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1167780 | How do I evaluate the integral of the Dirichlet function on [imath][0,1][/imath]?
Define [imath]f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if } x \in \mathbb{Q} \\ 0 & \mbox{if } \notin \mathbb{Q} \end{array} \right.[/imath] How to evaluate [imath]\int_0^1 f(x)\,dx[/imath] ? I have no idea how to solve it, but all I think is that: since between any two rational numbers there exist an irrational number and vice versa, so the number of rational and irrational numbers are same in the interval of [imath][0,1][/imath] (or [imath][0,1)[/imath] to be precise), and since the integral is equivalent to area under the function, so [imath]\int_0^1 f(x)\,dx[/imath] must be equal to [imath]\frac{1}{2}[/imath]. Is that correct? Thank you. | 437711 | Is Dirichlet function Riemann integrable?
"Dirichlet function" is meant to be the characteristic function of rational numbers on [imath][a,b]\subset\mathbb{R}[/imath]. On one hand, a function on [imath][a,b][/imath] is Riemann integrable if and only if it is bounded and continuous almost everywhere, which the Dirichlet function satisfies. On the other hand, the upper integral of Dirichlet function is [imath]b-a[/imath], while the lower integral is [imath]0[/imath]. They don't match, so that the function is not Riemann integrable. I feel confused about which explanation I should choose... |
1396878 | sum [imath]1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...[/imath]
find the sum if the following series converges: [imath]1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...[/imath] my attempt: The series can be grouped into difference of a series of odd terms and a series of even terms.. but how to find sum? | 1369198 | Baby Rudin claim: [imath]1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...[/imath] converges
This sequence is a rearrangement of the series [imath]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...[/imath]. Note that at this point in the text we do not have any theorem about the convergence of rearrangements. Let [imath]\{s_n\}[/imath] be the sequence of partials sums of the series then for [imath]n \ge 0[/imath] [imath]s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}[/imath] We can view it as the sequence(on [imath]n[/imath]) of partials sums of [imath]\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}[/imath] Where [imath]|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}[/imath]. By the comparison test [imath]s_{3(n+1)}[/imath] converges to some real [imath]\alpha[/imath]. But [imath]s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}[/imath] and [imath]s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} [/imath]hence we have a partition of [imath]\{s_n\}[/imath] into subsequences which tend to [imath]\alpha[/imath] and this implies [imath]s_n \rightarrow \alpha[/imath]. Is my proof correct? Any alternative solutions are appreciated. |
407828 | The sum [imath]1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots)[/imath] does not exist.
What are the argument(s) that I can use proving that [imath]1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots)[/imath] does not exist. The question was: Find a arrangement of [imath]\sum\frac{(-1)^{n-1}}{n}[/imath] for which the new sum is not exist(even not [imath]+\infty[/imath] or [imath]-\infty[/imath]) | 2606035 | To Prove [imath]\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k}\right).[/imath]
Prove That [imath]\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k}\right).[/imath] My Approach I was solving: [imath]\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}= \left(1+\frac 1 3 + \frac 1 5 +\ldots \right)-\left(\frac 1 2 + \frac 1 4 +\ldots \right) =\sum_{n=1}^{\infty}\frac{1}{2n-1}- \sum_{n=1}^{\infty}\frac{1}{2n}[/imath] Searching On Mathstack i found This I know that [imath] \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}[/imath] = [imath]\ln2[/imath] But i don't know how to prove it using this method. |
1396446 | Calculating [imath]\int_0^{\infty} \frac{\log^2(1 - e^{-x})\:x^5}{e^x - 1} \: dx [/imath]
I am having trouble calculating the following improper integral: [imath]\displaystyle \int\limits_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx [/imath] Can someone give me a way that I can calculate this? | 1367624 | Compute this integral
[imath] \displaystyle \int_{0}^{\infty}{\frac{{log}^{2}(1-{e}^{-x}){x}^{5}}{{e}^{x}-1} dx} [/imath] What I have done - [imath] \displaystyle I(k) = \int_{0}^{\infty}{\frac{{x}^{5}}{{e}^{x}{(1-{e}^{-x})}^{k}}}[/imath] And then, [imath] I''(1) [/imath] will be our integral. Now, we need to compute [imath] I(k) [/imath] [imath] \displaystyle I(k) = \int_{0}^{\infty}{\sum_{m=0}^{\infty}{\binom{k+m-1}{m} {e}^{-(m+1)x}{x}^{5}}} [/imath] [imath] \displaystyle I(k) = \sum_{m=0}^{\infty}{\binom{k+m-1}{m} \frac{\Gamma(6)}{{(m+1)}^{6}}} [/imath] Now, I computed [imath] \displaystyle I''(1) = 120 \left(\sum_{m=0}^{\infty}{\frac{{({H}_{m}^{(1)})}^{2} - {H}_{m}^{(2)}}{{(m+1)}^{6}}}\right)[/imath] Or, in style, [imath] \displaystyle 120\left({s}_{h}(2,6) - {\sigma}_{h}(2,6)\right) [/imath] I want to ask if the above computation is correct or not. EDIT I am updating the answer now. We can go by solving Euler sums from formula 22 as was suggested by Marco Cantarini here and get the form as given by Marco(watch out his answer). Then, his form can be expanded by hand as - [imath] 14\zeta(8) - 5\zeta(6)\zeta(2) - 6\zeta(5)\zeta(3) - 3{\zeta(4)}^{2} + \zeta(4){\zeta(2)}^{2} + {\zeta(3)}^{2}\zeta(2) [/imath] which on simplification gives a really nice form - [imath] 20{\pi}^{2}{\zeta(3)}^{2} - 720\zeta(3)\zeta(5) + 61{\pi}^{2}\zeta(6) [/imath] |
1397144 | Prove that a finite [imath]p-[/imath]group [imath]G[/imath] with a unique subgroup [imath]H[/imath] of index [imath]p[/imath] is cyclic.
I have the following question: Prove that a finite [imath]p-[/imath]group [imath]G[/imath] with a unique subgroup [imath]H[/imath] of index [imath]p[/imath] is cyclic. Since the subgroup is characteristic it is normal. I showed that either the center of the group [imath]Z(G)[/imath] is a subgroup of [imath]H[/imath] or [imath]G=Z(G)H[/imath] and the index [imath]Z(G)/(Z(G)\cap H)[/imath] is [imath]p[/imath]. But I couldn't continue from this. Any help would be great. | 296986 | Abelian [imath]p[/imath]-group with unique subgroup of index [imath]p[/imath]
Let [imath]G[/imath] be a finite abelian [imath]p[/imath]-group with a unique subgroup [imath]H[/imath] of index [imath]p[/imath]. It is a fact that [imath]G[/imath] is cyclic. This can be deduced from the classification theorem for finite abelian groups by writing [imath]G[/imath] as a product of cyclic groups and noting that [imath]G[/imath] does not have a unique subgroup of index [imath]p[/imath] unless there is only one such group. I am wondering if there is a nice proof of this result which does not use the classification theorem. Does anyone know of one? |
1397160 | G is finite group. Need to proof that exists natural k that [imath]g^k = e[/imath]
How do I prove that in a finite group G, for each element in G there is natural power (say [imath]k[/imath]) which depends on g,such that [imath]g^k=e[/imath] ? I need to show the existence and the dependence on which [imath]g[/imath] I choose. I tried write it that way, but I don't have any direction in the proof: [imath]G\:=\:\left|n\right|\::\:G=\left\{e,\:g,\:g^2,\:...\:,\:g^{n-1}\right\}[/imath] Can anybody give me any direction of thinking ? | 64575 | Prove that every element of a finite group has an order
I was reading Nielsen and Chuang's "Quantum Computation and Quantum Information" and in the appendices was a group theory refresher. In there, I found this question: Exercise A2.1 Prove that for any element [imath]g[/imath] of a finite group, there always exists a positive integer [imath]r[/imath] such that [imath]g^r=e[/imath]. That is, every element of such a group has an order. My first thought was to look at small groups and try an inductive argument. So, for the symmetric groups of small order e.g. [imath]S_1, S_2, S_3[/imath] the integer [imath]r[/imath] is less than or equal to the order of the group. I know this because the groups are small enough to calculate without using a general proof. For example, in [imath]S_3[/imath] there is an element that rearranges the identity [imath]\langle ABC \rangle[/imath] element by shifting one character to the left e.g. [imath]s_1 = \langle BCA \rangle[/imath]. Multiplying this element by itself produces the terms [imath]s_1^2 = \langle CAB \rangle[/imath]; and [imath]s_1^3 = \langle ABC \rangle[/imath] which is the identity element, so this element is equal to the order of the group, which is three. I have no idea if this relation holds for [imath]S_4[/imath] which means I am stuck well before I get to the general case. There's a second question I'd like to ask related to the first. Is the order or period of any given element always less than or equal to the order of the group it belongs to? |
1397458 | Does every even degree polynomial in the expansion of [imath]\exp x[/imath] have no real roots?
Does every even degree polynomial in the expansion of [imath]\exp x[/imath] have no real roots? I tried the first few values and it seems like it... Is this a known result? | 1397428 | The number of real roots of [imath]1+x/1!+x^2/2!+x^3/3! + \cdots + x^6/6! =0[/imath]
Attempts so far: Used Descartes signs stuff so possible number of real roots is [imath]6,4,2,0[/imath] tried differentiating the equation [imath]4[/imath] times and got an equation with no roots hence proving that above polynomial has [imath]4[/imath] real roots. But using online calculators I get zero real roots. Where am I wrong? |
1398017 | [imath]\displaystyle\int_{0}^{\pi} \frac{\sin^2 x}{a^2 - 2ab\cos x +b^2} \,\mathrm dx[/imath]
An integration: [imath]\int_{0}^{\pi} \frac{\sin^2 x}{a^2 - 2ab\cos x +b^2} \,\mathrm dx[/imath] I am stuck with this definite integral. Will putting [imath]\,\cos x = z\,[/imath] help out here? | 1038263 | Calculation of [imath]\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\mathrm dx[/imath]
Calculate the definite integral [imath] I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx [/imath] given that [imath]a>b>0[/imath]. My Attempt: If we replace [imath]x[/imath] by [imath]C[/imath], then [imath] I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;\mathrm dC [/imath] Now we can use the Cosine Formula ([imath]A+B+C=\pi[/imath]). Applying the formula gives [imath] \begin{align} \cos C &= \frac{a^2+b^2-c^2}{2ab}\\ a^2+b^2-2ab\cos C &= c^2 \end{align} [/imath] From here we can use the formula [imath]\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}[/imath] to transform the integral to [imath] \begin{align} I &= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;\mathrm dC\\ &= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;\mathrm dC\\ &= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;\mathrm dC \end{align} [/imath] Is my process right? If not, how can I calculate the above integral? |
1398196 | To evaluate [imath]\int_{0}^\pi \frac {\sin^2 x}{a^2-2ab \cos x + b^2}\mathrm dx[/imath]?
How do we evaluate [imath]\displaystyle\int_{0}^\pi \dfrac {\sin^2 x}{a^2-2ab \cos x + b^2}\mathrm dx[/imath] ? I tried substitution and some other methods , but its not working ; please help . | 2875167 | Evaluate [imath]\int_0^\pi \frac{\sin^2x}{a^2+b^2-2ab\cos x}\mathrm dx[/imath]
We have to evaluate- [imath]\int_0^\pi \frac{\sin^2x}{a^2+b^2-2ab\cos x}\mathrm dx[/imath] I tried applying the property [imath]\displaystyle\int_0^{2a}f(x)\,\mathrm dx=\int_0^a f(x)+f(2a-x)[/imath] and proceeding I got [imath]\frac{a^2+b^2}{2a^2b^2}\frac{\pi}{2}-\frac{2(a^2+b^2)}{4a^2b^2}\int_0^{\pi/2}\frac{(a^2-b^2)^2\sec^2x}{(a^2-b^2)^2\sec^2x+4a^2b^2\tan^2x}\mathrm dx[/imath] Things are getting messed up after this. Is there a better way out? |
1398331 | Prove that if [imath]E[X|\sigma(Y)] = Y[/imath] and [imath]E[Y|\sigma(X)] = X[/imath] then [imath]X = Y[/imath] almost surely.
Prove that if [imath]E[X|\sigma(Y)] = Y[/imath] and [imath]E[Y|\sigma(X)] = X[/imath] then [imath]X = Y[/imath] almost surely. This is my idea: By assumption, [imath]Y = E[X|\sigma(Y)] = E\left\lbrack E[Y|\sigma(X)]|\sigma(Y)\right\rbrack[/imath], and I would like to show that this equals [imath]E[Y|\sigma(X)][/imath] which equals [imath]X[/imath]. If I can show that [imath]E[Y|\sigma(X)][/imath] is [imath]\sigma(Y)[/imath] measurable, then the proof is finished. But I am not sure how to do this. Any help would be greatly appreciated. | 385109 | Question about Conditional Expectation
I have a seemingly trivial question regarding conditional expectation. Consider [imath]x[/imath] and [imath]y[/imath] be two integrable random variables on Probability space (X, [imath]\Sigma[/imath], [imath]P[/imath]) such that [imath]E(X|Y) =_{a.s} Y[/imath] and [imath]E(Y|X) =_{a.s} X[/imath] Show that [imath]X=_{a.s} Y[/imath] The thing looks like a simple proof result, but I don't know how to move on after showing that[imath]\int_{B} Y = \int_{B} X[/imath] for all [imath]B \in \sigma(x)[/imath] and [imath]B \in \sigma(y)[/imath] Now any hint on how to proceed? Thank you so much in advance |
1398398 | [imath]M=I\times J [/imath] for some [imath]I,J[/imath]
Let [imath]R,S[/imath] be two rings with identity. Prove that every ideal of [imath]R\times S[/imath] is of the form [imath]I \times J[/imath] where [imath]I[/imath] is an ideal of [imath]R[/imath] and [imath]J[/imath] is an ideal of [imath]S[/imath] . Obviously [imath]I \times J[/imath] is an ideal of [imath]R\times S[/imath]. Conversely let us assume that [imath]M[/imath] is an ideal of [imath]R\times S[/imath]. To show [imath]M=I\times J [/imath]. How to proceed? | 734476 | Ideals in direct product of rings
I am trying to solve this problem: Let [imath] R_1,...,R_n[/imath] be rings with identity. Every ideal of [imath]R=\prod_{i=1}^n R_i[/imath] is of the form [imath]\prod_{i=1}^n I_i[/imath] where [imath] I_i[/imath] is an ideal of [imath]R_i[/imath]. The first part is clearly if [imath]J[/imath] an ideal of [imath]R[/imath] then [imath]J\subset \prod_{i=1}^n I_i[/imath], but i can complete the rest..any help.. |
1398674 | Determining the image of a function
I was given a function that says: What is the image of the function [imath]F: \Bbb Z \times \Bbb N \rightarrow \Bbb R[/imath] given by [imath]f(a,b) = \frac{(a-4)}{7b}[/imath] I need help really understanding how to find an image. I did a few questions where it said to make the question = b, but that was dealing with only one variable and this question has a and b. | 1396495 | Define the image of the function [imath]f :\Bbb Z \times \Bbb N →\Bbb R[/imath] given by [imath]f(a, b) = \frac{a−4}{7b}[/imath]?
[imath]\Bbb Z[/imath] - integers [imath]\Bbb N[/imath] - natural numbers (starting from 1) [imath]\Bbb R[/imath] - real numbers I believe the answer is the set of real numbers ([imath]\Bbb R[/imath]), seeing as [imath]b[/imath] will not equal [imath]0[/imath] as the set of natural numbers start from [imath]1[/imath]. Thoughts? |
1398917 | Finding the sum of this series: [imath]1+\frac 12 + \frac 13 + \cdots + \frac 1{50}[/imath]
I need to find the sum of this series: [imath]1+\frac 12 + \frac 13 + \cdots + \frac 1{50}[/imath] Please help me find the sum of this series. | 644791 | Sum of Harmonic series: [imath] 1+\frac {1}{2}+\frac {1}{3} + \frac {1}{4} + \cdots + \frac {1}{50}[/imath]
How to find the sum of this series : [imath] 1+\frac {1}{2}+\frac {1}{3} + \frac {1}{4} + \cdots + \frac {1}{50}[/imath] |
1197732 | Prove that [imath]HK[/imath] is a subgroup iff [imath]HK=KH[/imath].
Let [imath]H[/imath] and [imath]K[/imath] be subgroups of a group [imath]G[/imath], and let [imath]HK=\{hk: h \in H, k \in K\}[/imath], [imath]KH=\{kh: k \in K, h \in H\}[/imath]. How can we prove that [imath]HK[/imath] is a subgroup iff [imath]HK=KH[/imath]? | 2418894 | If [imath]G[/imath] is a group, [imath]H \leq G[/imath] and [imath]K \leq G[/imath], then [imath]HK \leq G \iff HK = KH[/imath].
PROBLEM Let [imath]H[/imath] and [imath]K[/imath] be subgroups of a group [imath]G[/imath]. Prove that [imath]HK = \{ab | a \in H, b \in K\}[/imath] is a subgroup of [imath]G[/imath] if and only if [imath]HK = KH[/imath]. WHAT I KNOW (1) If [imath]H = \{e, \alpha, \alpha^2\}[/imath] and [imath]K = \{e, \beta, \beta^2\}[/imath] are subgroups of [imath]S_4[/imath], one can show that [imath]HK \neq KH[/imath]. We can compute [imath]HK[/imath] and [imath]KH[/imath] to verify that: In general, the product of two subgroups of a group [imath]G[/imath] is not a subgroup of [imath]G[/imath]. (2) If [imath]H = \{h_1, h_2, \ldots, h_r\}[/imath] and [imath]K = \{b_1, b_2, \ldots, b_p\}[/imath] are subgroups and one of [imath]H[/imath] or [imath]K[/imath] is normal in [imath]G[/imath], then (a) [imath]HK = KH[/imath], and (b) [imath]HK[/imath] is a subgroup of [imath]G[/imath]. (3) If [imath]H[/imath] and [imath]K[/imath] are normal subgroups of [imath]G[/imath], so also is [imath]HK[/imath]. QUESTION How do I then solve the problem at hand, given that what I know seems to require that at least one of [imath]H[/imath] or [imath]K[/imath] is normal in [imath]G[/imath]? That is, how does one prove the statement in the PROBLEM section in full generality (without requiring that one of [imath]H[/imath] or [imath]K[/imath] is normal in [imath]G[/imath])? |
1399498 | Is [imath]\log(-1)[/imath] equal to [imath]-\log(-1)[/imath]
I thought it should be because if the logarithmic identities hold then, [imath]-\log(-1)=\log(-1^{-1})=\log(-1)[/imath] But [imath]\log(-1)=i*\pi[/imath] and [imath]-\log(-1)=-i*\pi[/imath] | 25414 | Is it standard to say [imath]-i \log(-1)[/imath] is [imath]\pi[/imath]?
I typed [imath]\pi[/imath] into Wolfram Alpha and in the short list of definitions there appeared [imath] \pi = -i \log(-1)[/imath] which really bothered me. Multiplying on both sides by [imath]2i[/imath]: [imath] 2\pi i = 2 \log(-1) = \log(-1)^2 = \log 1= 0[/imath] which is clearly false. I guess my error is [imath]\log 1 = 0[/imath] when [imath]\log[/imath] is complex-valued. I need to use [imath]1 = e^{2 \pi i}[/imath] instead. So my question: is it correct for WA to say [imath]\pi = -i \log(-1)[/imath]? Or should be they specifying "which" [imath]-1[/imath] they mean? Clearly [imath] -1 = e^{i \pi}[/imath] is the "correct" value of [imath]-1[/imath] here. |
1343066 | Obstruction to the splitness of an exact sequence of holomorphic vector bundles
In the book of S. Kobayashi, hyperbolic complex spaces, there is the lemma (3.A.3) on the page 153, section Royden's extension lemma, whose statement is Every exact sequence of holomorphic vector bundles [imath]0\to E'\to E\to E''\to 0[/imath] on a Stein manifold splits. The author's argument is that the obstruction to a splitting lies in [imath]H^1(X,\mathrm{Hom}(E'',E'))[/imath] which is zero since [imath]X[/imath] is a Stein manifold. Could someone explain what obstruction means and why it lies in this cohomology group? Thank you very much. | 1345658 | Obstruction to the splitness of an exact sequence of holomorphic vector bundles
This question was asked here, but I think there will not be an answer, so let me recopy the question on Mathoverflow. In the book of S. Kobayashi, hyperbolic complex spaces, there is the lemma (3.A.3) on the page 153, section Royden's extension lemma, whose statement is Every exact sequence of holomorphic vector bundles [imath]0\to E'\to E\to E''\to 0[/imath] on a Stein manifold splits. The author's argument is that the obstruction to a splitting lies in [imath]H^1(X,\mathrm{Hom}(E'',E'))[/imath] which is zero since [imath]X[/imath] is a Stein manifold. Could someone explain what obstruction means and why it lies in this cohomology group? Thank you very much. |
1399662 | Summing up the elements of a finite field.
I want to show that [imath]\sum_{x\in \mathbb F_{q}}x^i=0[/imath] if [imath]q-1[/imath] does not divide [imath]i[/imath]. Can someone give me a hint? | 1268419 | Sum of elements of a finite field
Let [imath]F[/imath] be a finite field and [imath]i[/imath] an integer. Calculate the sum of all the elements of [imath]F[/imath], each raised to the [imath]i[/imath]th power. My approach so far: Let [imath]F=(0,1,\alpha,\alpha^2,...,\alpha^{p^n-1})[/imath], where [imath]p[/imath] is a prime number and let [imath]\sigma=1+\sum_{k=1}^{p^n-1}(α^k)^i[/imath] be the desired sum. Since [imath]F[/imath] is a ring, each element has an additive inverse. Thus, the trivial case of [imath]i=1[/imath] results in [imath]\sigma=1+[\alpha+(-\alpha)+\alpha^2+(-\alpha^2)+\cdots+\alpha^{p^n-1}+(-\alpha^{p^n-1})]=1[/imath]. We can also calculate the trivial case of [imath]i=0[/imath], where [imath]\sigma=1+p^n-1=p^n[/imath] In the general case, let [imath]i=t (\text{mod } p^n)[/imath]. Then: [imath]i=t+mp^n[/imath], where [imath]t,m[/imath] are integers. We take into account the fact that the order of the multiplicative group of the finite field is [imath]p^n[/imath], so each element raised to the [imath]p^n[/imath] equals [imath]1[/imath]. Thus [imath]σ=1+α^t+α^{2t}+\cdots+α^{t(p^n-1)}[/imath]. This is the sum of a geometric progression, plus [imath]1[/imath]. Thus, [imath]σ=1+α(α^{p^n-1})/α-1[/imath] = [imath] 1[/imath]+([imath]α^{p^n}-α[/imath])/[imath]{\alpha-1}[/imath]. But according to Lagrange's theorem, for every [imath]\alpha\in F[/imath], the polynomial [imath]x^{p^n}-x[/imath] is zero. It follows that [imath]\sigma=1[/imath]. An important question is raised: Is it true that the characteristic of the finite field is [imath]p[/imath] if its order is [imath]p^n[/imath]? If so, the case for [imath]i=0[/imath] leads to [imath]σ=0≠1[/imath] and all cases have been considered? |
1399457 | Listing all elements of a set
I was given a question like the following: Let [imath]A = \Bbb Z[/imath], [imath]B = [-1,\pi][/imath] , [imath]C=(2,7)[/imath]. List all Elements of [imath]A \cap (B^c \cap C)[/imath]. I do not really understand how to got about this problem. I understand [imath]\cap[/imath] means intersection, but I have trouble reading the question; for instance, why place brackets between [imath]B^c \cap C[/imath]? | 1396523 | Let [imath]A = \mathbb{Z}[/imath], [imath]B = [−1, \pi][/imath], [imath]C = (2, 7)[/imath]. List all elements of [imath]A \cap (B^c \cap C)[/imath].
After working it out on a number line, I got: [imath]\{4, 5, 6\}[/imath]. As it stands, the expression contains the integers that do not belong to the set [imath]B[/imath] that cross into [imath]C[/imath]. This would result in [imath]4, 5, 6[/imath]. Number [imath]7[/imath] would not be included because of the soft brackets surrounding [imath]C = (2, 7)[/imath]. Is this correct? |
1399921 | If [imath]A^n = B[/imath] and I know [imath]B[/imath], can I find [imath]A[/imath]?
Suppose that [imath]A[/imath] and [imath]B[/imath] are invertible, [imath]p \times p[/imath] matrices. If [imath]A^n = B[/imath] and I know all of the entries in [imath]B[/imath], can I find an [imath]A[/imath] for some or all integers [imath]n \ge 0[/imath]? How many solutions for [imath]A[/imath] exist? If I'm thinking correctly, then [imath]A = B * (A^{-1})^{n-1},[/imath] but this is sort of self referential. Thanks! | 460354 | Nth roots of square matrices
Is there a general method (which can be implemented by hand) to finding the [imath]n[/imath]-th roots of [imath]2 \times 2[/imath] matrices? Is there a similar method for a general [imath]m \times m[/imath] matrix? (for [imath]n > 1[/imath] and [imath]n\in\mathbb{Z}[/imath]) |
1392976 | Simplifying [imath]\sum_{i=1}^{n-2}i(n-1-i)[/imath]
I have been trying to simplify [imath]\sum_{i=1}^{n-2}i(n-1-i)[/imath] i.e - remove the summation, put it in polynomial form Since [imath]i[/imath] is the changing variable, I don't think this is possible. I also know that the simplified answer should be in polynomial form with highest degree 3, as this stems from a nested loop time complexity problem in programming. Is it possible if we rewrite this summation, as it has a kind of symmetrical property. The second half of the summation is identical to the first half. | 1113556 | How to show that [imath]\sum_{k=1}^n k(n+1-k)=\binom{n+2}3[/imath]?
While thinking about another question I found out that this equality might be useful there: [imath]n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6[/imath] To rewrite it in a more compact way: [imath]\sum_{k=1}^n k(n+1-k)=\frac{n(n+1)(n+2)}6.[/imath] This equality is relatively easy to prove: [imath]\sum_{k=1}^n k(n+1-k)= (n+1)\sum_{k=1}^n k - \sum_{k=1}^n k^2 = (n+1) \frac{n(n+1)}2 - \frac{n(n+1)(2n+1)}6 = n(n+1) \left(\frac{n+1}2-\frac{2n+1}6\right) = n(n+1)\frac{3(n+1)-(2n+1)}6 = \frac{n(n+1)(n+2)}6.[/imath] (We only used the known formulas for the sum of the first [imath]n[/imath] squares and the sum of the first [imath]n[/imath] numbers.) Are there some other nice proofs of this equality? (Induction, combinatorial arguments, visual proofs, ...) EDIT: Now I found another question which asks about the same identity: Combinatorial interpretation of a sum identity: [imath]\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}[/imath] (I have tried to search before posting. But the answers posted here so far gave me some new ideas for good keywords to search which lead me to finding that question.) The questions are, in my opinion, not exact duplicates since the other question asks specifically about combinatorial proofs and my question does not have that restriction. But I agree that this is a very minor distinction. In any case, if you think that one of them should be closed as a duplicate, then you can vote to close. I will refrain from voting to close/reopen on this question. (If one of the two questions is voted to be a duplicate of the other one, they probably cannot be merged, since the summation variables are off by one.) |
1399701 | Does [imath](p(0) \wedge (P(n) \implies P(n-1))) \implies P(n) \forall n\leq 0[/imath]?
Does [imath](p(0) \wedge (P(n) \implies P(n-1))) \implies P(n) \forall n\leq 0[/imath]? In other words, what I'm asking is, can I use the axiom of induction for negative numbers? Why/why not? E: This is not a duplicate from that other post, I ask nowhere about having a negative argument for the base case. | 1209845 | Is induction valid when starting at a negative number as a base case?
I'm reading the text Discrete and Combinatorial Mathematics by Grimaldi, and he puts forth the theorem of the principle of mathematical induction as such: Let [imath]S(n)[/imath] denote an open mathematical statement that involves one or more occurrences of the arable [imath]n[/imath] which represents a positive integer. [imath]\textbf{a)}[/imath] If [imath]S(1)[/imath] is true, and [imath]\textbf{b)}[/imath] if whenever [imath]S(k)[/imath] is true ( for some particular, but arbitrarily chosen, [imath]k \in\mathbb{Z^+}[/imath]), then [imath]S(k+1)[/imath] is true, then [imath]S(n)[/imath] is true for all [imath]n \in \mathbb{Z^+}[/imath]. He then states that: All that is needed is for the open statement [imath]S(n)[/imath] to be true for some first element [imath]n_{0}\in \mathbb{Z}[/imath] so that the induction process has a starting place. We need the truth of [imath]S(n_{0})[/imath] for our basis step. The integer [imath]n_{0}[/imath] could be [imath]5[/imath] just as well as [imath]1[/imath]. [imath]\textbf{It could even be zero or negative}[/imath] , because the set [imath]\mathbb{Z^+}[/imath] is in union with [imath]\{0\}[/imath] or any finite set of negative integers is well ordered. The negative part confuses me specifically. Won't this mean that the theorem can be false or would require changes both in premises and conclusion, since if the base case [imath]n_{0}<0[/imath] there could be some integer [imath]m<k[/imath] for instance, such that [imath]S(m)[/imath] is false? Also, can anyone direct me or provide an example of an induction proof that uses a negative base case? |
1400321 | Is the number [imath]0.1234567891011121314\ldots[/imath] a rational or irrational number?
Is the number [imath]0.1234567891011121314\ldots[/imath] a rational or irrational number? The number has a very clear pattern but however in order for the number to be a rational number it would have to be written as a/b. The normal tricks of writing it as an equation and solving [imath]3.3333333... = x \\ 10*3.333333 = 10x\\ 33.333333 = 10x\\ 30 + x = 10x\\ 30=9x\\ x=30/9 \\= 10/3[/imath] does not seem to work here | 480078 | Show that the sequence [imath]1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0,2,1,\cdots[/imath] isn't periodic
Show that the sequence [imath]\{a_n\}_{n\in \mathbb{N}} = \{x: x=[/imath] the nth decimal digit of Champernowne's constant[imath]\}[/imath] is not periodic. For those who don't know what Champernowne's constant is, it's the following number: [imath]C_{10} = 0.123456789101112131415161718192021...[/imath] Well, before someone says that [imath]\{a_n\}_{n\in \mathbb{N}}[/imath] is definitely aperiodic because Champernowne's constant is irrational(and even transcendental), I have to say that I'm well aware of the fact that a number is rational if and only if the sequence of its digits is periodic, but I'm looking for another proof that the sequence of the digits of [imath]C_{10}[/imath] is aperiodic that doesn't use anything but a smart way of showing that [imath]a_{n+k} = a_n[/imath] can't hold for any [imath]k\in \mathbb{N}[/imath]. I've observed that by grouping the terms of a sequence in a clever way we can show that no such [imath]k[/imath] can exist for some sequences. For example I created the following sequence: [imath]1,0,0,1,1,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,1,1,1,1,1,\cdots[/imath] The sequence is created this way: First you put one [imath]1[/imath], then two [imath]0[/imath]'s, and in the nth step you add either [imath]1[/imath]'s or [imath]0[/imath]'s n times depending on whether n is odd or even respectively. It's easy to show this sequence is not periodic, because no matter how large [imath]k[/imath] is you can always find [imath]n[/imath] such that [imath]a_n = 1[/imath] but [imath]a_{n+k}=0[/imath] or [imath]a_n=0[/imath] but [imath]a_{n+k}=1[/imath]. That proves the sequence is not periodic and as a corollary it' proved that the real number [imath]0.1001110000111110000001111111\cdots[/imath] is irrational. After showing that I've been trying to apply this technique of grouping to the sequence in problem without much success so far. I think if the periodicity of a sequence [imath]\{a_n\}[/imath] is denoted by [imath]\Omega(\{a_n\})[/imath] then if we define a new sequence [imath]\bar{a}_m(n) := \bar{a}_{m,n} = [a_n]_m[/imath] then [imath]\Omega(\{\bar{a}_{m,n}\}) \mid \Omega({a_n})[/imath]. This shouldn't be very hard to prove I think. Now, to show that [imath]1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0,2,1,\cdots[/imath] is not periodic it's sufficient to prove that the sequence [imath]1,0,1,0,1,0,1,0,1,1,0,1,1,1,0,1,0,1,0,1,\cdots[/imath] is not periodic. But I haven't progressed much from here :/ |
1400337 | [imath]a \in G[/imath] commutes with all its conjugates iff [imath]a[/imath] belongs to an abelian normal subgroup of [imath]G[/imath]
Let [imath]a\in G[/imath], where [imath]G[/imath] is a group. Prove that [imath]a[/imath] commutes with each of its conjugates in [imath]G[/imath] if and only if [imath]a[/imath] belongs to an abelian normal subgroup of [imath]G[/imath]. This is what I did: [imath]"⟹"[/imath] First, I thought of [imath]<a>[/imath] to be that normal subgroup containing a. But if it is true, I am not sure how to prove it. Then I thought of another subgroup, namely [imath]C_G(a)[/imath]. I conclude from the hypothesis given that [imath]Cl_G(a) ⊂ C_G(a)[/imath] And I know that [imath]|Cl_G(a)| = |G : C_G(a)|[/imath] by G acting transitively on [imath]Cl(a)[/imath] by conjugation. But I don't know how to get that [imath] C_G(a)[/imath] is normal in [imath]G[/imath]. I am thinking also of a normal subgroup of [imath]G[/imath] , say [imath]K[/imath] containing [imath]a[/imath], so [imath]<a>[/imath] [imath]\cap[/imath] [imath]K[/imath] and [imath] C_G(a)[/imath] [imath]\cap[/imath] [imath]K[/imath] are subgroups of G inheriting abelianity from [imath]<a>[/imath] and [imath] C_G(a)[/imath] respectively, and normality from [imath]K[/imath]. [imath]"\Leftarrow"[/imath] Let [imath]a ∈ H[/imath] s.t. [imath]H[/imath] is an abelian normal subgroup of G. So [imath]ah=ha[/imath], ∀ [imath]h∈H[/imath]. Since [imath]H[/imath] is normal in [imath]G[/imath], hence [imath]g^{-1}hg ∈ H[/imath], ∀ [imath]g∈G[/imath] So [imath]g^{-1}ag ∈ H[/imath], ∀ [imath]g∈G[/imath] Which means [imath]Cl_G(a)[/imath] [imath]\subseteq[/imath] [imath] H[/imath] Therefore, a commutes with each of its conjugates in G. Could someone correct me? | 1386718 | Prove that [imath]a[/imath] commutes with each of its conjugates in [imath]G[/imath] iff [imath]a[/imath] belongs to an abelian normal subgroup of [imath]G[/imath].
Let [imath]a[/imath] be an element of a group G. Prove that [imath]a[/imath] commutes with each of its conjugates in [imath]G[/imath] iff [imath]a[/imath] belongs to an abelian normal subgroup of [imath]G[/imath]. My Try: I proved the backward direction. For forward one, given [imath]g\in G, a(gag^{-1})=(gag^{-1})a[/imath]. I was trying to prove that [imath]a\in Z(G)[/imath]. In order to prove that, I must show that [imath]ag=ga[/imath]. But failed. Am I on the right track? Can anybody please give me a hint? I want to try it myself. |
1384321 | Prove inequality [imath]x^sy^{1-s} \leq sx + (1-s)y[/imath]
Given [imath]s \in (0,1)[/imath], prove [imath]x^sy^{1-s} \leq sx + (1-s)y[/imath] for [imath]x,y > 0[/imath] Tried some algebraic manipulations but I'm guessing I need to use some trick. Any suggestions, hints? | 594135 | Prove [imath]a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b > 0,\; 0 < \alpha < 1[/imath]
I have no idea how to do this. Any help would be appreciated. The chapter I'm on is about differentiation and the mean value theorem. Prove [imath]a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b > 0,\; 0 < \alpha < 1[/imath] |
1401057 | Question about complex numbers
Proof that if [imath]z = 1[/imath], then [imath]|z-w| = |1- \overline{w}z|[/imath], [imath]\forall w \in[/imath] [imath]\mathbb{C}[/imath] My attempt below: [imath](z-w)\overline{(z-w)} = |z-w|^2[/imath] [imath](z-w)\overline{(z-w)} = (z-w)(\overline{z}-\overline{w}) = |z|^2 -z\overline{w} -\overline{z\overline{w}} +|w|^2 = |z|^2 - 2Re(z\overline{w}) + |\overline{w}|^2*1 = 1 - 2Re(z\overline{w}) + |\overline{w}|^2*|z|^2 = 1 - 2Re(z\overline{w}) + |\overline{w}z|^2 = 1 - 2Re(\overline{w}z) + |\overline{w}z|^2[/imath] [imath]|z-w|^2 = 1 - 2Re(\overline{w}z) + |\overline{w}z|^2[/imath] I was trying identfiy a remarkable product, but I didn't have success. Someone can help me with this question? | 198582 | complex number question involving modulus
Let [imath]z, w[/imath] be distinct complex numbers. Show that if [imath]|z| = 1[/imath] or [imath]|w| = 1[/imath], then [imath]\frac{|z − w|}{|1 − z^*w|} = 1[/imath] [Hint: Note that [imath]|a|^2 = aa^*[/imath].] Hey guys, couldn't get my thinking cap on for this question. Any helpful input? Will appreciate it! |
1401176 | Multiple order axioms independence
Let [imath]T[/imath] be a theory, let [imath]L[/imath] be its language, let [imath]A[/imath] be its set of axioms and let [imath]P_0 \in L[/imath] be a property. [imath]P_0[/imath] could be : Consequence of [imath]A[/imath] The negation of a consequence of [imath]A[/imath] Independent of [imath]A[/imath] Assume that the property [imath]P_1[/imath] : "[imath]P_0[/imath] is independent of [imath]A[/imath]" is also in [imath]L[/imath]. Can [imath]P_1[/imath] be itself independent of [imath]A[/imath] ? For instance, assume [imath]P[/imath] versus [imath]NP[/imath] is independent of the axioms (let's say ZFC), could it be the case that this independence is itself independent of the axioms ? This could explain why nobody already solved this problem. More generally, assume that [imath]P_{i+1}[/imath] : "[imath]P_i[/imath] is independent of [imath]A[/imath]" is also in [imath]L[/imath]. Does it exist a property [imath]P_0[/imath] such that [imath]\forall i.\ P_i[/imath] is independent of [imath]A[/imath] ? If [imath]A[/imath] is empty, any property is independent of the axioms, in particular all the [imath]P_i[/imath] are independent of the axioms whatever is [imath]P_0[/imath]. If [imath]A[/imath] is the axioms of Presburger arithmetic, none of the property is independent of the axioms, but I am not sure that we can express "[imath]P_i[/imath] independent of [imath]A[/imath]" in the language of the theory. What happens when A is ZFC, Peano axioms, Euclide axioms, or any other usual theory ? | 17212 | Is there a statement whose undecidability is undecidable?
We know there are statements that are undecidable/independent of ZFC. Can there be a statement S, such that (ZFC [imath]\not\vdash[/imath] S and ZFC [imath]\not\vdash[/imath] ~S) is undecidable? |
1401351 | Scalar product on Lie algebra of compact Lie group
I am studying Differential Geometry and I am facing with a lemma in which there is a step that I do not understand. In particular, let [imath]G[/imath] be a connected compact Lie group, is used "[imath]\langle\ \cdot , \cdot \rangle[/imath] any [imath]Ad(G)[/imath]-invariant scalar product (or equivalently a non-degenerate semi-positive bilinear form) on [imath]\mathfrak{g}[/imath], the Lie algebra of [imath]G[/imath] ". I can't find any such of scalar product. When a connected compact Lie group [imath]G[/imath] has such a scalar product? I tried with the Killing form that is semi-definite negative bilinear form, that is non-degenerate iff the lie algebra is semisimple, but how I can prove that a compact connected Lie group have semisimple Lie algebra? I tried also with a proposition (3.6.2) in "lie groups", Duistermaat-Kolk, saying that the killing form of a lie algebra [imath]\mathfrak{g}[/imath] is negative on the complement of the center [imath]\mathfrak{z}[/imath] of [imath]\mathfrak{g}[/imath]. Then if I prove that [imath]G[/imath] is centerless, so is [imath]\mathfrak{g}[/imath] and the opposite of the killing form is an [imath]Ad(G)[/imath]-invariant scalar product on [imath]\mathfrak{g}[/imath]. [imath]\underline{NOTE}[/imath]: The lemma I have to do states "Two maximal tori of a compact connected Lie group [imath]G[/imath] are coniugate", so we can suppose that [imath]G[/imath] is non-abelian. | 319593 | Which Lie groups have Lie algebras admitting an Ad-invariant inner product?
I am trying to answer the following question: Which Lie groups have a Lie algebra admitting an [imath]\text{Ad}[/imath]-invariant inner product? First of all, all compact Lie groups satisfy this condition because one can average an arbitrary inner product on the algebra over the group. This means that all Lie groups that have the same Lie algebra as some compact group also satisfy the condition. Indeed, the group is then locally isomorphic to a compact group, and taking "the same" inner product on the Lie algebra, we see that this inner product is invariant under all [imath]\text{Ad}_g[/imath] for [imath]g[/imath] in some neighbourhood of the identity, but this neighbourhood generates the whole group, from which the claim follows. Another approach would be to show that an inner product is [imath]\text{Ad}[/imath]-invariant precisely if it is [imath]\text{ad}[/imath]-invariant (like here, although there may be an issue with connectedness of the group) and conclude that the Lie algebra completely characterizes whether or not an invariant inner product exists. I suspect these are all the Lie groups that answer the question, because Wikipedia claims this (the second property, at "this property characterizes compact Lie algebras"). However, the reference given is just "SpringerLink", which is of no use. Could somebody confirm that the answer to the above question is "all Lie groups with a compact Lie algebra"? I have consulted multiple sources, but none of them completely establish the result I'm looking for. For example, Lie groups beyond an introduction by Knapp establishes what I already know (that it works for groups with a compact Lie algebra), and establishes the converse only in case the inner product is actually the Killing form. |
1401891 | Gauss lemma for arbitrary commutative ring
Part (iv) of exercise #2 for chapter 1 in Atiyah and Macdonald's book Introduction to Commutative Algebra asserts that if [imath]f, g \in A[x][/imath] are primitive then [imath]fg[/imath] is primitive. We know that this is true when the coefficient ring [imath]A[/imath] is a UFD in which case it is the statement of Gauss' lemma. I have tried hard to show it for arbitrary ring as the problem says but did not manage to do that. So my question is: Though I do not doubt the author, is Gauss lemma really valid for arbitrary commutative rings with unity and can I have some hint? | 413788 | Product of two primitive polynomials
I'm having troubles with one of the problems in the book Introduction to Commutative Algebra by Atiyah and MacDonald. It's on page 11, and is the last part of the second question. Given [imath]R[/imath] a commutative ring with unit. We say [imath]f = \sum\limits_{i=0}^n r_ix^i \in R[x][/imath] is primitive if [imath]\langle r_0,r_1,...,r_n\rangle = R[/imath], i.e., the ideal generated by the coefficients of [imath]f[/imath] is [imath]R[/imath]. Prove that [imath]fg[/imath] is primitive iff [imath]f[/imath] and [imath]g[/imath] are both primitive. The [imath]\Rightarrow[/imath] part is easy. Say, [imath]f = \sum\limits_{i=0}^n r_ix^i[/imath], [imath]g = \sum\limits_{i=0}^m s_ix^i[/imath]; then [imath]fg = \sum\limits_{i=0}^{m+n} c_ix^i[/imath], where [imath]c_k = \sum\limits_{i + j = k}r_is_j[/imath]. Since [imath]fg[/imath] is primitive, there exists a set of [imath]\{\alpha_i\} \subset R[/imath], such that [imath]\sum\limits_{i=0}^{m+n} \alpha_ic_i = 1[/imath], to prove [imath]f[/imath] is primitive, I just need to write all [imath]c_i[/imath]'s in terms of [imath]r_i[/imath]'s, and [imath]s_j[/imath]'s, then group all [imath]r_i[/imath] accordingly, rearranging it a little bit, and everything is perfectly done. And the proof of the primitivity of [imath]g[/imath] is basically the same. The [imath]\Leftarrow[/imath] part is just so difficult. Say [imath]f = \sum\limits_{i=0}^n r_ix^i[/imath], [imath]g = \sum\limits_{i=0}^m s_ix^i[/imath] are both primitive, then there exists [imath]\{\alpha_i\}; \{\beta_i\} \subset R[/imath], such that [imath]\sum\limits_{i=0}^{n} \alpha_ir_i = 1[/imath], and [imath]\sum\limits_{i=0}^{m} \beta_is_i = 1[/imath]. At first, I thought of multiplying the two together, but it just didn't work. So, I'm stuck since I cannot see any way other than multiplying the two sums together. I hope you guys can give me a small push on this. Thanks very much in advance, And have a good day. |
1401726 | How to prove the Riemann hypothesis holds for the first non-trivial zero?
The Riemann hypothesis states that all non-trivial zeros of the Riemann zeta function [imath]\zeta(z)[/imath] lie on the critical line [imath]\Re(z)=1/2[/imath]. The MathWorld page on this topic mentions that the hypothesis has been verified for the first ten trillion [imath](10^{13})[/imath] zeros. Unfortunaly I have never seen an explicit proof that the first non-trivial zero [imath]\rho_1\approx0.5000000000...+i\,14.1347251417...[/imath] lies exactly on the critical line. Could you please show me a proof of this? | 7783 | Proving a known zero of the Riemann Zeta has real part exactly 1/2
Much effort has been expended on a famous unsolved problem about the Riemann Zeta function [imath]\zeta(s)[/imath]. Not surprisingly, it's called the Riemann hypothesis, which asserts: [imath] \zeta(s) = 0 \Rightarrow \operatorname{Re} s = \frac1 2 \text{ or } \operatorname{Im} s = 0 .[/imath] Now there are numerical methods for approximating [imath]\zeta(s)[/imath], but as I understand, no one knows any exact values except at even integers, which include the trivial zeroes for which [imath]\operatorname{Im} s = 0 [/imath]. So I've always wondered: For the rest, how does one prove [imath]\operatorname{Re} s = 1/2[/imath] exactly? (All I know that seems vaguely useful is the argument principle, which I'm not sure helps here, but I'd be happy to learn about other techniques that aren't too far advanced.) Edit: Looking over this again, I found out I missed the closed-form values at odd negative integers. This doesn't affect the question but seemed worth correcting. Thanks to the people who contributed. |
1401903 | Prove by induction: for [imath]n \ge 0[/imath], [imath]\frac{(2n)!}{n!2^n}[/imath] is an integer
Another prove by induction question: for [imath]n \ge 0[/imath], [imath]\frac{(2n)!}{n!2^n}[/imath] is an integer Base step: [imath]n = 0[/imath] [imath]\frac{(2 \times 0)!}{0! \times 2^0} = \frac{0!}{1 \times 1} = 1[/imath] Induction step: please help | 1352477 | Show [imath]\frac{(2n)!}{n!\cdot 2^n}[/imath] is an integer for [imath]n[/imath] greater than or equal to [imath]0[/imath]
Show [imath]\frac{(2n)!}{n!\cdot 2^n}[/imath] is an integer for [imath]n[/imath] greater than or equal to [imath]0[/imath]. Could anyone please help me with this proving? Thanks! |
1400195 | Finding the Frenet frame
I am trying to find the Frenet frame of the following curve: [imath]\zeta(t)=\left(\frac13(1+t)^{3/2},\frac{1}3(1-t)^{3/2},\frac{t}{\sqrt2}\right)[/imath] How do I do this? Is there a straightforward way from the curvature and Torsion? I can't find the definition anywhere. | 1402308 | Finding the curvature and normal vector for an arclength curve
I am trying to find the curvature and normal vector for [imath]\alpha(t) = \left(\frac13 (1+t)^{3/2}, \frac13 (1-t)^{3/2}, \frac{t}{\sqrt2}\right)[/imath] [imath]\alpha'(t) = \left(\frac12 (1+t)^\frac12 , - \frac 12 (1-t)^\frac12, \frac{1}{\sqrt2}\right)[/imath] [imath]\|\alpha'(t)\|=\sqrt{\frac14 (1+t) + \frac 14(1-t) + \frac12}=1[/imath] This is an Arclength parameterization, so we don't need chain rule for the frenet equations: Curvature is given by [imath]\|\alpha''(t)\|[/imath] [imath]\alpha''(t) = \left(\frac14 (1+t)^{-\frac12} , \frac 14 (1-t)^{-\frac12}, 0\right)[/imath] [imath]=\frac14 \left((1+t)^{-\frac12} , (1-t)^{-\frac12}, 0\right)[/imath] [imath]k(t)= \|\alpha''(t)\|=\frac14 \sqrt{(1+t)^{-1} + (1-t)^{-1}}[/imath] The normal vector [imath]n(t)=\frac{\alpha''(t)}{k(t)}= \left(\frac{1}{\sqrt{-\frac{2}{t-1}}}, \frac{1}{\sqrt{\frac{2}{t+1}}},0\right)[/imath] isn't orthogonal to [imath]\alpha'(t)[/imath] though? What have I done wrong? |
1401186 | Help solving a basic polynomial problem
[imath](3x-1)^7=a_7x^7+a_6x^6+a_5x^5+...+a_1x+a_0[/imath] Now it is required to find the value of [imath]a_7+a_6+a_5+...+a_1+a_0[/imath] Please give me some hints to solve this problem.Thanks a lot in advance. | 1082271 | Given that[imath](3x-1)^7=a_7x^7+a_6x^6+a_5x^5+...+a_1x+a_0[/imath], find [imath]a_7+a_6+a_5+a_4+...+a_1+a_0[/imath]
Given that[imath](3x-1)^7=a_7x^7+a_6x^6+a_5x^5+...+a_1x+a_0[/imath]find [imath]a_7+a_6+a_5+a_4+...+a_1+a_0[/imath] Is there anyway to do the question without using binomial theorem and expanding the expression on the LHS?? |
1402645 | If [imath]2xy[/imath] divides [imath]x^2+y^2-x[/imath], prove that [imath]x[/imath] is a perfect square
This problem is from ( BMO Exam1991 ). I tried to solve but it was difficult. The problem is: If [imath] x^{2} + y^{2} - x [/imath] is a multiple of [imath] 2xy [/imath] where [imath]x[/imath] & [imath]y[/imath] are integers, prove that [imath]x [/imath] is a perfect square. Any help or hint please. | 663283 | Prove that [imath]2xy\mid x^2+y^2-x[/imath] implies [imath]x[/imath] is a perfect square.
Prove that [imath]2xy\mid x^2+y^2-x[/imath] implies [imath]x[/imath] is a perfect square. My work: [imath]2xy\mid x^2+y^2-x \implies x^2+y^2-x=2xy\cdot k[/imath] So,[imath]x^2+y^2+2xy-x=(x+y)^2-x=2xy \cdot (k+1)[/imath] And,[imath]x^2+y^2-2xy-x=(x-y)^2-x=2xy \cdot (k-1)[/imath] I found that for [imath]x,y[/imath] both odd, no solution exists. For [imath]x[/imath] even, and [imath]y[/imath] odd,no solution exists. Solution exists only for [imath]x[/imath] odd, [imath]y[/imath] even and [imath]x[/imath] even and [imath]y[/imath] even solution exists. Cannot do anything more. Please help! |
1402957 | Solving Frobenius minimization with linear algebra
Given that [imath]A[/imath] is a symmetric matrix, find [imath]X[/imath] that solves [imath]\mathop {\min }\limits_X {\left\| {A - X{X^T}} \right\|_F}[/imath] I think that the problem can be solved using eigenvalue or singular value decomposition technics. [imath]XX^T=A[/imath] seems an obvious solution, but the problem is that [imath]XX^T[/imath] is positive semidefinite, while [imath]A[/imath] may not be, although they are both symmetric. At this point I am thinking about taking eigenvalue decomposition of A, then replacing the negative eigenvalues in the middle diagonal matrix with 0-s (denote the resulting matrix by [imath]\bar{A}[/imath]). But am having difficulties to understand why [imath]XX^T=\bar{A}[/imath] gives the [imath]X[/imath] with minimal distance from A. | 1402537 | Minimizing Frobenius norm of a special type
Given that [imath]A[/imath] is a symmetric matrix, find [imath]X[/imath] that solves [imath]\mathop {\min }\limits_X {\left\| {A - X{X^T}} \right\|_F}[/imath] I think that the problem can be solved using eigenvalue or singular value decomposition technics. [imath]XX^T=A[/imath] seems an obvious solution, but the problem is that [imath]XX^T[/imath] is positive semidefinite, while [imath]A[/imath] may not be, although they are both symmetric. At this point I am thinking about taking eigenvalue decomposition of A, then replace the negative eigenvalues in the middle diagonal matrix with 0-s (denote the resulting matrix by [imath]\bar{A}[/imath]). But am having difficulties to understand why [imath]XX^T=\bar{A}[/imath] gives the [imath]X[/imath] with minimal distance from A. |
1401970 | Survival probability of 1D Random Walker
For a 1 D random walk on [imath]Z[/imath] axis, starting at [imath]z=0[/imath], equal probability to go to right or left, what is the probability that during the first k steps the walker's position remains [imath]z\leq m[/imath]? This is also called the survival probability. I can write is as a sum of first passage time probabilities, but I am looking for a closed form. Thanks, | 4234 | Random walk [imath]< 0[/imath]
Suppose [imath]{X_t}[/imath] is a random walk with mean zero. (either discrete or continuous time) Fix a time [imath]T[/imath]. What is: [imath]P[X_t < 0 \text{ for all } t \leq T][/imath]? In words, what's the probability the random walk from [imath]0[/imath] to [imath]T[/imath] has never been above [imath]0[/imath]? Thank you. |
1402679 | Proving that all integers are even or odd
I know that [imath]\mathbb{Z}[/imath] is a group under addition with a multiplication defined. I have just the definition of even and odd integers: [imath]n[/imath] is even if [imath]n = 2k[/imath] for some integer [imath]k[/imath] and [imath]n[/imath] is odd if [imath]n = 2k+1[/imath] for some integer [imath]k[/imath]. Using just this I am wondering how to prove that all integers are either even or odd. That is, how can I prove that given an integer [imath]n[/imath], [imath]n[/imath] must be even or odd? My problem with this is that it seems so simple. I know that one can divide an integer by [imath]2[/imath] and the remainder will be [imath]0[/imath] or [imath]1[/imath]. Using this, it is clear that the even and odd integers make up everything. But how can I prove it without using this fact about remainders and such? I guess one could also use facts about prime numbers, but I am looking for a proof that just uses the definition of odd and even. | 166756 | Proving that an integer is even if and only if it is not odd
There is this question, but the definition of "even" and "odd" that I am using uses integers instead of just natural numbers; i.e., An integer [imath]n[/imath] is even iff there is some integer [imath]k[/imath] such that [imath]n=2k[/imath]. An integer [imath]n[/imath] is odd iff there is some integer [imath]k[/imath] such that [imath]n=2k+1[/imath]. Here is what I have so far: First we show that an integer [imath]n[/imath] is even or odd. We first use induction on the positive integers. For the base case, [imath]1=2\cdot0+1[/imath] so we are done. Now suppose inductively that [imath]n[/imath] is even or odd. If [imath]n[/imath] is even, then [imath]n=2k[/imath] for some [imath]k[/imath] so that [imath]n+1=2k+1[/imath] (odd). If [imath]n[/imath] is odd, then [imath]n=2k+1[/imath] for some [imath]k[/imath] so that [imath]n+1=2(k+1)[/imath] (even). This closes the induction, so every [imath]n\in\mathbf{Z}^+[/imath] is even or odd. Now we show every [imath]n\in\mathbf{Z}^-[/imath] is even or odd. Let [imath]n\in\mathbf{Z}^-[/imath]. Then [imath]n=-k[/imath] for some [imath]k\in\mathbf{Z}[/imath] (I think this follows immediately from most definitions of the integers.). Suppose [imath]k[/imath] is even. Then [imath]k=2j[/imath] for some [imath]j[/imath] so that [imath]n=-k=-2j=2(-j)[/imath] (even). Now suppose [imath]k[/imath] is odd. Then [imath]k=2j+1[/imath] for some [imath]j[/imath] so that [imath]n=-k=-(2j+1)=-2j-1=-2j-1+1-1=-2j-2+1=2(-j-1)+1[/imath] (odd). For [imath]0[/imath], note that [imath]0=2\cdot0[/imath] (even). Now we show that [imath]n\in\mathbf{Z}[/imath] cannot be both even and odd. Suppose for the sake of contradiction that [imath]n\in\mathbf{Z}[/imath] is both even and odd. Then there are integers [imath]k,j[/imath] such that [imath]n=2k=2j+1[/imath]. This implies that [imath]2(k-j)=1[/imath] (like in the referenced question). So we must show that [imath]1[/imath] cannot be even in order to complete the proof. This is where I am having trouble. I know that if I let [imath]f:\mathbf{Z}\to\mathbf{Z};x\mapsto2x[/imath] be a function, then [imath]f[/imath] is increasing so since [imath]f(0)=0[/imath] and [imath]f(1)=2[/imath] and [imath]0<1<2[/imath], there is no integer [imath]m[/imath] such that [imath]f(m)=1[/imath]. But this seems complicated so I was wondering if there was an easier way to do this. So my real question is: how can I show that [imath]1[/imath] is not even? (This is not homework.) |
1403445 | Prove that [imath]\int_0^{a}{\int_x^a{t^{-1}f(t)dt}} = \int_0^a{f(x)dx}[/imath]
I got stuck on this problem from Real Analysis by Folland. Can anybody give me any hints on how to solve this? If [imath]f[/imath] is Lebesgue integrable on [imath](0, a)[/imath] and [imath] g(x) = \int_x^a{t^{-1}f(t)dt} [/imath] then [imath]g[/imath] is integrable on [imath](0, a)[/imath] and [imath] \int_0^a{g(x)dx} = \int_0^a{f(x)dx} [/imath] | 637155 | How do we prove [imath]\int_I\int_x^1\frac{1}{t}f(t)\text{ dt}\text{ dx}=\int_If(x)\text{ dx}[/imath]
Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be Borel-measurable and Lebesgue-integrable over [imath]I:=(0,1)[/imath]. Further, let [imath]\;\;\;\;\;\;\;\;\;\;g : I\to \mathbb{R}\;,\;\;\; \displaystyle x \mapsto\int_x^1\frac{1}{t}f(t)\text{ dt}[/imath] I want to show that [imath]g[/imath] is integrable over [imath]I[/imath] and that it holds [imath]\;\;\;\;\;\;\;\;\;\;\displaystyle\int_Ig(x)\text{ dx}=\int_If(x)\text{ dx}[/imath] Proof: [imath]\;\;\;[/imath] Please note that [imath]g[/imath] is well-defined, because [imath]f[/imath] is Lebesgue-integrable over [imath]I[/imath]. So, by Fubini's theorem [imath]g[/imath] is integrable and it holds: [imath]\;\;\;\;\;\;\;\;\;\;\displaystyle\int_Ig(x)\text{ dx}=\displaystyle\int_I\int_x^1\frac{1}{t}f(t)\text{ dt}\text{ dx}=\displaystyle\int_x^1 \int_I\frac{1}{t}f(t)\text{ dx}\text{ dt}=\displaystyle\int_x^1 \frac{1}{t}f(t)\text{ dt}=\cdots[/imath] How do we integrate [imath]\int_x^1 \frac{1}{t}f(t)\text{ dt}?[/imath] |
95702 | A condition for [imath]\hat f[/imath] to be integrable
Let [imath]f \in L^1 (\mathbb R^n)[/imath]. Suppose that [imath]f[/imath] is continuous at zero and that the fourier transform [imath]\hat f[/imath] of [imath]f[/imath] is non-negative. Does this imply that [imath]\hat f \in L^1[/imath] (and hence, by the inversion theorem, that [imath]f[/imath] is continuous)? If so, how could I go about proving it? | 193114 | A positive "Fourier transform" is integrable
Let [imath]f\in L^1_\mathbb C(\mathbb R^n)[/imath]. I once read, in one of my old exam, that if [imath]\hat{f}(\mathbb R^n)\subset\mathbb R_+[/imath], then [imath]\hat{f}\in L^1(\mathbb R^n)[/imath]. As far as I remember, the professor gave an identity using mollifiers but I'm not 100% sure. Any hints on how to prove that ? Thanks! |
1404602 | Upper and lower bounds for [imath]S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.[/imath]
For a positive integer [imath]n[/imath] let [imath]S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.[/imath] Then which of the following are true. (a) [imath]S(100)\leq 100[/imath]. (b) [imath]S(100)>100[/imath]. (c) [imath]S(200)\leq 100[/imath]. (d) [imath]S(200)>100[/imath]. My attempt For the upper bound [imath]\begin{align} S(n) &= 1 + \left( \frac 12 + \frac 13 \right) + \left( \frac 14 + \frac 15 + \frac 16 + \frac 17 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}+1} + \cdots + \frac 1{2^n-1} \right) \\ &< 1 + \left( \frac 12 + \frac 12 \right) + \left( \frac 14 + \frac 14 + \frac 14 + \frac 14 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}} + \cdots + \frac 1{2^{n-1}} \right) \\ &= \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} \\ &= n. \end{align}[/imath] So we get [imath]S(n) < n[/imath] (for [imath]n > 1[/imath]), and in particular [imath]S(100) < 100[/imath]. Now I did not understand how to calculate a lower bound, or if there is any other method by which we can solve this. | 1366267 | Prove this inequality: [imath]\frac n2 \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac1{2^n - 1} \le n[/imath]
[imath]\dfrac{n}{2} \le \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n - 1} \le n [/imath] I've Tried for hours but didn't got any striking idea. I don't have any efforts to show rather than induction. Please Try If you 've non-inductive proof. |
1406209 | Number theory: Is this argument correct and cube question
First of all, is this argument correct? Suppose [imath]m < n[/imath] are integers. Then for every [imath]k \in \mathbb{N}[/imath] [imath]m + k < n + k.[/imath] What I did: Suppose that [imath]m + k \ge n + k[/imath], the by the cancelation law [imath]m\ge n[/imath]. What gives us a contradiction. Is this right? Another problem is, I have to show that [imath]2^n + 1[/imath] is not a cube for every [imath]n\ge 0[/imath]. I think that I have to use induction, but I don't know how. Thanks! | 947306 | Prove that [imath]2^n +1[/imath] in never a perfect cube
Prove that [imath]2^n +1[/imath] in never a perfect cube I've been thinking about this problem, but I don't know how to do it. I know that if [imath]m^3=2^n+1[/imath], then [imath]m[/imath] should be an odd number, but I 'm not able to get to a contradiction. |
1406639 | matrix multiplication questions
[imath]A[/imath] and [imath]B[/imath] are two matrices, when is [imath](A-B)(A+B)=A^2 - B^2[/imath] | 170241 | When is matrix multiplication commutative?
I know that matrix multiplication in general is not commutative. So, in general: [imath]A, B \in \mathbb{R}^{n \times n}: A \cdot B \neq B \cdot A[/imath] But for some matrices, this equations holds, e.g. A = Identity or A = Null-matrix [imath]\forall B \in \mathbb{R}^{n \times n}[/imath]. I think I remember that a group of special matrices (was it [imath]O(n)[/imath], the group of orthogonal matrices?) exist, for which matrix multiplication is commutative. For which matrices [imath]A, B \in \mathbb{R}^{n \times n}[/imath] is [imath]A\cdot B = B \cdot A[/imath]? |
1407003 | Sum of [imath]\sum_{k=1}^n\sin{k\theta}[/imath]
I have to calculate the sum of this series [imath] \sum_{k=1}^n\sin{k\theta} [/imath] I tried solving it like this [imath] \sum_{k=1}^n\sin{k\theta} = \operatorname{Im}\sum_{k=1}^ne^{(i\theta)k} [/imath] I recognized it as a geometric sum, but to get the sum I first had to check [imath] \left|e^{i\theta}\right| = 1 [/imath] which means that the geometric sum is divergent. Apparently that is not the right answer, but I can't find the spot where I made the mistake. Could you help me finding it? | 2159092 | Proof for [imath]z\neq 0[/imath] , z is any complex number?
[imath]\sum _{k=1}^{n}\mathrm{sin}kz=\frac{\mathrm{sin}\frac{n+1}{{2}}\cdot \mathrm{sin}\frac{nz}{2}}{\mathrm{sin}\frac{z}{2}}[/imath] Proof for [imath]z\neq 0[/imath] |
1407222 | [imath]G[/imath] of finite order [imath]2p[/imath] ([imath]p[/imath] is prime). Prove that [imath]G[/imath] abelian.
I have a group [imath]G[/imath] of order [imath]2p[/imath], where [imath]p>2[/imath] and prime. The additional thing that I also know, that [imath]\exists a\in Z(G)\mid O(a)=2[/imath]. I need to prove that G is abelian. But first, before that, tell me please, isn't here (in my next direction of thinking any objection to the given information). So, I look at two sub-groups of [imath]G[/imath] (the order isnt prime so they exist and their order devides the order of the group) of order [imath]p[/imath], [imath]A,B\le G\mid O(a)=O(b)=p[/imath] They are cyclic because of their order (is prime). And their elements are the total elements in [imath]G[/imath] (because they are sub-groups and there are totally [imath]2p[/imath] elements, and here I look at different sub-groups with [imath]p[/imath] elements each). So left to prove that any [imath]a\in A[/imath] abelian with any [imath]b\in B[/imath]. Now I think about the second data that given. The center of [imath]G[/imath] is in [imath]G[/imath], so there is [imath]g\in G\mid g^2=e[/imath]. But must [imath]g\in A[/imath] or [imath]g\in B[/imath], but their order is [imath]p[/imath] and [imath]2[/imath] does not divide any of them. Isn't that an objection ? Otherwise, which of the step is wrong, what can't I conclude from what I wrote? | 1403792 | A group of order 2p (p prime) and other conditions - prove abelian.
I have G where [imath]|G|=2p[/imath] ; p is prime. [imath]\exists a\in Z\left(G\right);\:a^2=e[/imath]. I need to prove that G is abelian. Now, let's translate it into math. To prove that G is abelian, is in other words ti prove that [imath]Z\left(G\right)=G[/imath], but [imath]Z\left(G\right)\in G[/imath], so if I prove that \left|Z\left(G\right)\right|=\left|G\right|, I've done. Now, by lagrange's theorem, [imath]\left|Z\left(G\right)\right|[/imath] can be 2, p or 2p (which I want to prove). It cant be p, because there is [imath]a\in Z\left(G\right)[/imath], which means [imath]a\in G[/imath] so [imath]a^{2}=e[/imath], therefore [imath]2 : |Z(G)|[/imath]. So I'm left with 2 and 2p. But as I see, theoretical the center can be only 2 elements: [imath]\left\{e,a\right\}[/imath], because all the data I have that there is there some [imath]a[/imath], but [imath]a^{-1}=a[/imath] because [imath]a^{2}=e[/imath], so I have for sure 2 elements. I dont see how can i prove that there is at least one another abelian element in Z(G), because the existence of two fits perfect eather, as I can see. Can somebody give me a hint ? Can this exercize be solved like that or Am I in the wrong direction from the first place ? Can somebody give me a hint? |
1407649 | If [imath]\text{gcd}(a,561)=1[/imath], then [imath]a^{560}=1\mod 561[/imath]
We have the factorization [imath]561=3*11*17[/imath]. Because [imath]\text{gcd}(a,561)=1[/imath], there are integers [imath]x,y[/imath] such that [imath]ax+561y=1[/imath]. So [imath]ax=1\mod 561[/imath]. Since the gcd is [imath]1[/imath], we have [imath]a\not \in 3\mathbb{Z}, 11\mathbb{Z}, 17\mathbb{Z}[/imath]. Therefore, [imath]a^{560}\not \in 3\mathbb{Z}, 11\mathbb{Z}, 17\mathbb{Z}[/imath]. So [imath]b^{560}=1\mod 561[/imath]. Is my reasoning correct? I have seen a problem that was similar to this that used Fermat's Little Theorem, but that can only be used when the number we mod out by is prime. | 22461 | Verifying Carmichael numbers
I'm trying to understand a solution I was given in a tutorial regarding a problem with Carmichael numbers and I was wondering if you guys can help clarify things: A composite number [imath]m[/imath] is called a Carmichael number if the congruence [imath]a^{m-1} \equiv 1 \pmod{m}[/imath] is true for every number a with [imath]\gcd(a,m) = 1[/imath]. Verify that [imath]m = 561 = 3 \times 11 \times 17[/imath] is a Carmichael number. Solution given: Apply Fermat's Little Theorem to each prime divisor of [imath]m[/imath]: \begin{align*} a^2 &\equiv 1 \pmod{3}\\ a^{10} &\equiv 1 \pmod{11}\\ a^{16} &\equiv 1 \pmod{17} \end{align*} This somehow then implies that [imath]a^{80} \equiv 1 \pmod{561}[/imath] then accordingly [imath]a^{560} \equiv 1 \pmod{561}[/imath]. I am lost as to how the 3 congruences imply [imath]a^{80} \equiv 1 \pmod{561}[/imath] ([imath]80 = \mathrm{LCM}(2,10,16)[/imath]). Can somebody clarify this for me? Thanks! |
1407169 | Similar Matrices and their Jordan Canonical Forms
Let [imath]A[/imath] and [imath]B[/imath] be two matrices in [imath]M_n[/imath]. Is the following ture: [imath]A[/imath] and [imath]B[/imath] are similar [imath]\iff[/imath] [imath]A[/imath] and [imath]B[/imath] have the same jordan canonical form. Could someone explain? | 1017581 | Are two matrices similar iff they have the same Jordan Canonical form?
Are two matrices similar if and only if they have the same Jordan Canonical form? Does the Jordan form have to have ordered eigenvalues? For example, if [imath]\lambda_1[/imath] and [imath]\lambda_2[/imath] are eigenvalues of [imath]A[/imath], are [imath]\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}[/imath] and [imath]\begin{pmatrix}\lambda_2&0\\0&\lambda_1\end{pmatrix}[/imath] both Jordan forms of [imath]A[/imath]? |
739145 | How to solve [imath](1-\tan^2x)\sec^2x+2^{\tan^2x}=0[/imath]
I am stuck the following problem : Solve the following problem: [imath](1-\tan^2x)\sec^2x+2^{\tan^2x}=0[/imath] Taking [imath]p=\tan^2x,[/imath] we get [imath](1-p^2)+2^p=0[/imath]. Now,I do not know how to go from here. Simple observation says [imath]x=2n\pi \pm \frac{\pi}{3}[/imath],satisfies the equation and so it will be the solution but I do not know how to get there. Can someone help? | 1559146 | Solve for x : If [imath]( 1 + \tan x ) ( 1 - \tan x ) ( \sec^ 2 x ) + 2 ^ { \tan^ 2 x } = 0[/imath]
If [imath]( 1 + \tan x ) ( 1 - \tan x ) ( \sec^ 2 x ) + 2 ^ { \tan^ 2 x } = 0[/imath] , then find the number of solutions in the interval [imath]( -\pi/2 , \pi/2 )[/imath]. I did solve , but I get 2 solutions instead of 4 as stated in answer. My solution : [imath]( 1 + \tan x ) ( 1 - \tan x ) ( \sec^2 x ) + 2 ^ { \tan^ 2 x } = 0 \\ ( 1 - \tan^2 x ) ( 1 + \tan^2 x ) + 2 ^ { \tan^ 2 x } = 0 \\ 1 - \tan^4 x + 2 ^ { \tan^ 2 x } = 0[/imath] Let [imath]\tan^2 x = z[/imath] [imath]\implies 2 ^ z = z^2 - 1[/imath] By hit and trial ( putting random integers ) I got [imath]z = 3[/imath] so then [imath]x = n\pi \pm \pi/3[/imath]. In [imath]( -\pi/2 , \pi/2 )[/imath] , I got , [imath]-\pi/3[/imath], [imath]\pi/3[/imath] only i.e. 2 solutions . But the answer in the book says , there are 4 . |
942158 | Necessary and sufficient conditions for when spectral radius equals the largest singular value.
One well known fact about matrix norms is the following: If [imath]\lambda_1\geq \dots\geq \lambda_n[/imath] are eigenvalues of a square matrix [imath]A[/imath], then: [imath]\frac{1}{||A^{-1}||} \leq |\lambda|\leq ||A||[/imath] If we take our matrix norm to be the matrix 2-norm, and recall that the matrix 2 norm gives us the largest singular value, i.e. [imath]||A||_2=\sigma_1[/imath], then the upper bound implies: [imath]|\lambda_1|\leq \sigma_1[/imath] My question is: are there necessary and sufficient conditions for when equality above holds? I vaguely remember something like: [imath]|\lambda_1|=\sigma_1[/imath] iff [imath]\lambda_1[/imath] is non-defective, i.e. its algebraic multiplicity equals its geometric multiplicity. However, I have not been able to find a reference or prove this. Can someone point me to a reference or provide a proof. Thank you in advance for your time. | 650632 | Quick question: matrix with norm equal to spectral radius
For [imath]A\in \mathcal{M}_n(\mathbb{C})[/imath], define: the spectral radius [imath] \rho(A)=\max\{|\lambda|:\lambda \mbox{ is an eigenvalue of } A\} [/imath] and the norm [imath] \|A\|=\max_{|x|=1}|A(x)| [/imath] where |.| is the Euclidean norm on [imath]\mathbb{C}^n[/imath]. Problem: Find all [imath]A\in \mathcal{M}_n(\mathbb{C})[/imath] such that [imath]\rho(A)=\|A\|[/imath]. Thank you very much! |
1407820 | Prove complete metric space for [imath]I=]0,\infty[[/imath] with [imath]d(x,y)=\lvert\ln(x)-\ln(y)\rvert[/imath]
Let [imath]I=]0,\infty[[/imath] equipped with the metric [imath]d(x,y)=\lvert\ln(x)-\ln(y)\rvert[/imath], [imath]\forall x,y \in I[/imath]. Prove that [imath](I,d)[/imath] is complete. Any help, and thanks in advance. | 1371285 | Proving that given metric space is complete: [imath]X := (0,\infty)[/imath] and [imath]d:=|\ln(x)-\ln(y)|[/imath]
Given the metric space [imath](X,d)[/imath] with [imath]X := (0,\infty)[/imath] and [imath]d:=|\ln(x)-\ln(y)|[/imath], how can I show that [imath](X,d)[/imath] is complete? I need to prove that any Cauchy sequence converges, so: If [imath](x_n)[/imath] is a Cauchy sequence in X, then it follows for all [imath]\epsilon \gt 0[/imath] that there exists an [imath]n_0 \in \mathbb{R} : \forall n,m \leq n_0: d(x_n,x_m) \lt \epsilon[/imath]. I couldn't find a direct way to prove this, I guess an indirect approach might go as follows: Let [imath](x_n)[/imath] be a Cauchy sequence in X and assume that it does not converge, then it follows that there exists an [imath]\epsilon > 0[/imath] such that for an arbitrary [imath]x \in X : \forall n \in \mathbb{N}: d(x_n - x) \geq \epsilon[/imath] and this would contradict the fact that [imath](x_n)[/imath] is a cauchy sequence. I'm assuming that this is incorrect since I didn't even use the given metric and this proof would mean that no cauchy sequence converges. So how can I prove that this metric space is complete? And in general is there a way how to approach these completeness proofs? |
1408080 | [imath]A[/imath] is an invertible matrix such that the sum of each row is [imath]1[/imath]
Let [imath]A[/imath] be an [imath]10\times 10[/imath] invertible matrix with real entries such that the sum of each row is [imath]1[/imath]. Then [imath]A[/imath]. The sum of entries of each row of the inverse of [imath]A[/imath] is [imath]1[/imath] [imath]B[/imath]. The sum of entries of each column of the inverse of [imath]A[/imath] is [imath]1[/imath] [imath]C[/imath]. The trace of the inverse of [imath]A[/imath] is non-zero [imath]D[/imath]. None of the above I don't see how to proceed or how to use the sum of the elements of of rows. Please help. I know option [imath]A[/imath] is correct and option [imath]C[/imath] is not. Why is option [imath]B[/imath] not correct? | 1304871 | The inverse of a matrix in which the sum of each row is [imath]1[/imath]
Let [imath]A[/imath] be an invertible [imath]10\times 10[/imath] matrix with real entries such that the sum of each row is [imath]1[/imath]. Then choose the correct option. The sum of the entries of each row of the inverse of [imath]A[/imath] is [imath]1[/imath]. The sum of the entries of each column of the inverse of [imath] A[/imath] is [imath]1[/imath]. The trace of the inverse of [imath]A[/imath] is non-zero. None of the above. If the matrix is given we can find its inverse but how can we find its inverse if the matrix itself not given? Any idea on how to find the answer? |
1408483 | [imath]\forall a[P(a)\implies Q(a)]\wedge \forall a[Q(a)\implies P(a)]\stackrel{?}{\equiv} \forall a[[P(a)\implies Q(a)]\wedge [Q(a)\implies P(a)]][/imath]
I'm reading: Devlin's Joy of Sets. He gives the definition of the axiom of extensionality: The definition of subset: And then there is this exercise: Rewriting it, I'd have: \begin{eqnarray*} {(x=y)}& \iff &{(x\subseteq y)\wedge (y\subseteq x)} \\ {}&\iff &{\forall a[a\in x\implies a\in y] \wedge \forall a[a\in y\implies a\in x]} \end{eqnarray*} Are the following two expressions equivalent: [imath]\forall a[a\in x\implies a\in y] \wedge \forall a[a\in y\implies a\in x] \\ \stackrel{?}{\equiv}\\ \forall a([a\in x\implies a\in y] \wedge [a\in y\implies a\in x])[/imath] Intuitively, I guess they are. But I'm not sure. | 268750 | Distributivity of quantifiers over logical connectives
I'm given the following expresssion: [imath] \forall a [\phi(a) \to \psi(a)] \wedge \forall a [\psi(a) \to \phi(a)] [/imath] that I wish to logically reduce to: [imath] \forall a [\phi(a) \leftrightarrow \psi(a)] [/imath] The only area I'm uncertain about is showing that universal quantification is distributive over conjunction, as trivial as it seems. Rather, I'm not sure where to find theorems or lemmas related to the topic. The set theory text from which this comes provides no insight and the introductory logic text I own also doesn't touch on the matter, at least in any great detail. |
496595 | Let [imath]\lim_{n\to\infty}a_n[/imath] = a, [imath]\lim_{n\to\infty}b_n[/imath] = b. We need to find [imath]\lim_{n\to\infty}\frac{a_nb_1+a_{n-1}b_2+...+a_1b_1}{n}[/imath].
Let [imath]\lim_{n\to\infty}a_n[/imath] = a, [imath]\lim_{n\to\infty}b_n[/imath] = b. We need to find [imath]\lim_{n\to\infty}\frac{a_nb_1+a_{n-1}b_2+...+a_1b_1}{n}[/imath]. As far as I get it, the answer is [imath]ab[/imath]. I've tried this to prove it: as both [imath]a_n[/imath] and [imath]b_n[/imath] have a limit, they also both have a supremum. Let [imath]C_1[/imath] and [imath]C_2[/imath] be the supremua. Then by the properties of supremum for any given [imath]е[/imath] there are [imath]k, l[/imath] such that [imath]\frac{a_nb_1+a_{n-1}b_2+...+a_1b_n}{n}<C_1C_2<(a_k+e)(b_l+e)[/imath] And I feel like there is some tiny step from here to the solution as the above is more or less [imath]ab[/imath], but I just can't get there. Would you mind giving some hints? | 725914 | Find the limit [imath]\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}[/imath]
When [imath]\lim_{n\to\infty} x_n = a[/imath], and [imath]\lim_{n\to\infty} y_n = b[/imath], find the limit, [imath]\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}.[/imath] Thank you for your help in advance. |
1406388 | What is [imath]\varphi(0)[/imath]?
[imath]\varphi[/imath] is Euler's totient function. My question is: When/if [imath]\varphi[/imath] is defined at [imath]0[/imath], what is it usually defined as? Is there a "most natural" or more commonly accepted definition of [imath]\varphi(0)[/imath]? This is a soft question, because it's of course a matter of convention. Here are three possible definitions, with some justification. Definition 1: [imath]\boldsymbol{\varphi(0) = 2}[/imath]. Note that for [imath]n \ge 1[/imath], [imath]\varphi(n) = \left| \left( \mathbb{Z} / n \mathbb{Z} \right)^* \right|[/imath], the number of units mod [imath]n[/imath]. Plugging in [imath]n = 0[/imath], we get [imath] \varphi(0) = \left| \left( \mathbb{Z} / 0 \mathbb{Z} \right)^* \right| = \left| \left( \mathbb{Z} \right)^* \right| = \left| \{-1, 1\} \right| = 2. [/imath] Definition 2: [imath]\boldsymbol{\varphi(0) = 0}[/imath]. If [imath]a \mid b[/imath], then [imath]\varphi(a) \mid \varphi(b)[/imath]. To preserve this property for [imath]b = 0[/imath], we need that [imath]\varphi(a) \mid \varphi(0)[/imath] for all [imath]a[/imath], implying [imath]\varphi(0) = 0[/imath]. WolframAlpha returns [imath]\varphi(0) = 0[/imath]. However, the Wolfram Math World page explains: By convention, [imath]\phi(0)=1[/imath], although the Wolfram Language defines EulerPhi[0] equal to 0 for consistency with its FactorInteger[0] command. which gives us Definition 3: [imath]\boldsymbol{\varphi(0) = 1}[/imath]. Does anyone know the reason for this convention? More observations Any choice of [imath]\varphi(0)[/imath] is consistent with the multiplicativity of [imath]\varphi[/imath]. [imath]\varphi(mn) = \varphi(m) \varphi(n) \frac{d}{\varphi(d)}[/imath], where [imath]d = \gcd(m,n)[/imath], implies [imath]\varphi(0) = 0[/imath]. This supports definition 2. If [imath]\varphi(n) = \sum_{ab = n} a \mu(b)[/imath], then note that when [imath]ab = 0[/imath], [imath]a = 0[/imath] or [imath]b = 0[/imath], and since [imath]\mu(0) = 0[/imath], [imath]a \mu(b) = 0[/imath]. So this is [imath]\sum 0 = 0[/imath], again agreeing with definition 2. | 1226241 | What is the Euler Totient of Zero?
Wolfram MathWorld defines the Euler Totient function as follows: The totient function phi(n), also called Euler's totient function, is defined as the number of positive integers <=n that are relatively prime to (i.e., do not contain any factor in common with) n, where 1 is counted as being relatively prime to all numbers By this definition would [imath]\phi(0)=0[/imath]? |
1409439 | Is the set containing just zero a mathematical field?
Consider the set [imath]\left\lbrace0\right\rbrace[/imath] together with the usual operations of addition and multiplication. Is this set together with these operations a field? I know that one of the requirements of a field is that for every nonzero element [imath]\alpha[/imath] of the set, there exists a unique element [imath]\beta[/imath] such that [imath]\alpha\cdot\beta[/imath] is equal to the multiplicative identity (in the case of my set the multiplicative identity is zero). Since there are no nonzero elements in my set, this should be true, right? | 634783 | What's the rationale for requiring that a field be a [imath]\boldsymbol{non}[/imath]-[imath]\boldsymbol{trivial}[/imath] ring?
The title pretty much says it all. Of course, one answer (IMO unsatisfactory) to such questions goes something like "a definition is a definition, period." In my experience, mathematical definitions are rarely completely arbitrary. Therefore I figure there must be a good reason for insisting that a field be a non-trivial ring (among other things), but it's not obvious to me. I realize that a trivial ring would make for a very boring field, but it is also a rather boring ring, and yet it is not disallowed as such. EDIT: I found a hint of a rationale in the statement that "the zero ring ... does not behave like a finite field," but I could not find in the source for this assertion exactly how the zero ring fails to behave like a finite field. |
1408990 | Prove that [imath]\sqrt{3}[/imath] is not a rational number
There is a similar question however that question asks why [imath]3 |p^2[/imath]. Here the question is about [imath] 3 | p^2 \rightarrow 3 | p[/imath]. It is a simple exercise (1.2.1) from Abbot's "Understanding Analysis". [imath]\nexists p,q \in \mathbb{N} : \left(\dfrac{p}{q}\right)^2 = 3[/imath] [imath]p[/imath] and [imath]q[/imath] have no common factors, otherwise they would cancel each other out. Contradiction: [imath]p^2 = 3q^2[/imath] The troublesome part for me is From this, we can see that [imath]p^2[/imath] is a multiple of 3 and hence [imath]p[/imath] must also be a multiple of 3. Why is that? The proof goes on with [imath]p = 3 r[/imath] [imath]q^2 = 3r^2 \rightarrow q = 3 \lambda[/imath]. It concluses now that [imath]q[/imath] and [imath]p[/imath] have common factors, which invalidates the original statement. How can I prove [imath]p^2 = 3q^2 \rightarrow p = 3 r[/imath]? If this is not true, then I don't see how the contradiction proves the statement, since it only invalidates it for such [imath]p,q[/imath] that have common factors. What if [imath]p^2 = 3q^2[/imath] and [imath]p \ne 3 r[/imath]? | 220663 | Please explain this step in proving the square root of 3 is irrational
Assume that [imath]3 = \frac{p^2}{q^2}[/imath] So, [imath] 3 q^2 = p^2[/imath] So [imath]p^2[/imath] is divisible by [imath]3[/imath]. How we can conclude this? |
1408642 | Combinatorial Proof for Binomial Identity: [imath]\sum_{k = 0}^n \binom{k}{p} = \binom{n+1}{p+1}[/imath]
I am studying combinatorics and I came across the identity [imath]\sum\limits_{k=0}^n \binom kp =\binom {n+1}{p+1}.[/imath] I have read the algebraic proof and it does not appeal to me. Is there an elegant counting trick we can use to arrive at this identity? | 1522826 | Prove combinatorial Identity using a combinatorial argument.
I have proved by induction on k that : [imath] \binom{n}{n} + \binom{n+1}{n} + \binom{n+2}{n}+ ... + \binom{n+k}{n} = \binom{n+k+1}{n+1} [/imath] But now, I need to prove it using a combinatorial argument. Could someone give me a hint ? :/ Thank you ! |
1409591 | [imath]S_n\to S[/imath] implies [imath]\sigma_n\to S[/imath]
Let [imath](a_n)_{n\ge0}[/imath] be a sequence of complex numbers and [imath]S_n=\sum_{k=0}^na_k[/imath], set [imath]\sigma_n:=\frac{S_0+S_1+\dots+S_n}{n+1}[/imath], if [imath]\lim_{n\to\infty}S_n=S[/imath] then show that [imath]\lim_{n\to\infty}\sigma_n=S[/imath], prove also that one cannot deduce from the convergence of [imath]\sigma_n[/imath] the convergence of [imath]S_n[/imath] If [imath]S_n[/imath] is convergent then there is an [imath]N[/imath] s.t. for [imath]n>N[/imath], we have [imath]|S_n|<S+1[/imath] [imath]|\sigma_n|\le\frac{S_0+\dots+S_N}{n+1}+\frac{(n-N)(S+1)}{n+1}[/imath] but how can I show that the limit is the same for [imath]\sigma_n[/imath] ? the convergence is proven, I think if [imath]a_k=(-1)^k[/imath] then [imath]\sigma_n\to0[/imath] but [imath]S_n[/imath] diverges | 1401736 | If [imath]\lim_{n\to\infty}a_{n}=l[/imath], Then prove that [imath]\lim_{n\to\infty}\frac{a_{1}+a_2+\cdot..+a_n}{n}=l[/imath]
Given [imath]a_n[/imath] be a sequence and IF [imath]\lim_{n\to\infty}a_{n}=l[/imath], Then prove that [imath]\lim_{n\to\infty}\frac{a_{1}+a_2+\cdot..+a_n}{n}=l[/imath] I do not know how to do this. Can someone help me with this? Thanks ATTEMPT |
1409997 | Value of [imath]x \log x[/imath] at [imath]x=0[/imath]
What is the value of [imath]f(x) = x \ln (x)[/imath] at [imath]x= 0[/imath]? Is it [imath]0[/imath] or indeterminate? | 470952 | Limit of [imath]x \log x[/imath] as [imath]x[/imath] tends to [imath]0^+[/imath]
Why is the limit of [imath]x \log x[/imath] as [imath]x[/imath] tends to [imath]0^+[/imath], [imath]0[/imath]? The limit of [imath]x[/imath] as [imath]x[/imath] tends to [imath]0[/imath] is [imath]0[/imath]. The limit of [imath]\log x[/imath] as [imath]x[/imath] tends to [imath]0^+[/imath] is [imath]-\infty[/imath]. The limit of products is the product of each limit, provided each limit exists. Therefore, the limit of [imath]x \log x[/imath] as [imath]x[/imath] tends to [imath]0^+[/imath] should be [imath]0 \times (-\infty)[/imath], which is undefined and not [imath]0[/imath]. |
1408038 | Reference request about Thm which use Transversality to compute Homotopy Groups
I'm following the following notes, and my attention was caught by Theorem [imath]1.1.4[/imath]. I am unable to find any reference of the proof. Could you suggest me some books in which there is a proof of this theorem? | 1344027 | Is simply connectedness preserved after deleting a high codimension set
Suppose [imath]X[/imath] is a complex manifold of complex dimension [imath]n[/imath], [imath]Z[/imath] is a subvariety of complex codimension at least [imath]2[/imath]. Suppose [imath]\pi_1(X)=0[/imath], do we have [imath]\pi_1(X-Z)=0[/imath]? Do we have [imath]\pi_1(X-Z)=\pi_1(X)[/imath] in general? |
1409965 | Limits of cosine and sine
When [imath]\theta[/imath] is very small why [imath]\sin \theta[/imath] is similar to [imath]\theta[/imath] and [imath]\cos\theta[/imath] similar to [imath]1[/imath]? Is it related to limits or we can prove it simply by using diagrams? | 1234218 | Approximation of the Sine function near [imath]0[/imath]
What is the reason that for [imath]x<0.5[/imath], [imath]\sin(x)\approx x[/imath]? Are there more known properties of these kind for other trigonometry functions? |
1410626 | Find a [imath]4\times 3[/imath] matrix to make a [imath]3 \times 4[/imath] matrix invertible
The matrix [imath]P[/imath] is 3x4 given. Is there a 4x3 matrix [imath]Q[/imath] such that the product of [imath]QP[/imath] is invertible? If this can't happen can someone explain why?Thanks. Matrix P | 1410599 | Can the product of an [imath]4\times 3[/imath] matrix and a [imath]3\times 4[/imath] matrix be invertible?
I want to find the inverse of the product of [imath]2[/imath] non-square matrices, but is this even possible? |
962412 | when [imath]{\rm gcd} (a,b)=1[/imath], what is [imath]{\rm gcd} (a+b , a^2+b^2)[/imath]?
I want to prove above statement "what is [imath]{\rm gcd} (a+b , a^2+b^2)[/imath] when [imath]{\rm gcd}(a,b) = 1[/imath]" I've seen some proofs of it, but i couldn't find useful one. here is one of the proof of it. some proofs I want prove it using method like 2nd answer. (user 9413) but i can't understand why "if [imath]d | 2ab[/imath] then [imath]d = 1[/imath] or [imath]2[/imath] or [imath]d | a[/imath] or [imath]d|b[/imath]" (isn't Euclid's lemma which only holds when [imath]d[/imath] is prime ?) please show me proof step by step | 153125 | Prove [imath]\gcd(a+b,a^2+b^2)[/imath] is [imath]1[/imath] or [imath]2[/imath] if [imath]\gcd(a,b) = 1[/imath]
Assuming that [imath]\gcd(a,b) = 1[/imath], prove that [imath]\gcd(a+b,a^2+b^2) = 1[/imath] or [imath]2[/imath]. I tried this problem and ended up with [imath]d\mid 2a^2,\quad d\mid 2b^2[/imath] where [imath]d = \gcd(a+b,a^2+b^2)[/imath], but then I am stuck; by these two conclusions how can I conclude [imath]d=1[/imath] or [imath]2[/imath]? And also is there any other way of proving this result? |
1411380 | Where's the foolish part ? Prove that 0/0 = 2
I've been across this on the web: \begin{align} \frac{0}{0} & = \frac{100-100}{100-100} \\ & = \frac{10^2-10^2}{10(10-10)} \\ & = \frac{(10-10)(10+10)}{(10)(10-10)} \\ & = \frac{(10+10)}{10} \\ & = \frac{2}{1} \\ & = 2 \end{align} Of course this is wrong but I can't tell where exactly since there is no obvious (to me at least) atrocity... Is this in the first line [imath] \frac{0}{0} = \frac{100-100}{100-100}[/imath] ? | 1376989 | Does square difference prove that 1 = 2?
I was mathematically shown 1 = 2 by a function that states the following [imath]x^2-x^2 = x^2-x^2 [/imath] [imath]x(x-x)=(x-x)(x+x)[/imath] dividing by [imath](x-x)[/imath] we get... [imath]x=x+x[/imath] [imath] x=2x[/imath] [imath]1=2[/imath] I can see that mathematically he was right, but for sure that I was missing something as it doesn't make mathematical sense |
1411331 | If [imath](V,\|\cdot \|)[/imath] is a finite dimensional space, then all norms are equivalent.
I want to show that if [imath](V,\|\cdot \|)[/imath] is a finite dimensional space, then all norms are equivalent. I have shown that if [imath]\dim V=m[/imath] all norms [imath]\|x\|_p=\sqrt[p]{x_1^p+...+x_m^p}[/imath] are equivalent, but how to show that for an unspecified norm [imath]\mathcal N[/imath] on [imath]V[/imath], there is a [imath]p[/imath] and two constante [imath]A,B>0[/imath] such that [imath]A\|x\|_p\leq \mathcal N(x)\leq B\|x\|_p[/imath] for all [imath]x\in V[/imath] ? | 754159 | Equivalence of norms proof
This question is from a set of optional, much harder problems from my first year analysis course, but the subject material is norms on [imath]\mathbb R^K[/imath]. (c) Show that there exists a constant [imath]C > 0[/imath] such that for all [imath]\mathbf x ∈ \mathbb R^k[/imath], [imath]||\mathbf x||\le C||\mathbf x||_\infty[/imath] Hint: Use the standard unit vectors to write [imath]\mathbf x = x_1\mathbf e_1 +. . .+x_n \mathbf e_n[/imath], and use the triangle inequality. (d) Show that there also exists a constant [imath]c > 0[/imath] such that for all [imath]x ∈ \mathbb R^k[/imath], [imath]||\mathbf x||_\infty \le c||\mathbf x||[/imath] Hint: If it isn’t true, then you can find a sequence [imath]⟨\mathbf x_n⟩[/imath] such that [imath]||\mathbf x_n||[/imath] is bounded but [imath]c_n := ||x_n||_∞ → ∞[/imath]. (Why?) Think about the sequence [imath]⟨\mathbf x_n/c_n⟩[/imath]. The Bolzano-Weierstrass theorem may be of assistance. Note that the unspecified norm represents any norm for [imath]\mathbb R^k[/imath]. I've done part c), but am stuck on part d). I've had a look for proofs on other questions on MSE, but none of them seem relevant to the hint given, and frequently the terminology is too advanced for me. I tried considering the case if it isn't true, as suggested, but I can't see how to bound any sequence using that. Any help would be much appreciated. |
992722 | Show that [imath]|I_m-AB|=|I_n-BA|[/imath]
Let [imath]A[/imath] be an [imath]m\times n[/imath] matrix and [imath]B[/imath] an [imath]n\times m[/imath] matrix. Show that [imath] |I_m-AB|=|I_n-BA|. [/imath] I don't know where to start. | 17831 | Sylvester's determinant identity
Sylvester's determinant identity states that if [imath]A[/imath] and [imath]B[/imath] are matrices of sizes [imath]m\times n[/imath] and [imath]n\times m[/imath], then [imath] \det(I_m+AB) = \det(I_n+BA)[/imath] where [imath]I_m[/imath] and [imath]I_n[/imath] denote the [imath]m \times m[/imath] and [imath]n \times n[/imath] identity matrices, respectively. Could you sketch a proof for me, or point to an accessible reference? |
1412496 | Frattini subgroups and nilpotent groups: bijection?
I have been proved that the Frattini subgroup of a finite group is nilpotent. Now I am wondering: is the converse true? I mean, if [imath]G[/imath] is a finite nilpotent group, is there always a finite group [imath]H:\operatorname{Frat}(H)\cong G[/imath]? | 717619 | Does every finite nilpotent group occur as a Frattini subgroup?
The Frattini subgroup of a finite group is the intersection of its maximal subgroups. It is well-known that the Frattini subgroup of a finite group is nilpotent. My question is whether a kind of converse is true: Does every finite nilpotent group occur (up to isomorphism, of course) as the Frattini subgroup of some finite group? The problem reduces immediately to the case of non-abelian groups of prime power order, since the Frattini subgroup of a direct product of finite groups is the direct product of the Frattini subgroups of the factors, and it is not hard to show that any finite abelian group is the Frattini subgroup of a finite (abelian) group. I had hoped this might kick off an induction, but I've not been able to do anything with it. The first interesting case seems to be the non-abelian groups of order eight. A computer search of groups up to order [imath]2000[/imath] (not including groups whose order is a multiple of [imath]512[/imath], however, which could not complete on hardware available to me), turned up nothing for these two groups. I am not certain what I would expect the answer to this question to be! |
1412468 | Find [imath]\operatorname{min}(\omega\sqrt{2},\mathbb{Q})[/imath]
Let [imath]\omega[/imath] be a primitive third root of unity with [imath]K=\mathbb{Q}(\omega,\sqrt{2})[/imath]. Find [imath]\operatorname{min}(\omega\sqrt{2},\mathbb{Q})[/imath] Could anyone tell me how to find this? and generally which is the way to find similar problems with minimals? | 1391166 | Finding Galois group of [imath]K=\Bbb{Q}(\omega,\sqrt2)[/imath], showing that [imath]K=\Bbb{Q}(\omega\sqrt2)[/imath], and finding [imath]\operatorname{min}(\omega\sqrt2,\Bbb{Q})[/imath]
Let [imath]\omega[/imath] be a primitive third root of unity, and [imath]K=\mathbb{Q}(\omega,\sqrt{2})[/imath]. I found that the degree of [imath][K:\mathbb{Q}]=6[/imath]. How can I find the Galois group [imath]\operatorname{Gal}(K/\mathbb{Q})[/imath]? After this, could anyone give me any hints for showing that [imath]K=\mathbb{Q}(\omega\sqrt{2})[/imath] and finding [imath]\operatorname{min}(\omega\sqrt{2},\mathbb{Q})[/imath]? |
1412765 | Can Two Different Polynomials Agree on an open interval?
Question: For a high degree polynomial [imath]P_1[/imath] , can we have another polynomial [imath]P_2[/imath] that is a part of [imath]P_1[/imath] (or they agree on open interval)? TBN: This question is partially answered in :Overlapping Polynomials however I need to know if it is true for all polynomials of any degree. | 1412610 | Overlapping Polynomials
This question is related to this:https://math.stackexchange.com/questions/1412564/interpolating-polynomial-its-root We have [imath]P_3=P_2\cdot P_1[/imath],for three non-zero polynomials. The degree of each polynomial is at least 1. Question: Does [imath]P_1[/imath] overlap [imath]P_3[/imath] over more than one point? By overlap I do not mean intersect. Instead, I mean a part of the polynomials can lay on each other over more than one point. edit: [imath]P_1\neq P_2[/imath] |
1412790 | Result of [imath]\int \limits_{-\infty}^{+\infty}x^2\times\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x[/imath]
I would like to read a very thorough and explained calculation process for a couple of integrals. For the life of me I just can't figure out the result on my own, and no resource on the web were able to help me. First, a small question: Is a primitive of [imath]\exp\left(\frac{-x^2}{2}\right)[/imath], [imath]-x\times\exp\left(\frac{-x^2}{2}\right)[/imath] ? If so, what's the derivative ? Now the real question. I would like to find the result of: [imath]\int \limits_{-\infty}^{+\infty}\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x[/imath] And then, the result of: [imath]\int \limits_{-\infty}^{+\infty}x^2\times\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x[/imath] Supposedly both are equal to [imath]\sqrt{2\pi}[/imath], but there's no way I can get there on my own. | 66084 | Integral: [imath]\int_{-\infty}^{\infty} x^2 e^{-x^2}\mathrm dx[/imath]
I don't know how to evaluate it. I know there is one method using the gamma function. BUT I want to know the solution using a calculus method like polar coordinates. [imath]\int_{-\infty}^\infty x^2 e^{-x^2}\mathrm dx[/imath] I will wait for a solution. Thank you. |
1413147 | How to prove that G is a cyclic Group?
Suppose [imath]G[/imath] is a finite Abelian group and, [imath]\forall\ n\in \mathbb{N} [/imath], there exist at most [imath]n[/imath] elements in [imath]G[/imath] which satisfy [imath]x^n=1[/imath]. Prove [imath]G[/imath] is cyclic. Thanks for your help. | 1398328 | Show that [imath]G[/imath] is cyclic
Here is a problem from Herstein. Let G be a finite abelian group so that the equation [imath]x^n=e[/imath] has at most [imath]n[/imath] solutions in [imath]G[/imath] for every positive integer [imath]n[/imath]. Show that [imath]G[/imath] is cyclic. I will have to basically show that there exists an element [imath]g[/imath] in [imath]G[/imath] having order [imath]o(G)[/imath]. I showed that such an element exists if [imath]o(G)=p_1p_2...p_k[/imath] where [imath]p_i[/imath] are primes. But I could not proceed if [imath]o(G)=p_1^{a_1}p_2^{a_2}[/imath]. I feel that if I can prove this I will be able to extend. |
1413457 | Number of injective maps from one finite set to another
[imath]m\le n[/imath] be natural numbers. What is the number of injective maps from a set of cardinality [imath]m[/imath] to a set of cardinality [imath]n[/imath] [imath]?[/imath] I think it is the number of ways [imath]m[/imath] distinct elements can be chosen from [imath]n[/imath] elements and permuted i.e. [imath]{n!}\over {(n-m)!}[/imath] Is my formula correct? | 243500 | Number of Injective Maps
What is the number of injective maps from a set of cardinality [imath]m[/imath] into a set of cardinality [imath]n[/imath] [imath](m \leq n)[/imath]? |
1413583 | Finite ring having [imath]2^n-1[/imath] invertible elements
Let [imath]A[/imath] be a ring with [imath]0\neq1[/imath] having [imath]2^n-1[/imath] invertible elements and at most [imath]2^n-1[/imath] non-invertible elements. Then [imath]A[/imath] must be a field? How many elements does the ring need to have? Because [imath]2^n-1[/imath] is odd, the characteristic of the ring is [imath]2[/imath]. Thank you! | 489893 | A ring with few invertible elements
Let [imath]A[/imath] be a ring with [imath]0 \neq 1 [/imath], which has [imath]2^n-1[/imath] invertible elements and less non-invertible elements. Prove that [imath]A[/imath] is a field. |
1412977 | doubt about the solution to an induction problem exercise
I need to prove that [imath]5^n-1[/imath] is divisible by [imath]4[/imath], [imath]\forall n \in \mathbb{N}[/imath]. So for the inductive step I know that: [imath]5^{n+1} -1= 5\times5^n -1[/imath] but how do I get from there to: [imath](5^n -1) + 4\times 5^n?[/imath] (That is the solution to the exercise.) | 595233 | Prove by induction that [imath]5^n - 1[/imath] is divisible by [imath]4[/imath].
Prove by induction that [imath]5^n - 1[/imath] is divisible by [imath]4[/imath]. How should I use induction in this problem. Do you have any hints for solving this problem? Thank you so much. |
1413904 | Prove [imath]\cos(\sin x)>\sin(\cos x)[/imath]
Prove that [imath]\cos( \sin x)>\sin(\cos x), \forall x\in\mathbb{R}[/imath]. I have thought that we should consider their difference and show it is positive for all x, so: Let [imath]A=\cos\sin x-\sin\cos x=\cos\sin x-\cos\left( \frac{\pi}{2}-\cos x\right)[/imath] but I am unable to move it further. How can we prove that? | 351846 | Show that the equation [imath]\cos(\sin x)=\sin(\cos x)[/imath] has no real solutions.
The following problem was on a math competition that I participated in at my school about a month ago: Prove that the equation [imath]\cos(\sin x)=\sin(\cos x)[/imath] has no real solutions. I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s): [imath] \cos^2(\sin x)=\sin^2(\cos x)\\ 1-\cos^2(\sin x)=1-\sin^2(\cos x)\\ \sin^2(\sin x)=\cos^2(\cos x)\\ \sin(\sin x)=\pm\cos(\cos x)\\ [/imath] I then proceeded to split into cases and use the identity [imath]\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)[/imath] to get [imath] \sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\ [/imath] and the identity [imath]-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) [/imath] to get [imath] \sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\ [/imath] where [imath]y[/imath] is any integer. I argued that [imath]y=0[/imath] was the only value of [imath]y[/imath] that made any sense (since the values of sine and cosine remain between [imath]-1[/imath] and [imath]1[/imath]). Therefore, the above equations become [imath] \sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\ [/imath] and [imath] \sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\ [/imath] Then, by a short optimization argument, I showed that these last two equations have no real solutions. First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof? |
1413525 | What is the sum of this series? [imath]\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}[/imath]
I want to know what is the sum of this series: [imath]\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}[/imath] Mathematica does not help. | 1413541 | What is the sum of this series? [imath]\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}[/imath]
I want to know what is the sum of this series: [imath]\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}[/imath] [imath]B_n[/imath] are the Bernoulli numbers. Mathematica does not help. |
1413989 | a conceptual question on markov chain
Suppose [imath]\{X_n,n\ge 0\}[/imath] and [imath]\{Y_n,n\ge0\}[/imath] are two independent discrete-time markov chains (DTMC) with state space [imath]S=\{0,1,2,\ldots\}[/imath]. Prove or give a counterexample to: [imath]\{X_n+Y_n,n\ge 0\}[/imath] is also a DTMC. | 1365430 | When the sum of independent Markov chains is a Markov chain?
I try to find as much as possible cases, when the chain [imath]Z(t) = |X_1(t)-X_2(t)|[/imath] is Markov, where [imath]X_1(t)[/imath] and [imath]X_2(t)[/imath] are independent, discrete-time and space, preferably non-homogeneous Markov chains. I started to search for the sum of independent Markov chains and I found this statement in Stoyanov J. - Counterexamples in Probability (2ed., Wiley, 1997)(p.229, one can google it and find in google books): ... the sum of two Markov processes need not be a Markov process. Note, however, that that the sum of two independent Markov processes preserves this property. That seems very strange to me and I wish to find the proof or at least the statement elsewhere. EDIT: The way of thinking how it may look like for a sum of two independent Markov chains (If I want to prove that the sum of two independent Markov chains is again a Markov chain): Let [imath]Y(n) = X_1(n)+X_2(n)[/imath]. Then [imath]P(Y(n+1) = i_{n+1} \ | \ Y(n) = i_n, ..., Y(0)=i_0) =[/imath] [imath]= P(X_1(n+1)+X_2(n+1)=i_{n+1}\ | \ X_1(n)+X_2(n)=i_n,...,X_1(0)+X_2(0)=i_0)=[/imath] [imath]=/ (?) / = \sum_{j+k=i_{n+1}}P(X_1(n+1)=j,X_2(n+1)=k \ | \ \cdot)=[/imath] [imath]=/\text{X's are independent}/ = \sum_{j+k=i_{n+1}}P(X_1(n+1)=j \ | \ \cdot)\cdot P(X_2(n+1)=k \ | \ \cdot) = [/imath] [imath]= /\text{Markov property + (??) } / = \sum_{j+k=i_{n+1}}P(X_1(n+1)=j \ | \ X_1(n)+X_2(n)=i_n)\cdot P(X_2(n+1)=k \ | \ X_1(n)+X_2(n)=i_n)=[/imath] [imath]=P(X_1(n+1)+X_2(n+1)=i_{n+1} \ | \ X_1(n)+X_2(n)=i_n) = P(Y(n+1)=i_{n+1}\ | \ Y(n)=i_n)[/imath]. Here I am uncertain about (?) and (??) steps. |
1391509 | Finding the functional form of the Green Function for a parabolic diff eq.
I need help finding the functional form of the Green function G(x,t) for a parabolic equation (i.e. heat diffusion etc) [imath] \frac{\partial{}G(x,t)}{\partial{}t}=a\frac{\partial{}^2G(x,t)}{\partial{}x^2}+\delta(t)\delta(x)[/imath] Using this result, I'd like to write the general solution of the heat equation: [imath]\frac{\partial{}T(x,t)}{\partial{}t}=a\frac{\partial{}^2T(x,t)}{\partial{}x^2}+f(x,t) [/imath] where f is a known source function. I know that I need to solve for G using suitable integral transforms. I am not sure what transforms that I should be using to solve this problem and would appreciate some help. | 1396282 | Product of 2 dirac delta functions
Solve for a parabolic diff. eq: [imath] \frac{\partial{}G(x,t)}{\partial{}t}=a\frac{\partial{}^2G(x,t)}{\partial{}x^2}+\delta(t)\delta(x)[/imath] Using the result, write the general solution of: [imath]\frac{\partial{}T(x,t)}{\partial{}t}=a\frac{\partial{}^2T(x,t)}{\partial{}x^2}+f(x,t) [/imath] |
1414456 | [imath]A^k = I[/imath] implies diagonalizable?
If [imath]A[/imath] is a square complex matrix with [imath]A^k = I[/imath] (where [imath]I[/imath] is the identity matrix of the same size as [imath]A[/imath]) for some positive integer [imath]k[/imath], does it follow that [imath]A[/imath] is diagonalizable? | 473831 | If [imath]A^m=I[/imath] then A is Diagonalizable
Let [imath]A[/imath] be an [imath]n\times n[/imath] complex matrix. If [imath]A^m=I_n[/imath] for some positive integer [imath]n[/imath]. How to show that [imath]A[/imath] is diagonalizable? |
1414431 | Dirichlet's Test in convergence
Can Dirichlet's test be applied to establish the convergence of [imath]1 - \frac 1 2 - \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 - \dots[/imath] where the number of signs increase by one in each block? Can somebody give me some ideas or hints? | 753960 | Convergence of "alternating" harmonic series where sign is +, --, +++, ----, etc.
Exercise 11 from section 9.3 of Introduction to Real Analysis (Bartle): Can Dirichlet’s Test be applied to establish the convergence of [imath] 1 - \dfrac12 - \dfrac13 + \dfrac14 + \dfrac15 + \dfrac16 - \cdots [/imath] [imath]\qquad \qquad[/imath] where the number of signs increases by one in each ‘‘block’’? If not, use another method to establish the convergence of this series. Dirichlet's test cannot be used because the partial sums generated by (1, -1, -1, 1, 1, 1, ...) are not bounded. But we can group the terms of the series in the following way: [imath] 1 - \left(\dfrac12 + \dfrac13\right) + \left(\dfrac14 + \dfrac15 + \dfrac16\right) - \left( \dfrac17 + \dfrac18 + \dfrac19 + \dfrac{1}{10} \right) + \cdots \\ = \sum _{n=1}^{\infty}(-1)^{n+1}a_n [/imath] where [imath] (a_n) = \left(1, \left(\dfrac12 + \dfrac13\right), \left(\dfrac14 + \dfrac15 + \dfrac16\right), ... \right) [/imath] So by Leibniz's test, if the sequence [imath](a_n)[/imath] is decreasing and [imath]\lim{a_n} = 0[/imath] then the grouped series is convergent. I've shown that since we are grouping terms of the same sign it is sufficient to show the convergence of the grouped series. I've shown that [imath]\lim{a_n} = 0[/imath], but how do I show that [imath](a_n)[/imath] is decreasing? |
1414604 | Supremum of a functional on [imath]C^1([0,1])[/imath]
I've tried to solve the following problem (1.1.30 from Berkeley Problems in Mathematics, no solution is provided therein), but I'm not sure if my solution is correct. I wanted to ask you for your opinion. The problem: Let [imath]S[/imath] be the set of all real [imath]C^1[/imath] functions [imath]f[/imath] on [imath][0,1][/imath] such that [imath]f(0)=0[/imath] and [imath] (*)\int_0^1f'(x)^2\mathrm{d}x\leq 1 [/imath] Define [imath] J(f) = \int_0^1 f(x)\mathrm{d}x. [/imath] Prove that the function [imath]J[/imath] is bounded on [imath]S[/imath] and compute its supremum. Is there a function [imath]f_0\in S[/imath] at which [imath]J[/imath] attains its supremum? If so, give [imath]f_0[/imath]. My attempts: First note that for any [imath]y\in[0,1][/imath], [imath]\int_0^1 f'(x)^2\mathrm{d}x \geq \int_0^y f'(x)^2\mathrm{d}x[/imath]. Thus, using the Cauchy Schwartz inequality, [imath]1\geq \int_0^y f'(x)^2\mathrm{d}x \cdot\underbrace{\int_0^1 1^2\mathrm{d}x}_{\geq\int_0^y 1^2\mathrm{d}x}\geq \left(\int_0^y f'(x)\cdot 1\mathrm{d}x\right)^2=(f(y)-f(0))^2=f(y)^2 [/imath] which implies that [imath]|f(y)|\leq 1[/imath]. Therefore, [imath]J(f)\leq|J(f)|\leq1[/imath] so we have shown boundedness. The rest is still a bit shaky:\ Now the supremum of [imath]J[/imath] on [imath]S[/imath] is attained when the equality in [imath](*)[/imath] holds (why is that? I'm not sure I could show this). This is the case for [imath]f_0(x)=x[/imath]. Now [imath]J(f_0)=\frac{1}{2}[/imath]. My guess is that therefore [imath]J(f)\leq \frac{1}{2}[/imath] is the supremum and maximum. Thank you for your suggestions. | 465353 | Example of the equality of an inequality
This question is related to Daniel Fischer's answer here. Suppose [imath]f[/imath] is a real [imath]C^{1}[/imath] function on [imath][0, 1][/imath] such that [imath]f(0) = 0[/imath] and [imath]\int_{0}^{1}f'(x)^{2}\, dx \leq 1[/imath]. Then (essentially by Cauchy-Schwarz), we have [imath]\left|\int_{0}^{1}f(x)\, dx\right| \leq 2/3[/imath] as Daniel Fischer stated in his answer. My question is: Is there an example of a function such that we have equality in [imath]\left|\int_{0}^{1}f(x)\, dx\right| \leq 2/3[/imath]? I was thinking of maybe a smoothed version of a function which takes the value [imath]0[/imath] on [imath][0, 1/2)[/imath] and [imath]4/3[/imath] on [imath](1/2, 1][/imath] but that seems complicated to construct, moreover, the derivative in some neighbourhood of [imath]1/2[/imath] might be hard to control. |
1414728 | Finding all homomorphisms
How can I solve this question Find all group homomorphisms from [imath]\mathbb{Z_{5}}[/imath] into [imath]\mathbb{Z_{12}^{x}}[/imath]? | 762727 | Finding homomorphisms from [imath]\mathbb Z_{12}[/imath] to [imath]\mathbb Z_{6}[/imath].
Find all homomorphisms from [imath]\mathbb Z_{12}[/imath], the cyclic group of order [imath]12[/imath], to [imath]\mathbb Z_6[/imath]. For each homomorphism [imath]f\colon \mathbb Z_{12}\to \mathbb Z_6[/imath], determine the kernel [imath]\ker(f)[/imath] and the image [imath]f(\mathbb Z_{12})[/imath]. I've already used the fact that the generators of [imath]\mathbb Z_{12}[/imath] are [imath]\bar 1, \bar 5, \bar 7, \bar 11[/imath] and the generators of [imath]\mathbb Z_6[/imath] are [imath]\bar 1[/imath] and [imath]\bar 5[/imath]. To get the homomorphisms which lead to the [imath]\mathbb Z_6[/imath] outputs of [imath](\bar 0, \bar 5, \bar 4, \bar 3, \bar 2, \bar 1, \bar 0, \bar 5, \bar 4, \bar 3, \bar 2, \bar 1)[/imath] and [imath](\bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5, \bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5)[/imath] but I don't know where to go from here to obtain the rest. |
1414778 | Evaluate [imath]\int \dfrac{1}{\sqrt{x-1}+\sqrt{x}+\sqrt{x+1}} \ \mathrm{d}x[/imath]
Evaluate [imath]\int \dfrac{1}{\sqrt{x-1}+\sqrt{x}+\sqrt{x+1}} \ \mathrm{d}x[/imath] I tried rationalizing the denominator by twice multiplying, but it didn't do any good. I also tried trig substitutions but that too didn't work out. Can anyone help please? | 1274212 | Integral [imath]\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}[/imath]
[imath]\int \frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}}[/imath] I tried substituting [imath]x=z^2[/imath] ... also [imath]x=\tan^2 \theta[/imath] ... but couldn't solve it either ways... if someone can help then it would be good. |
1414878 | Maximum degree of a polynomial
What is the maximum degree of a polynomial of the form [imath]\sum_{i=0}^n a_i x^{n-i}[/imath] with [imath]a_i = \pm 1[/imath] for [imath]0 \leq i \leq n, 1 \leq n[/imath], such that all the zeros are real? I have no idea where to start. | 1239878 | Only 12 polynomials exist with given properties
Prove that there are only 12 polynomials that have all real roots, and whose coefficients are all [imath]-1[/imath] or [imath]1[/imath]. (zero coefficients are not allowed, and constant polynomials do not count.) Two of them are obviously [imath]x-1[/imath] and [imath]x+1[/imath]. I tried applying Viete's theorems, but could not get anything useful. |
1414758 | Least Upper Bound Proof
I am looking for a proof for this: If [imath]S=(a,b)=\{x\in\mathbb R\, :\, a < x < b\}[/imath], then the least upper bound of [imath] S[/imath] is [imath]b[/imath] My professor was fumbling through an explanation and I got lost. | 1401307 | Proving that the supremum of [imath](a,b)[/imath] is [imath]b[/imath].
I was trying to prove it this way. Let [imath]a,b\in\mathbb{R}[/imath] with [imath]a<b[/imath]. Let [imath] I=(a,b)[/imath]. As for all [imath]x\in I[/imath],[imath]x<b[/imath], we get [imath]b[/imath] is an upper bound. let [imath]u=\sup I[/imath], then [imath]u\leq b[/imath]. If [imath]u=b[/imath] it's done. Otherwise, let's suppose [imath]u<b[/imath]. Furthermore, for all [imath]x\in I[/imath], [imath]a<x[/imath] and [imath]x\leq u[/imath], so [imath]a<u[/imath], then [imath]u\in I[/imath]... Can I get somewhere this way? |
1414641 | Are vectors [imath]e_i[/imath] linearly independent if and only if the matrix [imath]A[/imath] is nonsingular?
Let [imath]V[/imath] be an inner product space over [imath]\mathbb{R}[/imath] or [imath]\mathbb{C}[/imath], and let [imath]e_1, e_2, \dots, e_n[/imath] be a collection of vectors in [imath]V[/imath], not necessarily orthonormal. (Here, [imath]n[/imath] has nothing to do with the dimension of [imath]V[/imath].) Let [imath]\langle\cdot,\cdot\rangle[/imath] denote the inner product in [imath]V[/imath], and form the [imath]n \times n[/imath] matrix [imath]A[/imath] with entries [imath]\langle e_i, e_j\rangle[/imath] (where [imath]i, j = 1, \dots, n[/imath]). Are the vectors [imath]e_i[/imath] linearly independent if and only if the matrix [imath]A[/imath] is nonsingular? My thoughts on the problem so far. Let [imath]F[/imath] be [imath]\mathbb{R}[/imath] or [imath]\mathbb{C}[/imath] and [imath]f[/imath] the linear map [imath]F^n \to V[/imath] which sends the [imath]i[/imath]th standard basis vector to [imath]e_i[/imath]. I want to show that, if [imath]v \in F^n[/imath] is nonzero, then [imath]Av[/imath] is nonzero. I've observed that [imath]v^t Av = \langle f(v), f(v)\rangle[/imath] (which is positive). But I don't know what to do next. Can someone help me out? | 36580 | Gram matrix invertible iff set of vectors linearly independent
Given a set of vectors [imath]v_1 \cdots v_n[/imath], the [imath]n\times n[/imath] Gram matrix [imath]G[/imath] is defined as [imath]G_{i,j}=v_i \cdot v_j[/imath] Due to symmetry in the dot product, [imath]G[/imath] is Hermitian. I'm trying to remember why [imath]|G|=0[/imath] iff the set of vectors are not linearly independent. |
1414759 | Integrating Factor of [imath](axy^2+by)dx+(bx^2y+ax)dy=0[/imath].
[imath](axy^2+by)dx+(bx^2y+ax)dy=0[/imath] I have asked this question before too, but i wish to know the method for evaluating the integrating factor which is [imath]\dfrac {1} {(a-b)(x^2y^2-xy)}.[/imath] So far i know: [imath]M(x,y)dx + N(x,y)dy=0[/imath] is said to be a perfect differential when [imath]\frac{\partial (M(x,y))}{\partial y}=\frac{\partial (N(x,y))}{\partial x}[/imath]. In this case the equation is not exact and integrating factor is multiplied to make the differential equation exact. Is there any other method to solve this differential equation? | 1413266 | Exact Differential Equations [imath](axy^2+by)dx+(bx^2y+ax)dy=0[/imath].
[imath]M(x,y)dx + N(x,y)dy=0[/imath] is said to be a perfect differential when [imath]\frac{\partial (M(x,y))}{\partial y}=\frac{\partial (N(x,y))}{\partial x}[/imath] Let [imath]M_y=\frac{\partial (M(x,y))}{\partial y}[/imath] and [imath]N_x=\frac{\partial (N(x,y))}{\partial x}[/imath]. In case if: [imath]\frac{M_y-N_x}{N(x,y)}[/imath] is only a function of x only (say [imath]f(x)[/imath]) then it has a integrating factor [imath]e^{\int f(x)dx}[/imath]. But if [imath]\frac{M_y-N_x}{N(x,y)}[/imath] is a constant then what will be the integrating factor? Say for this problem: [imath](axy^2+by)dx+(bx^2y+ax)dy=0[/imath] Is there any other method to solve this differential equation? |
668307 | What are functions with the property [imath]f(f(x)) = x[/imath] called?
Do functions which, when composed with themselves, are equivalent to the identity function (i.e. functions for which [imath]f(f(x)) = x[/imath] in general) have a name and if so, what is it? Additionally, am I correct in saying that a such function has a splinter of two, or is it perhaps splinter of size 2 or something else entirely? Or could I say that a such function has an orbit of size 2? | 85854 | What's the name for the property of a function [imath]f[/imath] that means [imath]f(f(x))=x[/imath]?
I can think of several examples of functions such that twice application of the function is equivalent to no application of it. Additive inverse Multiplicative inverse Fourier transform Complex conjugation Any group built up from [imath]\mathbb{Z}_2[/imath], applying (one of) the [imath]\mathbb{Z}_2[/imath] parts' operation. "Idempotent" came to mind, but that's wrong. It means [imath]f(f(x)) = f(x)[/imath], not [imath]f(f(x))=x[/imath]. What is the word for this "flip-flop" property? |
157245 | Is the sum and difference of two irrationals always irrational?
If [imath]x[/imath] and [imath]y[/imath] are irrational, is [imath]x + y[/imath] irrational? Is [imath]x - y[/imath] irrational? | 1491971 | subtraction of two irrational numbers to get a rational
Say you have a number like [imath]\pi[/imath] or e. Is it possible to subtract another number from it and end up with a rational number? I mean I guess you could write an equation like [imath]\pi-x=3[/imath] But could there ever be a solution for x (that we could know and write out)? Correction: Any number besides the irrational number itself. Damn math ppl are too quick.. |
1415871 | Sufficient conditions for this function being linear
Let [imath]f[/imath] be a real-valued function for which, for every real [imath]x,y[/imath]: [imath]f(x+y) = f(x)+f(y)[/imath] Does this imply that [imath]f[/imath] is a linear function ([imath]f(x)=a\cdot x[/imath])? If [imath]f[/imath] is differentiable, I think the answer is yes. Take the derivative of [imath]f(x+y)[/imath] by [imath]x[/imath]: [imath]f'(x+y)=f'(x)[/imath]. This is true for all [imath]y[/imath], hence [imath]f'[/imath] is a constant function. Additionally, [imath]f(0)=f(x+0)-f(x)=0[/imath]. Hence [imath]f(x)=ax[/imath]. (is this correct?) If [imath]f[/imath] is continuous, I think the answer is also yes. First, note that [imath]f(0)=0[/imath]. Let [imath]a=f(1)[/imath]. By addition, for every integer [imath]x[/imath], [imath]f(x)=a\cdot x[/imath]. This also must be true for every rational [imath]x[/imath]. Hence, by continuity, it must be true for all [imath]x[/imath]. (is this correct?) What is the answer for general [imath]f[/imath]? | 385586 | Do there exist functions satisfying [imath]f(x+y)=f(x)+f(y)[/imath] that aren't linear?
Do there exist functions [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] such that [imath]f(x+y)=f(x)+f(y),[/imath] but which aren't linear? I bet you they exist, but I can't think of any examples. Furthermore, what hypotheses do we need to put on [imath]f[/imath] before no such functions exist? I feel continuity should be enough. |
1415722 | Kirschenhofer Ramanujan functional equations part I(alternative form)
Ramanujan analyzed [imath]\sum _{k=1}^{\infty } \frac{e^{-k x}}{e^{-2 k x}+1}=\sum _{k=1}^{\infty } \frac{\pi \operatorname{sech}\left(\frac{\pi ^2 k}{x}\right)}{2 x}+\frac{\pi }{4 x}-\frac{1}{4}[/imath] it is possible to show the following identity,numerically it seem equal [imath]\sum _{n=1}^{\infty } \frac{\pi \left(\text{sech}\left(\frac{\pi ^2 n}{2 x}\right)-\text{sech}\left(\frac{\pi ^2 n}{x}\right)\right)}{2 x}-\frac{1}{4}=\sum _{n=1}^{\infty } \frac{1}{2} (-1)^n \text{sech}(n x)[/imath] | 1415694 | Kirschenhofer Ramanujan functional equations
Ramanujan analyzed [imath]\sum _{k=1}^{\infty } \frac{e^{-k x}}{e^{-2 k x}+1}=\sum _{k=1}^{\infty } \frac{\pi \operatorname{sech}\left(\frac{\pi ^2 k}{x}\right)}{2 x}+\frac{\pi }{4 x}-\frac{1}{4}[/imath] it is possible to show the following identity,numerically it seem equal [imath]\sum _{n=1}^{\infty } \frac{\pi \csc \left(\frac{\pi (x+2 i \pi n)}{4 x}\right) \sec \left(\frac{\pi (x+2 i \pi n)}{4 x}\right)}{4 x}=\sum _{k=1}^{\infty } \frac{\pi \operatorname{sech}\left(\frac{\pi ^2 k}{x}\right)}{2 x}[/imath] |
1415915 | [imath](a,b) \mathbin\# (c,d)=(a+c,b+d)[/imath] and [imath](a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)[/imath]. Multiplicative inverse?
Let [imath]r\in \mathbb{R}[/imath] and let [imath]0\neq s \in \mathbb{R}[/imath]. Define operations [imath]\#[/imath] and [imath]\&[/imath] on [imath]\mathbb{R}[/imath] x [imath]\mathbb{R}[/imath] by [imath](a,b) \mathbin\#(c,d)=(a+c,b+d)[/imath] and [imath](a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)[/imath] as addition and multiplication, respectively. Do these operations form a field? I have shown that every axiom of the field is satisfied except for the "existence of multiplicative inverses" axiom. This is where I need help. We either need to show that this axiom fails and so we don't have a field or that the axiom [imath]aa^{-1}=1[/imath] is satisfied for every [imath]a\neq 0 \in \mathbb{R}[/imath] The multiplicative identity in this problem is [imath](1,0)[/imath]. Any solutions or hints are greatly appreciated. | 1416478 | Define multiplication as [imath](a,b) \boxdot (c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)[/imath] where [imath]a,b,c,d,r\in \mathbb{R}[/imath] and [imath]0\neq s\in \mathbb{R}[/imath].
Define multiplication as [imath](a,b) \boxdot (c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)[/imath] where [imath]a,b,c,d,r\in \mathbb{R}[/imath] and [imath]0\neq s\in \mathbb{R}[/imath]. The multiplicative inverse is [imath](1,0)[/imath]. I need to show that every ordered pair other than [imath](0,0)[/imath] has a multiplicative inverse. How on earth can I show that this property is satisfied (or produce a counterexample)? I'm trying to do this in the most elementary way possible. I do not have a background in abstract algebra and this problem should be done using basic algebra with an understanding of fields. |
1411035 | Let [imath]q[/imath] be a prime integer. Show that for each [imath]x∈GF(q)[/imath] there exist elements [imath]r[/imath] and [imath]s[/imath] in [imath]GF(q)[/imath] satisfying [imath]x=r^2+s^2[/imath].
Let [imath]q[/imath] be a prime integer. Show that for each [imath]x∈GF(q)[/imath] there exist elements [imath]r[/imath] and [imath]s[/imath] in [imath]GF(q)[/imath] satisfying [imath]x=r^2+s^2[/imath]. I'm stuck on this problem. Any solutions or hints are greatly appreciated. | 1140122 | Let [imath]p[/imath] be a prime integer. Show that for each [imath]a ∈ GF(p)[/imath] there exist elements [imath]b[/imath] and [imath]c[/imath] of [imath]GF(p)[/imath] satisfying [imath]a = b^2 + c^2[/imath].
Let [imath]p[/imath] be a prime integer. Show that for each [imath]a ∈ GF(p)[/imath] there exist elements [imath]b[/imath] and [imath]c[/imath] of [imath]GF(p)[/imath] satisfying [imath]a = b^2 + c^2[/imath]. I got some ideas like to show the elements of the form [imath]b^2 + c^2[/imath] is asubfield so they are all the field but I dont know how to prove that too, so I ask for some help ;) |
1417073 | A question on spectrum
Let [imath]A,B \in {C^{n \times n}}[/imath] and [imath]{\sigma (A + B)}[/imath] is spectrom of [imath](A+B)[/imath]. Suppose [imath]M = \left\{ {\lambda \in C:\lambda \in \sigma (A + B),\left\| B \right\| \le \varepsilon } \right\}[/imath] [imath]F(A) = \left\{ {{v^*}Av:v \in {C^n},{v^*}v = 1} \right\}[/imath] [imath]K = \left\{ {\lambda \in C:dist(\lambda ,F(A)) \le \varepsilon } \right\}[/imath] ,where [imath]{dist(\lambda ,F(A))}[/imath] denotes the distance between [imath]\lambda [/imath] and [imath]F(A)[/imath]. Why does [imath]M \subseteq K[/imath]? | 1416309 | A question on numerical range
Let [imath]A,B \in {C^{n \times n}}[/imath] and [imath]{\sigma (A + B)}[/imath] is spectrum of [imath](A+B)[/imath]. Suppose [imath]M = \left\{ {\lambda \in C:\lambda \in \sigma (A + B),\left\| B \right\| \le \varepsilon } \right\}[/imath] [imath]F(A) = \left\{ {{v^*}Av:v \in {C^n},{v^*}v = 1} \right\}[/imath] [imath]K = \left\{ {\lambda \in C:dist(\lambda ,F(A)) \le \varepsilon } \right\}[/imath] ,where [imath]{dist(\lambda ,F(A))}[/imath] denotes the distance between [imath]\lambda [/imath] and [imath]F(A)[/imath]. Why does [imath]M \subseteq K[/imath]? |
1417358 | How to prove the sum of squares larger than 1/n without induction?
known that: [imath]1\geq R_1 \geq R_2 \geq \dots \geq R_n \geq 0[/imath] and [imath]\sum_{i=1}^n R_i=1[/imath] To prove: [imath]\sum_{i=1}^n R_i^2 \geq \frac{1}{n}[/imath] Using induction, the problem can be easily proved. I'd like to know is there any other ways can prove this? | 1249986 | Prove QM-AM inequality
[imath]\dfrac{x_1^2+ x_2^2 + \cdots + x_n^2}n \geq \left(\dfrac{x_1+x_2+\cdots+x_n}n\right)^2[/imath] I don't think AM, GM can be used here. And simple expansion doesn't help too. What should I do? |
1266564 | What can I say of an [imath]m[/imath]-dimensional submanifold [imath]S[/imath] of an [imath]m[/imath]-dimensional manifold [imath]M[/imath]?
I consider a differentiable manifold [imath]M[/imath] of dimension [imath]m[/imath]. Let be [imath]S[/imath] a submanifold of [imath]M[/imath] of the same dimension [imath]m[/imath]. What can I say about [imath]S[/imath]? I have tried to prove that [imath]S[/imath] is open but I get stuck. My idea was to use the inverse function theorem to the inclution [imath]i:M \rightarrow S[/imath]. Any help will be appreciated. Thanks! | 84577 | Open subsets in a manifold as submanifold of the same dimension?
An open set in an [imath]n[/imath]-manifold is clearly a submanifold of the same dimension as its containing manifold (see open manifolds). Now, given an [imath]n[/imath]-manifold [imath]M[/imath], is it true that a set, to be the underlying set of a submanifold of [imath]M[/imath] with dimension [imath]n[/imath], must be open? |
1416937 | Another example of a non-Hausdorff topological space in which all compact subsets are closed
One example of a non-Hausdorff topological space in which all compact subsets are closed is the co-countable topology on an uncountable set, as demonstrated here. It was claimed (as a now-deleted answer to the above question) that the compact complement topology on [imath]\mathbb{R}[/imath] was another example, but this was proven incorrect here. Can someone provide an additional example of a non-Hausdorff space in whihc all compact subsets are closed? | 530638 | Compactness and Hausdorffness
Is it possible to have a topological space having all compact subsets closed, but the space itselt is not Hausdorff? I couldn't find any counterexample. I tried [imath]\mathbb R[/imath] with cofinite, point inclusion and point exclusion topologies, but these didn't work. Any suggestion in this context will be helpful for me. |
1417555 | I am issues with proving the following problem: [imath]f^{-1}(f(A)) ⊃ A[/imath]
I am unsure as to where to start with this problem. The way I read it is that [imath]f^{-1}(f(A)) ⊃ A[/imath] means that [imath]A[/imath] is a subset of the preimage of the image of [imath]A[/imath]. But I am unsure. | 347880 | Proving that [imath]C[/imath] is a subset of [imath]f^{-1}[f(C)][/imath]
More homework help. Given the function [imath]f:A \to B[/imath]. Let [imath]C[/imath] be a subset of [imath]A[/imath] and let [imath]D[/imath] be a subset of [imath]B[/imath]. Prove that: [imath]C[/imath] is a subset of [imath]f^{-1}[f(C)][/imath] So I have to show that every element of [imath]C[/imath] is in the set [imath]f^{-1}[f(C)][/imath] I know that [imath]f(C)[/imath] is the image of [imath]C[/imath] in [imath]B[/imath] and that [imath]f^{-1}[f(C)][/imath] is the pre-image of [imath]f(C)[/imath] into [imath]A[/imath]. Where I'm stuck is how to use all of this information to show/prove that [imath]C[/imath] is indeed a subset. Do I start with an arbitrary element (hey, let's call it [imath]x[/imath]) of [imath]C[/imath]? and then show that [imath]f^{-1}[f(x)][/imath] is [imath]x[/imath]? I could use a little direction here... Thanks. |
1417228 | Does [imath]S(n)[/imath] contain infinite many primes?
Denote [imath]p_j := j\text{th prime}[/imath] and [imath]S(n)\:=\sum_{j=1}^n p_j[/imath] (The sum of the first [imath]n[/imath] primes). Is it known whether [imath]S(n)[/imath] is prime for infinite many [imath]n[/imath]? OEIS gives the sum of the prime numbers upto [imath]10^{18}[/imath], which is [imath]12,212,914,292,949,226,570,880,576,733,896,687[/imath], being a [imath]35[/imath]-digit prime. The relative low growth rate of [imath]S(n)[/imath] (of order [imath]n^2\ln n[/imath]) is another heuristic argument that the answer to the given question should be yes. | 636479 | Conjecture: the sequence of sums of all consecutive primes contains an infinite number of primes
Starting from 2, the sequence of sums of all consecutive primes is: [imath]\begin{array}{lcl}2 &=& 2\\ 2+3 &=& 5 \\ 2+3+5 &=& 10 \\ 2+3+5+7 &=& 17 \\ 2+3+5+7+11 &=& 28 \\ &\vdots& \end{array} [/imath] If the [imath]n^\text{th}[/imath] prime is [imath]P_n[/imath], then we can write [imath]S_n=\sum_{i=1}^n P_n[/imath]. I conjecture that the sequence [imath]S_n[/imath] contains an infinite number of primes. I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof. Computationally, it can be verified that [imath]S_{13,932}=998,658,581=P_{50,783,012},[/imath] and the sequence shows no sign of slowing down. Can you prove the sequence [imath]S_n[/imath] contains an infinite number of primes? |
1418231 | Prove √2 exists by Archimedean Axiom
I am trying to prove the existence of the square root of 2. The proof: Let [imath]S=\{x \in \mathbb{R} ∣x \ge 0, x^2 < 2\}.[/imath] I understand the proof of LUB, [imath]\alpha[/imath] and so I am at the step where [imath]\alpha^2=2.[/imath] I know that we are to prove by contradiction so we state let [imath]\alpha^2 <2[/imath] and [imath]\alpha^2 >2[/imath]. Now my instructor wants us to use the Archimedean Axiom [imath]1/n < \epsilon[/imath]. [imath](\alpha^2 + 1/n)^2[/imath] then what..... | 1415235 | Prove the existence of the square root of [imath]2[/imath].
I am trying to prove the existence of the square root of [imath]2[/imath]. I have some steps with a very vague explanation and I would like to clarify. The proof: Let [imath]S=\{x\in\mathbb R\mid x\geqslant 0 \text{ and } x^2<2\}.[/imath] I understand the proof of LUB, ∝ and so I am at the step where [imath]\alpha^2=2[/imath]. I know that we are to prove by contradiction so we state let [imath]\alpha^2 <2[/imath] and [imath]\alpha^2 >2[/imath]. Now my instructor wants us to use the Archimedean Axiom [imath]1/n = \varepsilon[/imath]. [imath](\alpha^2 + 1/n)^2[/imath] then what..... |
1413680 | We all use mathematical induction to prove results, but is there a proof of mathematical induction itself?
I just realized something interesting. At schools and universities you get taught mathematical induction. Usually you jump right into using it to prove something like [imath]1+2+3+\cdots+n = \frac{n(n+1)}{2}[/imath] However. I just realized that at no point is mathematical induction proved itself? What's the mathematical induction's proof? Is mathematical induction proved using mathematical induction itself? (That would be mind blowing.) | 1783734 | I'm having trouble understanding why inductive proofs are logical
I am new to Mathematics, reading books in my free time. I have recently learned about proving Mathematical propositions by induction. I am having a bit of trouble understanding the process and why it is logical. I have learned that if you have a statement [imath]A[/imath], and you assume [imath]A_{n}[/imath] is true, then you can prove [imath]A[/imath] is true for all numbers greater than [imath]n[/imath] by showing that [imath]A_{n+1}[/imath]. I don't understand this because from my point of view it looks like you only proved [imath]A[/imath] is true for [imath]n[/imath] and [imath]n + 1[/imath]. Maybe I am missing something, or maybe I need a formal introduction to logic. I never read anything about philosophy or logic before learning Mathematics, but if reading about logic would help me understand proofs better I would be glad to. |
1418398 | Find [imath]\lim_{n \rightarrow \infty } 2^{n} \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}[/imath]
[imath]\lim_{n \rightarrow \infty } 2^{n} \sqrt{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}[/imath] where the 2 inside the roots appear n times. For example if n = 2 : [imath]2^{2} \sqrt{2-\sqrt{2}}[/imath] I discovered this. Has this been already developed/made before? | 428046 | Find [imath]\lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}[/imath].
Find [imath]\displaystyle \lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}[/imath]. By geometry method, I know that this is [imath]\pi[/imath]. But is there algebraic method to find this ? Thank you. |
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