qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
1046465 | proving [imath](A \rightarrow B \vee C) \rightarrow ((A\rightarrow B) \vee (A\rightarrow C))[/imath]in hilbert system(HP)
I'm looking for [imath](A \rightarrow B \vee C) \rightarrow ((A\rightarrow B) \vee (A\rightarrow C))[/imath] in Hilbert system | 1046362 | proving [imath] (A \rightarrow B \vee C )\rightarrow((A\rightarrow B) \vee (A\rightarrow C))[/imath]
I'm looking for a way to prow [imath] (A \rightarrow B \vee C )\rightarrow((A\rightarrow B) \vee (A\rightarrow C))[/imath] from the following axioms and rules [imath]\vdash A \rightarrow A[/imath] [imath]\vdash A \wedge B \rightarrow A[/imath] [imath]\vdash A \wedge B \rightarrow B[/imath] [imath]\vdash A \rightarrow A \vee B [/imath] [imath]\vdash B \rightarrow A \vee B [/imath] [imath]A, B \vdash A\wedge B[/imath] [imath]A, A \rightarrow B \vdash B[/imath] [imath]\vdash A\wedge (B \vee C) \rightarrow (A\wedge B)\vee (A\wedge C)[/imath] [imath]\vdash A \rightarrow (B \rightarrow B)[/imath] [imath]\vdash (A\rightarrow B)\wedge (A\rightarrow C) \rightarrow(A\rightarrow B\wedge C)[/imath] [imath]\vdash (A\rightarrow C)\wedge (B\rightarrow C) \rightarrow(A\vee B \rightarrow C)[/imath] [imath]\vdash (A\rightarrow B)\wedge (B\rightarrow C) \rightarrow(A\rightarrow C)[/imath] [imath](A\rightarrow B) \vee E,(C\rightarrow D) \vee E \vdash((B\rightarrow C)\rightarrow(A\rightarrow D)) \vee E[/imath] [imath]A \vee C, (A\rightarrow B)\vee C\vdash B\vee C[/imath] [imath]A\rightarrow B, C\rightarrow D\vdash(B\rightarrow C)\rightarrow (A\rightarrow D)[/imath] |
1039990 | Find [imath]\int_0^2 \arctan(\pi x)-\arctan(x)\, \mathrm dx[/imath]
Find [imath]\int_0^2 \arctan(\pi x)-\arctan(x)\, \mathrm dx[/imath] The hint is also given : Re-write the Integrand as an Integral I think we have to Re-write this single integral as a double integral and then do something. Any one any ideas | 1005422 | Evaluating [imath]\int_0^2(\tan^{-1}(\pi x)-\tan^{-1} x)\,\mathrm{d}x[/imath]
Hint given: Write the integrand as an integral. I'm supposed to do this as double integration. My attempt: [imath]\int_0^2 [\tan^{-1}y]^{\pi x}_{x}[/imath] [imath]= \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \mathrm{d}x} {y^2+1}[/imath] [imath]= \int_2^{2\pi} \int_{\frac{y}{\pi}}^2 \frac { \mathrm{d}x \mathrm{d}y} {y^2+1}[/imath] [imath]= \int_2^{2\pi} \frac { [x]^2_{\frac{y}{\pi}} \mathrm{d}y } { y^2+1}[/imath] [imath]= \int_2^{2\pi} \frac { 2- {\frac{y}{\pi}} \mathrm{d}y } { y^2+1}[/imath] Carrying out this integration, I got, [imath]2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}][/imath] But I'm supposed to get [imath]2[\tan^{-1} 2 \pi - \tan^{-1} 2] - \frac {1}{2 \pi} [\ln \frac {1+4 {\pi}^2} {5}]+ [\frac {\pi-1}{2 \pi}] \ln 5[/imath] Can someone please explain where I'm wrong? I've failed to pinpoint my mistake. Thank you. |
1048265 | binomial coefficient and recurrence relation
any hints on how to solve the recurrence relations for the following binomial coefficient \begin{equation} {n \choose k}=\begin{cases} 1, & \text{if [imath]k\in\{0,n\}[/imath]}.\\ {n-1 \choose k-1}+{n-1 \choose k}, & \text{otherwise}. \end{cases} \end{equation} for [imath] 0\ge k \le n [/imath] | 545089 | Proving an Combination formula [imath] \binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}[/imath]
Proving an Combination formula [imath]\displaystyle \binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}[/imath] [imath]\bf{My Try}[/imath]::[imath]\displaystyle{\binom{n-1}{k}+\binom{n-1}{k-1}=}[/imath] [imath]\displaystyle{\frac{\left(n-1\right)!}{k!\left(n-k-1\right)!}+\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k\right)!}=}[/imath] [imath]\displaystyle{\frac{\left(n-1\right)!}{k\,\left(k-1\right)!\left(n-k-1\right)!}+\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k-1\right)!\,\left(n-k\right)}}=[/imath] [imath]\displaystyle \frac{(n-1)!}{(k-1)!\cdot (n-k-1)!}\left(\frac{1}{k}+\frac{1}{n-k}\right) = \binom{n}{k}[/imath] My Question is How can i prove using combinational argument Help Required Thanks |
1046502 | A question on Kernel.
Suppose [imath]K(x,y)[/imath] is a symmetric kernel and [imath]K(x,.), K(.,y)[/imath] are in [imath]L^1(\Omega)[/imath]. Let [imath]\phi\in L^{2}(\Omega)[/imath], where [imath]\Omega[/imath] everywhere is a domain in [imath]\mathbb{R}^n[/imath]. Can [imath]\int K(x,y)\phi(y)dy[/imath] belong to [imath]L^2[/imath]?. In other words can an inequality of the type [imath]||\int K(x,y)\phi(y)dy||_{2}\leq||K||_{L^1{\Omega\times\Omega}}||\phi||_{2}[/imath] exist?. | 1047820 | Does an integral operator with a symmetric integrable kernel have to be bounded on [imath]L^2[/imath]?
Suppose [imath]K(x,y)[/imath] is a symmetric kernel. Let [imath]\phi\in L^2(\Omega)[/imath], where [imath]\Omega[/imath] everywhere is a domain in [imath]R^n[/imath]. Can [imath]\int_{\Omega}K(x,y)\,\phi(y)\,dy[/imath] belong to [imath]L^2[/imath]? In other words can an inequality of the type [imath] \left\|\int_{\Omega}K(x,y)\,\phi(y)\,dy\right\|_{2} \leq \|K\|_{L^1(\Omega\times\Omega)} \|\varphi\|_{2} [/imath] exist? |
1048702 | Integral with [imath]e[/imath] , [imath] \int e^{x^3}\;dx[/imath]
[imath] \int e^{x^3}\;dx[/imath] so, i'm searching for answer this question, i think that this not is too easily, but i think this integral not exist solution undefined, this integral would be easy if had the derivative out but i'snt happen | 270721 | How to evaluate the integral [imath]\int e^{x^3}dx [/imath]
How to evaluate the integral [imath]\int e^{x^3}dx \quad ?[/imath] I've tried to set [imath]t=x^3[/imath], but it seems to be a blind alley; I don't know what to do with [imath]\int\frac{e^t}{3\sqrt[3]{t^2}}dt[/imath]. |
1048566 | Diophantine solutions to [imath]x^y-y^x=1[/imath]
[imath]x^y-y^x=1[/imath] for [imath]x,y\in\Bbb Z[/imath] and [imath]x,y>1[/imath] is clearly a special case of very well known Catalan's conjecture (now resolved). It seems to be very limited special case, but I was told by someone today that solution to this, appears simple, problem, is in fact equivalent to all of Catalan's conjecture. I had some hard time believing this, and I still doubt that this is true. When graphing solutions to this equation among real numbers it clearly looks like [imath]y[/imath] is bounded below 2 for all [imath]x[/imath]. I tried to employ some analytical tools to show this last fact, but I didn't manage to get anywhere. My question is, is this special case really equivalent to full Catalan's conjecture? If it's not, is there some elementary way of showing that it's only integer solution is [imath](3,2)[/imath]? Thanks in advance. | 1042371 | Number theory: [imath]x^y + 1 = y^x[/imath]
Today a friend told me the equality: [imath]2^3 + 1 = 3^2[/imath], and I wondered if there exist more solutions to the general problem [imath]x^y + 1 = y^x[/imath] where [imath]x[/imath] and [imath]y[/imath] are integers. Some research led me to the result for [imath]x^y = y^x[/imath], which has no integer solutions except [imath]x = 2[/imath] and [imath]y = 4[/imath] (assumed [imath]x\neq y[/imath]). Is this a related result, or do more integer solutions exist? |
1048642 | Evaluate [imath]\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x}\text{d}x[/imath]
Evaluate [imath] \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} \text{d}x[/imath] where [imath]n\in\mathbb{N}[/imath] This one is another intriguing question from my worksheet. I'm only allowed to use elementary methods and high school math. However, I can't see a way to do this without derivative under the integral (which is not allowed). Please Help! Thanks. | 1034156 | Ways to prove [imath] \int_0^\pi dx \frac{\sin^2(n x)}{\sin^2 x} = n\pi[/imath]
In how many ways can we prove the following theorem? [imath]I(n):= \int_0^\pi dx \frac{\sin^2(n x)}{\sin^2 x} = n\pi[/imath] Here [imath]n[/imath] is a nonnegative integer. The proof I found is by considering [imath]I(n+1)-I(n)[/imath], which can be reduced to [imath] g(n):= \int_0^\pi dx \frac{\sin(2 n x) \cos x}{\sin x}. [/imath] I then proved that [imath]g(n)=g(n+1)[/imath], whence [imath]g(n) = g(1) = \pi[/imath]. This completes the proof. I was wondering if there is a more direct way to prove it. By 'direct' I mean without deriving auxiliary recursions. |
1045443 | Prove that series converge
Show if series [imath]\displaystyle \sum_{n=1}^{\infty}a_n[/imath] has a positive terms and converge then the series [imath]\displaystyle \sum_{n=1}^{\infty}\frac{\sqrt{a_n}}{n}[/imath] converge too. I don't have idea what to use here, I'd be greatful for any hints | 1048143 | Suppose [imath]\sum{a_n}[/imath] converges. How do I prove that [imath]\sum{\frac{\sqrt{a_n}}{n}}[/imath] converges
Suppose [imath]\sum{a_n}[/imath] converges. How do I prove that [imath]\sum{\frac{\sqrt{a_n}}{n}}[/imath] converges. So I know that just because [imath]\sum{a_n}[/imath] converges, do not mean I can say anything about the converges of its square root. So I know that I can prove that if [imath]\sum{\sqrt{a_n}}[/imath] converges then [imath]\sum{\frac{\sqrt{a_n}}{n}}[/imath] converges by Abel's Test, but I do not know where to start for the case where [imath]\sum{\frac{\sqrt{a_n}}{n}}[/imath] diverges |
1049500 | To find all Ring homomorphisms from [imath]\mathbb Z_m[/imath] to [imath]\mathbb Z_n[/imath]
How to find all Ring homomorphisms from [imath]\mathbb Z_m[/imath] to [imath]\mathbb Z_n[/imath] (with the usual ring structure ) ? | 263063 | The number of ring homomorphisms from [imath]\mathbb{Z}_m[/imath] to [imath]\mathbb{Z}_n[/imath]
I face the problem of finding how many non-trivial ring or group homomorphisms there are from [imath]\mathbb{Z}_m[/imath] to [imath]\mathbb{Z}_n[/imath], where [imath]m<n[/imath]. Is there any general formula? At the moment, I want to know how many ring homomorphisms there are when [imath]m=12,n=28[/imath]. Please help. |
1050037 | Permutations with repeated elements
Say you have n digits and you want to see how many different numbers you can form with these digits (i.e. all possible numbers, not only the ones that are n digits long). These digits however aren't unique. How would one approach such a question? There wouldn't be any problem if you considered only the numbers that are 8 digits long, but unfortunately that's not the case. For example: how many numbers can you make with the following 8 digits: [imath]1, 2, 2, 3, 3, 3, 0, 0[/imath]. For 8 digits the answer would be quite simple: [imath]\frac{8!}{1!2!3!2!}[/imath] I also figured the answer above includes the answer for 7 and 6 digits because of the two zeros. But I don't know how to continue for 5 etc.. digits. | 1048536 | How many different numbers can you make with the following digits?
How many different numbers can you make with the following eight digits? [imath]1, 2, 2, 3, 3, 3, 0, 0[/imath] The problem I encounter is how to include numbers that aren't 8 digits long? For instance the number 12. I know the answer for 8 digit numbers (assuming zero can be a starting digit): [imath]\frac{8!}{2!3!2!}[/imath] But I can't seem to figure out how to include all the other numbers that aren't 8 digits long. |
292423 | Proving the sum of the first [imath]n[/imath] natural numbers by induction
I am currently studying proving by induction but I am faced with a problem. I need to solve by induction the following question. [imath]1+2+3+\ldots+n=\frac{1}{2}n(n+1)[/imath] for all [imath]n > 1[/imath]. Any help on how to solve this would be appreciated. This is what I have done so far. Show truth for [imath]N = 1[/imath] Left Hand Side = 1 Right Hand Side = [imath]\frac{1}{2} (1) (1+1) = 1[/imath] Suppose truth for [imath]N = k[/imath] [imath]1 + 2 + 3 + ... + k = \frac{1}{2} k(k+1)[/imath] Proof that the equation is true for [imath]N = k + 1[/imath] [imath]1 + 2 + 3 + ... + k + (k + 1)[/imath] Which is Equal To [imath]\frac{1}{2} k (k + 1) + (k + 1)[/imath] This is where I'm stuck, I don't know what else to do. The answer should be: [imath]\frac{1}{2} (k+1) (k+1+1)[/imath] Which is equal to: [imath]\frac{1}{2} (k+1) (k+2)[/imath] Right? By the way sorry about the formatting, I'm still new. | 1079191 | Understanding the proof that the sum of the first [imath]n[/imath] natural numbers is [imath]\frac{1}{2}n(n+1)[/imath]
I'm reading the following book: http://www.cs.princeton.edu/courses/archive/spring10/cos433/mathcs.pdf. In page 26, they attempt to prove the following theorem using Induction: For all [imath]n \in \mathbb{N}[/imath]: [imath]1 + 2 + \cdots + n = \frac{n(n+1)}{2}[/imath] In order to prove the theorem we need to prove the following statements: [imath]P(0)[/imath] is true. For all [imath]n \in \mathbb{N}[/imath], [imath]P(n)[/imath] implies [imath]P(n + 1)[/imath]. Proving the first one is easy: [imath]P(0) = 0 = 0 * (0 + 1) / 2[/imath] [imath]= 0 = 0 * 1 / 2[/imath] [imath]= 0 = 0 / 2[/imath] [imath]= 0 = 0[/imath] In order to prove the second, they state the following: [imath]1+2+\cdots + n + (n+1) = \frac{n(n+1)}{2} + (n+1) \tag{1}[/imath] [imath] = \frac{(n+1)(n+2)}{2} \tag{2}[/imath] I don't understand how they get from (1) to (2) and how that proves that proves that for all [imath]n \in \mathbb{N}[/imath], [imath]P(n)[/imath] implies [imath]P(n + 1)[/imath] is true. I'm clearly missing something in the process. Can someone clear the fog for me? |
1050588 | Proof by induction of the formula for [imath]2^0+2^1+2^2+...+2^n[/imath]?
[imath]2^0+2^1+2^2+...+2^n[/imath] for [imath]n ∈ \mathbb{N} \cup \{0\}[/imath]. I made a conjecture that this is [imath]2^{n+1} - 1[/imath]. Now I have to prove it by induction. I tested the base case where it's equal to zero, and it worked. Then I did "assume [imath]k∈ \mathbb{N} \cup \{0\}[/imath] and [imath]2^0+2^1+2^2+...+2^k[/imath]". I got marked down for saying "Let [imath]n=k+1[/imath]", and did a bunch of math for it to become [imath]2^{k+2} -1[/imath] and I was apparently wrong >_>? | 658992 | Proving the geometric sum formula by induction
[imath]\sum_{k=0}^nq^k = \frac{1-q^{n+1}}{1-q}[/imath] I want to prove this by induction. Here's what I have. [imath]\frac{1-q^{n+1}}{1-q} + q^{n+1} = \frac{1-q^{n+1}+q^{n+1}(1-q)}{1-q}[/imath] I wanted to factor a [imath]q^{n+1}[/imath] out of the second expression but that 1- is screwing it up... |
1050983 | Simple Combinatorial Proof
[imath] \sum_{i=0}^n {n \choose i} i = n2^{n-1} [/imath] I am having trouble formulating a combinatorial proof. An algebraic proof is quite simple, where one expands [imath](1 + x)^{n}[/imath] and then takes the derivative. Any help with the combinatorial proof would be greatly appreciated. | 388587 | Combinatorial proof of [imath]\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}[/imath].
Prove that [imath]\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}[/imath] I can't find counting interpretations for either of the sides. A hint of "if [imath]S[/imath] is a subset of [imath]\{1, . . . , n\}[/imath] and [imath]S^\prime[/imath] is its complement then [imath]|S| + |S^\prime| = n[/imath]" was also given, but I still don't know how to begin. |
1051358 | linear algebra diagonalization
[imath]A =\begin{pmatrix}a &b \\ c & d \end{pmatrix}[/imath] show that: [imath]A[/imath] is diagonalizable if [imath](a-d)^2 +4bc > 0[/imath] [imath]A[/imath] is not diagonalizable if [imath](a-d)^2 + 4bc\leq 0[/imath] | 1046515 | Diagonalisable or not?
Let [imath]A = \begin{pmatrix}a & b\\c & d\end{pmatrix}.[/imath] Show that 1) [imath]A[/imath] is diagonalisable if [imath](a - d)^2 + 4 bc > 0[/imath] 2) [imath]A[/imath] is not diagonalisable if [imath](a - d)^2 + 4 bc < 0[/imath] |
1051466 | Integration-Can anyone give the technique t0 integrate this type of problems?
[imath] \int_0^\pi \sqrt{4\sin^2x-4\sin x+1}\;dx [/imath] Can anyone give the technique t0 integrate this type of problems? I didn't do this type of problems earlier. | 1042761 | Most efficient way to integrate [imath]\int_0^\pi \sqrt{4\sin^2 x - 4\sin x + 1}\,dx[/imath]?
[imath]\int_0^\pi \sqrt{4\sin^2 x - 4\sin x + 1}\,dx[/imath] Please help with this. I cannot do this problem in a definite way. |
1051431 | Convergent sequence?
Why does the [imath]\lim \sum{n^{(1/n)}-1}[/imath] diverge as [imath]n\rightarrow \infty[/imath]? I suspected it would converge as [imath]\lim {n^{(1/n)}}=1[/imath] as [imath]n\rightarrow \infty[/imath] but computation show otherwise. So, now I am trying to show the partial sums are unbounded but didnt get far. | 416627 | Convergence of the series [imath]\sqrt[n]n-1[/imath]
Let [imath]a_n=\sqrt[n]n-1[/imath]. Does [imath]\sum_{n=1}^\infty a_n[/imath] converge? |
1051634 | Is [imath]\mathbb{Q}\times \mathbb{Q}[/imath] a field?
Is [imath]\mathbb{Q} \times \mathbb{Q}[/imath] a field ? We know that [imath]\mathbb Q[/imath] is a field. But, if we think of [imath]\mathbb Q\times \mathbb Q[/imath], what would be the answer? | 229836 | Explain why [imath]\mathbb{Z \times Z}[/imath] and [imath]\mathbb{R \times R}[/imath] is not a field
Explain why [imath]\mathbb{Z \times Z}[/imath] and [imath]\mathbb{R \times R}[/imath] is not a field and that why any external direct sum of two fields cannot be a field. I believe it has much to do with the lack of every non-zero element having an inverse, however I am having difficulty seeing it. |
1051352 | Circles intersecting at A and B
Question: Two given circles intersect at A and B. A straight line through B meets the circles again at C and D. Prove that CD is greatest when it is parallel to the line joining the centres My attempt: I made this diagram on Paint.NET: Here CD is the required line parallel to the line [imath]MNOP[/imath] joining the centers N and O, and EF is another line not parallel to [imath]MNOP[/imath]. Both CD and EF pass through point B. I know I have to prove that [imath]CD > EF[/imath]. But I don't know how to solve this problem. And I need hints. | 458726 | problem on intersecting circles
If two non-congruent circles with centres [imath]A[/imath] and [imath]B[/imath] intersect at [imath]P[/imath] and [imath]Q[/imath], then prove that among all lines drawn through [imath]P[/imath] and terminated by the circumference of two circles, the line which is parallel to [imath]AB[/imath] is the greatest. |
1051790 | Showing components of sets in product space are equal
Let [imath]U[/imath] and [imath]V[/imath] be nonempty open sets in the product space [imath]X = \prod_{\alpha \in I}X_{\alpha}[/imath]. Show that there exists [imath]u \in U[/imath] and [imath]v \in V[/imath] such that for all but finitely many coordinates [imath]\alpha[/imath], the components [imath]u_{\alpha}[/imath] and [imath]v_{\alpha}[/imath] are equal. So [imath](U \times V) \in X[/imath] and [imath]U,V[/imath] are open. We want to show that there is some [imath]\alpha[/imath] such that [imath]u_{\alpha} = v_{\alpha}[/imath] where [imath]u_{alpha}[/imath] and [imath]v_{\alpha}[/imath] are in [imath]U,V[/imath] respectively. Yes? I'm not sure how we can use the information given here. | 1050820 | any two open sets in the product topology contain points that differ in finitely many components
For non empty open sets [imath]U[/imath] and [imath]V[/imath] in the product topology [imath] \prod_{\alpha \in I} X_\alpha[/imath] show that there are points [imath]x \in U[/imath] and [imath]y \in V[/imath] that only differ in finitely many coordinates. What I have so far (not much): Since [imath]U[/imath] and [imath]V[/imath] are open in the product topology, both [imath]U[/imath] and [imath]V[/imath] are the union of products of open sets that differ from their corresponding topologies in only finitely many elements. Even though the possible values that coordinates of points in [imath]U[/imath] can differ in only finitely many elements from possible values that coordinates of points in [imath]V[/imath] can take on, points in [imath]U[/imath] and [imath]V[/imath] can have infinitely many coordinates, so how would I go about finding two possibly infinitely long points that only differ in finitely many coordinates? |
1052146 | Prove that [imath]f(x) = 0[/imath] for all [imath]x ∈ R[/imath].
Let [imath]f(x)[/imath] be a continuous function such that [imath]f(r) = 0[/imath] for all rational numbers r. Prove that [imath]f(x) = 0[/imath] for all [imath]x ∈ R[/imath]. | 344821 | if [imath]f[/imath] is continuous on [imath]\mathbb{R}[/imath] and [imath]f(r)=0,r \in \mathbb{Q}[/imath], then [imath]f(x)=0,x \in \mathbb{R}[/imath]
Suppose that [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] is continuous on [imath]\mathbb{R}[/imath] and that [imath]f(r)=0[/imath] for all [imath]r \in \mathbb{Q}[/imath]. Prove that [imath]f(x)=0[/imath] for all [imath]x \in \mathbb{R}[/imath]. My attempt: Define a sequence [imath](x_n)[/imath] where [imath]x_n \in \mathbb{Q}[/imath] for all [imath]n \in \mathbb{N}[/imath] and assume that [imath](x_n) \rightarrow a \not\in \mathbb{Q}[/imath]. Since [imath]f[/imath] is continuous, we have [imath]\lim_n{f(x_n)}=f(a)=0[/imath]. Since [imath]a[/imath] is arbitrary irrational number, we have [imath]f(a)=0[/imath] for all [imath]a \not\in \mathbb{Q}[/imath]. Hence, we proved the statement. Is my proof valid? or is there any flaw ? |
1052390 | Property of subsequences of a Cauchy sequence
I'm going over an old homework question in my Analysis class that I missed. My professor's solution was confusing so I am looking for additional help. The question is: If [imath]\{p_n\}_{n \in \mathbb{N}} [/imath] is a cauchy sequence in a metric space, show that for any [imath]\epsilon > 0[/imath] there exists a subsequence [imath]p_{n_i}[/imath] such that [imath]d(p_{n_i}, p_{n_{i+1}}) < \frac{\epsilon}{2^{i+1}}[/imath]. My solution was to choose [imath]p_{n_1}[/imath] and [imath]p_{n_2}[/imath] such that [imath]d(p_{n_1}, p_{n_2}) < \frac{\epsilon}{4}[/imath] and then choose [imath]p_{n_j}[/imath] for [imath]j \geq 2[/imath] such that [imath]d(p_{n_j}, p_{n_{j+1}}) < \frac{\epsilon}{2^{j+1}}[/imath] which we can do since the sequence is Cauchy. If you could explain how to do it and why my solution doesn't work, that would be much appreciated. Thanks in advance. | 1030207 | If [imath](a_n)[/imath] is Cauchy it has a subsequence [imath](a_{n_i})[/imath] such that [imath]d(a_{n_{i+1}},a_{n_i})<2^{-i}[/imath] for all [imath]i[/imath].
Let [imath](X,d)[/imath] be a metric space and [imath](a_n)[/imath] a Cauchy sequence in [imath]X[/imath]. How to show that there exists [imath]n_1<n_2<\cdots[/imath] such that [imath]d(a_{n_{i+1}},a_{n_i})<2^{-i},[/imath] for all [imath]i\in\mathbb{N}[/imath]? I tried to use induction, but it doesn't work. The base case is OK, but if we have [imath]d(a_{n_2},a_{n_1})<2^{-1},\ldots,d(a_{n_k},a_{n_{k-1}})<2^{-(k-1)},[/imath] I don't see any reason for to exist [imath]n_{k+1}>n_k[/imath] such that [imath]d(a_{n_{k+1}},a_{n_k})<2^{-k}[/imath]. |
1049003 | Show [imath]E(X)= \int_{[0, \infty)} P[X>t] dt[/imath], without using density argument
For [imath]X[/imath], a positive random variable, use Fubini's theorem applied to [imath]\sigma[/imath]-finite measure to prove \begin{align*} E(X)= \int_{[0, \infty)} P[X>t] dt \end{align*} I found several references on the the site like this: Proving [imath]EX=\int_{0}^{\infty}(1-F(x))\, dx[/imath] when [imath]X\geq0[/imath] Expression for [imath]n[/imath]-th moment but all of them assume that [imath]X[/imath] has density or discrete is the are different proof that does not use this assumption. This is what I tried \begin{align*} \int_{[0, \infty)} P[X>t] dt&= \int_{[0, \infty)} E(\mathsf{1}_{[X>t]}) dt=\int_{[0, \infty)} \int_{\Omega} \mathsf{1}_{[X>t]}dP dt\\ &=\int_{\Omega} \int_{[0, \infty)} \mathsf{1}_{[X>t]} dt dP=\int_{\Omega} \int_{[0, X)} 1 dt dP=\int_{\Omega} X dP=E[X] \end{align*} where the first step in second line is due to Fubinni's theorem | 578358 | Show that [imath]E(X)=\int_{0}^{\infty}P(X\ge x)\,dx[/imath] for non-negative random variable [imath]X[/imath]
Show that for a non-negative random variable [imath]X[/imath], [imath]\mathbb E(X)=\int_0^\infty \mathbb P(X\ge x) \, dx.[/imath] I started with [imath]\mathbb E(X)=\int_0^\infty x \, dF(x)=\int_0^\infty \int_0^x dt\,dF(x).[/imath] This is my try. |
132495 | Why don't you need the Axiom of Choice when constructing the "inverse" of an injection?
Suppose [imath]f:X\rightarrow Y[/imath] is a surjection and you want to show that there exists [imath]g:Y\rightarrow X[/imath] s.t. [imath]f\circ g=\mathrm{id}_Y[/imath]. You need the AC to show this. However, suppose [imath]f[/imath] is a injection and you want show there is [imath]g[/imath] s.t. [imath]g\circ f=\mathrm{id}_X[/imath]. Then, according to my textbook, you don't need the AC to show this. This is counterintuitive to me, because it's like you need a special axiom to claim that an infinite product of big sets is nonempty, while you don't need one to claim that an infinite product of singleton sets is nonempty, which seems smaller than the former. So why don't you need the AC to show the latter? EDIT: [imath]X[/imath] should be nonempty. EDIT 2: I realized (after asking this) that my question mostly concerns whether the AC is needed to say that an infinite product of finite sets is nonempty, and why. | 1792980 | Axiom of Choice and LEFT inverse
I am aware of why the Axiom of Choice is equivalent to the the statement that every surjection splits. However, I don't see why we don't also need AC to show that every injection splits. In particular, for any injection [imath]f[/imath], we need to pick some element in the domain of [imath]f[/imath] to which elements in the codomain of [imath]f[/imath] are sent in the case that those elements are not in the image of [imath]f[/imath]. This shows moreover that left inverses are not in general unique. How can we do this without AC? EDIT: My real issue here is in the comparison of the proof of "[imath]f: A \to B[/imath] is a surjection implies [imath]f[/imath] has a right inverse" with the proof "[imath]f: A \to B[/imath] is an injection implies [imath]f[/imath] has a left inverse", and the use of the Axiom of Choice therein. In particular, one apparently need not invoke AC in the injection proof to select in advance a single element [imath]a_0 \in A[/imath] for which to send [imath]b \in B[/imath] in the case that [imath]b \notin f(A)[/imath]. But then, in the same way, in the surjection proof surely one may select for each [imath]b \in B[/imath] a single element [imath]a_b \in f^{-1}[\{b\}][/imath] which then determines a function [imath]g: B \to A[/imath] such that [imath]fg = \text{id}[/imath], all without AC? We know each [imath]f^{-1}[\{b\}][/imath] is not empty, so pick an element in [imath]f^{-1}[\{b\}][/imath] (in the same way we did so in the injection proof by picking [imath]a_0 \in A[/imath]) and send [imath]b[/imath] to that element. |
923355 | Is the series [imath]\sum_{1}^{\infty}\frac{1}{p_{j}}[/imath] where [imath]p_{j}[/imath] is the [imath]j[/imath]th prime convergent?
Does the series [imath]\sum_{n=1 }^{\infty}1/p_{j} [/imath] of reciprocal primes converge? Experimentally, it seems convergent. | 1539980 | Divergence of the series of the inverses of the prime numbers
Can you help me prove that the [imath]\sum \limits _{p \text{ prime}} \dfrac 1 p[/imath] is divergent? |
1053390 | Investigate whether there exists a function [imath]f : [a,b]\rightarrow R[/imath] that is continuous and that takes exactly twice each of its values.
Let a < b. Investigate whether there exists a function [imath]f : [a,b]\rightarrow R[/imath] that is continuous and that takes exactly twice each of its values. | 652550 | A function takes every function value twice - proof it is not continuous
I want to prove the following nice statement I've found: A function [imath]f: [0,1] \rightarrow \mathbb{R}[/imath] takes every function value twice - proof it is not continuous I've already found an answer to my question but one thing is still not clear to my mind. The answer I found was here on math.StackExchange posted by Arthur: Whole post: "Assume that you have a function [imath]f[/imath] that does take every real value exactly twice. Let [imath]f(a) = f(b) = 0[/imath] with [imath]a < b[/imath]. Let [imath]c \in (a,b)[/imath] be given, and assume without loss of generality [imath]f(c) > 0[/imath]. Then for every [imath]0 < \epsilon < f(c)[/imath] there exist [imath]d \in (a,c)[/imath] and [imath]e \in (c,b)[/imath] such that [imath]f(d) = f(e) = \epsilon[/imath]. So on the intervals [imath][0,a)[/imath] and [imath](b,1][/imath] our function cannot go any higher than zero. That means that [imath]f[/imath] should take every positive real value on the interval [imath][a,b][/imath]. However the restriction of [imath]f[/imath] to [imath][a,b][/imath] is continuous from a closed interval to [imath]\mathbb{R}[/imath] and thus bounded. Contradiction." Now everything is clear here except for: "However the restriction of [imath]f[/imath] to [imath][a,b][/imath] is continuous from a closed interval to [imath]\mathbb{R}[/imath] and thus bounded. Contradiction." Well I think know what Arthur is telling there: Because of the boundary there is always at least one function value that can't be taken twice by the function, e.g. just like the peak of [imath]-x^2[/imath]. But that is exactly where I was stuck before I started researching that issue. Is it really possible to just say it the way Arthur did? I thought that I'd have to further somehow prove the statement that boundary leads to that contradiction. I hope that it is clear what I was trying to express here... Anyways as always - thank you for your help! FunkyPeanut |
1053488 | Define [imath] f:\mathbb C\rightarrow \mathbb C[/imath]
Define [imath] f:\mathbb C\rightarrow \mathbb C[/imath] by [imath]f(z)=\begin{cases}0 & \text{if } Re(z)=0\text{ or }Im(z)=0\\z & \text{otherwise}.\end{cases}[/imath] Then the set of points where [imath]f[/imath] is analytic is: (a) [imath]\{z:Re(z)\neq 0[/imath] and [imath]Im(z)\neq 0\}[/imath], (b) [imath]\{z:Re(z)\neq 0[/imath] or [imath]Im(z)\neq 0\}[/imath], (c) [imath]\{z:Re(z)\neq 0\}[/imath], (d) [imath]\{z:Im(z)\neq 0\}[/imath]. my think all the options are correct. because if consider option a) then the function [imath]f[/imath] will be [imath]f(z)=z[/imath] and it is a polynomial function. hence analytic. but for option b) if let [imath]Re(z)\neq 0\}[/imath] but [imath]\{z:Im(z)= 0\}[/imath] then the function [imath]f[/imath] will be [imath]f(z)=0[/imath] and it is also analytic. similarly we can say every option is correct. Am I right? please help me. | 256278 | [imath] f:\mathbb C\rightarrow \mathbb C[/imath] defined by [imath]f(z)=0[/imath], if [imath]Re(z)=0[/imath] or [imath]Im(z)=0[/imath] and [imath]f(z)=z[/imath], otherwise.
I was thinking about the following problem : Define [imath] f:\mathbb C\rightarrow \mathbb C[/imath] by [imath]f(z)=\begin{cases}0 & \text{if } Re(z)=0\text{ or }Im(z)=0\\z & \text{otherwise}.\end{cases}[/imath] Then the set of points where [imath]f[/imath] is analytic is: (a) [imath]\{z:Re(z)\neq 0[/imath] and [imath]Im(z)\neq 0\}[/imath], (b) [imath]\{z:Re(z)\neq 0[/imath] or [imath]Im(z)\neq 0\}[/imath], (c) [imath]\{z:Re(z)\neq 0\}[/imath], (d) [imath]\{z:Im(z)\neq 0\}[/imath]. I think [imath]f(z)=z[/imath] if [imath]f[/imath] is defined on the set [imath]\{z:Re(z)\neq 0[/imath] and [imath]Im(z)\neq 0\}[/imath] and in that case [imath]f[/imath] is analytic. But I am not sure about the other options. So, choice (a) is right. Am I going in the right direction? Please help. Thanks in advance for your time. |
1052883 | If [imath]X[/imath] and [imath]Y[/imath] are independent random variables, does it follow that [imath]X^2[/imath] and [imath]Y[/imath] are independent?
If [imath]X[/imath] and [imath]Y[/imath] are independent random variables, then can I say that [imath]X^2[/imath] and [imath]Y[/imath] are independent? | 707655 | Showing independence of random variables
When proving [imath]\bar x[/imath] and [imath]S^2[/imath] are independent in my noted it says that "functions of independent quantities are independent ". Can someone tell me how functions of independent quantities are independent happen? Also let [imath]X_1,X_2,X_3,\dots,X_n[/imath] be a random sample.And suppose we want to estimate parameter [imath]\theta[/imath] using [imath]T(x)[/imath] as an estimator. In this I have a expression as [imath]E\{ [ T(X)-E(T(X))][E(T(x)-\theta)]\}[/imath]. I want to know if I can say that in [imath][T(X)-E(T(X))][/imath] since [imath]E(T(X)[/imath] (expected value is a one particular fixed value) and [imath]T(X)[/imath] are independent of one another. Also the expression [imath][E(T(x)-\theta)][/imath] is independent of [imath]X[/imath]. Hence the two expressions [imath][T(X)-E(T(X))][/imath] and [imath][E(T(x)-\theta)][/imath] are independent of one another. So that I can use if [imath]a,b[/imath] are independent [imath]E(ab)=E(a)E(b)[/imath] form |
1054094 | Finding a group that spans a vector space of Polynomials
I am having problem finding a group that spans this vector space: [imath]S={ p(x) \in R_4[x]} \big| p(x)=p(x-1) [/imath] My idea was to substitute: [imath] a+bx+cx^{2}+dx^{3}+ex^{4}=a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}[/imath] But it led me nowhere. what can I do? Thanks in advance. | 1053465 | Finding a basis for a certain vector space of periodic polynomials
I am having a little bit of trouble solving an homework question. I found that [imath]S={ p(x) \in R_4[x]} \big| p(x)=p(x-1) [/imath] is a vector space. Now I need to find some set, K that holds [imath]{span(k)=S}[/imath] My idea was to substitute: [imath] a+bx+cx^{2}+dx^{3}+ex^{4}=a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}[/imath] But here I get something that seems a little bit weird, I'd like to look if I have any mistakes that I did not notice. and if I had mistakes, how to make my answer right. I calculated the right side of the equation and I got to : [imath]a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}=a+bx-b+cx^{2}-2cx+c+dx^{3}-3dx^{2}+3dx-d+ex^{4}-4ex^{3}+6ex^{2}-4ex+e= (a-b+c-d+e)+(b-2c+3d-4e)x+(c-3d+6e)x^{2}+(d-4e)x^{3}+ex^{4} [/imath] From here, [imath] (a-b+c-d+e)+(b-2c+3d-4e)x+(c-3d+6e)x^{2}+(d-4e)x^{3}+ex^{4}=a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}\Rightarrow [/imath] [imath](-b+c-d+e)+(-2c+3d-4e)x+(-3d+6e)x^{2}+(d-4e)x^{3}=0+0x+0x^{2}+0x^{3}+0x^{4}[/imath] And the solution to this equation is only whan [imath]a=b=c=d=e=0[/imath]. I think that I am maybe not in the right direction. Do you have any Idea how to find a set [imath]K\Rightarrow span(K)=S[/imath] ? Thanks in advance. |
1054262 | Finding the uniformising parameter of a DVR
I am looking to find the intersection number of the affine plane curves [imath]F=Y^2-X^3+X[/imath] and [imath]G=(X^2+Y^2)^3-4X^2Y^2[/imath], and I need to do it with the order function of the local ring of [imath]F[/imath] at the origin. So my attempt goes as follows: [imath]k[X,Y] \twoheadrightarrow \Gamma(F)=k[X,Y]/(F) \hookrightarrow \mathcal{O}_{(0,0)}(F)[/imath] [imath] G \mapsto \bar{G} = G\mod (F) \mapsto \frac{\bar{G}}{1} [/imath] Then I get the intersection number as [imath]I((0,0),F\cap G) = \text{Ord}_{(0,0)}^F(\bar{G})[/imath]. For the map onto the coordinate ring of [imath]F[/imath], I wrote [imath]X\mapsto x, Y\mapsto y : y^2=x^3-x[/imath]. From there I thought that [imath]\bar{G} = (x^2+(x^3-x))^3 - 4x^2(x^3-x)[/imath]. The map into the local ring (which I know should be a DVR as the origin is a simple point of [imath]F[/imath]) is a natural injection, so I thought I should be able to write [imath]\bar{G} = u\cdot t^n[/imath] as one can do in a DVR. I can't! I'm not sure what is going wrong. | 1053352 | Intersection Number of [imath]B = Y^2 - X^3 + X[/imath] and [imath]F = (X^2 + Y^2)^3 - 4X^2Y^2[/imath] using the fact [imath]I(P,F \cap B) = ord_P^B(F) [/imath].
Firstly, I apologise that this question is specific to two polynomials. I understand that this post will not help a lot of people, and for that I am sorry. I will make it up to you all by anwsering lots more questions in the future! However, as my lecturers enjoy time off on the weekend, I am unable to ask this to anyone else who would know what the heck I'm talking about. Sorry if my notation is not widely used, I can elaborate more if required. Please excuse my poor Latex too. Let: [imath]B = Y^2 - X^3 + X[/imath], and [imath]F = (X^2 + Y^2)^3 - 4X^2Y^2[/imath]. I wish to find the intersection number of [imath]B[/imath] and [imath]F[/imath] at [imath]P = (0,0)[/imath] i.e [imath]I(P,F \cap B). [/imath] However, I want to use the following proposition to do this: If [imath]P[/imath] is simple on [imath]B[/imath] then [imath]I(P,F \cap B) = ord_P^B(F) [/imath]. I will omit proving that [imath]P[/imath] is simple on [imath]B[/imath] as I hope it is clear to see it is. I know that [imath]I(P,F \cap B) > M_p(F) * M_p(B) = 1 * 4 [/imath], where [imath]M_P(G)[/imath] is the multiplicity of [imath]P[/imath] on [imath]G[/imath]. Now, consider the Image of [imath]X[/imath] and [imath]Y[/imath]: [imath]K[X,Y] \longrightarrow \frac{K[X,Y]}{Y^2 - X^3 + X} \longrightarrow O_P(Y^2 - X^3 + X)[/imath] [imath]X \longrightarrow x[/imath] [imath]Y \longrightarrow y[/imath] Where lower case denotes the image. Then [imath]ord_P^{Y^2 - X^3 + X}((X^2 + Y^2)^3 - 4X^2Y^2)[/imath] Can be found by substituting [imath]y^2 = x^3 - x[/imath] into [imath](x^2 + y^2)^3 - 4x^2y^2[/imath] Which gives: [imath]ord_P^{Y^2 - X^3 + X}(x^9+3x^8 -5x^6 -4x^5 +3x^4 +3x^3)=3[/imath] i.e [imath]I(P,F \cap B) = 3[/imath], a contradiction to [imath]I(P,F \cap B) > 4[/imath] as above. The correct answer I know is 6, so where have I gone wrong? |
1054456 | Polar cones' property
I am trying to prove: [imath]A \subseteq B \implies B^\circ \subseteq A^\circ[/imath] where [imath]A^\circ[/imath] is polar cone of [imath]A[/imath] ([imath]A[/imath] convex cone) and [imath]B^\circ[/imath] is polar cone of [imath]B[/imath] ([imath]B[/imath] convex cone) | 677540 | Proof of properties of dual cone
Show that if C [imath]\subseteq[/imath] D then [imath]D^*[/imath] [imath]\subseteq[/imath] [imath]C^*[/imath] where * is dual cone operation. Can somebody explain it. |
1009066 | Density function for RV X
The density function for a random variable X is given in terms of a constant c. Find the value of c. What is the corresponding distribution function? Sketch both the density and the distribution functions. Finally, find the probabilities. 5.1 [imath]f(x)=0[/imath] for [imath]x<0[/imath] and [imath]f(x)=\frac{c}{(x+1)^4}[/imath] for [imath]0<x[/imath] [imath]P(X>4)[/imath] [imath]P(X<2)[/imath] [imath]P(1\le\ X<3)[/imath] I get c=3 by taking the integral, and then I am a little confused as to why [imath]F(x)=1-\frac{1}{(x+1)^3}[/imath] Is that because in solving for c, I ended up with [imath]1=\frac{c}{3(x+1)^3}[/imath], so substituting c in I get [imath]1=\frac{1}{(x+1)^3}[/imath]? And, then [imath]P(X>4)=1-P(X\le\ 4)[/imath] [imath]=1-(1-{1}{(1+4)^3})[/imath] [imath]=1/125[/imath] [imath]P(X<2)=1-{1}{(1+2)^3}[/imath] [imath]=26/27[/imath] I guess I am a little confused about what F(x) and f(x) actually MEAN? And thus that leads to my confusion about what the differences between > and < and greater than or equal to are, etc etc in terms of the equations. The ones above I solved by comparing to the book but I still don't understand what they mean really. So as a result, I don't really get how to solve [imath]P(1\le\ X<3)[/imath]. | 1008534 | Density function for RV
The density function for a random variable X is given in terms of a constant [imath]c[/imath]. Find the value of [imath]c[/imath]. What is the corresponding distribution function? Sketch both the density and the distribution functions. Finally, find the probabilities. 5.1 [imath]f(x)=0[/imath] for [imath]x<0[/imath] and [imath]f(x)=\frac{c}{(x+1)^4}[/imath] for [imath]0>x[/imath] [imath]P(X>4)[/imath] [imath]P(X<2)[/imath] [imath]P(1\le\ X<3)[/imath] I get [imath]c=3[/imath] by taking the integral, and then I am a little confused as to why [imath]F(x)=1-\frac{1}{(x+1)^3}[/imath] Is that because in solving for [imath]c[/imath], I ended up with [imath]1=\frac{c}{3(x+1)^3}[/imath], so substituting c in I get [imath]1=\frac{1}{(x+1)^3}[/imath]? And, then [imath]P(X>4)=1-P(X\le\ 4)[/imath] [imath]=1-(1-{1}{(1+4)^3})[/imath] [imath]=1/125[/imath] [imath]P(X<2)=1-{1}{(1+2)^3}[/imath] [imath]=26/27[/imath] I guess I am a little confused about what [imath]F(x)[/imath] and [imath]f(x)[/imath] actually MEAN? And thus that leads to my confusion about what the differences between [imath]>[/imath] and [imath]<[/imath] and greater than or equal to are, etc etc in terms of the equations. The ones above I solved by comparing to the book but I still don't understand what they mean really. So as a result, I don't really get how to solve [imath]P(1\le\ X<3)[/imath]. 5.2 [imath]f(x)=ce^x[/imath] for [imath]x<0[/imath] and [imath]f(x)=ce^{-x}[/imath] for [imath]0>x[/imath] 5.3 [imath]f(x)=0[/imath] for [imath]x<-1[/imath] or [imath]x>2[/imath] and [imath]f(x)=cx^2[/imath] for [imath]-1<x<2[/imath] |
1048160 | Find the root of C
Can u help me to find a root for C (except c = 0) in below equation. [imath]ce^{-c}-{10\over5}(1-e^{-c})^2=0[/imath] by expanding this I got, [imath]ce^{-c}-2 + 4 e^{-c}-2e^{-2c}=0[/imath] now grouping, [imath](c+4)e^{-c}-2-2e^{-2c}=0 \rightarrow (1) [/imath] let [imath] e^{-c} = x[/imath] then [imath] c = -logx[/imath] Substituting these values in (1), [imath]x^2 + (logx - 4) x +1 = 0[/imath] now if I apply formula to find the root of quadratic equation, (log x -4) term is coming inside the square root and making it complex to find the root, Am I proceeding it right? Is there any other way to find the root of this equation. If I plot a graph for this the curve is cutting x axis at 0.49, which is one of the root. How to arrive this mathematically? | 1040907 | Find the positive root of the equation [imath]ce^{-c}-2(1-e^{-c})^2=0[/imath]
Can you help me find a root for [imath]c[/imath] in the equation below? [imath]ce^{-c}-{10\over5}(1-e^{-c})^2=0[/imath] By expanding this I got, [imath]ce^{-c}-2 + 4 e^{-c}-2e^{-2c}=0[/imath] now grouping, [imath](c+4)e^{-c}-2-2e^{-2c}=0 \tag{1} [/imath] Let [imath] e^{-c} = x[/imath] then [imath] c = -\log x[/imath] Substituting these values in (1), [imath]x^2 + (\log x - 4) x +1 = 0[/imath] Now if I apply formula to find the root of quadratic equation, [imath](\log x -4)[/imath] term is coming inside the square root and making it complex to find the root. Am I proceeding right? Is there any other way to find the root of this equation? If I plot a graph for this the curve is cutting x axis at [imath]0.49[/imath], which is one of the root. How to arrive at this mathematically? ? |
1046313 | Differential equation [imath]x'=x^2- \frac{2}{t^2}[/imath]
[imath]x'=x^2- \frac{2}{t^2}[/imath] Hmm I have not idea how to do it... Maybe any smart substitution could work? | 982149 | Solve differential equation, [imath]x'=x^2-2t^{-2}[/imath]
Solve differential equation: [imath]x'=x^2- \frac{2}{t^2}[/imath] Maybe is it sth connected with homogeneous equation? I have no idea how to solve it. |
1055604 | Proof entire function is constant
[imath]f[/imath] is an entire function, suppose: [imath]\lim_{z\to\infty}\frac{\Re f}{z} =0[/imath] Can we proof [imath]f[/imath] is constant? Thanks! | 446124 | Proving that a function has a removable singularity at infinity
I'm having trouble with the following exercise from Ahlfors' text (not homework) "If [imath]f(z)[/imath] is analytic in a neighborhood of [imath]\infty[/imath] and if [imath]z^{-1} \Re f(z) \to0[/imath] as [imath]z \to \infty[/imath], show that [imath]\lim_{z \to \infty} f(z)[/imath] exists. (In other words, the isolated singularity at [imath]\infty[/imath] is removable) Hint: Show first, by use of Cauchy's integral formula, that [imath]f = f_1 +f_2[/imath] where [imath]f_1(z) \to 0[/imath] for [imath]z \to \infty[/imath] and [imath]f_2(z)[/imath] is analytic in the whole plane. My attempt: [imath]f[/imath] is analytic in the exterior of some disk [imath]\{|z|\geq R_0 \}[/imath]. For large enough [imath]|z|[/imath] the circle [imath]C_1(z)[/imath] around [imath]z[/imath] with radius one lies in the domain of analyticity, and we can apply Cauchy's integral formula: [imath]f(z)=\frac{1}{2\pi i}\oint_{C_1(z)}\frac{f(\zeta)}{\zeta-z} \mathrm{d} \zeta=\frac{1}{2 \pi i} \oint_{C_1(z)}\frac{\Re f(\zeta)}{\zeta-z} \mathrm{d} \zeta+\frac{1}{2\pi} \oint_{C_1(z)}\frac{\Im f(\zeta)}{\zeta-z} \mathrm{d} \zeta. [/imath] I want to say that the last expression is exactly the decomposition [imath]f_1+f_2[/imath]. Indeed, using Cauchy's estimate (and the fact about the real part) one can show that the first integral tends to zero for large [imath]|z|[/imath], but I can't understand how to define the other one for all [imath]z \in \mathbb C[/imath]. In fact, even if I have such a decomposition, I don't think that it will suffice: For instance [imath]f(z)=f_1(z)+f_2(z)[/imath] with [imath]f_1(z)=0,f_2(z)=\exp(z)[/imath] has an essential singularity at [imath]\infty[/imath]. Thank you for any directions. |
1055557 | Prove that [imath]x=0.1234567891011\cdots[/imath] is irrational
Prove that [imath]x=0.1234567891011\cdots[/imath] is irrational Proof: we argue by contradiction.suppose x is rational. then its decimal expansion ultimatetly periodic. Lets p denote the perid of this expansion. Now consider a block [imath]B=000 \cdots0[/imath] [imath] p[/imath] times since any integer of the form [imath]10^k[/imath] with [imath] k \geq p[/imath] contains [imath]p[/imath] consecutive [imath]0[/imath]'s . this block must accure infinitely often in thedecimal expansion of [imath]x[/imath] by ourassumption that this expansionis ultimately periodic with period [imath]p[/imath] , this implies that [imath]B[/imath] must be the repeatingperiod block. Which means that the sequence consists of all [imath]0[/imath]'s from some point onwards. but this clearly contradictions the construction of the sequence. This proof is correct? If it is not, what's the issue. | 935389 | How to prove that a number is irrational
We write all postive whole integers after the comma, how do we prove that this is an irrational number? ([imath]0.1234567891011121314...[/imath]) |
1047844 | Rational or irrational
We know that [imath]0.12345\cdots = \frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\frac{4}{10^4}+\frac{5}{10^5}\cdots = \sum_{k=1}^{\infty}\frac{k}{10^k}[/imath] prove that [imath]0.12345\cdots[/imath] is irrational . I though if let [imath]0.1345\cdots[/imath] be rational like proof of [imath]\mathbb e[/imath]. also I compute that series NOTE :if [imath]|x| < 1[/imath] then [imath]\sum_{k=1}^{\infty}x^k=\frac{1}{1-x} \implies \sum_{k=1}^{\infty}kx^k= \frac{x}{(1-x)^2}[/imath] [imath]\sum_{k=1}^{\infty}\frac{k}{10^k}=\sum_{k=1}^{\infty}k\left(\frac{1}{10}\right)^k= \frac{\frac{1}{10}}{\left(1-\frac{1}{10}\right)^2}=\frac{10}{81}=0.1234567912346679\cdots[/imath] which is rational. please give me hints. I don't want answer buddies. thanks a lot. | 1443877 | How to prove that [imath]x=0.1234567891011\dots [/imath] is irrational?
I'm in [imath]9th[/imath] class and I was wondering how to solve this problem. I only know how to prove that [imath]0.1010010001\dots[/imath] is irrattional. |
1055684 | Splitting field of [imath]X^6+X^3+1[/imath]
Let [imath]f(X)=X^6+X^3+1\in\mathbb Q[X][/imath]. I need to find a splitting field [imath]L[/imath] for this polynomial and the degree of [imath][L:\mathbb Q][/imath]. [imath]f[/imath] is irreducible with [imath]f(X+1)[/imath] and [imath]p=3[/imath] to apply Eisenstein. I know that then [imath]\mathbb Q[X]/(f(X))[/imath] is a field in which [imath]f[/imath] has a root (how can I find out what the root is?). I'm supposed to inductively find more splitting fields and adjoin them in the end to get [imath]L[/imath], but how is that done concretely? | 133079 | Splitting field of [imath]x^6+x^3+1[/imath] over [imath]\mathbb{Q}[/imath]
I am trying to find the splitting field of [imath]x^6+x^3+1[/imath] over [imath]\mathbb{Q}[/imath]. Finding the roots of the polynomial is easy (substituting [imath]x^3=t[/imath] , finding the two roots of the polynomial in [imath]t[/imath] and then taking a 3-rd root from each one). The roots can be seen here [if there is a more elegant way of finding the roots it will be nice to hear] Is is true the that the splitting field is [imath]\mathbb{Q}((-1)^\frac{1}{9})[/imath] ? I think so from the way the roots look, but I am unsure. Also, I am having trouble finding the minimal polynomial of [imath](-1)^\frac{1}{9}[/imath], it seems that it would be a polynomial of degree 9, but of course the degree can't be more than 6...can someone please help with this ? |
1054716 | Direct product of groups and isomorphism
Let [imath]A, B, C[/imath] three groups such that [imath]A \times C \cong B \times C[/imath]. I already know that if [imath]A, B[/imath] and [imath]C[/imath] are abelian and finite, then [imath]A \cong B[/imath]. I think this result does not hold anymore if they are not supposed to be finite or abelian. Is it possible to give a counterexample ? Thank you very much for the help ! | 2193 | Cancellation of Direct Products
Given a finite group [imath]G[/imath] and its subgroups [imath]H,K[/imath] such that [imath]G \times H \cong G \times K[/imath] does it imply that [imath]H=K[/imath]. Clearly, one can see that this doesn't work out for all subgroups. Is there any condition by which this can remain true. |
1055772 | How prove this integral [imath]\int_{0}^{1}\frac{(x-1)dx}{(x+1)\ln{x}}=\ln{\frac{\pi}{2}}[/imath]
show that [imath]I=\int_{0}^{1}\dfrac{(x-1)dx}{(x+1)\ln{x}}=\ln{\left(\dfrac{\pi}{2}\right)}[/imath] I say this integral background,today I have use this wolf play some function,and Suddenly I found this interesting problem I try solve this by hand, [imath]\dfrac{x-1}{(x+1)\ln{x}}=(1-\dfrac{2}{x+1})\dfrac{1}{\ln{x}}[/imath] then It's ugly then [imath]I=\int_{0}^{1}\dfrac{1}{\ln{x}}dx-\int_{0}^{1}\dfrac{2}{\ln{x}}d\ln{(1+x)}[/imath] | 280105 | Evaluating [imath]\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}[/imath]
Some time ago I came across to the following integral: [imath]I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}[/imath] What are the hints on how to compute this integral? |
1056029 | How do I prove this by induction?
thank you for taking the time to help me with the question. I am struggling to use proof by induction for this formula: [imath]\sum_{k=0}^{n}k\times k! = (n + 1)! - 1[/imath] So far, I came up with: [imath]S(n) = (n + 1)! – 1[/imath] Base case: When [imath]n = 0[/imath], [imath](n + 1)! - 1 = (1)! - 1 = 0 = S(0)[/imath], thus the base case is proved. Inductive step: Suppose S[imath](n) = (n + 1)! – 1[/imath], and n > 1 Then we have... [imath]S(n + 1) = ((n + 1) + 1)! - 1[/imath] I don't understand what is required of the proof by induction. What am I supposed to do? | 917367 | Using induction to prove that [imath]\sum_{r=1}^n r\cdot r! =(n+1)! -1[/imath]
Use induction to prove that [imath]\displaystyle\sum_{r=1}^n r\cdot r! =(n+1)! -1[/imath] I first showed that the formula holds true for [imath]n=1[/imath]. Then I put n as [imath]k[/imath] and got an expression for the sum in terms of [imath]k[/imath]. I then found the sum till the [imath](k+1)[/imath]th term by adding the [imath](k+1)[/imath]th term to both sides of the equation and compared it to the sum expression I get by plugging [imath]k+1[/imath] into the sum expression given in the question - they don't match. |
1056244 | The first countable space
Let [imath]X[/imath] uncountable set and let [imath](X,τ_{cf})[/imath] is a topological space such that [imath]τ_{cf}[/imath]: I mean complement finite topology why this topological spaces isn't first countable space ? | 546123 | Is the cofinite topology on an uncountable set first countable?
Let [imath]$X$[/imath] be any uncountable set with the cofinite topology. Is this space first countable? I don't think so because it seems that there must be an uncountable number of neighborhoods for each [imath] x \in X[/imath]. But I am not sure if this is true. |
1056265 | Evaluate [imath]\int\int_SF.dS[/imath] where [imath]F=(xz,yz,x^2+y^2)[/imath]
Evaluate [imath]\int\int_SF.dS[/imath] where [imath]F=(xz,yz,x^2+y^2)[/imath] Where [imath]S[/imath] is the closed surface obtained from the surfaces [imath]x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0[/imath] on the top and the bottom and [imath]z=4-\sqrt{x^2+y^2}[/imath] on the side. How we can solve such integrals?I need your help. Edit: without using the Gauss Divergence theorem. | 1056279 | How to evaluate [imath]\int\int_SF.dS[/imath] where [imath]F=(xz,yz,x^2+y^2)[/imath]?
Evaluate [imath]\int\int_SF.dS[/imath] where [imath]F=(xz,yz,x^2+y^2)[/imath] Where [imath]S[/imath] is the closed surface obtained from the surfaces [imath]x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0[/imath] on the top and the bottom and [imath]z=4-\sqrt{x^2+y^2}[/imath] on the side. How we can solve such integrals?I need your help. Without using the Gauss Divergence theorem. |
1056038 | Probability of exactly one empty box when n balls are randomly placed in n boxes.
Each of [imath]n[/imath] balls is independently placed into one of [imath]n[/imath] boxes, with all boxes equally likely. What is the probability that exactly one box is empty? (Introduction to Probability, Blitzstein and Nwang, p.36). The number of possible permutations with replacement is [imath]n^n[/imath] In order to have one empty box, we need a different box having [imath]2[/imath] balls in it. We have [imath]\dbinom{n}{1}[/imath] choices for the empty box, [imath]\dbinom{n-1}{1}[/imath] choices left for the box with [imath]2[/imath] balls, and [imath](n-2)![/imath] permutations to assign the remaining balls to the remaining boxes. Result: [imath]\frac{\dbinom{n}{1} \dbinom{n-1}{1} (n-2)!}{n^n}[/imath] Is this correct? | 1024258 | Probability: [imath]n[/imath] balls into [imath]n[/imath] holes with exactly one hole remaining empty
The question is: n balls are distributed into n holes at random. What is the probability that exactly one hole remains empty. I came up with [imath]P\left(A\right)=\frac{\:\dbinom{n\:}{1}\:\dbinom{n\:-1}{1}\:\left(n-2\right)!}{n^n}[/imath] but was told I’m way way off. Yes, I do realize I’m awful at probability. Thanks for the help. |
601113 | [imath]X[/imath] is homeomorphic to [imath]X\times X[/imath] (TIFR GS [imath]2014[/imath])
Question is : Suppose [imath]X[/imath] is a topological space of infinite cardinality which is homeomorphic to [imath]X\times X[/imath]. Then which of the following is true: [imath]X[/imath] is not connected. [imath]X[/imath] is not compact [imath]X[/imath] is not homeomorphic to a subset of [imath]\mathbb{R}[/imath] None of the above. I guess first two options are false. We do have possibility that product of two connected spaces is connected. So, [imath]X\times X[/imath] is connected if [imath]X[/imath] is connected. So I guess there is no problem. We do have possibility that product of two compact spaces is compact. So, [imath]X\times X[/imath] is compact if [imath]X[/imath] is compact. So I guess there is no problem. I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false. So, only thing I have problem with is third option. I could do nothing for that third option.. I would be thankful if some one can help me out to clear this. Thank you :) | 1571824 | What can be said about topological properties of a space [imath]X[/imath] that is homeomorphic to [imath]X\times X[/imath] [imath]?[/imath]
Given that [imath]X[/imath] is a topological space with infinite cardinality and [imath]X[/imath] is homeomorphic to [imath]X\times X[/imath]. (A) [imath]X[/imath] cannot be connected. (B) [imath]X[/imath] cannot be compact. (C) [imath]X[/imath] cannot be homeomorphic to a subspace of [imath]\mathbb R[/imath]. (D) None of the above. If I take [imath]\mathbb Q[/imath] with discrete topology then [imath]\mathbb Q[/imath] satisfies the given condition while contradicting option (C). Now for options (A) and (B), I know if [imath]X[/imath] is connected (compact) then [imath]X\times X[/imath] will be connected (compact). But I cannot think of any examples of connected or compact spaces that are homeomorphic to their two-fold product. Can anyone provide with some examples? |
1056517 | Proving that if a set [imath]A[/imath] is infinite then necessarily [imath]|A|\geq|\mathbb{N}|[/imath]
A set [imath]A[/imath] is set to be infinite if it is not finite, i.e. if there exists no [imath]n\in\mathbb{N}[/imath] such that [imath]|A|=n[/imath], meaning there exists a bijection [imath]A\leftrightarrow\{1,\dotsc,n\}[/imath]. How do I prove that for any such set there necessarily exists an injective function [imath]f:\mathbb{N}\to A[/imath], meaning there are no infinite cardinalities under countability? | 10085 | Why is [imath]\omega[/imath] the smallest [imath]\infty[/imath]?
I am comfortable with the different sizes of infinities and Cantor's "diagonal argument" to prove that the set of all subsets of an infinite set has cardinality strictly greater than the set itself. So if we have a set [imath]\Omega[/imath] and [imath]|\Omega| = \aleph_i[/imath], then (assuming continuum hypothesis) the cardinality of [imath]2^{\Omega}[/imath] is [imath]|2^{\Omega}| = \aleph_{i+1} > \aleph_i[/imath] and we have [imath]\aleph_{i+1} = 2^{\aleph_i}[/imath]. I am fine with these argument. What I don't understand is why should there be a smallest [imath]\infty[/imath]? Is there a proof (or) an axiom that there exists a smallest [imath]\infty[/imath] and that this is what we address as "countably infinite"? or to rephrase the question "why can't I find a set whose power set gives us [imath]\mathbb{N}[/imath]"? The reason why I am asking this question is in "some sense" [imath]\aleph_i = \log_2(\aleph_{i+1})[/imath]. I do not completely understand why this process should stop when [imath]\aleph_i = \aleph_0[/imath]. (Though coming to think about it I can feel that if I take an infinite set with [imath]\log_2 \aleph_0[/imath] elements I can still put it in one-to-one correspondence with the Natural number set. So Is [imath]\log_2 \aleph_0 = \aleph_0[/imath] (or) am I just confused? If [imath]n \rightarrow \infty[/imath], then [imath]2^n \rightarrow \infty[/imath] faster while [imath]\log (n) \rightarrow \infty[/imath] slower and [imath]\log (\log (n)) \rightarrow \infty[/imath] even slower and [imath]\log(\log(\log (n))) \rightarrow \infty[/imath] even "more" slower and so on). Is there a clean (and relatively elementary) way to explain this to me? (I am totally fine if you direct me to some paper (or) webpage. I tried googling but could not find an answer to my question) |
1056738 | How to prove this polynomial has an imaginary root?
How can we show that the polynomial [imath]a_nx^n + a_{n-1}x^{n-1} + a_3x^3 + x^2 + x + 1 = 0[/imath], where [imath]a_i\in \Bbb R[/imath], [imath]i=3,...,n[/imath] has an imaginary root? | 1027168 | Real roots of a polynomial of real co-efficients , with the co-efficients of [imath]x^2 , x[/imath] and the constant term all [imath]1[/imath]
Can all the roots of the polynomial equation (with real co-efficients) [imath]a_nx^n+...+a_3x^3+x^2+x+1=0[/imath] be real ? I tried using Vieta's formulae [imath]\prod \alpha=\dfrac {(-1)^n}{a_n}[/imath] , [imath](\prod \alpha )(\sum \dfrac 1 \alpha)=(-1)^{n-1}\dfrac 1 a_n[/imath] , so [imath]\sum \dfrac1\alpha = -1[/imath] but this is not going anywhere , please help . |
1033717 | When is a holomorphy ring a PID?
I will use the notation and language of Stichtenoth, Algebraic Function Fields and Codes. Let [imath]F[/imath] be a function field over a finite field [imath]\mathbb F_q[/imath], [imath]S[/imath] a non empty set of places (possibly infinite) and [imath]O_S[/imath] the holomorphy ring of [imath]S[/imath], i.e., [imath]O_S:=\bigcap_{P\in S} O_P[/imath] (being [imath]O_P[/imath] the valuation ring of the place [imath]P[/imath]). Give necessary and sufficient conditions for [imath]O_S[/imath] to be a Principal Ideal Domain. Many thanks in advance! G. | 1044330 | When a holomorphy ring is a PID?
I will use the notation and language of Stichtenoth, Algebraic Function Fields and Codes. Let [imath]F[/imath] be a function field over a finite field [imath]\mathbb F_q[/imath], [imath]S[/imath] a non empty set of places (possibly infinite) and [imath]O_S[/imath] the holomorphy ring of [imath]S[/imath], i.e., [imath]O_S:=\bigcap_{P\in S} O_P[/imath] (being [imath]O_P[/imath] the valuation ring of the place [imath]P[/imath]). Give necessary and sufficient conditions for [imath]O_S[/imath] to be a Principal Ideal Domain. Many thanks in advance! G. |
1056380 | Function continuous at irrationals and discontinuous at rationals
Q: Given the function [imath]f(x)=\sum_{n=1}^\infty f_n(x)[/imath], where [imath]f_n(x)= \left\{ \begin{array}{lr} 0; \;\;if \;x< r_n \\ \displaystyle \frac{1}{2^n}; x\geq r_n \end{array} \right. [/imath] Here [imath]r_n[/imath] is an enumeration of [imath]\mathbb{Q}[/imath]. Show that [imath]f(x)[/imath] is contionuous at irrationals and discontinuous at rationals. Here is my work. I proved that each [imath]f_n(x)[/imath] is continuous at irrationals. But now stuck in proving [imath]f(x)[/imath] continuous at irrationals. Can anyone please help me? | 1054922 | About the continuity of the function [imath]f(x) = \sum\limits_k2^{-k}\mathbf 1_{q_k \leq x}[/imath]
Let [imath]q: \mathbb{N} \to \mathbb{Q}[/imath] be a bijection and denote the image of [imath]k \in \mathbb{N}[/imath] by [imath]q_k[/imath]. Let [imath]f: \mathbb{R} \to (0,1)[/imath], [imath] f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} [/imath] Show that: 1) [imath]f[/imath] is strictly increasing 2) [imath]f[/imath] is continuous at [imath]x[/imath] iff [imath]x \in \mathbb{R} \setminus \mathbb{Q}[/imath]. My work: Let [imath]x, y \in \mathbb{R}[/imath], [imath]x < y[/imath]. Then there exists [imath]q_n \in \mathbb{Q}[/imath] s.t. [imath]x < q_n < y[/imath]. Now [imath] f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} < \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} + 2^{-n} \leq \underset{q_k \leq y}{\sum_{k \in \mathbb{N}}} 2^{-n} = f(y) [/imath] so [imath]f[/imath] is strictly increasing. I think my proof of 1) is ok, but how can I prove 2)? |
1057732 | Question about [imath]\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}[/imath].
In several places on this site the sum [imath]\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin(n)}{n}[/imath] has been discussed as a generalized alternating series, which therefore converges. I am curious about the sum [imath]\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}[/imath], which I am sure does not converge. Here is my "proof". Steps 2 and 3 clearly needs some clarification. Step 1) Define [imath]A\subset\mathbb{R}[/imath] by [imath]A:=[\pi/4,3\pi/4]\cup[5\pi/4,7\pi/4]\cup\ldots[/imath]. For any [imath]x\in A[/imath], [imath]|\sin(x)|\geq\sqrt{2}/2[/imath]. Step 2) Since [imath]\pi[/imath] is irrational, roughly half the integers are in [imath]A[/imath]. That is, if we define [imath]\#_A(n)[/imath] to be the cardinality of [imath]\{1,\ldots,n\}\cap A[/imath], [imath]\displaystyle\lim_{n\to\infty}\dfrac{\#_A(n)}{n}=\dfrac{1}{2}[/imath]. Step 3) Therefore [imath]\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}>\sum_{n>0,n\in A}\dfrac{|\sin(n)|}{n}>\sum_{n>0,n\in A}\dfrac{\sqrt{2}/2}{n}\approx\dfrac{1}{2}\sum_{i=1}^n\dfrac{\sqrt{2}/2}{n}[/imath]. Step 4) Since the last sum in Step 3 diverges, the sum [imath]\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}[/imath] diverges. I beleive this is more or less intuitively right, but I do not really know the theory to make rigorous Steps 2 and 3. | 1040918 | Does [imath]\sum\limits_{k=1}^\infty |\sin(ak)/k|[/imath] converge?
Does [imath]\sum_{k=1}^\infty |\sin(ak)/k|[/imath] converge for all [imath]0<a<\pi[/imath]? I do not think so since for [imath]a=\pi /2[/imath]: [imath]\sum_{k=1}^\infty\left\vert\frac{\sin(ak)}{k}\right\vert=\sum_{k=0}^\infty (2k+1)^{-1}=\infty.[/imath] But is this true for all [imath]a[/imath]? |
1057736 | Asymptotic for degree
How can I find asymptotic for [imath]\chi(n)[/imath], if [imath]\chi^{\chi^\chi} = n[/imath]. Is here self-qualification estimation? I tried to take the logarithm of both sides, but to nothing has come. | 1055588 | asymptotic of [imath]x^{x^x} = n[/imath]
How find the asymptotic behavior for [imath]x(n)[/imath] if [imath]x^{x^x} = n[/imath]? I supposed that [imath]x = O(\log\log{n})[/imath] and took logarithm two times. So I get [imath]x = O(\frac{\log\log{n}}{\log\log\log{n}})[/imath] Is it right? How to improve this estimate? |
1058059 | Induction Proof with a [imath]\neq[/imath] 1
For [imath]a \neq 1[/imath] and [imath]n>0[/imath], [imath](1-a^{n+1})/(1-a)=1+a+a^2+...+a^n[/imath] How do you prove this by induction? | 1126962 | Demonstration of sum of powers of [imath]2[/imath]
Theorem : For every natural number [imath]p[/imath]: [imath]\sum^p_{i=0} 2^i = 2^{p+1}-1[/imath] I trieed to demonstrate the theorem using induction Demonstration : [imath]1)[/imath] If we have [imath]p=0[/imath] then we get [imath]2^0=2^{0+1}-1[/imath] that is always true. [imath]2)[/imath] Supposing that the first statement is true then we get that [imath]\sum^{p+1}_{i=0} 2^i = 2^{p+2}-1[/imath] Now we know that: [imath]\sum^{p+1}_{i=0} 2^i - \sum^p_{i=0} 2^i = 2^{p+1}[/imath] and if that is true, must be also true that: [imath]2^{p+2}-1-(2^{p+1}-1) = 2^{p+1}[/imath] [imath]2^{p+2}-1-(2^{p+1}-1)=2^{p+2}-2^{p+1}=2^{p+1}(2^1-2^0)=2^{p+1}[/imath] QED Is this a valid demonstration? |
1058172 | Graph Theory triangle (3 colors)
Show that if the edges of [imath]K_n[/imath] are colored with [imath]n[/imath] different colors, then there must be a triangle where all three edges have distinct colors. So, I want to use induction on [imath]n[/imath] where [imath]n[/imath] is the number of vertices in the graph then the base case would be [imath]n=3[/imath] ([imath]K_3[/imath] graph) because you must have a triangle with three distinct edge colors. Then, I know if I draw it with [imath]n=4[/imath] and so on so forth it is true, but I'm having trouble putting it more eloquently. | 797829 | Proving a triangle with different edge colors exists in a graph.
This is again some homework translated (hopefully not too badly) from my book The graph [imath]K_{n}[/imath] is colored using [imath]n[/imath] different colors, in a way that each color is used at least once. Prove that there exists a triangle with its edges colored in different colors. |
1058640 | Expectation of random varible with normal distribution composed with exponential
I am trying to find [imath]\mathbb{E}(e^{-X})[/imath] where [imath]X[/imath] is a random variable with a general normal distribution. I end up with [imath](2\pi \sigma)^{-\frac{1}{2}} \int_{-\infty}^{\infty} e^{-x}e^{\left(\frac{-(\mu-x)^2}{2\sigma^2}\right)}dx [/imath] Tried some transformations and the polar coordinate trick used to derive values of [imath](2\pi \sigma)^{-\frac{1}{2}}\int e^{(\frac{-(\mu-x)^2}{2\sigma^2})}dx [/imath] but haven't been able to come up with anything explicit. Can anyone point me in the right direction? I am in despair... do I need to use a numerical method? If it helps, here is what the integral boils down to [imath](2\pi \sigma)^{-\frac{1}{2}}\int \exp\left(-\frac{x(2\sigma - 2\mu +x)}{2\sigma^2}\right)dx [/imath] | 1056502 | Proof of a Gaussian Integral property
I'm working through some old integrals and I found one that's interesting. I can't quite remember how it's proved, so if someone could set me off in the right direction, it would be really helpful. Thanks! [imath]\int_{\mathbb{R}} e^{-ax^2 + bx} dx = \sqrt{ \dfrac{\pi}{a}} e^{\frac{b^2}{4a}}[/imath] I've tried change of variable, but I'm not sure this is the right approach. |
1058960 | Sequences or 'chains' of adjoint functors
Suppose we have (some categories and some functors such that) [imath]F_1[/imath] is left adjoint to [imath]G_1[/imath], [imath]G_1[/imath] left adjoint to [imath]F_2[/imath], [imath]F_2[/imath] left adjoint to [imath]G_2[/imath]. Will [imath]F_1[/imath] then be equal to [imath]F_2[/imath] (and [imath]G_1[/imath] to [imath]G_2[/imath])? Or can there exist arbitrarily long 'chains' of adjoint functors that do not trivialize? | 92023 | Adjoint pairs, triplets and quadruplets
Often we have adjoint pairs [imath](A, B)[/imath] (meaning [imath]A[/imath] is left adjoint to [imath]B[/imath]). Sometimes we have adjoint triplets [imath](A,B,C)[/imath] (meaning [imath]A[/imath] is left adjoint to [imath]B[/imath] and [imath]B[/imath] is left adjoint to [imath]C[/imath]. No adjoint relation between [imath]A[/imath] and [imath]C[/imath] obviously, since they have the same source and target). So the first question is: can we have quadruplets [imath](A,B,C,D)[/imath] ? (meaning that [imath]C[/imath] is also left adjoint to [imath]D[/imath]). Second question: It would be then possible that we have [imath](A,D)[/imath] besides [imath](A,B)[/imath]. But I think not, because I think that it is not possible to have a functor [imath]A[/imath] with two different right adjoints, [imath]B[/imath] and [imath]D[/imath]. Is this correct? |
610448 | Conditional probability intuition.
Kolmogorov definition - Wikipedia According to the Kolmogorov definition, given two events A and B, [imath]P(A|B) = \frac{P(A \cap B)}{P(B)}[/imath] . But I dont really understand why this is true. Why isn't [imath] P(A|B) = P(A \cap B) [/imath]?? Somebody please help me get an intuition why the Kolmogorov definition is true?? | 294580 | Intuition behind the Definition of Conditional Probability (for 2 Events)
What is some intuitive insight regarding the conditional probability definition: [imath]P(A\mid B) = \large \frac{P(A \cap B)}{P(B)}[/imath] ? I am looking for an intuitive motivation. My textbook merely gives a definition, but no true development of that definition. Hopefully that's not too much to ask. |
1059667 | Let [imath]F|K[/imath] be a field extension and [imath]a \in F [/imath] such that [imath][K(a):K][/imath] is odd integer
Let [imath]F|K[/imath] be a field extension and [imath]a \in F[/imath] such that [imath][K(a):K][/imath] is odd integer, then prove that [imath]K(a)=K(a^2)[/imath]. | 665322 | Do we have [imath]K(\beta)=K(\beta^2)[/imath] for field extension of odd degree?
Let [imath]K\subseteq K(\beta)[/imath] be a finite field extension of odd degree. Does this imply [imath]K(\beta)=K(\beta^2)[/imath]? |
790927 | Is there a function [imath]f\colon\mathbb{R}\to\mathbb{R}[/imath] such that every non-empty open interval is mapped onto [imath]\mathbb{R}[/imath]?
I wonder whether there is a function [imath]f\colon\Bbb R\to\Bbb R[/imath] with the folowing characteristic? for every two real numbers [imath]\alpha,\beta,\alpha\lt\beta[/imath], [imath]\{f(x):x\in(\alpha,\beta)\}=\Bbb R[/imath] I can't say such a function does not exist, neither can I construct a example Thanks a lot! | 186427 | Function whose image of every open interval is [imath](-\infty,\infty)[/imath]
How to find a function from reals to reals such that the image of every open interval is the whole of R? Is there one which maps rationals to rationals? |
1060896 | Let [imath]R=M_n(D)[/imath], [imath]D[/imath] is a division ring. Prove that every [imath]R-[/imath]simple module is isomorphic to each other.
Let [imath]R=M_n(D)[/imath], [imath]D[/imath] is a division ring. Prove that every [imath]R-[/imath]simple module is isomorphic to each other. I need some hints to prove it. Thank you very much. | 1011168 | Are all simple left modules over a simple left artinian ring isomorphic?
I've a basic question. [imath]R[/imath] is a simple left artinian ring. I want to show that all simple left [imath]R[/imath]-modules are isomorphic. A simple [imath]R[/imath]-module [imath]M[/imath] is isomorphic to [imath]R/J[/imath] where [imath]J[/imath] is a maximal left ideal of [imath]R[/imath]. Does this imply that [imath]M\cong Ra[/imath] for some [imath]a\in R[/imath], [imath]a\neq 0[/imath]? Many thanks. |
1060925 | Maximal ideal of the ring of all real valued continuous functions on [imath][0,1][/imath]
Let [imath]R[/imath] be the ring of all real valued continuous functions on [imath][0,1][/imath] and [imath]M[/imath] be a maximal ideal of [imath]R[/imath]. Then how to prove that [imath]\exists c \in [0,1] [/imath] such that [imath]M:=\{f \in R:f(c)=0\}[/imath] ? | 701637 | Ideal in a ring of continuous functions
Let [imath]R[/imath] be the ring of all continuous real valued functions on the unit interval [imath][0,1][/imath] (with pointwise operations), and let [imath]I[/imath] be a proper ideal of [imath]R[/imath]. Show that there exists [imath]λ\in [0,1][/imath] such that [imath]I\subseteq M_{λ}= \left\{f \in R\mid f(λ)=0\right\}.[/imath] Can't really get anywhere with this one. Appreciate the help. |
1050291 | Show that quotient rings are not isomorphic
I've been given a homework problem that requires me to show that the rings [imath]\mathbb{C}[x,y]/(y - x^2)[/imath] and [imath]\mathbb{C}[x,y]/(xy-1)[/imath] are not isomorphic. This is my attempt at a solution: For [imath]\mathbb{C}[x,y]/(y - x^2)[/imath], we can parametrize in the following way: [imath]x = t[/imath] and [imath]y = t^2[/imath]. Then this ring is isomorphic to [imath]\mathbb{C}[t][/imath]. For [imath]\mathbb{C}[x,y]/(xy-1)[/imath], we can parametrize [imath]x = t[/imath] and [imath] y = 1/t[/imath]. Then this is isomorphic to [imath]\mathbb{C}[t, 1/t][/imath]. But [imath]\mathbb{C}[t, 1/t][/imath] is not isomorphic to [imath]\mathbb{C}[t][/imath]. Am I on the right track? If not, any helpful hints? | 2156617 | [imath]K[X,Y]/(XY-1)[/imath] not isomorphic to [imath]K[T][/imath]
Why is [imath]K[X,Y]/(XY-1)[/imath] not isomorphic to [imath]K[T]?[/imath] As a hint I should think about units. But I don't get a helpful idea. |
1061719 | Stuck on Induction Proof
Let [imath]0< a_1 < b_1[/imath] and define [imath]a_{n+1} = \sqrt{a_nb_n}[/imath], and [imath]b_{n+1}[/imath] = [imath]{a_n + b_n}\over2[/imath]. Use induction to show that [imath]a_n<a_{n+1}<b_{n+1}<b_n[/imath]. Also, prove [imath]a_n[/imath] and [imath]b_n[/imath] have limits and that they are equal. I've already proven the base case, I just don't know where to go from here. | 583857 | Question about arithmetic–geometric mean
We have two sequences: [imath]a_{n+1}=\sqrt{a_nb_n}[/imath] [imath]b_{n+1}=\frac{a_n+b_n}{2}[/imath] I need to prove that those are making Cantor's Lemma.(At the end I shold get that: [imath]\lim_{n\to \infty}a_n=\lim_{n\to \infty}b_n[/imath] by Cantor's Lemma) Any ideas how? Thank you. |
1063098 | [imath]2n+1[/imath] real numbers with equal sums
Suppose we have [imath]2n+1[/imath] real numbers. If we remove any of these numbers we can separate the remaining [imath]2n[/imath] numbers in two groups of [imath]n[/imath] numbers with equal sums. Show that all these numbers are equal. | 1002267 | After removing any part the rest can be split evenly. Consequences?
Let [imath]S[/imath] be a finite collection of real numbers not necessarily distinct . If any element of [imath]S[/imath] is removed then the remaining real numbers can be divided into two collections with same size and same sum ; then is it true that all elements of [imath]S[/imath] are equal ? ( I know the result is true if the "reals" are replaced by "integers" ) Please help . |
1063226 | 1:1 between N and P(N)
For each natural number [imath]n\in\mathbb{N_0}[/imath] list all the sets with at most [imath]n[/imath] elements and with maximal element as and including [imath]n[/imath], in numerical order. Now all the elements of [imath]P(N)[/imath] will be listed as n gets larger. Label each set of natural numbers with another natural number. Why is the above not a 1:1 correlation between N and P(N)? The sequence starts with the empty set, then [imath]\{1\}, \{2\}, \{1,2\}, \{3\}, \{1,3\}, \{2,3\}, \{1,2,3\}, \{4\}, \{1,4\}, \{2,4\}, \{3,4\}, \{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}[/imath] … | 1031256 | Do Cantor's Theorem and the Schroder-Bernstein Theorem Contradict?
I am confused as to how Cantor's Theorem and the Schroder-Bernstein Theorem interact. I think I understand the proofs for both theorems, and I agree with both of them. My problem is that I think you can use the Schroder-Bernstein Theorem to disprove Cantor's Theorem. I think I must be doing something wrong, but I can't figure out what. Can someone tell me where I am going wrong? Here's how I think I can prove that the set of all natural numbers has the same cardinality as its power set. Cantor's Theorem says that the power set of any set A has greater cardinality than A. Specifically, the set of all natural numbers, [imath]\mathbb{N}[/imath], is countably infinite, while its power set [imath]P(\mathbb{N})[/imath] is uncountably infinite. The Schroder-Bernstein Theorem says that, for two sets [imath]A[/imath] and [imath]B[/imath], if there exist one-to-one functions [imath]f:A\mapsto B[/imath] and [imath]g:B \mapsto A[/imath], then sets [imath]A[/imath] and [imath]B[/imath] have the same cardinality. I will try to use this to prove that the set of all natural numbers N has the same cardinality as its power set [imath]P(\mathbb{N})[/imath]. I need to prove that there exist one-to-one functions [imath]f:\mathbb{N}\mapsto P(\mathbb{N})[/imath] and [imath]g:P(\mathbb{N})\mapsto \mathbb{N} [/imath] [imath]f:\mathbb{N}\mapsto P(\mathbb{N})[/imath] Define [imath]f:\mathbb{N}\mapsto P(\mathbb{N})[/imath], [imath]n\mapsto f(n)=\{n\}[/imath] For each output [imath]f(n)[/imath], there can only be one n corresponding to [imath]f(n)[/imath], so [imath]f[/imath] is a one-to-one function. [imath]g:P(\mathbb{N})\mapsto \mathbb{N} [/imath] Define [imath]g:P(\mathbb{N})\mapsto \mathbb{N} [/imath] as follows Each element of [imath]P(\mathbb{N})[/imath] is a subset of [imath]\mathbb{N}[/imath], as represented by [imath]\{{a}_{1},{a}_{2},\dots ,{a}_{k}\}[/imath] where [imath]{a}_{1},{a}_{2},\dots ,{a}_{k}[/imath] are elements of [imath]\mathbb{N}[/imath] For each element [imath]s=\{{a}_{1},{a}_{2},\dots ,{a}_{k}\}[/imath] of [imath]P(\mathbb{N})[/imath], the elements of [imath]s=\{{a}_{1},{a}_{2},\dots ,{a}_{k}\}[/imath] are ordered from lowest to highest, so that [imath]{a}_{1}<{a}_{2}<\dots <{a}_{k}[/imath]. The order of a set does not matter, so this can be done without creating a new set. For each element [imath]s=\{{a}_{1},{a}_{2},\dots ,{a}_{k}\}[/imath] of [imath]P(\mathbb{N})[/imath], [imath]g(s)={{p}_{1}}^{{a}_{1}} *{{p}_{2}}^{{a}_{2}} * ... * {{p}_{k}}^{{a}_{k}}[/imath], where [imath]{p}_{1},{p}_{2},...,{p}_{k}[/imath] are each the [imath]k[/imath]th prime number. Ex. [imath]{p}_{1}=2, {p}_{2}=3, {p}_{3}=5,\dots[/imath] [imath]P(\mathbb{N})[/imath] contains the empty set, so [imath]g(\emptyset)=0.[/imath] The set of all prime numbers is countably infinite, so it has the same cardinality as the largest element of [imath]P(\mathbb{N})[/imath], which is [imath]\mathbb{N}[/imath]. Each output [imath]g(s)[/imath] is the prime factorization of some number n. Each number n has a unique prime factorization, so there is only one s in [imath]P(\mathbb{N})[/imath] whose [imath]g(s)[/imath] is the prime factorization of n. Since the elements of [imath]s=\{{a}_{1},{a}_{2},\dots ,{a}_{k}\}[/imath] are ordered from lowest to highest, [imath]{a}_{1}[/imath] is the only element of [imath]s[/imath] that can be [imath]0[/imath]. Thus, [imath]{p}_{1}=2[/imath] is the only prime number that can have an exponent of 0 in the output. Thus, it is impossible for any set [imath]s[/imath] in [imath]P(\mathbb{N})[/imath] to have the same output [imath]g(s)[/imath] as a set of different length in [imath]P(\mathbb{N})[/imath] . In other words, it is impossible to add or remove an element of s to create a new set [imath]{s}_{2}[/imath] so that [imath]g(s)=g({s}_{2})[/imath]. Every element [imath]s[/imath] of [imath]P(\mathbb{N})[/imath] has a unique output [imath]g(s)[/imath], so [imath]g:P(\mathbb{N})\mapsto \mathbb{N} [/imath] is a one-to-one function. So [imath]f:\mathbb{N}\mapsto P(\mathbb{N})[/imath] and [imath]g:P(\mathbb{N})\mapsto \mathbb{N} [/imath] are both one-to-one functions. So [imath]\mathbb{N}[/imath] and [imath]P(\mathbb{N})[/imath] have the same cardinality. I'm fairly certain that both Cantor's Theorem and the Schroder-Bernstein Theorem are correct, so where does my proof go wrong? |
1063286 | Show that [imath]\mathbb{R}^{\mathbb{R}} = U_{e} \oplus U_{o}[/imath]
A function [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] is called even if [imath]f(-x) = f(x)[/imath] [imath] \forall x \in \mathbb{R} [/imath] A function [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] is called odd if [imath]f(-x) = -f(x)[/imath] [imath] \forall x \in \mathbb{R} [/imath] Let [imath]U_{e}[/imath] denote the set of real-valued even functions on [imath]\mathbb{R}[/imath] and let [imath]U_{o}[/imath] denote the set of real valued odd functions on [imath]\mathbb{R}[/imath]. Show that [imath]\mathbb{R}^{\mathbb{R}} = U_{e} \oplus U_{o}[/imath] This question from Q24 of "Linear Algebra Done Right" 3rd Edition - Chp 1.4 (Subspaces). The author of the book does minimal work with functions during the chapter. I have no idea how to approach the question. | 1266022 | How to verify that [imath]\mathbb R^\mathbb R[/imath] is a direct sum of the Ue and Uo
The [imath]U_e[/imath] and [imath]U_o[/imath] denote the set of all real-valued even/odd function on [imath]\mathbb R[/imath] respectively. |
1063194 | "Converse" Schur's lemma
For representations over an algebraically closed field one can formulate Schur's lemma in the following form: Every endomorphism of irreducible representation is of the form [imath]\lambda\cdot id[/imath] I wonder, if the converse is true, i.e if all endomorphisms of representation are scalar then it is irreducible. It's easy to see that such representation must be indecomposable, but what about irreducibility?(case of non-algebraically closed field is also interesting) | 542380 | the converse of Schur lemma
I am interested in the converse of the following form of Schur's lemma: Lemma. (Schur) A group [imath]G[/imath], a [imath]\mathbb{C}[/imath]-vector space [imath]V[/imath] and a homomorphism [imath]D : G \rightarrow \operatorname{GL}(V)[/imath] are given. Suppose that [imath]D[/imath] is a irreducible reprsentation. If a linear map [imath]T : V \rightarrow V[/imath] commutes with [imath]D(g)[/imath] for all [imath]g \in G[/imath], then [imath]T[/imath] is some scalar multiple of the identity operator. Edit) What I want to prove or disprove is the converse of the above: Suppose that the representation [imath]D : G \rightarrow \operatorname{GL}(V)[/imath] is reducible. Then there is a linear map [imath]T : V \rightarrow V[/imath] which is not a scalar multiple of the identity such that [imath]T D(g) = D(g) T[/imath], [imath]\forall g\in G[/imath]. When [imath]G[/imath] is a finite group or a compact Lie group, this trivally holds, because every reducible representation of [imath]G[/imath] is decomposable (i.e. fully reducible); if [imath]W \leq V[/imath] is a subspace invariant under [imath]D : G \rightarrow \operatorname{GL}(V)[/imath], then we can find an invariant subspace [imath]U \leq V[/imath] satisfying [imath]V = W \oplus U[/imath], and [imath]T=\pi_W+2\pi_U[/imath] commutes with every [imath]D(g)[/imath] but is not a scalar multiple of the identity. ([imath]\pi[/imath]: projection operators) QUESTION. But what if [imath]G[/imath] is a general group, which admits reducible but indecomposable representations? Does the converse of Schur lemma still hold? |
1063583 | Show that the number [imath]n[/imath] is divisible by [imath]7[/imath]
How can I prove that [imath]n = 8709120[/imath] divisible by [imath]7[/imath]? I have tried a couple methods, but I can't show that. Can somebody please help me? | 879251 | Divisibility by 7 rule, and Congruence Arithmetic Laws
I have seen other criteria for divisibility by 7. Criterion described below present in the book Handbook of Mathematics for IN Bronshtein (p. 323) is interesting, but could not prove it. Let [imath]n = (a_ka_{k-1}\ldots a_2a_1a_0)_{10} = \displaystyle{\sum_{j=0}^{k}}a_{k-j}10^{k-j}[/imath]. The expression [imath] Q_{3}^{\prime}(n) = (a_2a_1a_0)_{10} - (a_5a_4a_3)_{10} + (a_8a_7a_6)_{10} -\ldots [/imath] are called alternating sum of the digits of third order of [imath]n[/imath]. For example, [imath] Q_{3}^{\prime}(123456789) = 789-456+123=456 [/imath] Proposition: [imath]7 | n \ \Leftrightarrow \ 7 | Q_{3}^{\prime}(n)[/imath]. proof. ?? Thanks for any help. |
1063573 | Fourier transform of a function with sine
I don't know how to compute the Fourier tranform of this function: [imath]f(x) = \frac{\sin \pi a x}{\pi x}[/imath] I know that [imath]\frac{\sin \pi a x}{\pi x} = \frac{e^{i \pi a x} - e^{- i \pi a x}}{2i \pi x}[/imath] Then I plug this function in the formula for the Fourier transform and I get: [imath]\int_{\mathbb{R}}\frac{e^{i \pi a x} - e^{- i \pi a x}}{2i \pi x} e^{-i s x} dx =[/imath] [imath]= \frac{1}{2 \pi i } \int_{\mathbb{R}} \frac{e^{ix( \pi a - s)} - e^{- i x( a \pi +s)}}{x} dx=[/imath] [imath]=\frac{1}{2 \pi i } \int_{\mathbb{R}} \frac{e^{ix( \pi a - s)}}{x} dx - \int_{\mathbb{R}} \frac{e^{- i x( a \pi +s)}}{x} dx[/imath] What should I do know? We can get rid of [imath]i[/imath] in the exponent by changing variables, but that doesn't help much | 757772 | a Fourier transform (sinc)
let [imath]K(u) = \frac{\sin(u)}{\pi u}[/imath] show that Fourier transform of [imath]K[/imath] is [imath] \hat{K}(\omega) = \textbf{1}_{|\omega|\leq 1} [/imath] Some help would be appreciated |
1063859 | Proof with chromatic index.
Prove that for [imath]G = (V,E) [/imath] [imath] \chi(G) \chi(\overline{G}) \ge |V| [/imath] Please for some advices. Thanks in advance. | 1000088 | Prove [imath]\chi(G)\chi(\bar{G}) \geq n[/imath] for chromatic number of graph and its complement
Let us denote by [imath]\chi(G)[/imath] the chromatic number, which is the smallest number of colours needed to colour the graph [imath]G[/imath] with [imath]n[/imath] vertices. Let [imath]\bar{G}[/imath] be the complement of [imath]G[/imath]. Show that (a) [imath]\chi(G)+\chi(\bar{G}) \leq n+1[/imath] (b) [imath]\chi(G)\chi(\bar{G}) \geq n[/imath] I was able to prove (a) using induction. Any hints on proving (b)? |
1064171 | A problem on conituous first derivative function
Suppose [imath]g\in C^1 [a,b][/imath]. Prove that for all [imath]\epsilon > 0[/imath], there is [imath]\delta > 0[/imath] such that [imath]|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{< \epsilon }[/imath] for all points [imath]c,d \in [a,b][/imath] with [imath]0 <|d-c|< \delta[/imath] First, I don't know what [imath]|{g'(c)} - {{g(d)-g(c)} \over {d-c}}|[/imath] mean. Does [imath]g\in C^1 [a,b][/imath] imply, [imath]g'(c)[/imath] is zero? Also how does [imath]g'(c) = \lim_{d\rightarrow c}{{g(d)-g(c)} \over {d-c}}[/imath] apply to the question? To prove the statement in the question, I guess I need to have [imath]||f(d)-f(c)|| \in \epsilon[/imath] whenever [imath]||d-c|| < \delta[/imath], for [imath]c,d \in [a,b][/imath]. Then how do I show [imath]|{g'(c)} - {{g(d)-g(c)} \over {d-c}}|[/imath] = [imath]||f(d)-f(c)||[/imath]. Hints please! Thanks! | 1063845 | For a [imath]C^1[/imath] function, the difference [imath]|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |[/imath] is small when [imath]|d-c|[/imath] is small
Suppose [imath]g\in C^1 [a,b][/imath]. Prove that for all [imath]\epsilon > 0[/imath], there is [imath]\delta > 0[/imath] such that [imath]|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{< \epsilon }[/imath] for all points [imath]c,d \in [a,b][/imath] with [imath]0 <|d-c|< \delta[/imath] First, I don't know what [imath]C^1 [a,b][/imath] means. Some ideas: By Mean value theorem, [imath]g'(c) = {{g(b)-g(a)} \over {b-a}} [/imath] since [imath]c\in[a,b][/imath]. To show [imath]|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{< \epsilon }[/imath] whenever [imath]0 <|d-c|< \delta[/imath] for [imath] c,d \in [a,b][/imath]. I guess I have to use the definition of limit of continuous function. But, I don't know how to connect all of these ideas. |
1064190 | Some complex variable problem
Problem. Suppose that [imath]f[/imath] has an isolated singularity at the point [imath]a[/imath], and [imath]f'/f[/imath] has a first-order pole at [imath]a[/imath]. Prove that [imath]f[/imath] has either a pole or a zero [imath]a[/imath] This is a problem from a past qualifying exam in the institution I'm attending now which might appear in the final exam on tomorrow.(so if asking this is a problematic, then let me know, I'll delete it.)Because [imath]f[/imath] should have pole or removable or essential singularity at [imath]a[/imath], I checked that if [imath]f[/imath] has removable singularity at [imath]a[/imath] then [imath]f(a)[/imath] has to be [imath]0[/imath] to satisfy the condition in the problem, and also checked that if [imath]f[/imath] has pole at [imath]a[/imath] then [imath]f[/imath] satisfies the condition in the problem. So, I was trying to show that [imath]f[/imath] can't have essential singularity at [imath]a[/imath], and tried to use Casorati-Weiestrass theorem to derive some contradiction and it didn't work. I also tried to just plug in Laurent series expanded at [imath]a[/imath] in [imath]f'/f[/imath] and do calculation to eliminate some coefficients with negative indices but calculation would result infinite system of linear equations so I gave up doing it. Therefore, I'm stuck and please give me some hint. | 626656 | Not an essential singularity
The following is an old qualifying exam that I cannot solve: Suppose [imath]f[/imath] is holomorphic on the punctured unit disk and [imath] \frac{f '}{f} [/imath] has a simple pole at [imath]0[/imath]. Show that [imath]f[/imath] does not have an essential singularity at [imath]0[/imath]. |
1064153 | A very quick way to prove a set is measurable.
All examples of non-measurable subset of [imath]\mathbb{R}[/imath] (in the Lebesgue sense) seem to need the axiom of choice in some way or the other. Hence, can we say: The set [imath]A\subseteq \mathbb{R}[/imath] is measurable because the axiom of choice was not used to define it. Is that a valid argument? Or does there exist non-measurable sets that do not require the axiom of choice? | 142499 | Are sets constructed using only ZF measurable using ZFC?
Suppose [imath]S[/imath] is a subset of [imath]\mathbb{R}[/imath] which can be defined without using the axiom of choice, i.e. which can be proved to exist using only the axioms of ZF. Does it follow that [imath]S[/imath] is measurable? We know that ZF + "All subsets of [imath]\mathbb{R}[/imath] are Lebesgue measurable" is consistent (assuming ZF is), but the claim doesn't follow from this alone, since there could be a proof that [imath]S[/imath] is not measurable which uses choice. |
1064755 | Questions about the dihedral group [imath]D_8[/imath]
Consider the dihedral group [imath]D_8[/imath] of order [imath]16[/imath]. Consider [imath]D_8[/imath] with the presentation [imath]D_8=\{r^i s^j : i=0,...,7; j=0,1; r^8=s^2=e; sr=r^7s=r^{-1}s\}[/imath], where [imath]\{e\}, \{rs, r^3s, r^5 s, r^7s\}[/imath] and [imath]\{s, r^2s , r^4s, r^6s\}[/imath] are three of its conjugacy classes (here [imath]r[/imath] is a rotation and [imath]s[/imath] is a reflection). 1) Find the remaining conjugacy classes. 2) Find the centre and class equation for [imath]D_8[/imath]. 3) Find the subgroups of order [imath]4[/imath] and explain which is normal and why. | 556508 | Conjugate class in the dihedral group
List all the conjugate classes in the dihedral group of order [imath]2n[/imath] and verify the class equation. The dihedral group is generated by two elements [imath]r[/imath] and [imath]s[/imath]. The order of [imath]r[/imath] is two since [imath]r^2=e[/imath] and [imath]s[/imath] is [imath]n[/imath] since [imath]s^n = e[/imath]. And I know all elements can be produced as either [imath]s^k[/imath] or [imath]rs^k[/imath]. How can I list all the conjugacy class? |
1061705 | Evaluate [imath]\int_0^{2\pi} \frac{\sin^2\theta}{5+4\cos\theta}\,\mathrm d\theta[/imath]
Evaluate [imath]\int_0^{2\pi} \frac{\sin^2\theta}{5+4\cos(\theta)}\mathrm d\theta[/imath] This is the final question on my review for my final exam tomorrow, and I will be honest and say that I have no clue how to begin problem. Any hints in the direction of how to solve this would be helpful. Following from @Adam Hughes, [imath]-\dfrac{\pi}{2}[f'(0)+Res_2][/imath]When I took the derivative of [imath]f(z)=\dfrac{(z^2-1)^2}{2z^2+5z+2}[/imath] and evaluated for 0, I got [imath]-\dfrac{5}{4}[/imath] Now, [imath]Res_2[/imath] | 1065284 | Evaluating [imath]\int_{0}^{2\pi } \frac{\sin^{2} (x) }{5+4\cos(x)}\,\mathrm dx[/imath]
[imath]\int_{0}^{2\pi } \frac{\sin^{2} (x) }{5+4\cos(x)}\,\mathrm dx[/imath] I am having trouble parsing the square of sine in the numerator. Could someone provide some hint? Thanks. |
1064853 | Abelian transitive subgroup of [imath]S_n[/imath]
Let [imath]A[/imath] be an abelian and transitive subgroup of [imath]S_n[/imath] ([imath]A[/imath] acts on [imath]\left\{ 1,2,\ldots,n\right\}[/imath] naturally). Prove that [imath]A[/imath] is cyclic and generated by cycle permutation of length [imath]n[/imath]. I've proved that [imath]\left|A\right|=n[/imath], but couldn't show [imath]A[/imath] contains such cycle.. | 128098 | Show that any abelian transitive subgroup of [imath]S_n[/imath] has order [imath]n[/imath]
Can anybody tell me what is known about the classification of abelian transitive groups of the symmetric groups? Let [imath]G[/imath] be a an abelian transitive subgroup of the symmetric group [imath]S_n[/imath]. Show that [imath]G[/imath] has order [imath]n[/imath]. Thanks for your help! |
1059843 | Proving [imath]A^2 = 0[/imath] given [imath]A^5 = 0[/imath]
I have a class question where I must prove [imath]A^2 = 0[/imath] given [imath]A^5 = 0[/imath] with A being a 2x2 matrix. I though that I could simply say that as [imath]A^5 = 0[/imath] then [imath]A^2 \cdot A^3 = 0 \implies A^2 = 0[/imath] as [imath]A^2 = A\cdot A[/imath] and [imath]A^3 = A\cdot A\cdot A[/imath] [imath]\implies A = 0 \implies A^2 = 0[/imath]. Could someone verify that my proof is valid and not just a circular argument (which I feel it may be) Thanks | 1053886 | Show that for a [imath]2\times 2[/imath] matrix [imath]A^2=0[/imath]
Assuming that given a [imath]2\times2[/imath] matrix [imath]A[/imath] with the property [imath]A^3=0[/imath] show that [imath]A^2=0[/imath]. Okay so this question is messing with me. When it says [imath]A^3=0[/imath] do you think it means the determinant? How would one approach this? Any help would be much appreciated |
1065009 | understanding the commutator of dihedral group
Let [imath]$G=D_{2n}=⟨x,y|x^2=y^n=e, $[/imath] [imath]yx=xy^{n-1}⟩[/imath] I need to find [imath]$G'$[/imath] [ the commutator of G] now I understand that [imath]$G'$[/imath] is the subgroup generated from [imath] U=xyx^{-1}y^{-1} , [/imath] [imath]$ \ \forall x,y \in G$[/imath] So, [imath]$U=$[/imath] whats the strategy here ? | 593542 | Commutator subgroup of a dihedral group.
I have a few questions concerning an example of the commutator subgroups in the dihedral group. This example is found on pg.171 of Abstract Algebra by Dummit and Foote. Let [imath]D_{2n}=\langle r,s |r^n=s^2=1, s^{-1}rs=r^{-1}\rangle[/imath]. Since [imath][r,s]=r^{-2}[/imath] we have that [imath]\langle r^{-2} \rangle = \langle r^2 \rangle \le D'_{2n}[/imath]. Furthermore, [imath]\langle r^2\rangle \trianglelefteq D_{2n}[/imath] and the images of [imath]r[/imath] and [imath]s[/imath] in [imath]D_{2n} / \langle r^2 \rangle[/imath] generate this quotient. What exactly is meant by the image of [imath]r[/imath] and [imath]s[/imath]? They (is this referring to [imath]r[/imath] and [imath]s[/imath]?) are commuting elements of order [imath]\le 2[/imath] (I know [imath]s[/imath] is of order [imath]2[/imath] but [imath]r[/imath] is of order [imath]n[/imath]??) so the quotient is abelian and [imath]D'_{2n} \le \langle r^2 \rangle[/imath]. (I thought the quotient was abelian due to the already established properties of [imath]\langle r^2 \rangle[/imath] i.e it is normal and a subgroup of the commutator subgroup.) |
851833 | Completeness of the set of convergent sequences
It's a problem from the book "Topology of Metric Spaces", written by Kumaresan: "Show that the set [imath]\textbf{c}[/imath] of convergent sequences in the Normed Linear Space of all bounded real sequences under the sup norm [imath]\|\|_\infty[/imath] is complete. Hint: Enough to show that [imath]\textbf{c}[/imath] is closed. If x = [imath](x_n)[/imath] is a limit point of [imath]\textbf{c}[/imath], it suffices to show that [imath]x[/imath] is Cauchy." I got a little bit confused here... What I tried to do: I know that to show that a metric space is complete, I have to show that every Cauchy sequence is convergent there. About the hint: I don't understand why it is enought to show that [imath]\textbf{c}[/imath] is closed, and what does it mean to say that [imath](x_n)[/imath] is a limit point of [imath]\textbf{c}[/imath]? Does that mean that there is a squence [imath](y_k) \in \textbf{c}[/imath] such that [imath](y_k) \to (x_n)[/imath]? If [imath](y_k) \to (x_n)[/imath], given [imath]\epsilon > 0[/imath], there are [imath]n_0, k_0 \in \mathbb{N}[/imath] such that [imath]\|y_k-x_n\|_\infty<\epsilon \forall k,n \geq \max\{n_0, k_0\}=m_0[/imath] As we have that [imath](y_n) \in \textbf{c}[/imath], we have that [imath](y_n) \to y \in \mathbb{R}[/imath]. Because of the uniqueness of the limit, [imath](x_n) \to y[/imath], so [imath](x_n) \in \textbf{c}[/imath], therefore, [imath]\textbf{c}[/imath] is closed. Now, let [imath](y_{n_k}) \in \textbf{c}[/imath] be a Cauchy sequence. Well, [imath](y_{n_k})=(y_{n_1},y_{n_2},\dots,y_{n_j},\dots)[/imath] is a sequence of sequences. Knowing that [imath](y_{n_k})[/imath] is Cauchy's, we know that, given [imath]\epsilon > 0, \exists j_0 \in \mathbb{N}[/imath] such that [imath]\forall m, p \geq j_0[/imath] [imath]\|y_{n_m}-y_{n_p}\|_\infty < \epsilon[/imath] I don't know what to do from here. My idea is to show that, since every term [imath]y_{n_j}[/imath], of the sequence [imath](y_{n_k})[/imath], belongs to [imath]\textbf{c}[/imath], therefore, every term [imath]y_{n_j}[/imath] is convergent, we have that [imath](y_{n_k})[/imath] converges for a convergent sequence [imath](y_n)[/imath]. So, since [imath](y_n)[/imath] is convergent, [imath](y_n) \in \textbf{c}[/imath], we have that every Cauchy sequence in [imath]\textbf{c}[/imath] is convergent, meaning that [imath]\textbf{c}[/imath] is complete. But I don't know how to do that, and I don't know if I understood the problem correctly, so I don't know if what I did before is correct... I'd be really grateful if someone could give a hint, a solution, a coment about anything I wrote here! :) | 2462682 | Prove that [imath]c_0[/imath] is complete by showing that [imath]c_0[/imath] is closed in [imath]l_\infty[/imath].
Question: How do I show that [imath]c_0[/imath], which is shorthand for [imath]C_0(\mathbb{N})[/imath], is closed in [imath]l_\infty[/imath]? I've tried to work with the fact that if [imath](f_n)[/imath] is a sequence in [imath]c_0[/imath] converging to [imath]f \in l_\infty[/imath], [imath]|f(k)|\leq|f(k) - f_n(k)| + |f_n(k)|[/imath] and that I can choose n so that [imath]|f(k) - f_n(k)|[/imath] is small independent of k. However, I don't get very far.. Thanks in advance! |
1065036 | Prove that if [imath]A[/imath] is a set with infinite cardinality, then there exists a bijection between [imath]A[/imath] and [imath]A^2[/imath].
Prove that if [imath]A[/imath] is a set with infinite cardinality, then there exists a bijection between [imath]A[/imath] and [imath]A^2[/imath]. I know that there exists a bijection between [imath]\Bbb{R}[/imath] and [imath]\Bbb{R^2}[/imath], but I don't think that technique can be generalized. I think the proof of this may be highly non-trivial. Any help would be great. | 577022 | Cardinal of [imath]X^2[/imath]
The axiom of choice is equivalent to the following statement: if [imath]X[/imath] is an infinite set then the cardinal of [imath]X^2[/imath] is the same as that of [imath]X[/imath]. Is there an elementary proof of this statement? |
1065368 | How to prove that a union of cardinals is a cardinal
I have this question: Let [imath]\omega_1[/imath] the least uncountable cardinal, and for all [imath]n \in \omega[/imath], [imath]n \geq 1[/imath]. Let [imath]\omega_{n+1}[/imath] the least cardianal greater than [imath]\omega_n[/imath]. Show that [imath]\bigcup_{n \in \omega}\omega_n[/imath] is a cardinal. My try: Let [imath]f_n : \omega_n \rightarrow A_n [/imath] a bijectionfor all [imath]n \in \omega[/imath]. And let [imath]F_n :\bigcup_{n \in \omega}\omega_n \rightarrow \bigcup_{n \in \omega}A_n[/imath]. [imath]F(x) = f_n(x)[/imath], where [imath]n = min\{ n: x \in A_n\}[/imath]. An d prove that [imath]F[/imath] is a bijection. But I'm not have sucess. | 1064019 | Union of uncountable cardinals
I have to prove the following: Let [imath]\omega_1[/imath] be the least uncountable cardinal and, for each [imath]n \in \omega[/imath] (here, [imath]\omega[/imath] is the set of natural numbers denoted as an ordinal/cardinal), [imath]n\ge1[/imath], let [imath]\omega_{n+1}[/imath] be the least cardinal that is greater than [imath]\omega_n[/imath]. Show that [imath]\bigcup_{n\in\omega}\omega_n[/imath] is a cardinal. I suppose the proof has something to do with transfinite induction/recursion, but I don't know exactly how I'm supposed to use it. Any thoughts? |
23443 | Function [imath]\mathbb{R}\to\mathbb{R}[/imath] that is continuous and bounded, but not uniformly continuous
I found an example of a function [imath]f: \mathbb{R}\to\mathbb{R}[/imath] that is continuous and bounded, but is not uniformly continuous. It is [imath]\sin(x^2)[/imath]. I think it's not uniformly continuous because the derivative is bigger and bigger as [imath]x[/imath] increases. But I don't know how to prove this is uniformly continuous. Is [imath]\sin(x^2)[/imath] uniformly continuous then? if it isn't, can you guys think of any other examples? thanks | 218971 | Prove that the function[imath]\ f(x)=\sin(x^2)[/imath] is not uniformly continuous on the domain [imath]\mathbb{R}[/imath].
If I want to prove that the function[imath]\ f(x)=\sin(x^2)[/imath] is not uniformly continuous on the domain [imath]\mathbb{R}[/imath], I need to show that: [imath]\exists\epsilon>0[/imath] [imath]\forall\delta>0[/imath] [imath]\exists{x,y}\in\mathbb{R}\ : |x-y|<\delta[/imath] and [imath]|\sin(x^2) - \sin(y^2)|\geq\epsilon[/imath]. So let's take [imath]\epsilon = 1[/imath]. Then I want [imath]|\sin(x^2)-\sin(y^2)|\ge1[/imath]. That's the case if [imath]\sin(x^2)[/imath]=0 and [imath]\sin(y^2)=\pm1[/imath]. Thus [imath]x^2=n\pi[/imath] and [imath]y^2=n\pi + \frac{1}{2}\pi[/imath]. Now I'm stuck on expressing x and y, which I want to express in [imath]\delta[/imath], to ensure that [imath]|x-y|<\delta[/imath]. Thanks in advance for any help. |
1065505 | Prove that [imath]T,S[/imath] are simultaneously diagonalizable iff [imath]TS=ST[/imath].
Definition: We say that [imath]S,T[/imath] are simultaneously diagonalizable if there's a basis, [imath]B[/imath] which composed by eigen-vectores of both [imath]T[/imath] and [imath]S[/imath] Show that [imath]S,T[/imath] are simultaneously diagonalizable iff [imath]ST=TS[/imath]. I tried both directions, but couldn't get much further. I'd be glad for help. Thanks. | 236212 | Prove that simultaneously diagonalizable matrices commute
Two [imath]n\times n[/imath] matrices [imath]A, B[/imath] are said to be simultaneously diagonalizable if there is a nonsingular matrix [imath]S[/imath] such that both [imath]S^{-1}AS[/imath] and [imath]S^{-1}BS[/imath] are diagonal matrices. a) Show that simultaneously diagonalizable matrices commute: [imath]$AB = BA$[/imath]. b) Prove that the converse is valid, provided that one of the matrices has no multiple eigenvalues. Is every pair of commuting matrices simultaneously diagonalizable? My attempt: a) Let [imath]M=S^{-1}AS[/imath] and [imath]P=S^{-1}BS[/imath], then it follows that [imath]A= S^{-1}MS[/imath] and [imath]B=S^{-1}PS[/imath]. Thus [imath]AB =S^{-1}MSS^{-1}PS = S^{-1}MPS = S^{-1}PMS[/imath] (is it wrong that I switched MP to PM?) [imath]=S^{-1}PSS^{-1}MS=BA[/imath]. b) How can I do this? |
1065640 | Prove that if [imath]M[/imath] is an [imath]R-[/imath] projective left module then [imath]M/IM[/imath] is an [imath]R/I-[/imath] projective left module.
Let [imath]I[/imath] ba a two-sided ideal of a ring [imath]R[/imath] and [imath]M[/imath] be an [imath]R-[/imath] left module. Prove that if [imath]M[/imath] is an [imath]R-[/imath] projective left module then [imath]M/IM[/imath] is an [imath]R/I-[/imath] projective left module. It is easy to see that [imath]M/IM[/imath] has the structure of [imath]R/I-[/imath]module. I need some ideas to prove that [imath]M/IM[/imath] is an [imath]R/I-[/imath] projective left module. Thank you very much. | 813076 | Proving that P/PJ is a projective right module over R/J
If P is a projective right module over a ring R and J is a two sided ideal of R. Prove that P/PJ is a projective right module over R/J . My idea was trying to proof that " [imath]M[/imath] is an [imath]R[/imath]-module iff [imath]M[/imath] is an module over [imath]R/J[/imath] " where J is an ideal of R . |
1065671 | How to approach an equation of the form [imath]f(x)=y[/imath] where [imath]f[/imath] is recursively defined?
This is a homework problem so I am not looking for someone to solve it for me. I would like to know how I should approach this problem, or in what direction I should research to figure it out myself. For [imath]y \in \mathbb N[/imath], [imath]y > 1[/imath], find [imath]x[/imath] for which [imath]f(x) = y[/imath]. Given: [imath]f(0)=1[/imath] [imath]f(1)=1[/imath] [imath]f(2)=2[/imath] [imath]f(2x)=f(x)+f(x+1)+x[/imath] [imath]f(2x+1)=f(x)+f(x−1)+1[/imath] | 1060154 | Help with a recurrence with even and odd terms
I have the following recurrence that I've been pounding on: [imath] a(0)=1\\ a(1)=1\\ a(2)=2\\ a(2n)=a(n)+a(n+1)+n\ \ \ (\forall n>1)\\ a(2n+1)=a(n)+a(n-1)+1\ \ \ (\forall n\ge 1) [/imath] I don't have much background in solving these things, so I've been googling around looking for different techniques. Unfortunately, I haven't been able to correctly apply them yet. So far I've tried: Characteristic equations Generating functions Plugging it into Wolfram Alpha Telescoping Observation I'm sure that one or more of these is the way to go, but my biggest problem right now is figuring out how to deal with the idea that there are different equations for odd and even values of [imath]n[/imath]. I'm not necessarily looking for a solution (though it would be gratefully accepted), but any nudges in the right direction would be much appreciated. To follow up, it turned out that the speed problem described in my comment, below was related to the Decimal implementation in Python 2.7. Switching to Python 3.3 (and Java) made everything many orders of magnitude better. |
1065822 | does derivative of convergent function go to 0?
My question: if [imath]f[/imath] is differentiable and [imath]\lim_{x \to \infty} f(x) = M[/imath], does this imply that [imath]\lim_{x \to \infty} f'(x) = 0[/imath]? My thinking: there exists [imath]X[/imath] such that [imath]\forall x,y>X[/imath], [imath]|f(x)-f(y)|<\epsilon[/imath] by the Cauchy criterion. Thus [imath]\frac{|f(x)-f(y)|}{x-y} = f'(d)< \epsilon[/imath] for [imath]d[/imath] between [imath]x[/imath] and [imath]y[/imath]. I worry this won't hold if [imath]x[/imath] is extremely close to [imath]y[/imath] though. | 211063 | Does the derivative of a continuous function goes to zero if the function converges to it?
Physicist here. I am puzzled by a question: looking at a continuous function [imath]g :\mathbb{R} \rightarrow \mathbb{R}[/imath] that goes to zero at infinity, I am interested in the behavior of its derivative [imath]\beta = g'[/imath]. Precisely, does it go to zero too? By writing on paper it looks like: [imath] \beta(+ \infty)=\text{lim}_{x\rightarrow \infty}\text{lim}_{h\rightarrow 0} \frac{g(x+h)-g(x)}{h}[/imath] And if I can invert the two limits, I get what I expect: [imath]\beta(+\infty)=0[/imath], so I was curious about the hypothesis behind this permutation. Is the requirement of continuity enough? Thanks. |
1066144 | Improper integral: [imath]\int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}[/imath]
I've tried many ways, but it seems that it didn't work: [imath] \int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)} = \int_0^1\frac{x^{n-1}}{(x+1)(2x+1)\cdots(nx+1)}\mathrm dx = \cdots [/imath] Any help would be appreciated! | 114155 | Computing [imath] \int_{0}^{\infty} \frac{1}{(x+1)(x+2)...(x+n)} \mathrm dx [/imath]
I would like to compute: [imath] \int_{0}^{\infty} \frac{1}{(x+1)(x+2)...(x+n)} \mathrm dx [/imath] [imath] n\geq 2[/imath] So my question is how can I find the partial fraction expansion of [imath] \frac{1}{(x+1)(x+2)...(x+n)} \; ?[/imath] |
1066471 | how many solution can be found of the form [imath]A \pmod{X} = B[/imath]
[imath]A[/imath] and [imath]B[/imath] are given, How many [imath]X[/imath] can be found to make the following equation true? [imath] A \pmod{X} = B [/imath] Is there any formula? | 1066472 | Given a, b How many solutions exists for x, such that: [imath]a \bmod{x}=b [/imath]
Given [imath]a, b[/imath]. How many solutions exists for [imath]x[/imath], such that: [imath]a \bmod{x}=b [/imath] By example: [imath]a = 21[/imath] and [imath]b = 5[/imath] [imath]21 \bmod{8} = 21 \bmod{16} = 5[/imath] Then [imath]x[/imath] has 2 solutions |
1066676 | Eigenvalues of a nilpotent matrix can only be [imath]0[/imath]
Prove that the eigenvalues for a square Nilpotent matrix A can only be [imath]0[/imath]. Definition of nilpotent A [imath]^n[/imath]=[imath]0[/imath] n is a positive whole integer | 372370 | Prove that the only eigenvalue of a nilpotent operator is 0?
I need to prove that: if a linear operator [imath]\phi : V \rightarrow V[/imath] on a vector space is nilpotent, then its only eigenvalue is [imath]$0$[/imath]. I know how to prove that this for a nilpotent matrix, but I'm not sure in the case of an operator. How would I be able to relate [imath]\phi[/imath] to a matrix? Note: A nilpotent operator [imath]\phi[/imath] has been defined as an operator that satisfies [imath]\phi^{n} = 0[/imath] for some [imath]n \geq 1[/imath]. |
1066872 | concerning Graph Theory/subgraphs/ even degree
Given a simple Graph [imath]G=(V,E)[/imath] ([imath]V[/imath] vertices, [imath]E[/imath] edges) I have to show that there exists a distribution [imath]V= V_1 \cup V_2[/imath] of the vertices such that all vertices in the induced subgraphs [imath]G[V_1][/imath] and [imath]G[V_2][/imath] have even degree. I honestly have no idea how to solve this question. I tried several things (which I will list below) but all attempts have failed. induction: Induction on the number of vertices has proven to be extremely ugly for this question, so I dropped the idea really quick. color all circles with odd elements first. use that in [imath]G[/imath] there is an even amount of vertices with odd degree. I do not want a complete answer to this question (at least not for the moment). Does anyone have an idea/tip for this question which doesn't give away too much? I hope you can help me slinshady | 1065448 | A partition of vertices of a graph
I've got an example for this question, but there are many different possibilites and I don't know how to show this for all graphs. Has got anyone any advice how to begin ? Let [imath]G=(V,E)[/imath] be an undirect graph.Proof: There is a partition [imath]V = V_1 \cup V_2[/imath] of a verticesset, so all vertices in [imath]G[V_1][/imath] and [imath]G[V_2][/imath] have got an even degree.' I draw it on paper,but how mentioned it, it's not proven for all and I don't know how. Thank you for any advice. |
1066898 | Find a function in [imath]L^p(\mathbb{R})[/imath] only for [imath]p=4[/imath]
I'm having trouble with this problem from an old analysis qual: Find a function [imath]f[/imath] such that for [imath]p\in (1,\infty)[/imath], [imath]f[/imath] is in [imath]L^p(\mathbb{R})[/imath] only when [imath]p=4[/imath]. | 55170 | Is it possible for a function to be in [imath]L^p[/imath] for only one [imath]p[/imath]?
I'm wondering if it's possible for a function to be an [imath]L^p[/imath] space for only one value of [imath]p \in [1,\infty)[/imath] (on either a bounded domain or an unbounded domain). One can use interpolation to show that if a function is in two [imath]L^p[/imath] spaces, (e.g. [imath]p_1[/imath] and [imath]p_2[/imath],with [imath]p_1 \leq p_2[/imath] then it is in all [imath]p_1\leq p \leq p_2[/imath]). Moreover, if we're on a bounded domain, we also have the relatively standard result that if [imath]f \in L^{p_1}[/imath] for some [imath]p_1 \in [1,\infty)[/imath], then it is in [imath]L^p[/imath] for every [imath]p\leq p_1[/imath] (which can be shown using Hölder's inequality). Thus, I think that the question can be reduced to unbounded domains if we consider the question for any [imath]p>1[/imath]. Intuitively, a function on an unbounded domain is inside an [imath]L^p[/imath] space if it decrease quickly enough toward infinity. This makes it seem like we might be able to multiply the function by a slightly larger exponent. At the same time, doing this might cause the function to blow up near zero. That's not precise/rigorous at all though. So I'm wondering if it is possible to either construct an example or prove that this can't be true. |
1067264 | How to calculate the sum of the infinite series [imath]\sum_{n = 0}^\infty \frac{n}{2^\sqrt{n}}[/imath]
How do you calculate the sum of an infinite series like [imath] \sum_{n = 0}^\infty \frac{n}{2^\sqrt{n}}[/imath] //EDIT //Ignore I searched up how to find this with infinite geometric series solution which was [imath]\frac{a}{1-r}[/imath] [imath]a[/imath] being first value and [imath]r[/imath] being common ratio. But I could not find common ratio. Please give me common ratio of this series. //Ignore I have also tried limits like [imath] \lim_{a \to \infty} \sum_{n = 0}^a \frac{n}{2^\sqrt{n}} [/imath] but they don't actually give me a solid answer. Correction Prove this infinite series is convergent. | 1066475 | What is [imath]\sum_{n=1}^{\infty} \frac{n}{2^{\sqrt{n}}}[/imath]?
What is the value of [imath]\sum_{n=1}^{\infty} \frac{n}{2^{\sqrt{n}}}[/imath] It's clearly convergent but is it possible to calculate the sum? |
1067189 | Is [imath]\sup_{y\in Y}f(x,y)[/imath] measurable?
I have [imath]f(x,y)\geq 0[/imath] measurable in [imath]\mathcal{A}\times\mathcal{B}[/imath] where [imath](X,\mathcal{A})[/imath] and [imath](Y,\mathcal{B})[/imath] are measurable spaces. I don't think [imath]g(x)=\sup_{y\in Y}f(x,y)[/imath] is measurable in [imath]\mathcal{A}[/imath] if [imath]g(x)<\infty[/imath] for each [imath]x\in X[/imath]. My thought comes from the fact that one of sufficient conditions to make a supremum in [imath]y\in Y[/imath] be measurable if [imath]Y[/imath] is countable so there is a counterexample if I set [imath]Y[/imath] be uncountable to make [imath]g(x)[/imath] be not measurable. But I can't not do further. Please help me to construct the counterexample of [imath]g(x)[/imath]. | 650249 | Supremum of a product measurable function...
This is an interesting question that has stumped the entirety of my measure theory class, including the professor: Prove or disprove: Let [imath](X,\mathcal A)[/imath], [imath](Y,\mathcal B)[/imath] be measure spaces. Let [imath]f[/imath] be an [imath]\mathcal A[/imath] [imath]\times[/imath] [imath]\mathcal B[/imath] measurable nonnegative function, and let [imath]g(x)[/imath] [imath]=[/imath] [imath]sup_{y \in Y}f(x,y)[/imath], with [imath]g(x)<\infty[/imath] for all [imath]x[/imath]. Is [imath]g(x)[/imath] necessarily an [imath]\mathcal A[/imath]-measurable function? We all feel the answer is no, given that slices are measurable, and sups of measurable functions are only guaranteed to be measurable over a countable index. We think the correct answer is to start with a nonmeasurable set [imath]S[/imath] in [imath]X[/imath], and to try to build a set [imath]T[/imath] in [imath]Y[/imath] that makes [imath]S \times T[/imath] measurable in [imath]\mathcal A \times \mathcal B[/imath], but we suspect this is quite difficult with no further guidance. Any ideas? |
1067379 | Prove or disprove If f is continuous and g is dicontinuous then f + g is discontinuous
Let [imath]A\subseteq \mathbb{R}[/imath], function [imath]f\colon A\to \mathbb{R}[/imath] continuous at [imath]a\in A[/imath], and [imath]g\colon A \to \mathbb{R}[/imath] is discontinuous at [imath]a\in A[/imath]. Prove or disprove that [imath]f+g[/imath] is discontinuous at [imath]a\in A[/imath]. A proof by definition is what I need. But if ones try with the other way that's okay. Help me. Thank you. | 995494 | If [imath]f[/imath] s discontinuous at [imath]x_0[/imath] but [imath]g[/imath] is continuous there, then [imath]f+g[/imath]?
Let [imath]x_0 \in \mathbb{R}[/imath] and [imath]f,g: \mathbb{R} \to \mathbb{R}[/imath] such that [imath]f[/imath] is discontinuous at [imath]x_0[/imath] but [imath]g[/imath] is continuous at [imath]x_0[/imath], then [imath]f+g[/imath] at [imath]x_0[/imath] is... My approach: After some graphical attempts to construct such a pathological case as given above I claim that the described scenario isn't possible. That is [imath]f \text{ disc. at } x_0 \text{ and } g \text{ cont. at } x_0 \implies f+g \text{ disc. at } x_0 [/imath] Proof: I was trying to come up with a contradiction, because if they happen to work out I find these kind of proofs to be most satisfying. Assume [imath]f+g[/imath] is continuous at [imath]x_0[/imath] that is [imath]\forall \epsilon > 0 ,\exists \delta_1 > 0 : \forall x \in \mathbb{R}, |x-x_0| < \delta_1 \implies |f(x)+g(x)-f(x_0)-g(x_0)| < \epsilon [/imath] Since this statement must be true for all [imath]\epsilon > 0[/imath] I could choose [imath]\epsilon:=1[/imath]. Next I know that also [imath]g[/imath] is continuous at said point [imath]x_0[/imath], thus [imath]\forall \epsilon > 0, \exists \delta_2 > 0 : \forall x \in \mathbb{R}, |x-x_0| < \delta_2 \implies |g(x)-g(x_0)| < \epsilon [/imath] and I have that [imath]f[/imath] is NOT continuous at [imath]x_0[/imath] therefore: [imath]\exists \epsilon^* > 0, \forall \delta>0: \exists x \in \mathbb{R}, |x-x_0| < \delta \wedge |f(x)-f(x_0)| \geq \epsilon [/imath] My problem is to bring all these three statements nicely together. Mainly I get confused by the quantors. In order to match all three statements I assume I have to choose [imath]\epsilon > 0[/imath] such that the last statement is met, so say [imath]\epsilon:= \epsilon^*>0[/imath], so statement 1 and 2 are still true for said [imath]\epsilon[/imath] I also need to choose [imath]\delta[/imath]. Lets say [imath]\delta = \min{(\delta_1,\delta_2)}[/imath] now I should be able to work with all three statements at the same time right? But then I stumble with my argumentation. Let [imath]|x-x_0|< \delta[/imath], then we obtain [imath] |f(x)+g(x)-f(x_0)-g(x_0)| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)| < - \epsilon + \epsilon =0 [/imath] which means that [imath]|f(x)+g(x)-f(x_0)-g(x_0)| <0[/imath] and I hope that this is a contradiction because [imath]|.| \geq 0[/imath]. Please comment (or answer) if I am right or absolutely on the wrong path. I'd also appreciate it if you criticize my formulation. |
1067926 | Polynomial P= Q² + R²
I want to show that every polynomial [imath]P[/imath] [imath](\geq0)[/imath] can be written with two other polynomials [imath]Q[/imath] and [imath]R[/imath]. [imath]P=Q^2+R^2[/imath] | 1012733 | Prove that a positive polynomial function can be written as the squares of two polynomial functions
Let [imath]f(x)[/imath] be a polynomial function with real coefficients such that [imath]f(x)\geq 0 \;\forall x\in\Bbb R[/imath]. Prove that there exist polynomials [imath]A(x),B(x)[/imath] with real coeficients such that [imath]f(x)=A^2(x)+B^2(x)\;\forall x\in\Bbb R[/imath] I don't know how to approach this, apart from some cases of specific polynomials that turned out really ugly. Any hints to point me to the right direction? |
1068037 | Find all positive integers [imath]n[/imath] such that [imath]\phi(n) + \tau(n) > n[/imath].
How do I find all positive integers [imath]n[/imath] such that [imath]\phi(n) + \tau(n) > n[/imath]? I attempted using the formulas for [imath]\phi(n)[/imath] and [imath]\tau(n)[/imath], but I feel this approach is kind of handwavy... | 1065608 | Relatively Prime Numbers and Number of Divisors of n
Find all positive integers [imath]n[/imath] such that [imath]\phi(n) + \tau(n) > n[/imath]. I'm not sure how to deal with such abstract functions... help? |
454843 | Equation of a function after rotation of the plane
I have a function [imath]\mathbb{R}\to\mathbb{R}[/imath], [imath]y=f(x)[/imath]. I want to find its equation [imath]y'=g(x')[/imath] when the basis [imath](x,y)[/imath] is rotated by [imath]a[/imath]. [imath](x',y')= R_a[(x,y)][/imath], where [imath]R_a[/imath] is the rotation of angle [imath]a[/imath]. I assume that f is symétric (f(x)=f(-x)) and that [imath]g[/imath] is still a function after the rotation ! I thought naively that I could rotate by -a, compute f, and then rotate by a. But this is wrong when [imath]f[/imath] is linear and commutes with [imath]R_a[/imath] ! More precisely, I thought that [imath]g(x)= P_y \circ R_a (0,f \circ P_x \circ (R_{-a}[(x,0)]))[/imath], where [imath]P_x[/imath] and [imath]P_y[/imath] are the projectors [imath]R^2 \rightarrow R[/imath] on the two coordinates Thank you for your help. | 17246 | Is there a way to rotate the graph of a function?
Assuming I have the graph of a function [imath]f(x)[/imath] is there function [imath]f_1(f(x))[/imath] that will give me a rotated version of the graph of that function? For example if I plot [imath]\sin(x)[/imath] I will get a sine wave which straddles the [imath]x[/imath]-axis, can I apply a function to [imath]\sin(x)[/imath] to yield a wave that straddles the line that would result from [imath]y = 2x[/imath]? |
1068339 | Prove that [imath]\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0[/imath] in the [imath]\epsilon[/imath]-[imath]\delta[/imath] way
Given: [imath]\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0[/imath] How to prove this limit using the [imath]\epsilon[/imath]-[imath]\delta[/imath] way? (the biggest problem is to find [imath]\delta[/imath]) | 1068087 | Proving a limit of a trigonometric function: [imath]\lim_{x \to 2/\pi}\lfloor \sin \frac{1}{x} \rfloor=0[/imath]
[imath]\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0[/imath] I need to prove the limit of this using the [imath]\epsilon - \delta [/imath] way but I don't know how to find [imath]\delta[/imath] when I'm given a trigonometric function I know only how to do it with polynomial functions |
1068516 | How to find [imath]\int \sec^{3} \ dx[/imath]
I am stuck trying to find [imath]\int \sec^3{x} \ dx.[/imath] Here is my attempt using integration by parts: [imath]\int \sec^3{x} \ dx = \sec{x}\tan{x} - \int \tan^2{x}\sec{x} \ dx.[/imath] At this point, I am stuck. How can I continue from here? | 1021188 | The proper and easiest way of doing an integral with derivative?
I have this integral: [imath]\int{\sec^3x\,\mathrm dx}[/imath] I don't understand how I would solve this. Google and YouTube videos don't help me understand much, other than just giving the answer. Is it possible to explain step-by-step how this would be solved, assuming that this is the first time I'm seeing it? |
1065656 | Extreme point of unit balls, the complex case
I've been trying to determine what are the extreme point of the unit balls of [imath]\ell^1[/imath] and [imath]C[0,1][/imath]. I think that I cracked the real case (I got for [imath]\ell^1[/imath]: [imath]\{e_n\}_{n\in \mathbb N}\bigcup\{-e_n\}_{n\in \mathbb N}[/imath], and for [imath]C[0,1][/imath] the constant functions [imath]f=1[/imath] and [imath]f=-1[/imath]). I had a little trouble figuring out what should be the answer in case the field is [imath]\mathbb C[/imath]. I think it should be (for [imath]\ell^1[/imath]) all the [imath]\{e_n\}_{n\in \mathbb N}\bigcup\{-e_n\}_{n\in \mathbb N}\{ie_n\}_{n\in \mathbb N}\bigcup\{-ie_n\}_{n\in \mathbb N}[/imath], and (for [imath]C[0,1][/imath]) the functions [imath]f=1, f=-1, f=i, f=-i[/imath]. But I can also see it go to (for [imath]\ell^1[/imath]) all the [imath]\{ae_n:|a|=1\}[/imath], and (for [imath]C[0,1][/imath] all the constant functions [imath]f=a[/imath] with [imath]|a|=1[/imath]. I would like some guidness if possible. Knowing the right answer could help, as well as the intuition behind and maybe some tips on how to get there. Thanks!! | 1066196 | Extreme point of unit balls, over [imath]\mathbb C[/imath]
I've been trying to determine what are the extreme point of the unit balls of [imath]\ell^1[/imath] and [imath]\mathcal{C}[0,1][/imath]. I think that I cracked the real case (I got for [imath]\ell^1[/imath]: [imath]\{e_n\}_{n\in \mathbb N}\bigcup\{−e_n\}_{n \in \mathbb N}[/imath], and for [imath]\mathcal{C}[0,1][/imath] the constant functions [imath]f=1[/imath] and [imath]f=−1[/imath]). I had a little trouble figuring out what should be the answer in case the field is [imath]\mathbb{C}[/imath]. I think it should be (for [imath]\ell^1[/imath]) all the [imath]\{e_n\}_{n\in \mathbb N}\bigcup\{−e_n\}_{n \in \mathbb N}\bigcup\{ie_n\}_{n\in \mathbb N}\bigcup\{−ie_n\}_{n \in \mathbb N}[/imath], and (for [imath]\mathcal{C}[0,1][/imath]) the functions [imath]f=1,f=−1,f=i,f=−i[/imath]. But I can also see it go to (for [imath]\ell^1[/imath]) all the [imath]\{ae_n:|a|=1\}[/imath], and (for [imath]\mathcal{C}[0,1][/imath] all the constant functions [imath]f=a[/imath] with [imath]|a|=1[/imath]. I would like some guidness if possible. Knowing the right answer could help, as well as the intuition behind and maybe some tips on how to get there. Thanks!! |
1067713 | Primitive roots of [imath]2^{16} + 1[/imath]
I have a primitive root [imath] \alpha [/imath] of a number [imath] p = 2^{16} + 1[/imath]. How can I show if [imath] \alpha^{3} [/imath] and [imath]\alpha^{14}[/imath] are primitive roots as well? | 134109 | [imath]g^k[/imath] is a primitive element modulo [imath]m[/imath] iff [imath]\gcd (k,\varphi(m))=1[/imath]
I'd really love your help with the following one: : Let [imath]g[/imath] a primitive element modulo [imath]m[/imath], [imath]g^{\varphi(m)} \equiv 1\pmod{m}[/imath]. I need to prove that [imath]g^k[/imath] is a primitive element modulo [imath]m[/imath] iff [imath]\gcd (k,\varphi(m))=1[/imath]. First I tried to assume the left part, since [imath]g^k[/imath] is a primitive element so [imath]g^k \equiv 1\pmod{m}[/imath], so that [imath]k \neq \varphi(m) [/imath]. so there is a [imath]r[/imath] such that [imath]r| \varphi(m)[/imath], [imath]r|k[/imath]. if [imath]k=r[/imath] so [imath]\varphi(m)=kt[/imath] for some [imath]t[/imath], [imath]t < \varphi(m)[/imath] so that [imath]1 \equiv g^{\varphi(m)}=g^{kt}=(g^{k})^t[/imath] so [imath]g^k[/imath] is not an primitive element. if [imath]k \neq r[/imath] so there are [imath]a,b[/imath] such that [imath]\varphi(m)a+kb=r[/imath] for [imath]r\gt 1[/imath], what can I do next? I succeeded the other direction. Thanks a lot |
1068879 | How to derive this interesting identity for [imath]\log(\sin(x))[/imath]
I saw on SE that: [imath]\log(\sin x)=-\log(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n} \phantom{a} (0<x<\pi)[/imath] This is an extremely useful identity, as it helps solve: [imath]\int_{0}^{\pi} \log(\sin(x)) dx[/imath] But how is it derived? From Taylor series, power series? How do I get this? Even if someone can start me off that would be great. | 292468 | Fourier series of Log sine and Log cos
I saw the two identities [imath] -\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2) [/imath] and [imath] -\log(\cos(x))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}+\log(2) [/imath] here: twist on classic log of sine and cosine integral. How can one prove these two identities? |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.