [imath]x\in {\mathbb{R}}[/imath] I can show continuity by using uniform convergence and the Weierstrass M test. But I don't know how to do differentiability. Any help would be highly appreciated. Thank you. | 1203743 | [imath]\sum_{n=1}^{\infty}\frac{\sin (nx^2)}{1+n^3}[/imath] represents a differentiable function
Show that the following series of function defines a continuous differentiable function function in [imath]\mathbb R[/imath]. [imath]\sum_{n=1}^{\infty}\frac{\sin (nx^2)}{1+n^3}.[/imath] We have , [imath]|f_n(x)|=\left|\frac{\sin (nx^2)}{1+n^3}\right|\le \frac{1}{1+n^3}\le \frac{1}{n^3}=M_n \text{ (say) }[/imath] As, [imath]\sum M_n[/imath] is convergent so, the given series is uniformly convergent. Also , each [imath]f_n(x)[/imath] is a continuous function in [imath]\mathbb R[/imath]. So, the given series converges to a continuous function , say [imath]f(x)[/imath]. But how we can show that [imath]f(x)[/imath] is differentiable function ? Please help... |
2611055 | Derivative of [imath]\sin_dx[/imath] when x is measured in degree
Find the derivative of [imath]\sin_d x,[/imath] if [imath]\sin_d x[/imath] is the sine of [imath]x[/imath] where [imath]x[/imath] is taken to be in degrees. please help me out solving this math question | 214912 | Derivative of the sine function when the argument is measured in degrees
I'm trying to show that the derivative of [imath]\sin\theta[/imath] is equal to [imath]\pi/180 \cos\theta[/imath] if [imath]\theta[/imath] is measured in degrees. The main idea is that we need to convert [imath]\theta[/imath] to radians to be able to apply the identity [imath]d/dx \sin x = \cos x [/imath]. So we need to express [imath] \sin \theta[/imath] as [imath] \sin_{deg} \theta = \sin(\pi \theta /180), [/imath] where [imath]\sin_{deg}[/imath] is the [imath]\sin[/imath] function that takes degrees as input. Then applying the chain rule yields [imath] d/d\theta [ \sin(\pi\theta/180)] = \cos(\pi \theta/180) \pi/180 = \frac{\pi}{180}\cos_{deg}\theta. [/imath] Is this derivation formally correct? |
2610297 | Finding minimum value of [imath]\sqrt{x^2+y^2}[/imath]
Find the minimum value of [imath]\sqrt{x^2+y^2}[/imath], given [imath]15x+8y=120[/imath]. My attempt: From [imath]15x+8y=120[/imath], I get [imath]y=\frac{120-15x}{8}[/imath]. I substitute this value into [imath]\sqrt{x^2+y^2}[/imath], getting [imath]\sqrt{\frac{289x^2-3600x+14400}{64}}[/imath]. I am stuck here as [imath]289x^2-3600x+14400[/imath] cannot be square-rooted. | 427843 | If [imath] 5x+12y=60[/imath] , what is the minimum of [imath]\sqrt{x^2+y^2}[/imath]?
I know this can be easily done by solving for [imath]y[/imath] and substituting, so that you only have to find the minimum value of the parabola [imath]\large x^2 + \left ( \frac {60-5x}{12} \right)^2[/imath] using standard techniques, but is there a less messy way to do this using inequalities? I tried various things such as [imath]AM-GM[/imath], but I don't get anywhere. One thing which I did notice about the restriction is that it is of the form [imath]ax+by=ab[/imath], but I don't know how to make any use of that. Thanks. |
2609905 | Dense set on unit circle
I have a task. Prove that [imath](\cos(n\alpha), \sin(n\alpha))[/imath], with [imath]n\in\mathbb{N}[/imath] is dense on unit circle.([imath]\alpha[/imath] chosen such that our set is infinite ) We want to show, that for every element's neighbourhood there is an element of that that is in the neighborhood. We know that there all points are different (there is no cycle on the circle). But why can we use pigeonhole principle to prove that if we divide the circle in [imath]k[/imath] arcs then there exists two elements which a closer than [imath]\frac{2\pi}{k}[/imath]? EDIT: Thanks everyone. Now everything is clear. This question is [SOLVED] | 1722526 | Let [imath]q \in \mathbb C[/imath], [imath]\|q\|=1[/imath] and [imath]q^n \neq 1, \forall n \in \mathbb N[/imath]. Show that [imath]\{q^n: n \in \mathbb N\}[/imath] is dense in [imath]S^1[/imath]
Let [imath]q \in \mathbb C[/imath], [imath]|q|=1[/imath] and [imath]q^n \neq 1, \forall n \in \mathbb N[/imath]. Show that [imath]\{q^n: n \in \mathbb N\}[/imath] is dense in [imath]S^1[/imath]. My attempt: As [imath](q^n)[/imath] is limited, there is subsequence [imath](q^{n_j})[/imath] convergent. So given [imath]\epsilon > 0, \exists j_0 \in \mathbb N, \quad |q^{n_j} - q^{n_k}| < \epsilon, \quad \forall j,k \geq j_0[/imath]. Supose that [imath]n_j > n_k[/imath], then [imath]|q^{n_j} - q^{n_k}| = |q^{n_k}||q^{n_j - n_k} - 1| = |q^{n_j - n_k} - 1| < \epsilon, \quad \forall j, k \geq j_0[/imath]. So, there is a subsequence of [imath](q^n)[/imath] that converges to [imath]1[/imath]. Using this fact, how can I ensure that for all [imath]w \in S^1[/imath] there is a subsequence of [imath](q^n)[/imath] that converges to [imath]w[/imath]? |
2610560 | Find minimum [imath]x+y+z[/imath] given [imath]3x=4y=7z[/imath] and [imath]x, y, z \in \mathbb N^+[/imath]
Given that [imath]x, y, z[/imath] are positive integers and that [imath]3x=4y=7z[/imath] find the minimum value of [imath]x+y+z[/imath]. The options are: A) 33 B) 40 C) 49 D) 61 E) 84 My attempt: [imath]y=\frac{3}{4}x, z=\frac{3}{7}x[/imath]. Substituting these values into [imath]x+y+z[/imath], I get [imath]\frac{117}{28}x[/imath]. I have no idea how to continue. [imath]x[/imath] in this case would have to be 28, meaning that the sum is [imath]117[/imath], which is not one of the options | 532139 | Least Value Of [imath]x+y+z[/imath]
If [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are positive integer and [imath]3x=4y=7z[/imath], then What is the least possible value of [imath]x+y+z[/imath]? |
2611158 | Is the set of cardinalities totally ordered?
If there exists no injection [imath]X \rightarrow Y[/imath] does that mean that there exists a surjection [imath]Y \rightarrow X[/imath] ? If there exists no surjection [imath]X \rightarrow Y[/imath] does that mean that there exists an injection [imath]Y \rightarrow X[/imath] ? In other words, are two sets always comparable by size ? | 151904 | Does the assertion that every two cardinalities are comparable imply the axiom of choice?
If, for any two sets [imath]A[/imath] and [imath]B[/imath], Either [imath]|A|<|B|, |B|<|A|[/imath] or [imath]|A|=|B|[/imath] holds, does the axiom of choice holds? Why? |
2611556 | Show that if f(n) is O(g(n) and d(n) is O(h(n)), then f(n) + d(n) is O(g(n) + h(n))
I have just started an algorithm design and analysis class and I am completely lost. We seem to be moving thru material that I am unfamiliar with and my book is terrible. My instructor tried to help but I still do not understand. The question: Show that if f(n) is O(g(n)) and d(n) is O(h(n)), then the summation f(n) + d(n) is O(g(n) + h(n)) If someone could help me understand this proof and show me the methodology for proving something like this I would be so grateful. This isn't homework, but is similar to stuff we are doing in class and it's going right over my head. Give me the child's explanation if you could. I have used this as a reference: Suppose [imath]f(n)[/imath] is [imath]O(g(n))[/imath] and [imath]d(n)[/imath] is [imath]O(e(n))[/imath]. How do i use Big O to show that [imath]f (n)[/imath] + [imath]d(n)[/imath] is [imath]O(g(n) + e(n))[/imath]? but im still very confused. Thanks so much in advance. | 2118277 | Suppose [imath]f(n)[/imath] is [imath]O(g(n))[/imath] and [imath]d(n)[/imath] is [imath]O(e(n))[/imath]. How do i use Big O to show that [imath]f (n)[/imath] + [imath]d(n)[/imath] is [imath]O(g(n) + e(n))[/imath]?
Suppose [imath]f(n)[/imath] is [imath]O(g(n))[/imath] and [imath]d(n)[/imath] is [imath]O(e(n))[/imath]. How do I use Big O to show that [imath]f(n) + d(n)[/imath] is [imath]O(g(n) + e(n))[/imath]? |
1957812 | Radius of convergence, power series
So I'm stuck with the question: Find the radius of convergence for the power series: [imath]f(z) = \sum^{ \infty}_{j=0} 2^{j} z^{j^2}[/imath] My issue is that there are two solutions depending if the [imath]a_{n}[/imath] is a perfect square or not, how should I approach this problem? Thanks in advance! | 603337 | Radius of Convergence - [imath]\sum_{n=1}^{\infty}2^n x^{n^2}[/imath]
What is the radius of convergence of this power series here? [imath]\sum_{n=1}^{\infty}2^n x^{n^2}[/imath] |
2612252 | A confusing sequence of products
Define [imath]x_n=\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\left(1-\frac{1}{10}\right)^2\left(1-\frac{1}{15}\right)^2...\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2[/imath] for [imath]n>1[/imath] then [imath]\lim_{n\to\infty} x_n [/imath] is A)..[imath]\frac{1}{3}[/imath] B)..[imath]\frac{1}{9}[/imath] C)..[imath]\frac{1}{81}[/imath] D)..[imath]0[/imath] What I did is... [imath]X_n[/imath]=[imath](\prod_{i=2}^{i} \frac{i^2+i-2}{i(i+1)})^2[/imath]=[imath](\prod_{i=2}^{i} \frac{(i+2)(i-1)}{i(i+1)})^2[/imath]=[imath](\frac{4.5.6.....(i+2)()1.2.3.4....(i-1)}{(1.2.3.4...i)(2.3.4.5...(i+1))})^2=(\frac{1}{3}(1+\frac{2}{i}))^2[/imath] as [imath]i\to\infty[/imath] we get [imath]\lim_{x\to\infty} x_n=\frac{1}{9}[/imath] Am I correct or wrong? Any help would be appreciated. | 639302 | Limit [imath]\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2[/imath]
How can I find the following limit :[imath]\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\left(1-\frac{1}{10}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2[/imath]There are four options [imath](a)\frac1{3}(b)\frac1{9}(c)\frac1{81}(d)0[/imath] Thanks! |
2612489 | Proof that [imath](-1)\cdot a=(-a)[/imath]
I'm supposed to prove this. The answer key doesn't tell me anything more than to add zero: [imath](a+(-a))[/imath] to: [imath](-1)\cdot a[/imath] and use that: [imath]a=a\cdot 1[/imath] I have these axioms to my disposal: (A1) for all [imath]x,y[/imath] we have [imath]+y=y+x[/imath] (A2) for all [imath]x,y,z[/imath] we have [imath](x+y)+z=x+(x+z)[/imath] (A3) there exists a number [imath]0[/imath] so that for all [imath]x[/imath] we have [imath]x+0=x[/imath] (A4) for all [imath]x[/imath] there exists a number [imath]a[/imath] so that [imath]x+a=0[/imath] (M1) for all [imath]x,y[/imath]we have [imath]x*y=y*x[/imath] (M2) for all [imath]x,y,z[/imath] we have [imath](x*y)*z=x*(y*z)[/imath] (M3) there exists a number [imath]1=/=0[/imath] so that for all [imath]x[/imath] we have [imath]x(1/x)=1[/imath] (M4) for all [imath]x=/=0[/imath] there exists a number [imath]b[/imath] so that [imath]x(1/x)=1[/imath] (AM) for all [imath]x,y,z[/imath] we have [imath]z*(x+y)=z*x+z*y)[/imath] [imath]a ∈ R[/imath] | 1236015 | Show that [imath]a(-1) = (-1)a = -a [/imath].
In a ring [imath]R[/imath] with identity 1, show that [imath]a(-1) = (-1)a = -a \qquad\forall\, a \in R[/imath] I have started with [imath]a + (-a) = 0[/imath] but cant proceed from here. |
2612058 | What is the Number of Non Isomorphic Spanning trees of [imath]K_n[/imath]? (Complete graph of size n)
So i know that for [imath]K_6[/imath] the answer is 6, but the book didn't explain how it solved this and i can't either! so what is the Number of Non Isomorphic Spanning trees of [imath]K_n[/imath] ? The only post related to this i found was this but there was no clear answer on there. | 356300 | All nonisomorphic trees of order [imath]n[/imath]
I have two questions regarding spanning trees: Q[imath]1[/imath]. Is there any formula for the number of distinct trees of order [imath]n[/imath]? I don't mean labelled trees, just distinct trees. For example: for [imath]n=3[/imath] there's only [imath]1[/imath] distinct tree, for [imath]n=4[/imath] there are [imath]2[/imath] distinct trees, for [imath]n=6[/imath] there are [imath]6[/imath]. Q[imath]2[/imath]. Is there any algorithm to implement the desired output in Q[imath]1[/imath]? |
2612695 | Finding the inverse of a matrix
Given the matrix [imath]A∈M_3(Z_9)[/imath] [imath] \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 0 \\ 0 & 3 & 6 \\ \end{matrix} [/imath] The determinant is: [imath]\det(A)=18+18-24=3[/imath] So the matrix is invertible because the determinant is different from zero. The transposed matrix is: [imath] \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 3 \\ 3 & 0 & 6 \\ \end{matrix} [/imath] The matrix [imath]A'[/imath] is: [imath] \begin{matrix} 0 & 6 & 0 \\ 6 & 6 & 6 \\ 6 & 6 & 8 \\ \end{matrix} [/imath] So the inverse matrix is: [imath] \begin{matrix} 0 & 2 & 0 \\ 2 & 2 & 2 \\ 2 & 2 & 8/3 \\ \end{matrix} [/imath] Please tell me if my attempt is correct. | 2611225 | Invertible matrix and inverse matrix
Establish if the matrix [imath]A\in M_3 (\Bbb{Z}_9)[/imath] [imath] \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 0 \\ 0 & 3 & 6 \end{bmatrix} [/imath] is invertible, and if so, find the inverse matrix. |
2612605 | Algebraic Tensor product of Hilbert spaces
Let [imath]E[/imath], [imath]F[/imath] be complex Hilbert spaces. The algebraic tensor product of [imath]E[/imath] and [imath]F[/imath] is given by [imath]E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N},\;\;v_i\in E,\;\;w_i\in F \right\}.[/imath] In [imath]E \otimes F[/imath], we define [imath] \langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,t_j\rangle_2, [/imath] for [imath]\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F[/imath] and [imath]\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F[/imath]. The above sesquilinear form is an inner product in [imath]E \otimes F[/imath]. Why [imath](E \otimes F,\langle\cdot,\cdot\rangle)[/imath] is not a complete space? Thank you!! | 1046329 | Tensor Product: Hilbert Spaces
This question has been modified... Problem Given Hilbert spaces. In general, their algebraic tensor product isn't complete: [imath]\mathcal{H}\hat{\otimes}\mathcal{K}=\mathcal{H}\otimes\mathcal{K}\iff\dim\mathcal{H}<\infty\lor\dim\mathcal{K}<\infty[/imath] How to prove this from scratch? Attempt Choose orthonormal bases: [imath]\mathcal{S}\otimes\mathcal{T}:=\{\sigma\otimes\tau:\sigma\in\mathcal{S},\tau\in\mathcal{T}\}[/imath] One obtains some candidates: [imath]\sigma_k\otimes\tau_l\in\mathcal{S}\otimes\mathcal{T}:\quad\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l\quad\sum_{k=1}^\infty\frac{1}{kl}\delta_{kl}\sigma_k\otimes\tau_l[/imath] However the former one drops out: [imath]\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l=\left(\sum_{k=1}^\infty\frac{1}{k}\sigma_k\right)\otimes\left(\sum_{l=1}^\infty\frac{1}{l}\tau_l\right)[/imath] So it is not obvious at all wether the latter one works out! Reference Build-up on: Vector Spaces: Tensor Product |
2612860 | Derivative of [imath]n![/imath] (factorial)?
I have a theory that uses the gamma function: [imath]\Gamma(n)=\int_0^\infty x^{n-1}e^{-x} \space dx[/imath] Then I was inclined to think that perhaps the derivative is: [imath]x^{n-1}e^{-x}[/imath] But I'm not sure we can just drop the integral along with the bounds to get the derivative. Then I thought about taking the limit: [imath]\lim_{x\to\infty}x^{n-1}e^{-x}[/imath] But now we can't specify at what [imath]x[/imath] value we want to get the rate of change of. At this point I feel like I can't get any further on my own and would appreciate some insight. EDIT: Looking for derivative in terms of [imath]n[/imath] actually. | 300526 | Derivative of a factorial
What is [imath]{\partial\over \partial x_i}(x_i !)[/imath] where [imath]x_i[/imath] is a discrete variable? Do you consider [imath](x_i!)=(x_i)(x_i-1)...1[/imath] and do product rule on each term, or something else? Thanks. |
2613177 | Calculate integral [imath]\int_{0}^{2\pi} \frac{dx}{a^2\sin^2x+b^2\cos^2x}[/imath]
Any hints of how to approach solving this integral? [imath]\int_{0}^{2\pi} \frac{dx}{a^2\sin^2x+b^2\cos^2x}, \quad a,b>0[/imath] I tried substituting [imath]z=e^{ix}[/imath] but I get a rather complex form I can't do anything with. | 1752721 | Any hint to solve given integral [imath]\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}[/imath]?
Show that for [imath]ab>0[/imath] [imath]\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}[/imath] I'm not sure how to go about this. Any solutions or hints are greatly appreciated. |
2612655 | If [imath]\sum\limits_{i=1}^na_i=\prod\limits_{i=1}^na_i[/imath] for every [imath]n[/imath], identify [imath]\lim\limits_{n\to \infty}a_n[/imath]
Let [imath]\left(a_n\right)_{n \in\mathbb{N}} [/imath] denote a sequence of real numbers such that, for every [imath]n\geqslant1[/imath], [imath]\sum_{i=1}^na_i=\prod_{i=1}^na_i[/imath] Identify the limit [imath]\lim_{n\to \infty}a_n[/imath] What I have done: [imath]a_1-a_1=0 \\a_1+a_2-(a_1a_2)=0 \to a_1\cdot a_2(\dfrac{a_1}{a_2}+\dfrac{a_2}{a_1}-1)=0\\.\\.\\.\\a_1\cdot a_2\cdot \cdot \cdot a_n(\dfrac{a_1}{a_2\cdot \cdot \cdot a_n}+\dfrac{a_2}{a_1\cdot \cdot \cdot a_n}+...+\dfrac{a_n}{a_2\cdot \cdot \cdot a_{n-1}}-1)=0[/imath] Now what do I do ? | 2615849 | find the [imath]\lim_{n\to \infty}a_n=?[/imath]
Let [imath]a_1=a\neq1[/imath] and [imath]a_{n+1}=\dfrac{1}{S_n-1}+1[/imath] then find the [imath]\lim_{n\to \infty}a_n=?[/imath] such that : [imath]S_n=\sum_{i=1}^n a_i[/imath] [imath]a_1=a \\a_2=\dfrac{1}{a-1}+1 \\a_3=2a+\dfrac{1}{a-1}+1 \\ a_4=3a+2+\dfrac{2}{a-1}+\dfrac{a-1}{2a^2-2a+1} [/imath] now what do I do ? |
2613697 | [imath]Z[i]/(2+3i) \simeq Z/13Z[/imath] is my proof correct?
Here's my new attempt at showing [imath]Z[i]/(2+3i) \simeq Z/13Z[/imath]. proof: Let's define the natural homomorphism [imath]\phi: Z \to Z[i]/(2+3i)[/imath] where [imath]\phi(z)= z+ (2+3i)[/imath] It is easy to check that this is a ring homomorphism. Claim 1: [imath]Ker (\phi)= 13 Z[/imath] proof: We will show both inclusions. Let [imath]z\in 13Z[/imath] then [imath]z=13k[/imath] for some integer k. But [imath]13k= (2+3i)(2k-3ik)[/imath] and so [imath]z \in (2+3i)[/imath] and so [imath]z+(2+3i)=0+(2+3i)[/imath] as cosets and hence [imath]\phi(z)= 0 +(2+3i)[/imath] i.e. [imath]z \in ker(\phi)[/imath] and hence [imath]13Z \subset ker(\phi)[/imath] Now let's show the reverse inclusion. Let [imath]z\in ker(\phi)[/imath], then [imath]z \in (2+3i) \cap Z[/imath]. Then [imath]z= (2+3i)(a+bi)[/imath] for some integers a,b such that [imath]z\in Z[/imath], and so multiplying out we get [imath]z= (2a-3b) + (2b+3a)i[/imath] for some [imath]a,b[/imath] such that [imath]z[/imath] is an integer. Now for z to be an integer, the complex part must be [imath]0[/imath], hence [imath]2b=-3a[/imath]. Now 3 is a prime dividing [imath]2b[/imath] which implies 3 divides b. Similarly we have that 2 divides a. Hence [imath]b=3k[/imath] and [imath]a=2m[/imath] for some integers k and m. Substituting into [imath]2b=-3a[/imath] we obtain [imath]6k=-6m[/imath] and hence [imath]k= -m[/imath] and hence [imath]b=-3m[/imath] and [imath]a= 2m[/imath] Finally substituting into [imath]z=(2+3i)(a+bi)[/imath] we get [imath]z=(2+3i)(2m-3mi)= 13m[/imath] and hence [imath]z\in 13Z[/imath] and we are done. Claim 2: [imath]\phi[/imath] is surjective I have also done this part but I would like to make sure my first claim is correct first. Hence by the first isomorphism theorem, we have that [imath]Z/13Z \simeq Z[i]/(2+3i)[/imath] Thanks. | 2611693 | [imath]\mathbb{Z}[i] / (2+3i)[/imath] has 13 elements
I am self studying algebra, and have come across this exercise: Show that [imath]\mathbb{Z}[i]/(2+3i)[/imath] has 13 elements and is a field. My proof: Since Z[i] is a euclidian domain with euclidian norm function [imath]f(c+id)= c^2+d^2[/imath], for any element [imath]a+bi \in Z[i][/imath], we can "divide" it by [imath]2+3i[/imath] to get [imath]a+bi= q (2+3i) + r[/imath] where either [imath]r=0[/imath] or [imath]r=c+di[/imath] for some integers [imath]c,d[/imath] not both [imath]0[/imath] with [imath]c^2 + d^2=f(c+di) < f(2+3i)= 13[/imath]. But then [imath]a+bi[/imath] belongs to the coset [imath] r + (2+3i) [/imath]. So every element in the quotient ring belongs to some [imath]r + (2+3i)[/imath] for [imath]r=0[/imath] or [imath]r=c+di[/imath] with [imath]c^2 + d^2<13[/imath]. Now we can count all the possible pairs of integers satisfying this condition, ie [imath](0,0),(0,1),(0,2),(0,3), ..., (3,1)[/imath] to get 13 and so there are 13 elements in the quotient ring. Is this a correct argument? Also is there an elegant way of showing it is a field rather than showing its an integral domain by showing each element has an inverse by brute force? Thanks in advance! |
2613744 | How can I prove that a number of the form [imath]2^{2n}-1[/imath] is divisible by 3?
How can I prove that for each positive integer [imath]n[/imath] the number [imath]2^{2n}-1[/imath] is divisible by [imath]3[/imath]? [imath]3[/imath]? | 648273 | Prove by induction that [imath]2^{2n} – 1[/imath] is divisible by [imath]3[/imath] whenever n is a positive integer.
I am confused as to how to solve this question. For the Base case [imath]n=1[/imath], [imath](2^{2(1)} - 1)\,/\, 3 = 1[/imath], base case holds My induction hypothesis is: Assume [imath]2^{2k} -1[/imath] is divisible by [imath]3[/imath] when [imath]k[/imath] is a positive integer So, [imath]2^{2k} -1 = 3m[/imath] [imath]2^{2k} = 3m+1[/imath] after this, I'm not quite sure where to go. Can anyone provide any hints? |
2604252 | The general term of the sequence : 1, 1, -1, -1, 1, 1, ...?
Suppose we have the following sequence: {[imath]a_n[/imath]} such as: [imath]a_0 = 1, \\ a_1 = 1, \\ a_2 = -1, \\ a_3 = -1, \\ a_4 = 1, \\ a_5 = 1, \\ ... [/imath] How can we find the general term of this sequence? I tried using a trigonometric function e.g. [imath]\alpha \sin(x+\phi)[/imath], then we impose some constraints on [imath]\alpha[/imath] and [imath]\phi[/imath] to get that sequence, but I get lost, is there any clever way to find the general term? EDIT: The question is identified as duplicate, but that answer does not solve the question, because I am looking for a solution that does not involve floor function. | 2587188 | Generalizing a sequence.
Consider a sequence of the form: [imath]x[n]=-1,-1...,1,1,...,-1,-1...,1,1... \quad n>0[/imath] One can think of this as a square wave ([imath]\pm1[/imath]) with a 50 % duty cycle (coming from EE). For the simplest case of such a series i.e. [imath]-1,+1,-1,+1,-1...[/imath], a general equation is quite trivial ([imath]x[n]=-1^n[/imath]). Can we have e general equation for the other cases i.e. when there are multiple [imath]-1[/imath]'s followed by [imath]+1[/imath]'s ? |
2612180 | Find the dimension of a subspace and the orthogonality complement of another
I would like to the fastest way to find [imath]\dim(V \cap W^\bot)[/imath] where [imath]V[/imath] and [imath]W[/imath] are two subspaces of [imath]\Re^4[/imath]. [imath]V[/imath] is defined as the span of three vectors and [imath]W[/imath] the span of one vector.For example as in here - [imath] V = \left\langle\begin{pmatrix}2\\2\\2\\1\end{pmatrix}, \begin{pmatrix}5\\4\\1\\1\end{pmatrix}, \begin{pmatrix}2\\1\\1\\0\end{pmatrix}\right\rangle,\quad W = \left\langle\begin{pmatrix}1\\-3\\2\\1\end{pmatrix}\right\rangle [/imath] | 2609215 | The dimension of the intersection of two vector subspaces
We're given: [imath] V = Span\left\{ \begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}, \ \begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \ \begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix} \right\} \ \ \ \mathrm{and} \ \ \ W = Span \left\{\begin{bmatrix}1\\-3\\2\\1 \end{bmatrix} \right\} [/imath] and we're asked to find [imath]\mathrm{dim}(V\cap W^{\bot})[/imath] Here's my approach. First, by inspecting the basis of [imath]W[/imath], I managed to construct a basis for [imath]W^{\bot}[/imath], which is the following: [imath]W^{\bot} = Span \left\{\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\right\}[/imath] Then, for each vector [imath]\textbf{w}[/imath] in the basis of [imath]W^{\bot}[/imath], I tried to see if the system [imath]A\textbf{x}=\textbf{w}[/imath] was compatible or not. In this case, I found out that the system was only compatible with two of the vectors in [imath]W^{\bot}[/imath], thus indicating me that [imath]\mathrm{dim}(V \cap W^{\bot} ) = 2[/imath] (which is correct). What I do not get, however, is that since the dimension of the intersection is [imath]2[/imath], why aren't two of the vectors in the basis of [imath]V[/imath] orthogonal to the vector which spans [imath]W[/imath] (that was my initial approach, i.e try to see which vectors of the basis of [imath]V[/imath] are orthogonal to the vector that spans [imath]W[/imath]). Also, I was wondering if there was a simpler/quicker way of doing this. |
2614291 | Integral part of [imath]\sqrt{2018+\sqrt{2018+\sqrt{...+2018}}}[/imath]
I am tasked to find the integral part of [imath]\sqrt{2018+\sqrt{2018+\sqrt{...+2018}}}[/imath], where the number of [imath]2018[/imath] is [imath]2018[/imath]. My attempt: I came up with an upper bound, which was [imath]\sqrt{2018+\sqrt{2018+\sqrt{2018+...}}}[/imath], where there were infinite number of [imath]2018[/imath]. I let this be [imath]a[/imath], and then I have [imath]a^2=2018+a[/imath], giving me [imath]45.4249374...[/imath]. My problem here is that I have no idea what to set as a lower bound. Can anyone give me a hint to solve this? Alternatively, another approach could also be provided if it is more efficient. | 2600313 | Calculating [imath]\Big\lfloor\underbrace{\sqrt{2017+\sqrt{2017+...+\sqrt{2017}}}}_\text{2017 roots}\Big\rfloor[/imath]
Could you please help me computing [imath]\Big\lfloor\underbrace{\sqrt{2017+\sqrt{2017+...+\sqrt{2017}}}}_\text{2017 roots}\Big\rfloor[/imath]? |
2614458 | Proving that a set of 2016 natural numbers contain a non-empty set with a sum divisible by 2016
Prove: Each set which consists of 2016 natural numbers contains a non-empty set that has a sum divisible by 2016 My attempt: Use the pigeonhole principle. Let [imath]A[/imath] be a set of 2016 natural numbers. I'll split the answer to two different possibilities: [imath]A[/imath] contains at least one number which is divisible by 2016. In this case the claim is true. [imath]A[/imath] does not contain numbers that are divisible by 2016. In this case, we'll apply the pigeonhole principle. Holes: [imath]\{1,2,3,4,...,2016\}[/imath], A set of all possible remainders when dividing a number (which is not divisible by 2016) by 2016. In total there are 2015 holes. Pigeons: 2016 numbers. Each number will be divided by 2016 and put into the appropriate hole according to the that number's division remainder. By the pigeonhole principle, at least in one hole will be [imath]\lceil \frac{2016}{2015} \rceil[/imath]. objects. In this case, the claim is also true. Is my proof sufficient? Can I prove this not using 2 different possibilities, or maybe not using the pigeonhole principle? Edit: As stated in the comments and in the answers, my proof is false. The proof shows that there exists at least 2 numbers in the set [imath]A[/imath] with a difference divisible by 2016. It does not prove that there is a non-empty set with a sum divisible by 2016. | 2556550 | Consider a set [imath]A[/imath] of [imath]10[/imath] integers, [imath]A = \{a_1,a_2, \ldots ,a_{10}\}[/imath]. Prove that there is at least one subset of [imath]A[/imath] whose sum is divisible by [imath]10[/imath].
Consider a set A of 10 integers, [imath]A= \{a_1,a_2, \ldots ,a_{10}\}[/imath]. Prove that there is at least one subset of A whose sum is divisible by 10. Use the pigeon-hole principle. Hint: Consider the 10 sums [imath]s_1 = a_1, s_2 = a_1 + a_2, \ldots ,s_{10} = a_1 + a_2 + \ldots+a_{10}[/imath]. If one of these ten sums is divisible by [imath]10[/imath], we are done. So assume that none of these ten sums is divisible by [imath]10[/imath]. I'm pretty sure this is the proof but im not exactly sure what's [imath]b_l[/imath] and [imath]b_k[/imath] stand for? Let [imath]A=\{a_1,a_2,...,a_{10}\}[/imath] Define set of numbers [imath]b_n=\sum\limits_{i=1}^n a_n[/imath] i.e. the set [imath]\{a_1, (a_1+a_2),(a_1+a_2+a_3),\ldots ,(a_1+a_2+\ldots +a_{10})\}[/imath] If any element of the set is divisible by [imath]10[/imath], then we are done. Hence proving case [imath]1[/imath]. Case 2: Assume that no element of this set is disible by [imath]10[/imath]. Then, by Pigeon hole priciple, there exists some [imath]k<l[/imath], such that, [imath]b_l\equiv b_k[/imath] then [imath]b_l\equiv b_k (mod 10)[/imath] with [imath]l>k>0[/imath] Hence, [imath]\{b_k,b_{k+1}, \ldots,b_l\}[/imath] is a subset of [imath]A[/imath] whose sum is divisble by [imath]10[/imath]. Thus proving Case [imath]2[/imath]. |
2611964 | then the value of [imath] \frac{1-\vert a \vert^2}{\pi} \int_{\gamma} \frac{\vert dz \vert}{\vert z+a \vert^2} [/imath].
If [imath]a\in \mathbb{C}[/imath] with [imath]\vert a \vert <1[/imath] , then the value of [imath] \frac{1-\vert a \vert^2}{\pi} \int_{\gamma} \frac{\vert dz \vert}{\vert z+a \vert^2} [/imath] where [imath]\gamma[/imath] is the simple closed curve with [imath]\vert z \vert=1[/imath], taken with the positive orientation. I tried but I cant manipulate the denominator, I used [imath]z \bar z =\vert z \vert^2[/imath], Please help. | 2024356 | complex integration [imath]\frac{1-|a|^{2}}{\pi}\int_{L}\frac{|dz|}{{|z+a|}^{2}}[/imath]
How to solve complex integration [imath]\frac{1-|a|^{2}}{\pi}\int_{L}\frac{|dz|}{{|z+a|}^{2}}[/imath], where [imath]L[/imath] is the simple closed curve [imath]|z|=1[/imath] and [imath]a\in\mathbb{C}[/imath] with [imath]|a|<1.[/imath] [imath]z=e^{it}[/imath] is the parametrization of the circle [imath]|z|=1.[/imath]If we put values it becomes as [imath]\frac{1-|a|^{2}}{\pi}\int_{0}^{\pi}\frac{|dt|}{{|e^{it}+a|}^{2}}[/imath]. Now i don't know to proceed further. Please help. Thanks a lot. |
2615215 | Show that the element [imath]φ(a)\in G'[/imath] has also order d!
EXERCISE: Consider that [imath]Φ:G\rightarrow G'[/imath] homomorphism and [imath]a\in G[/imath] which has order d. Show that the element [imath]φ(a)\in G'[/imath] has also order d. I am sorry i could't appose any attempt here but i don't even know how to start.I have just started to examine homomorphism and i have to clear my mind on them! It would be very helpful if someone can give me some hints or a thourough solution to my problem. I want to learn how i can work with these problems as i don't have many experience with these type of problems! Thanks in advance! | 981297 | Order of an Element Under a Homomorphism
I was reading an example about describing homomorphisms and I'm having a bit of trouble with this one. Show there is no group homomorphism [imath]f : \mathbb{Z}_{10} \to \mathbb{Z}_{25}[/imath] such that [imath]f(1) = 3[/imath]. The first step of the proof is [imath]|1| = 10[/imath] in [imath]\mathbb{Z}_{10}[/imath], so [imath]|f(1)| \in\{ 1, 2, 5, 10\}[/imath]. Does this mean that [imath]|f(1)|[/imath] can be [imath]1[/imath], [imath]2[/imath], [imath]5[/imath] or [imath]10[/imath]? I see that [imath]|1| = 10[/imath] since [imath]1[/imath] is a generator of [imath]\mathbb{Z}_{10}[/imath], and that [imath]|f(1)|[/imath] seems to be a factor of [imath]10[/imath]. If [imath]f(1)[/imath] is the image of [imath]1[/imath] under [imath]f[/imath], how would I find the order of [imath]f(1)[/imath]? Thanks for your help. |
2615083 | Finding a homormorphism form [imath]\mathbb{Z}/4\mathbb{Z}[/imath] to [imath]\mathbb{Z}/6\mathbb{Z}[/imath]
I've got the following question: find a homormorphism form [imath]\mathbb{Z}/4\mathbb{Z}[/imath] to [imath]\mathbb{Z}/6\mathbb{Z}[/imath]. My work so far: The group [imath]\mathbb{Z}/N\mathbb{Z}[/imath] has a generator of 1, so: If we have a homomorphism from [imath]\mathbb{Z}/4\mathbb{Z}[/imath] to [imath]\mathbb{Z}/6\mathbb{Z}[/imath] which sends [imath]1 \in \mathbb{Z}/4\mathbb{Z}[/imath] to an [imath]m\in \mathbb{Z}/6\mathbb{Z}[/imath] with [imath]m \cdot 4=\bar{0}[/imath] , does this mean that if [imath]4m=\bar{0}[/imath] in [imath]\mathbb{Z}/6\mathbb{Z}[/imath] implies [imath]2m=\bar{0}[/imath] and thus [imath]m=0 \mod 2[/imath]? How should i finish this? Does this mean that there are three homomorphisms, namely you can send 1 t0 0, 2 and 4? I have read another question regarding the homomorphisms between [imath]\mathbb{Z}/n\mathbb{Z}[/imath] and [imath]\mathbb{Z}/m\mathbb{Z}[/imath], but when I tried to apply it, I got stuck | 982341 | Finding all homomorphisms between [imath]\mathbb{Z}/m\mathbb{Z}[/imath] and [imath]\mathbb{Z}/n\mathbb{Z}[/imath]
I want to find all group homomorphisms [imath]\varphi: \mathbb{Z}/ m \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}[/imath], with [imath]m[/imath] and [imath]n[/imath] natural numbers. Clearly [imath]\varphi(0)=0[/imath] since the identity in [imath]\mathbb{Z}/m\mathbb{Z}[/imath] must map to the identity in [imath]\mathbb{Z}/n\mathbb{Z}[/imath], and it follows that [imath]m \varphi(1)=0[/imath], so is [imath]mx \equiv 0 \bmod n[/imath]? This as far as I get. How can I find all the homomorphisms? |
2615860 | Which is/are true?
If [imath]A,B\in M_n(\mathbb{R})[/imath] such that [imath]AB=BA=O[/imath] and [imath]A+B[/imath] is invertible ,then select the correct statements [imath]\ldots[/imath] [imath](a)[/imath] rank(A)=rank(B); [imath](b)[/imath] rank(A)+rank(B)=[imath]n[/imath]; [imath](c)[/imath] nullity(A)+nullity(B)=[imath]n[/imath]; [imath](d)[/imath] [imath]A-B[/imath] is invertible. I've proved [imath](d)[/imath] is true. As If [imath]A[/imath] is a [imath]n \times n[/imath] matrix ,then it goes a vector space with [imath]n[/imath] dimension to an another vector space and similarily for [imath]B[/imath] but when [imath]BA=0[/imath] it means that every base from image of [imath]A[/imath] will be at kernel of [imath]B[/imath]. So [imath]B[/imath] at least has all of basis of image [imath]A[/imath], therefore null [imath]B \geq[/imath] rank [imath]A[/imath]. For [imath](a),(b),(c)[/imath] I took some examples and found [imath](a)[/imath] is not correct. But I'm unable to prove [imath](b),(c)[/imath]. Please help me to prove these. | 822590 | Which one about the rank and nullity is true
Let [imath]A[/imath] and [imath]B[/imath] be [imath]n[/imath] [imath]\times[/imath] [imath]n[/imath] real matrices such that [imath]AB[/imath]=[imath]BA[/imath]=[imath]0[/imath] and [imath]A+B[/imath] is invertible. Which of the following are always true? [imath]1[/imath]. rank [imath](A)[/imath]=rank [imath](B)[/imath]. [imath]2[/imath]. rank [imath](A)[/imath][imath]+[/imath]rank [imath](B)[/imath] =[imath]n[/imath] [imath]3[/imath]. nullity [imath](A)[/imath] [imath]+[/imath] nullity [imath](B)[/imath] = [imath]n[/imath] [imath]4[/imath]. [imath]A-B[/imath] is invertible. I am not getting how to do. Atleast it is clear that Rank[imath](A+B)[/imath]= [imath]n[/imath]. help me! |
2615119 | Show that there is a unique polynomial of degree at most [imath]2k+1[/imath] such that [imath]p^{[j]}(x_1)=a_j \text{ and } p^{[j]}(x_2)=b_j \text{ for } j=0,\dots, k.[/imath]
Let [imath]x_1,x_2 \in \mathbb R[/imath] and let [imath](a_0, \dots, a_k), (b_0, \dots, b_k)[/imath] be [imath](k+1)[/imath]-tuples of real numbers. Show that there is a unique polynomial of degree at most [imath]2k+1[/imath] such that [imath]p^{[j]}(x_1)=a_j \text{ and } p^{[j]}(x_2)=b_j \text{ for } j=0,\dots, k.[/imath] [Consider the polynomials [imath](x-x_1)^k (x-x_2)^{k+1}[/imath], [imath](x-x_1)^{k+1}(x-x_2)^k[/imath] and argue by induction.] Any help with this exercise? | 2612791 | Show that there is a unique polynomial of degree at most [imath]2n+1[/imath] such that [imath]q^{[k]}(x_1)=a_k,[/imath] [imath]q^{[k]}(x_2)=b_k[/imath] for [imath]k=0, \dots, n[/imath].
Let [imath]x_1, x_2 \in \mathbb R[/imath] and let [imath](a_0, a_1, \dots, a_n), (b_0, b_1, \dots, b_n)[/imath] be [imath](n+1)[/imath]-tuples of real numbers. Show that there is a unique polynomial of degree at most [imath]2n+1[/imath] such that [imath]q^{[k]}(x_1)=a_k,[/imath][imath]q^{[k]}(x_2)=b_k[/imath] for [imath]k=0, \dots, n[/imath]. Any hints on how to get started with this exercise? |
1159858 | Why we use [imath]\mathrm{Fn}(\kappa\times\lambda,2,\lambda)[/imath] to force [imath]2^\lambda\ge\kappa[/imath] instead of [imath]\mathrm{Fn}(\kappa\times \lambda,2)[/imath]
I am reading Kunen's Set Theory and I learn that, [imath]\operatorname{Fn}(\kappa\times\omega,2)[/imath] forces [imath]2^{\aleph_0}\ge|\kappa|[/imath], where [imath]\operatorname{Fn}(I,J,\lambda)=\{p:p\text{ function},\,\operatorname{dom}p\subset I,\,\operatorname{ran}p\subset J,\,|p|<\lambda\}[/imath] and [imath]\operatorname{Fn}(I,J)=\operatorname{Fn}(I,J,\omega)[/imath]. However, when forcing [imath]2^\lambda\ge\kappa[/imath] for cardinal [imath]\kappa[/imath] and regular cardinal [imath]\lambda[/imath], Kunen uses [imath]\operatorname{Fn}(\kappa\times\lambda,2,\lambda)[/imath] rather than [imath]\operatorname{Fn}(\kappa\times\lambda,2)[/imath]. As I think, [imath]\operatorname{Fn}(\kappa\times\lambda,2)[/imath] preserves cardinals and seems to add [imath]|\kappa|^{V[G]}[/imath] many subsets of [imath]\lambda[/imath] so I think we can use [imath]\operatorname{Fn}(\kappa\times\lambda,2)[/imath] to force [imath]2^\kappa\ge\lambda[/imath]. However, every reference I see uses [imath]\operatorname{Fn}(\kappa\times\lambda,2,\lambda)[/imath], not [imath]\operatorname{Fn}(\kappa\times\lambda,2)[/imath]. Is there a reason to use [imath]\operatorname{Fn}(\kappa\times\lambda,2,\lambda)[/imath] instead of [imath]\operatorname{Fn}(\kappa\times\lambda,2)[/imath]? Did I misunderstand something? Thanks for any help. | 838215 | Adding subsets of regular cardinals (Jech p. 226)
On p. 226 of his [imath]{\it Set}[/imath] [imath]{\it Theory}[/imath], Jech considers adding [imath]\lambda[/imath] many subsets of [imath]\kappa[/imath] to a ground model [imath]M[/imath]. He outlines a suitable partial order on the assumption that [imath]M[/imath] satisfies [imath]\kappa<\lambda[/imath], [imath]2^{<\kappa} = \kappa[/imath], and [imath]\lambda^\kappa = \lambda[/imath]. Any resulting generic extension, he claims, will preserve cardinals and satisfy [imath]2^\kappa =\lambda[/imath]. [imath]\DeclareMathOperator{\dom}{dom}\DeclareMathOperator{\rng}{rng}[/imath] It looks like we can get a generic extension with the same properties without assuming [imath]2^{<\kappa} = \kappa[/imath] (though we do assume that [imath]\kappa<\lambda[/imath] and [imath]\lambda^\kappa = \lambda[/imath]). In particular, let [imath]P[/imath] be the set of all [imath]p[/imath] such that: (i) [imath]\dom(p)\subseteq \lambda\times\kappa[/imath] and [imath]|\dom(p)|<\omega[/imath] (ii) [imath]\rng(p)\subseteq\{0,1\}[/imath] and let [imath]p[/imath] be stronger than [imath]q[/imath] iff [imath]p\supseteq q[/imath]. Since [imath]P[/imath] is c.c.c., cardinals will be preserved. In addition, we seem to have: [imath](2^\kappa)^{M[G]} \leq (|B|^\kappa)^{M} \leq (|P|^\kappa)^{M}\leq (\lambda^\kappa)^{M} \leq \lambda \leq (2^\kappa)^{M[G]}[/imath] So, [imath]\lambda = (2^\kappa)^{M[G]}[/imath]. Am I missing something? |
2613198 | limit of: [imath]\begin{align} \lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n} \end{align}[/imath]
How to get the limit of: [imath]\begin{align} \lim_{n\to \infty}\sqrt{n-2\sqrt{n}}-\sqrt{n} \end{align}[/imath] I have formed it to: [imath]\begin{align} \lim_{n\to \infty}\frac{n(\sqrt{n-2\sqrt{n}}-\sqrt{n})}{n} \end{align}[/imath] And tried to use L'Hopital but unfortunately I am not able to get a result with that: [imath]f'(x)=\frac{d}{dn}(n(\sqrt{n-2\sqrt{n}}-\sqrt{n}))= (\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left( \frac{1-\frac{1}{\sqrt{n}}}{2\sqrt{n-2\sqrt{n}}} - \frac{1}{2\sqrt{n}}\right)[/imath] [imath]g'(x)=\frac{d}{dn} n = 1[/imath] [imath]\begin{align} \lim_{n \to \infty}\frac{f'(x)}{g'(x)} &= \lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left( \frac{1-\frac{1}{\sqrt{n}}}{2\sqrt{n-2\sqrt{n}}} - \frac{1}{2\sqrt{n}}\right)}{1}} \\ &= \lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})+n\left( 0\right)}{1}} \\ &= \lim_{n \to \infty}{\frac{(\sqrt{n-2\sqrt{n}}-\sqrt{n})}{1}} \\ &= \lim_{n \to \infty}{\frac{(\sqrt{\frac{1}{n}-2\sqrt{\frac{1}{n}}}-\sqrt{\frac{1}{n}})}{\frac{1}{n}}} \\ &= 0\\ \end{align}[/imath] Which is wrong. Wolfram says that the limit is [imath]-1[/imath] I appreciate every help :) | 2614626 | Convergence of the sequence [imath] \sqrt {n-2\sqrt n} - \sqrt n [/imath]
Here's my attempt at proving it: Given the sequence [imath] a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} [/imath] To get rid of the square root in the numerator: \begin{align} \frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{\sqrt {n-2\sqrt n} + \sqrt n} &= \frac { {n-2\sqrt n} - \ n}{\sqrt {n-2\sqrt n} + \sqrt n} = \frac { {-2\sqrt n}}{\sqrt {n-2\sqrt n} + \sqrt n} \\&= \frac { {-2}}{\frac {\sqrt {n-2\sqrt n}} {\sqrt n} + 1} \end{align} By using the limit laws it should converge against: [imath] \frac { \lim_{x \to \infty} -2 } { \lim_{x \to \infty} \frac {\sqrt {n-2\sqrt n}}{\sqrt n} ~~+~~\lim_{x \to \infty} 1} [/imath] So now we have to figure out what [imath]\frac {\sqrt {n-2\sqrt n}}{\sqrt n}[/imath] converges against: [imath] \frac {\sqrt {n-2\sqrt n}}{\sqrt n} \leftrightarrow \frac { {n-2\sqrt n}}{ n} = \frac {1-\frac{2\sqrt n}{n}}{1} [/imath] [imath]{\frac{2\sqrt n}{n}}[/imath] converges to [imath]0[/imath] since: [imath] 2\sqrt n = \sqrt n + \sqrt n \leq \sqrt n ~\cdot ~ \sqrt n = n [/imath] Therefore [imath]~\lim_{n\to \infty} a_n = -1[/imath] Is this correct and sufficient enough? |
2616081 | Find a difficult integral
How to find the following intergal: [imath]\mathcal{I}_\text{n}:=\int_0^\infty\frac{x^\text{n}}{1+x^2}\space\text{d}x\tag1[/imath] I've no idea where to start, I've tried integration by parts but it lead not to a closed form. | 2530926 | Finding a difficult integral
Good afternoon, I need to find: [imath]\int_0^\infty\frac{x^n}{ax^2+bx+c}\space dx[/imath] I've no idea about any clue what so ever. |
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