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2493443 | What is the intuition behind why the integration of [imath]f(x) = x[/imath] for closed interval of negative to positive infinity diverges, rather than being zero?
I'm currently studying integrals in calculus. In teaching improper integrals over infinite intervals, I come across this example in my lecture notes, [imath]\int_{-\infty}^{\infty} x \ dx[/imath] Naturally, the integration is split into two halves, such as [imath]\int_{-\infty}^{0}x\ dx=-\infty[/imath] and [imath]\int_{0}^{\infty}x\ dx=\infty[/imath]. The lecture notes conclude that since both these improper integrals diverge, then so does the above improper integral, i.e. [imath]\Rightarrow\int_{-\infty}^{\infty}x\ dx\ \mathrm{diverges}[/imath] Intuitively, however, I would expect that the first improper integral, [imath]\int_{-\infty}^{\infty}x\ dx[/imath] should evaluate to [imath]0[/imath], given that, An intuitive meaning of the definite integral is the area under the curve The curve of [imath]y=x[/imath] gives a negative area for [imath](-\infty,0)[/imath] and positive area for [imath](0, \infty)[/imath] The curve of [imath]y=x[/imath] is symmetrical about [imath]y=0[/imath] So, my questions are Why is this not the case, and Is [imath]\int_{-\infty}^{\infty}x\ dx\ \mathrm{diverges}[/imath] even what I think it means, that [imath]\int_{-\infty}^{\infty}x\ dx=\infty[/imath]? | 2596476 | Can we give a definition of [imath]\int_{-\infty}^{\infty} xdx[/imath],or why this integral is divergent?
I know that [imath]\int_{-\infty}^{\infty} x dx= \int_{-\infty}^{0} xdx + \int_{0}^{\infty} xdx[/imath] and both of these integrals is divergent, so the previous integral is divergent too, but I think it is easy to see that [imath]\int_{-\infty}^{\infty} xdx=0[/imath], what is wrong what is wrong with my intuition? |
2564473 | How do you find the equation from the series expansion
How do you find the equation which is a Taylor series expansion of [imath]\displaystyle 1+\frac{2}{3^2}+\frac{3}{3^3} + \ldots + \frac{k}{3^k} \ldots[/imath] I believe the Taylor series expansion : [imath]\displaystyle \frac13+\frac{2}{3^2}+\frac{3}{3^3} + \ldots + \frac{k}{3^k} \ldots[/imath] is the expansion of [imath]\displaystyle \frac{x}{(1-x)^2}[/imath]. Where does the [imath]1[/imath] come from then? How do I find what the original equation is? | 2564431 | How to find the infinite sum
I need to find the infinite sum of the following series expansion [imath]1/3 + 2/3^2 + 3/3^3 + 4/3^4 + \dots + k/3^k + \dots[/imath] I know that [imath]x/(1 - x) = x + x^2 + x^3 + \dots + x^k + \dots[/imath] We need to find the [imath]x[/imath] value in order to find the infinite sum. What could the [imath]x[/imath] value be? I am not sure. |
2563699 | [imath]G[/imath] acts as a group of automorphisms on [imath]A[/imath], [imath]\textrm{Spec}(A^G)=G \backslash \textrm{Spec}(A)[/imath]
[imath]\DeclareMathOperator{\Spec}{Spec}[/imath] I'm starting to read about quotients of group schemes and am working through some basic exercises. This is from Algebraic Geometry and Arithmetic Curves by Qing Liu. I'm trying to figure out part (a). I want to show that if [imath]\mathfrak P_1, \mathfrak P_2[/imath] are primes of [imath]A[/imath] with [imath]\mathfrak P_1 \cap A^G = \mathfrak P_2 \cap A^G[/imath], then [imath]\mathfrak P_1 = \sigma \mathfrak P_2[/imath] for some [imath]\sigma \in G[/imath]. Since [imath]A[/imath] is integral over [imath]R:=A^G[/imath] by part (b), the problem becomes to show that for any prime [imath]\mathfrak p[/imath] of [imath]R[/imath], the group [imath]G[/imath] acts transitively on the primes in [imath]A[/imath] lying over [imath]\mathfrak p[/imath]. If we just stick to maximal ideals, I can solve the problem by modifying an argument from basic algebraic number theory: Solution when [imath]\mathfrak p[/imath] is a maximal ideal: Since [imath]R \subseteq A[/imath] is integral, every prime lying over [imath]\mathfrak p[/imath] is also maximal. Let [imath]\mathfrak P, \mathfrak Q[/imath] be distinct primes of [imath]A[/imath] lying over [imath]\mathfrak p[/imath]. Suppose that [imath]\mathfrak P \neq \sigma \mathfrak P[/imath] for any [imath]\sigma \in G[/imath]. Then [imath]\mathfrak P[/imath] and [imath]\sigma \mathfrak Q[/imath] are comaximal ideals for every [imath]\sigma \in G[/imath], and so by the Chinese remainder theorem there exists a solution [imath]x \in A[/imath] to the system [imath]x \equiv 0 \pmod{\mathfrak P}[/imath] [imath]x \equiv 1 \pmod{\sigma^{-1}\mathfrak Q} : \sigma \in G [/imath] Then [imath]\sigma(x) - 1 \in \mathfrak Q[/imath] for all [imath]\sigma[/imath], so [imath]\sigma(x)[/imath] is never in [imath]\mathfrak Q[/imath]. Hence neither is [imath]y := \prod\limits_{\sigma \in G} \sigma(x)[/imath]. But [imath]y \in \mathfrak P \cap A^G = \mathfrak p \subseteq \mathfrak Q[/imath], contradiction. [imath]\blacksquare[/imath] I had an idea of how to reduce to the case where [imath]\mathfrak p[/imath] is maximal. Let [imath]S = R - \mathfrak p[/imath]. Then the inclusion [imath]R \subseteq A[/imath] induces an injective ring homomorphism [imath]R_{\mathfrak p} = R\otimes_R R_{\mathfrak p} \rightarrow A \otimes_R R_{\mathfrak p} = S^{-1}A[/imath]. By tensoring with [imath]1_{R_{\mathfrak p}}[/imath], we still get an action of [imath]G[/imath] as a group of automorphisms of the ring [imath]S^{-1}A[/imath]. If I can show that [imath]R_{\mathfrak p} = (S^{-1}A)^G[/imath], then I will be in the same situation as before, with [imath]\mathfrak p R_{\mathfrak p}[/imath] a maximal ideal. From the diagram [imath]\begin{array} \textrm{Spec } A & \leftarrow & \Spec S^{-1}A \\ \downarrow & & \downarrow \\ \Spec A^G & \leftarrow &\Spec (A^G)_{\mathfrak p} \end{array}[/imath] with horizontal injections and vertical surjections, with the action of [imath]G[/imath] on the top right object being the restriction of the action on the top left, I'll get the result. | 2465914 | Action of finite group of automorphisms on Spec A
Question 2.3.20 in Qing Liu's Algebraic Geometry and Arithmetic Curves: Let [imath]A[/imath] be a ring, [imath]G[/imath] a finite group of automorphisms of [imath]A[/imath]. Let [imath] p: \mathrm{Spec}\ A \to \mathrm{Spec}\ A^{G} [/imath] be the morphism of affine schemes induced by the obvious inclusion. Show that, for [imath]x_{1}[/imath], [imath]x_{2} \in \mathrm{Spec}\ A[/imath], [imath] p(x_{1}) = p(x_{2}) \iff \exists \ \sigma \in G : \sigma(x_{1}) = x_{2}.[/imath] The '[imath]\impliedby[/imath]' direction is easy, but I'm having trouble with the other direction (although I have no doubt that it is also simple). My attempts so far have included assuming that there is no [imath]\sigma \in G[/imath] such that [imath]\sigma(x_{1}) = x_{2}[/imath], so that for each [imath]\sigma[/imath] there is [imath]g_{\sigma} \in x_{1}[/imath] with [imath]\sigma(g_{\sigma}) \notin x_{2}[/imath], and then trying to construct an element of [imath]p(x_{1})\backslash p(x_{2})[/imath] out of these [imath]g_{\sigma}[/imath], for example by taking [imath] g = \prod_{\sigma \in G}g_{\sigma} [/imath] and then [imath] h = \prod_{\sigma}\sigma(g) [/imath] but this appears to fail as I can't ensure that [imath]\sigma(g_{\tau}) \notin p(x_{2})[/imath] for [imath]\sigma \neq \tau[/imath]. Any hints would be much appreciated! |
2564292 | Find all functions from [imath]\mathbb{Z}[/imath] to [imath]\mathbb{Z}[/imath] such that [imath]f(x+y)=f(x)+f(y)[/imath].
Find all functions from [imath]\mathbb{Z}[/imath] to [imath]\mathbb{Z}[/imath] such that [imath]f(x+y)=f(x)+f(y)[/imath]. We showed these claims: Claim1. [imath]f(0)=0[/imath]. Claim2. [imath]f(-x)=-f(x)[/imath]. But, how should I show that there is no other [imath]g[/imath] function from [imath]\mathbb{Z}[/imath] to [imath]\mathbb{Z}[/imath] which satisfies [imath]g(x+y)=g(x)+g(y)[/imath]. | 2006468 | Find all functions [imath]f: \mathbb Z \to \mathbb Z[/imath] such that [imath]f(x+y)=f(x)+f(y)[/imath]
Let [imath]x=0\,\, f(x+y)=f(0+y)[/imath] every [imath]y \in \mathbb{Z}[/imath] So we get [imath]f(0)+f(y)=f(y)[/imath] then [imath]f(0)=0[/imath] [imath]0=f(0)=f(x(-x))=f(x)+f(-x)=0[/imath] Let [imath]f(1)=k[/imath] for some [imath]k \in \mathbb{Z}[/imath] [imath]f(n)=f(1+1+1+...+1) (n \text{ many } 1)[/imath] Β Β Β Β Β Β Β Β Β Β [imath]=f(1)+f(1)+...+f(1) (n \text{ many } f(1))[/imath] Β Β Β Β Β Β Β Β Β Β [imath]=n\cdot k[/imath] Thus [imath]f: \mathbb{Z} \to \mathbb{Z}[/imath] a function such that [imath]f(n)=nk[/imath] I found this but I must also show that there is no other function which satisfies [imath]f(x+y)=f(x)+f(y)[/imath] and I am stuck can somebody help me. |
2564679 | Show that there exists [imath]x \in X[/imath] with [imath]\|x\|=1[/imath] and [imath]f(x)=\|f\|[/imath] in a non-zero reflexive space
Let [imath]X \neq \lbrace 0 \rbrace[/imath] be a reflexive space, and let [imath]f \in X'[/imath]. Show that there exists [imath]x \in X[/imath] with [imath]\|x\|=1[/imath] and [imath]f(x)=\|f\|[/imath]. The definition of a reflexive space: [imath]X[/imath] is called reflexive if [imath]J_X(X)=X''[/imath] where [imath]J_X: X \to X''[/imath] defined by [imath]J_Xx(f)=f(x)[/imath] with [imath]x \in X[/imath] and [imath]f \in X'[/imath]. How to find this such [imath]x[/imath]? | 2558598 | For [imath]f[/imath] in dual space, there exists [imath]x[/imath] with norm 1 and [imath]f(x)=\|f\|[/imath] if space is reflexive (and nontrivial)
Let [imath]X\ne\{0\}[/imath] be a reflexive space and let [imath]f\in X^*[/imath], where [imath]X^*[/imath] is the dual of [imath]X[/imath]. I want to know: in general, does there exist an [imath]x\in X[/imath] with [imath]\|x\|=1[/imath], and [imath]f(x)=\|f\|[/imath], where [imath]\|f\|[/imath] is defined as [imath]\sup\{|f(x)|:x\in X,\|x\|=1\}[/imath]? I know this is true for [imath]\mathbb{R}^n[/imath] with the norm from the standard inner product, but I'm wondering if it is true in general. |
2563956 | Find all natural numbers [imath]n[/imath] such that [imath]2^n[/imath] divides [imath]3^n -1[/imath]
Find all natural numbers [imath]n[/imath] such that [imath]2^n[/imath] divides [imath]3^n -1[/imath] I think that the only solutions are [imath]n = 0,1,2,4[/imath], but I have no idea on how to prove it. I tried to write [imath]3^n-1[/imath] as [imath]1+3+3^2+...+3^{n-1}[/imath] and manipulate the sum but found my self at the equally hard problem of finding the power of two dividing [imath]3^k+1[/imath] | 1252716 | When does [imath]2^{n}[/imath] divide [imath]3^{n}-1[/imath]
The title says it all. For what natural numbers [imath]n[/imath] does [imath]2^n[/imath] divide [imath]3^n - 1[/imath]? By substituting values I can see that this happens for [imath]n = 1, 2, 4[/imath] but are there more? I am not able to prove/disprove. |
2562619 | On the volume of a non-right tetrahedron and the integral of a linear function over it
I've solved quite a bit of these questions with tetrahedrons with a 90 degree angle, but I am not sure how to approach a non-right angle tetrahedron. I am unsure how to find the triple integral's bounds. Calculation itself is no problem. [imath] \int_B (x + y + 2z) dV[/imath] where B is the tetrahedron with vertices [imath](0, 0, 0)[/imath], [imath](0, 2, 0)[/imath], [imath](1, 3, 1)[/imath] and [imath](1, 1, 0)[/imath]. Is the way of finding the bounds same for non-right angle tetrahedrons? | 225431 | Finding the volume of the tetrahedron.
Find the volume of the tetrahedron with the vertices [imath]P(1,1,1)[/imath], [imath]Q(1, 2, 3)[/imath], [imath]R(3, 1, 2)[/imath], and [imath]S(2, 3, 1)[/imath]. |
2563623 | Relation between Pascal's triangle and fibonacci series.
Accidentally, I had seen this relation. I tried to find the formula Here is my try:- [imath]f_0= {{0}\choose{0}} [/imath] [imath]f_1={{1}\choose{0}} [/imath] [imath]f_2={{2}\choose{0}}+ {{1}\choose{1}} [/imath] [imath]f_3= {{3}\choose{0}}+ {{2}\choose{1}} [/imath] [imath]f_4= {{4}\choose{0}}+ {{3}\choose{1}}+{{2}\choose{2}} [/imath] [imath]...[/imath] Generalizing, [imath]f_n=\sum_{s+r=n,s\ge r}{{s}\choose{r}}, 0\le s,r \le n [/imath] How to write the correct formula of n-th term. Why this pattern is coming. I could verify using calculation for upto n=10. How to give a rigorous proof?. | 859367 | Fibonacci Numbers Proof: [imath] f_n = \binom n0 + \binom{n-1}1 +\dots+ \binom{n-k}k[/imath]
Prove the following fibonacci sequence, which appear in Pascal's Triangle. I am not sure where to start on this, any pointers? [imath] f_n = {n\choose0} + {n-1\choose1} + ... + {n-k\choose k}[/imath] where [imath]\displaystyle k=\left\lfloor\frac{n}{2}\right\rfloor[/imath] |
2561790 | Set of [imath]2008[/imath] Integers Having Prime Divisors Less Than [imath]24[/imath]
A set [imath]L[/imath] consists of [imath]2008[/imath] integers, none of which have a prime divisor larger than [imath]24[/imath]. Prove that [imath]L[/imath] has four elements, the product of which is equal to the fourth power of an integer. | 1362056 | Olympiad question on Pigeonhole principle
Given a set [imath]M[/imath] of [imath]1985[/imath] distinct positive integers, none of which has a prime divisor greater than [imath]26[/imath], prove that [imath]M[/imath] contains at least one subset of four distinct elements, whose product is the fourth power of an integer. I couldn't even start. |
2143999 | Show that, if [imath]S_1[/imath] and [imath]S_2[/imath] are subsets of a vector space [imath]V[/imath] such that [imath]S_1 \subseteq S_2[/imath], then [imath]span(S_1) \subseteq span(S_2)[/imath].
Show that, if [imath]S_1[/imath] and [imath]S_2[/imath] are subsets of a vector space [imath]V[/imath] such that [imath]S_1 \subseteq S_2[/imath], then [imath]span(S_1) \subseteq span(S_2)[/imath]. A (hypothesis): [imath]S_1[/imath] and [imath]S_2[/imath] are subsets of a vector space [imath]V[/imath] such that [imath]S_1 \subseteq S_2[/imath]. B (conclusion): [imath]span(S_1) \subseteq span(S_2)[/imath] A1: Let [imath]S_2 = \{x^n, x^{n - 1}, ..., x^0\}[/imath] A2: [imath]span(S_2) = \{a_nx^n + a_{n - 1}x^{n - 1}+ ... + a_0 : a \in F \}[/imath] A3: Let [imath]S_1 = \{w^n, w^{n - 1}, ..., w^0 \}[/imath] where [imath]w = a_nx^n + a_{n - 1}x^{n - 1}+ ... + a_0[/imath]. [imath]\therefore[/imath] The elements of [imath]S_1[/imath] are linear combinations of [imath]S_2[/imath]. A4: [imath]span(S_1) \subseteq span(S_2)[/imath] [imath]Q.E.D.[/imath] I would greatly appreciate it if people could please take the time to review my proof for correctness. If there are any errors, then please explain why and what the correct solution is. | 1447177 | If X and Y are two sets of vectors in a vector space V, and if X [imath]\subset[/imath] Y, then is span X [imath]\subset[/imath] span Y?
If X and Y are two sets of vectors in a vector space V, and if X [imath]\subset[/imath] Y, then is span X [imath]\subset[/imath] span Y? If so, why is or isn't the span of X a subset of the span of Y? EDIT: Thank you for the hints! Here is the proof I came up with; please let me know if it is correct. The spanning set of X can be written as: Span(X) = {[imath]a_1 x_1 + a_2 x_2 + ... + a_n x_n[/imath]} where all [imath]x_i[/imath] are vectors and all [imath]a_i[/imath] are scalars. Since X[imath]\subset[/imath]Y then all [imath]x_i[/imath] are also in Y. Thus, Span(x) is a linear combination of vectors from Y So Span(x) [imath]\subset[/imath] Span(Y). |
2564538 | if p is odd prime such that [imath]p \equiv 1\,(\text{mod } 7)[/imath] then [imath]p \equiv 1\,(\text{mod } 14)[/imath]
Prove : if [imath]p[/imath] is odd prime such that [imath]p\equiv 1 \ (\bmod 7)[/imath] then [imath]p \equiv 1\ (\bmod 14)[/imath] my prove for it that let [imath]p[/imath] is odd prime and [imath]p\equiv 1 \ (\bmod 7)[/imath] so every odd prime number can be written as [imath]p\equiv 1 \ (\bmod 2)[/imath] so by multiple the [imath]p\equiv 1\ (\bmod 7)[/imath] and [imath]p\equiv 1\ (\bmod 2)[/imath] we get [imath]p\equiv 1\ (\bmod 14)[/imath] is that right please? | 2559664 | If [imath]p[/imath] is a prime such that [imath]pβ‘1\pmod 7[/imath] then [imath]p β‘1\pmod{14}[/imath]
Prove or disprove the following: 1) if [imath]p[/imath] is odd prime such that [imath]pβ‘1\pmod 7[/imath] then [imath]p β‘1\pmod{14}[/imath] 2) if [imath]p[/imath] is odd prime such that [imath]pβ‘4 \pmod 7[/imath] then [imath]p β‘4\pmod{14}[/imath] I think that both statement are true and have a similar proof 1) [imath]pβ‘1\pmod 7[/imath] so [imath]pβ‘8\pmod 7[/imath] then [imath]7|p-8[/imath] so [imath]p-8= 7 k[/imath] where [imath]k[/imath] is an integer so [imath]p=8+7k[/imath], so [imath]p β‘1\pmod {14}[/imath] It is that a correct proof? Thank you. |
2564012 | Product of a quotient space and the unit interval
If [imath]X[/imath] is a topological space and [imath]\sim[/imath] is an equivalence relation on [imath]X[/imath], I want to show that [imath]\tau: X \times I \to (X/{\sim}) \times I, \; (x,t) \mapsto (\pi(x),t)[/imath] (where [imath]I =[0,1][/imath] and [imath]\pi: X \to X/{\sim}[/imath] is the quotient map) is an identification. Obviously, [imath]\tau[/imath] is continuous. But how do I prove that [imath]\tau^{-1}(U)[/imath] is open iff [imath]U[/imath] is open? So far all of my attempts have failed since [imath]\pi[/imath] can't be assumed to be an open map. Thanks for help in advance. | 31697 | When is the product of two quotient maps a quotient map?
It is not true in general that the product of two quotient maps is a quotient maps (I don't know any examples though). Are any weaker statements true? For example, if [imath]X, Y, Z[/imath] are spaces and [imath]f : X \to Y[/imath] is a quotient map, is it true that [imath] f \times {\rm id} : X \times Z \to Y \times Z[/imath] is a quotient map? |
2564687 | Prove that [imath]f[/imath] is continuous-Proof-Verification.
Let [imath](X,d)[/imath] be a compact metric space. Let [imath]f:X\to X[/imath] be a function such that graph(f)=[imath](G(f))=\{(x,f(x):x\in X\}[/imath] is closed. Prove that [imath]f[/imath] is continuous. TRY: To show that [imath]f[/imath] is continuous enough to show that if [imath]x_n\to x[/imath] then [imath]f(x_n)\to f(x)[/imath]. Let [imath]x_n\to x[/imath] ;; since [imath]X[/imath] is compact so [imath]f(x_n)[/imath] has a convergent subsequence say [imath]f(x_{n_k})\to y[/imath]. Consider the corresponding terms in the sequence [imath]x_n[/imath] then [imath](x_{n_k },f(x_{n_k}))\to (x,y).[/imath] Now [imath](x_{n_k },f(x_{n_k}))\in G(f)[/imath] which is closed hence [imath](x,y)\in G(f))[/imath] which in turn implies that [imath]y=f(x)[/imath] So [imath]f(x_{n_k})\to f(x)[/imath]. But I need to show that [imath]f(x_n)\to f(x)[/imath] How can I show that??Please help me out. | 1769333 | Compact topological space with closed graph implies continuity
Let [imath]X,Y[/imath] be topological spaces and [imath]f:X \rightarrow Y[/imath] If [imath]Y[/imath] is compact and [imath]G(f)[/imath] (graph of [imath]f[/imath]) is closed, show that [imath]f[/imath] is continuous. I consider an open set in [imath]X[/imath], and as [imath]G(f)[/imath] is closed, [imath](X \times Y)-G(f)[/imath] is open. And I guess I need to do something here. Then I can get some open sets. Also, I notice that compact is something relate to "finite". I guess I can use the compactness to do finite intersection or union and get the result finally. But what should I do in the middle or I am not even on a right track. |
2564572 | Understanding [imath]1[/imath]-line proof that if [imath]V[/imath] is irreducible [imath]\mathbb{C}S_n[/imath]-module, then [imath]\operatorname{res}_{S_{n-1}}V[/imath] is multiplicity free?
I'm having trouble understanding a terse proof of a theorem. Theorem. If [imath]V[/imath] is an irreducible [imath]\mathbb{C}S_n[/imath]-module, then the restriction [imath]\operatorname{res}^{S_n}_{S_{n-1}}V[/imath] is multiplicity free. Proof: The centralizer of [imath]\mathbb{C}S_{n-1}[/imath] in [imath]\mathbb{C}S_n[/imath] is commutative. I know how to show that the centralizer above is commutative, but I don't see the connection with the restricted module being multiplicity free. I know that a criterion for a [imath]\mathbb{C}G[/imath]-module to be commutative is that its endomorphism algebra be commutative. Is [imath]\operatorname{End}_{S_{n-1}}(\operatorname{res}^{S_n}_{S_{n-1}}V)[/imath] seen to be commutative somehow using the fact that the centralizer is commutative? | 1220623 | Centralizer of [imath]\mathbb{C}[G][/imath] in [imath]\mathbb{C}[H][/imath]
I found this result, but can't understand how to prove. Let [imath]H[/imath] be a subgroup of [imath]G[/imath]. Then prove [imath]Z(\mathbb{C}[G],\mathbb{C}[H])[/imath] is commutative iff every irreducible [imath]G[/imath] module when restricted to [imath]H[/imath] splits into irreducible [imath]H[/imath] modules of multiplicity at most [imath]1[/imath]. Where [imath]Z(M,N)=\{m \mid mn=nm , \forall n\in N\}[/imath] |
2565246 | How to solve [imath]\int \frac{a}{(a^{2}+x^{2})^{3/2}}dx[/imath]
I tried the Euler substitution: [imath]t=x+\sqrt{x^{2}+a}[/imath] and I got [imath]\int \frac{4ta}{(t^{2}+a)^{2}}dt[/imath] and I don't know wnat now or if it is the right way. Thank you in advance. | 2222429 | Integrate: [imath]\int (x^2+a^2)^{-3/2} \cdot dx[/imath]
Integrate: [imath]\int (x^2+a^2)^{-3/2} \cdot dx[/imath] My Approach: [imath]\int (x^2+a^2)^{-3/2} \cdot dx[/imath] [imath]\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}[/imath] But this doesn't give the right answer. I showed this to my friend and he said [imath]d(x^2+a^2)[/imath] is not possible which makes sense since you can't take a small element of the form [imath](x^2+a^2)[/imath]. How can I then solve this integration without using trigonometry? |
2547140 | Proving trigonometric identity: [imath] \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} [/imath]
Prove that [imath] \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} [/imath] This is my working - [imath]\text{LHS} = \frac{\sin (y+x)}{\sin (y-x)} = \frac{\sin y \cos x + \cos y \sin x}{\sin y \cos x - \cos y \sin x} [/imath] I used the compound angle relations formula to do this step. However , now I'm stuck. I checked out the answer and the answer basically carried on my step by dividing it by [imath] \frac{\cos x \cos y}{\cos x \cos y} [/imath] If I'm not wrong , we cannot add in our own expression right? Because the question is asking to prove left is equal to right. Adding in our own expression to the left hand side will be not answering the question, I feel . | 668619 | Proving a trig identity: [imath]\frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B}[/imath]
I'm learning about trig identities, and I'm struggling to prove that two expressions are equal: [imath] \frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B} [/imath] How do you go about proving this? I know about compound angles - i.e. the sine, cosine and tangent of [imath](A \pm B)[/imath], but don't know how to apply it in this situation. |
2565861 | If [imath]f\in \mathcal{C}^1(\mathbb{R}^2)[/imath], then there exists [imath]g:[0,1]\to \mathbb{R}^2[/imath] such that [imath]f\circ g [/imath] is constant.
If [imath]f\in \mathcal{C}^1(\mathbb{R}^2)[/imath]. Prove that there exists a continuous one-one function [imath]g:[0,1]\to \mathbb{R}^2[/imath] such that [imath]f\circ g:[0,1]\to \mathbb{R}^2 [/imath] is constant. Would anybody give me some hint to do this problem? | 801644 | Show there exists [imath]g:(0,1)\to\mathbb{R}^2[/imath] such that [imath]h\circ g[/imath] constant, [imath]g[/imath] continuously differentiable and [imath]g[/imath] injective
Let [imath]h\in C^1(\mathbb{R}^2,\mathbb{R})[/imath]. Show: There exists [imath]g:(0,1)\to\mathbb{R}^2[/imath] such that [imath]h\circ g[/imath] constant, [imath]g[/imath] continuously differentiable and [imath]g[/imath] injective. Dini's theorem: Let [imath]\Omega \subseteq \mathbb{R}^2[/imath] open, [imath]f \in C^1(\Omega, \mathbb{R})[/imath] and [imath]p \in \Omega[/imath] with [imath]f(p) = 0[/imath]. If [imath]\frac{\partial f}{\partial y}(p) \neq 0[/imath], then there exist open intervals [imath]V_1, V_2[/imath] with [imath]p \in V_1 \times V_2 \in \Omega[/imath] and [imath]g \in C^1(V, \mathbb{R})[/imath] with [imath]\{(x,y) \in V_1 \times V_2 \mid f(x,y) = 0 \} = \{(x,g(x) \mid x \in V_1\}[/imath] and [imath]g[/imath] furthermore satisfies [imath]g' = -\left(\frac{\partial f}{\partial y}(x, g(x))\right)^{-1}\frac{\partial f}{\partial x}(x, g(x)).[/imath] Proof of existance of [imath]g[/imath]: Let without loss of generality [imath]\frac{\partial f}{\partial y}(p) \gt 0[/imath]. Then there exists [imath]\epsilon \gt 0[/imath] with [imath]\overline{B_\epsilon(p)} \subseteq \Omega[/imath] and [imath]\frac{\partial f}{\partial y} \gt 0[/imath] on [imath]\overline{B_\epsilon(p)}[/imath]. Let [imath]y^- = p - \frac{\epsilon}{2}[/imath] and [imath]y^+ = p + \frac{\epsilon}{2}[/imath]. Then there exists [imath]\delta \in (0, \epsilon)[/imath] with [imath]f(x, y^-) \lt 0 \lt f(x, y^+)[/imath] for [imath]x \in (p - \frac{\delta}{2}, p + \frac{\delta}{2}) =: V_1[/imath] and [imath](y^-, y^+) =: V_2[/imath]. For [imath]x \in V_1[/imath] there exists exactly one [imath]y \in V_2[/imath] with [imath]f(x,y) = 0[/imath] (mean value theorem). Set [imath]y(x) := g[/imath] and the claim follows. |
417272 | Mean value properties of [imath]f(t)=(\cos t,\sin t)[/imath]
I am stuck on the following question which I came across in a recent exam paper: Let [imath]f \colon [\pi,2 \pi] \to \Bbb R^2[/imath] be the function [imath]f(t)=(\cos t,\sin t)[/imath].Then which of the following are necessarily correct? The options are: There exists [imath]t_0 \in [\pi,2\pi][/imath] such that [imath]f'(t_0)=\frac{f(2 \pi)-f(\pi)}{\pi}[/imath] There exists no [imath]t_0 \in [\pi,2\pi][/imath] such that [imath]f'(t_0)=\frac{f(2 \pi)-f(\pi)}{\pi}[/imath] There exists [imath]t_0 \in [\pi,2\pi][/imath] such that [imath]||f(2 \pi)-f(\pi)|| \le \pi ||f'(t_0)||[/imath] [imath]f'(t)=(-\sin t,\cos t) \,\,\forall t \in [\pi,2\pi][/imath]. MY ATTEMPT: Clearly option 4 is correct but I am not sure how to prove/disprove for other options. Looking at options 1,2,3 I think mean value theorem has a role to play. Can someone explain me with some details. My edit: Option 3 appears to be true since [imath]||f(2 \pi)-f(\pi)||=||(1,0)-(-1,0)||=||(2,0)||=2[/imath] and [imath]\pi ||f'(t_0)||=\pi \,\,\text{since}\, ||f'(t)||=||(- \sin t,\cos t)||=1[/imath] and so [imath]||f(2 \pi)-f(\pi)|| \le \pi ||f'(t_0)||[/imath] is true. Option 1 makes no sense after putting the values of [imath]\pi, 2\pi[/imath] in [imath]f[/imath]. So option 2 appears to be correct. Am I going in the right direction? Is there any other elegant way of tackling this problem? | 365839 | continuous function from [imath][\pi,2\pi]\to \mathbb{R}^2[/imath]
[imath]f:I=[\pi,2\pi]\to \mathbb{R}^2[/imath] be given by [imath]f(t)=(\cos t,\sin t)[/imath] which of the following are necessarily correct? [imath]1[/imath]. [imath]\exists t_0\in I[/imath] such that [imath]f'(t_0)=\frac{1}{\pi}(f(2\pi)-f(\pi))[/imath] [imath]2[/imath]. [imath]\nexists t_0\in I[/imath] such that [imath]f'(t_0)=\frac{1}{\pi}(f(2\pi)-f(\pi))[/imath] [imath]3[/imath].[imath]\exists t_0\in I[/imath] such that [imath]\pi||f'(t_0)||\ge ||(f(2\pi)-f(\pi)||[/imath] [imath]4[/imath]. [imath]f'(t)=(-\sin t,\cos t)\forall t[/imath] [imath]4[/imath] is clearly true, [imath]1[/imath] is true by Mean Value Theorem as [imath]f[/imath] satisfies all the conditions of MVT? have no idea of [imath]2[/imath] and [imath]3[/imath], Thank you for help |
2565850 | How to solve Rudin Chapter 2 problem 11?
I have finished the entire problem except the 5th part. Roughly speaking the questions asks to show that over [imath]\mathbb{R}[/imath] prove/disprove that the distance function [imath] d_5(x,y) = \frac{|x-y|}{1 + |x-y|} [/imath] Is a metric. I have proven symmetry and I have proven that if [imath]d_5(x,y) = 0[/imath] that [imath]x = y[/imath] so all that remains to be shown is the triangle inequality namely : [imath] \frac{|x-y|}{1 + |x-y|}+ \frac{|y-z|}{1 + |y-z|} \ge \frac{|x-z|}{1 + |x-z|} [/imath] This of course can be cast more simply in terms of [imath]u,v[/imath] in [imath]\mathbb{R}[/imath] as [imath] \frac{|u|}{1 + |u|}+ \frac{|v|}{1 + |v|} \ge \frac{|u+v|}{1 + |u+v|} [/imath] This reminds me an awful lot of contest style inequalities but i can't for the love of god place which one this is (not that it should matter, I should ideally be able to just generate a proof with ease). My one angle of attack was to do case work on fixing [imath]u,v[/imath] to be both positive or both negative, or a mix of positive and negative. But is there a easier way to proceed than that? | 981684 | Distance function is in fact a metric
I know I should be able to show this, but for some reason I am having trouble. I need to show that [imath]d(x,y) = \frac{|x-y|}{1+|x-y|}[/imath] is a metric on [imath]\Bbb R[/imath] where [imath]|*|[/imath] is the absolute value metric. I am getting confused trying to show that the triangle inequality holds for this function. My friend also said that he proved that this distance function defines a metric even if you replace [imath]|*|[/imath] with any other metric. So I'd like to try and show both, but I cannot even get the specific case down first. Please help. |
2566099 | Question about localisation
Suppose [imath]A[/imath] be a commutative ring with unity. [imath]S[/imath] be a multiplicative closed subset of [imath]A[/imath]. Suppose it is given that [imath]A_S[/imath] is a local ring. What can we say about [imath]S[/imath]. Is it true that [imath]S[/imath] must be a complement of a prime ideal? Thank you. | 1409338 | Converse of "localization at a prime is local"
Suppose [imath]S^{-1}R[/imath] is the localization of a ring R at a multiplicative subset S, and is local. Must S be the complement of a prime ideal? |
2566394 | [imath]A[/imath] is normal if and only if [imath]trace(A^\ast A)=\sum_{i=1}^n|\lambda_i|^2.[/imath] where the [imath]\lambda_i[/imath] are the eigenvalues of [imath]A[/imath].
Let [imath]A[/imath] complex matrix of [imath]n\times n[/imath]. Show that [imath]A[/imath] is normal if and only if [imath]trace(A^\ast A)=\sum_{i=1}^n|\lambda_i|^2.[/imath] where the [imath]\lambda_i[/imath] are the eigenvalues of [imath]A[/imath]. I try using the trace properties and the definition of normal matrix...I use that [imath]A[/imath] is similar to unitary matrix...but i not can prove the exercise. I wait that can you help me. Thanks! | 1030016 | Show that [imath]A\in\mathbb{C}_n[/imath] is normal [imath]\iff[/imath] [imath]tr(A*A) = \sum_{i = 1}^n|\lambda_i|^2[/imath], where [imath]\lambda_1,...,\lambda_n[/imath] are the eigenvalues of [imath]A[/imath].
Title restated: Show that [imath]A\in\mathbb{C}_n[/imath] is normal [imath]\iff[/imath] [imath]tr(A^*A) = \sum_{i = 1}^n|\lambda_i|^2[/imath], where [imath]\lambda_1,...,\lambda_n[/imath] are the eigenvalues of [imath]A[/imath]. This question comes from "Matrices and Linear Transformations" by Charles Cullen. I'm studying for an exam and am trying to do some problems but I'm having trouble with this one. Any help would be appreciated. Thank you. |
2566591 | How do we find the average value of the following function?
How do we find the average value of this function with the intervals [imath][0,1][/imath]? [imath]f(t)=10(e^{5t}-1)[/imath] | 2565155 | Find the average value of the function [imath]f(t) = 10(e^{5t} - 1)[/imath] over the interval [imath][0, 1][/imath].
I have the following solution for my question but I am not sure if I have done the steps right: |
2566176 | If [imath]A \in M_5(\mathbb R)[/imath] and [imath]A^2-4A-I=0[/imath] find [imath](a_1-\frac{1}{a_1})+\ldots+(a_5-\frac{1}{a_5})[/imath]
Let [imath]A \in M_5(\mathbb R)[/imath] and [imath]A^2-4A-I=0[/imath], moreover we consider [imath]a_1[/imath], [imath]a_2[/imath], [imath]\ldots[/imath] [imath]a_5[/imath] be eigen value of [imath]A[/imath], I gonna find [imath](a_1-\frac{1}{a_1})+\ldots+(a_5-\frac{1}{a_5})[/imath], while I have no idea how to proceed, any piece of advice would be much appreciated | 2200369 | A question about eigenvalues of [imath]5\times5[/imath] matrix
[imath]A\in M_5({R})[/imath] and [imath]A^2-4A-I=0[/imath] If [imath]a_1,a_2,a_3,a_4,a_5[/imath] are eigenvalues of [imath]A[/imath] what is value of [imath]\left(a_1-\dfrac{1}{a_1}\right)+\left(a_2-\dfrac{1}{a_2}\right)+\cdots+\left(a_5-\dfrac{1}{a_5}\right)[/imath] I tried to work with [imath]A^2-4A-I=0[/imath] but get nothing because [imath]A_{5\times5}[/imath] characteristic polynomial is something like [imath]A^5+k_1A^4+...[/imath] Any hint will be appreciated . |
2567374 | Proving [imath]\arctan(\mathrm{e}^{-v}) = \text{arccot}(\mathrm{e}^v)[/imath]
Is this a correct way to prove that [imath] \arctan(\mathrm{e}^{-v}) = \text{arccot}(\mathrm{e}^v) [/imath] [imath]\tan x = \frac{1}{\mathrm{e}^v} [/imath] Turn [imath]\tan x[/imath] into [imath]\sin x[/imath] and [imath]\cos x[/imath] [imath]\frac{\sin x}{\cos x} = \frac{1}{\mathrm{e}^v}[/imath] Multiply [imath]\cos x[/imath] [imath] \sin x = \frac{\cos x}{\mathrm{e}^v} [/imath] [imath] \mathrm{e}^v \sin x = \cos x[/imath] [imath] \mathrm{e}^v = \frac{\cos x}{\sin x}[/imath] Multiply by [imath]\mathrm{e}^v[/imath] and divide by [imath]\sin x[/imath] [imath] \mathrm{e}^v = \cot x[/imath] [imath]\text{arccot}(\mathrm{e}^v) = x[/imath] | 304399 | Are [imath]\mathrm{arccot}(x)[/imath] and [imath]\arctan(1/x)[/imath] the same function?
In my textbook it asks for me to: Prove that there is no constant [imath]C[/imath] such that [imath]\text{arccot}(x) - \text{arctan}(\frac{1}{x}) = C [/imath] for all [imath]x \ne 0[/imath]. Explain why this does not violate the zero-derivative theorem. But I believe I have found such a [imath]C[/imath], i.e. [imath]C =0[/imath]! I even asked WolframAlpha (http://www.wolframalpha.com/input/?i=arccot%28x%29+-+arctan%281%2Fx%29) which corroborates my answer. This question appears in Apostol's Calculus Volume I, Second Edition: Exersize 6.22-11b Edit: Mathematica's definition of arccot is different from the one in my textbook. Apostol's arccot maps a real number into [imath](0, \pi)[/imath] while Mathematica's maps a real number into [imath](-\pi/2, \pi/2)[/imath] Here they are super-imposed: http://www.wolframalpha.com/input/?i=integral%28-1%2F%281%2Bx%5E2%29%29+%2B+pi%2F2%3B+arccot |
2567321 | How to construct an auxiliary function in this mean value theorem question?
Let [imath]f(x)[/imath] be a real-valued function that is thrice differentiable on [imath][a,b][/imath], prove there exists a [imath]\xi \in (a,b)[/imath] such that [imath]f(b)=f(a)+\frac{1}{2}(b-a)(f'(a)+f'(b))-\frac{1}{12}(b-a)^3f'''(\xi)[/imath] I failed to establish a fine auxiliary function and the Taylor formula seems useless here. Please help me with your fabulous auxiliary function :) | 312429 | Trapezoid rule error analysis
How can I prove that the max error of the trapezoid rule for the integral [imath]\int_{a}^{b}{f(x)\, \mathrm{d}x} [/imath] is: [imath]\Delta=-\frac{1}{12n^2}f''(c)(b-a)^3 \text{for } c \in (a,b) \ ?[/imath] I know that to obtain that result first you have to prove that [imath]\exists \;c \in (a,b); \int_{a}^{b}f(x)\,dx = \frac{b - a}{2}\{f(a) + f(b)\} - \frac{1}{12}f''(c)(b-a)^3[/imath] But I'm stuck here, I tried using the mean value theorem but got nowhere. Anyone got any ideas? If it helps: [imath]\forall x_0 \in (a,b) \;\exists\;\xi_0 \in (a,b);\; f(x_0) - p(x_0) = f''(\xi_0)\frac{(x_0 - a)(x_0 - b)}{2}[/imath], where p(x) is the linear function that interpolates f(x) in the points a and b ([imath]p(x) = f(a) + \frac{f(b) - f(a)}{b-a}(x-a)[/imath]) |
2567787 | [imath]\frac{\int_0^1x^n\sqrt{1-x^2}}{\int_0^1x^{n-2}\sqrt{1-x^2}}[/imath]
I want to compute the following ratio: [imath]\frac{\int_0^1x^n\sqrt{1-x^2}}{\int_0^1x^{n-2}\sqrt{1-x^2}}[/imath] As I found from wolframe it is [imath]\frac{n-1}{n+2}[/imath]. I am trying to solve the integral using the following substitution: [imath]f(x) = x^{n-2}\sqrt{1-x^2}[/imath] [imath]\frac{\int_0^1x^2f(x)}{\int_0^1f(x)}[/imath] But I am sticking here. | 2166595 | given [imath]I_{n}=\int^1_{0}x^n\sqrt{1-x^2}dx[/imath], then finding value of [imath] \frac{I_{n}}{I_{n-2}}[/imath]
given [imath]\displaystyle I_{n}=\int^{1}_{0}x^n\sqrt{1-x^2}dx[/imath], then finding value of [imath]\displaystyle \frac{I_{n}}{I_{n-2}}[/imath] Attempt: put [imath]x=\sin \theta[/imath] and [imath]dx = \cos \theta[/imath] [imath]\displaystyle I_{n} = \int^{\frac{\pi}{2}}_{0}\sin^{n}\theta \cdot \cos^2 \theta d \theta = \int^{\frac{\pi}{2}}_{0}\sin^{n}\theta (1-\sin^2 \theta)d \theta [/imath] could some help me how to solve it, thanks |
2568347 | Describe all abelian groups of order 24 up to isomorphism
I know every finite abelian group is the direct sum of cyclic groups, each of prime power order. So since [imath]24 = 2^3 \times 3, 24 \neq 120[/imath] does that just give us [imath]\mathbb{Z}_{24}, \mathbb{Z}_8 \times \mathbb{Z}_3, \mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3,[/imath] and [imath]\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3[/imath]? Is that the full answer or am I missing something? | 583958 | Find all possible abelian groups of order [imath]120[/imath].
Find all possible abelian groups of order [imath]120[/imath]. If someone could walk me through how to do this, that would be great. |
2565858 | How to express the eigenvalue of [imath]AB[/imath] in tems of eigenvalues of [imath]A[/imath] and [imath]B[/imath]
My friend has asked me to find out the relationship between the eigenvalues of [imath]AB[/imath] with eigenvalues of [imath]A[/imath],[imath]B[/imath].Where [imath]A,B[/imath] are any two matrices. My try:Let [imath]A,B[/imath] are two matrices with [imath]v[/imath] be a common eigenvector.Then [imath]Av=x_1v[/imath] and [imath]Bv=x_2v[/imath],[imath]x_1,x_2[/imath] are the eigenvalues of [imath]A[/imath] and [imath]B[/imath] respectively.Then [imath]ABv=A(x_2v)=x_2(Av)=x_2x_1v[/imath]......So [imath]x_1x_2[/imath] is also the eigenvalue of [imath]AB[/imath]. If [imath]A,B[/imath] are [imath]2*2[/imath] matrices let [imath]x_3,x_4,x[/imath] are the other eigenvalues of [imath]A,B,AB[/imath] respectively.As we know [imath]det(AB)=det(A)det(B)[/imath],and the determinant of matrix is the product of it's eigenvalues; I get [imath]x_1x_2x=x_1x_3x_2x_4[/imath][imath]=>x=x_3x_4[/imath].So each eigenvalue of [imath]AB[/imath] is the product of suitable pair of eigenvalues from [imath]A[/imath] and [imath]B[/imath].If I generalize this result I can conclude that if the [imath]n*n[/imath] matrices [imath]A,B[/imath] has same [imath](n-1)[/imath] linearly independent common eigenvalues then each of the eigenvalues of [imath]AB[/imath] is the product of suitable pair of eigenvalues of [imath]A[/imath] and [imath]B[/imath]. Can I extend this result in more general case?Is the condition "[imath]A[/imath] and [imath]B[/imath] has [imath]n-1[/imath] common linearly independent eigenvectors" is necessary?I will be happy if anyone give me any result almost familer to that result with another condition or any beautiful result concerning eigenvalues of [imath]AB[/imath]. | 716990 | Eigenvalues of Matrices and Eigenvalue of product of Matrices
If [imath]n\times n[/imath] matrix [imath]A[/imath] has eigenvalues [imath]1,-1[/imath] and [imath]n\times n[/imath] matrix [imath]B[/imath] also has eigenvalues [imath]1,-1[/imath], can I then say something about eigenvalues of [imath]AB[/imath] and [imath]BA[/imath]? |
2568581 | Prve that [imath]x^4+x^2+x+1[/imath] is irreducible over [imath]\mathbb{Q}[/imath].
I try Eisenstein's Criterion, replacing [imath]x[/imath] as [imath]x+1[/imath], [imath]x+2[/imath] or [imath]x-1[/imath]. But they are the forms we need. | 841006 | Is this polynomial irreducible in [imath]\mathbb{Q}[/imath]?
this is a really easy question but I cant find an answer; In need to see if [imath]x^4+x^2+x+1[/imath] is an irreducible polynomial over [imath]\mathbb{Q}[/imath] |
1688392 | Find the Change of Basis Matrix
Consider the 2-dimensional vector subspace [imath]V = \{(x, y, z) : x + y + z = 0\}[/imath] of [imath]\mathbb{R}^3[/imath] , and two bases: [imath]\alpha_1 = (1,β1,0)[/imath], [imath]\alpha_2 = (1,0,β1)[/imath], and [imath]\beta_1 = (0,1,β1)[/imath], [imath]\beta_2 = (1,1,β2)[/imath]. Find the change-of-basis matrix [imath]A[/imath]. I (foolishly) tried the standard way from linear algebra to find it and I found that [imath]A[/imath] is a 2*2 matrix [imath]\begin{bmatrix} -1 &1 \\ 2&-1 \end{bmatrix}[/imath], which doesn't make sense since the vectors in [imath]V[/imath] are of [imath]\mathbb{R}^3[/imath] any help? | 1688528 | Change of basis matrix for [imath]B_2\{(1,-1,0),(1,0,-1)\}[/imath] to [imath]B_1\{(0,1,-1),(1,1,-2)\}[/imath]
Change of basis matrix for [imath]B_2\{a_1=(1,-1,0),a_2=(1,0,-1)\}[/imath] to [imath]B_1\{b_1=(0,1,-1),b_2=(1,1,-2)\}[/imath], where [imath]B_{1,2}[/imath] are bases of [imath]V=\{(x,y,z)|x+y+z=0\}[/imath]. I haven't practiced that for a very long time and the definitions are a rather formal, total and loaded. I think I got the idea right, but I think I should have it checked: I know that [imath]b_1=-a_1+a_2,b_2=-a_1+2a_2[/imath]. Hence, for [imath]P=\begin{pmatrix}-1&-1\\1&2\\ \end{pmatrix}[/imath], [imath][b_1]_{B_1}=P[b_1]_{B_2}[/imath] and the same goes for [imath]b_2[/imath]. How do I arrive at the requested matrix from here? What about the standard basis? Is [imath]B_2[/imath] the standard basis here? I am confused, and could use some help. |
2569313 | Find all matrices [3x3] that commute with given matrix
Find all [imath]3\times 3[/imath] matrices that commute with [imath]A =\left( \begin{array}{cc} a_1 & 0 & 0\\ 0 & a_2 & 0\\ 0 & 0 & a_3\end{array} \right)[/imath] My progress: I know that a I need to find a matrix such that [imath]AX = XA[/imath]. However I'm getting stuck when: [imath]AX =\left( \begin{array}{cc} a_1x_{11} & a_1x_{12} & a_1x_{13}\\ a_2x_{21} & a_2x_{22} & a_2x_{23}\\ a_3x_{31} & a_3x_{32} & a_2x_{33}\end{array} \right)[/imath] [imath]XA =\left( \begin{array}{cc} a_1x_{11} & a_2x_{12} & a_3x_{13}\\ a_1x_{21} & a_2x_{22} & a_3x_{23}\\ a_1x_{31} & a_2x_{32} & a_3x_{33}\end{array} \right)[/imath] The answer has been given as: [imath]\left( \begin{array}{cc} b_1 & 0 & 0 \\ 0 & b_2 & 0 \\ 0 & 0 & b_3 \end{array} \right)[/imath] I don't understand how they're getting that form. Can someone please explain? | 1697991 | Do Diagonal Matrices Always Commute?
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix and let [imath]\Lambda[/imath] be an [imath]n \times n[/imath] diagonal matrix. Is it always the case that [imath]A\Lambda = \Lambda A[/imath]? If not, when is it the case that [imath]A \Lambda = \Lambda A[/imath]? If we restrict the diagonal entries of [imath]\Lambda[/imath] to being the equal (i.e. [imath]\Lambda = \text{drag}(a, a, \dots, a)[/imath]), then it is clear that [imath]A\Lambda = AaI = aIA = \Lambda A[/imath]. However, I can't seem to come up with an argument for the general case. |
2566402 | Prove that the set of [imath]n\times n[/imath] symmetric matrices is a subspace of [imath]M_{n\times n}(R)[/imath]?
Let H be the set of all symmetric [imath]n\times n[/imath] matrices: [imath]H = \{A β M_{n\times n}(R) | A^T = A\}[/imath]. Prove that H is a subspace of [imath]M_{n\times n}(R)[/imath]. | 12531 | Show that the set of all symmetric, real matrices is a subspace, determine the dimension
Question: Let [imath]V \subset M(n,n,\mathbb{R})[/imath] be the set of all symmetric, real [imath](n \times n)[/imath] matrices, that is [imath]a_{ij} = a_{ji}[/imath] for all [imath]i,j[/imath]. Show that [imath]V[/imath] is a subspace of [imath]M(n,n,\mathbb{R})[/imath] and calculate dim[imath](V)[/imath]. My attempt so far: First part: To show that [imath]V[/imath] is a subspace I need to show: (a) [imath] 0 \in V[/imath] and (b) $\forall A,B \in V: (i) A + B \in V (ii) \lambda A \in V$ For (a) I would say: Let [imath]a_{ij} \in 0[/imath](this should represent a zero matrix, is that how to write it?) [imath]a_{ij} = 0 = a_{ji} \Rightarrow 0 \in V[/imath] For (b) I am actually confused since I would first think: both a [imath](2 \times 2)[/imath] matrix and a [imath](3 \times 3)[/imath] matrix belong to [imath]V[/imath] but addition of matrices of different size is undefined [imath]\Rightarrow[/imath] [imath]V[/imath] is not closed under addition [imath]\Rightarrow[/imath] [imath]V[/imath] is not a subspace of [imath]M(n,n,\mathbb{R})[/imath]... what am I missing here? (To start I don't really understand the notation [imath]M(n,n,\mathbb{R})[/imath]... what exactly does the [imath]\mathbb{R}[/imath] represent there?). Disregarding my confusion I would still try to show (b), but my mathematical notation is still lacking competence... Is the following somewhat clear? Would anyone ever use "[imath]\in[/imath]" to denote "is an element of matrix"? (i)Let [imath]a_{ij},a_{ji} \in A[/imath] and [imath]b_{ij}, b_{ji} \in B[/imath]. Let [imath]A,B \in V[/imath] [imath]\Rightarrow a_{ij} = a_{ji}, b_{ij} = b_{ji}[/imath] [imath]A + B = C \Rightarrow c_{ij} = (a_{ij}+b_{ij}) = (a_{ji} + b_{ij}) = (a_{ij} + b_{ji}) = c_{ji} = (a_{ji} + b_{ji})[/imath] [imath]\Rightarrow C \in V[/imath] (ii) Let [imath]A\in V, \lambda \in \mathbb{R}[/imath]. Let [imath]a_{ij},a_{ji} \in A[/imath]. [imath]\Rightarrow a_{ij} = a_{ji}[/imath] [imath]\lambda \cdot A = A'[/imath] with [imath]\lambda a_{ij} = \lambda a_{ji} \Rightarrow A' \in V[/imath] Second part: I feel that I understand the answer... For an [imath](n \times n)[/imath] matrix, the diagonal length [imath] = n[/imath] and these are the elements which have no counterpart and are not critical to the symmetry. When these elements are subtracted from the total[imath](n^{2})[/imath], half of the remainder can be independently selected and the other half will follow as a result. Therefore I think it makes sense to write that dim[imath](V) = n + \frac{n^{2}-n}{2}[/imath]. Is this correct? If so, given the context of the exercise, how could I make my answer more acceptable? |
2568814 | Picard Iteration: Convergence of system
I want to prove that, for any [imath]t[/imath], a solution exists in the interval [imath][0,T][/imath], when [imath]T>0[/imath]. [imath]x'(t)=A(t)x(t)[/imath] My question is quite similar to this one Picard iteration (general), but with one small difference. My [imath]A(t)[/imath] is a function of [imath]t[/imath], whereas in the other question [imath]A[/imath] seems to be constant. Picard iteration: [imath]x^{[k+1]}=x_0+\int^t_0A(Ο)x^{[k]}(Ο) dΟ[/imath]. Where [imath]x_0[/imath] is arbitrary. I want tor prove that [imath]x^{[k+1]}[/imath] converges on the given interval and that it satisfies the equation below. [imath]|x^{[k+1]}(t)-x^{[k]}(t)|\le |x_0| [/imath][imath]\frac{Ο^{k+1}(t)}{(k+1)!}[/imath] where [imath]Ο(t)=\int^t_0||A(Ο)||dΟ[/imath]. I'm not sure how to account for the variant function [imath]A(t)[/imath] in my proof, so I was hoping someone could show me what it look like | 2553408 | First Order Time-Variant System : Picard Method
I looking at a time variant first order system. I am trying to prove that a sequence of functions [imath]x^{[k]}(t)[/imath] generated using Picard iterations converges uniformly on some interval [imath][0,T][/imath]. Given: [imath]\dot{x}(t)=A(t)x(t)[/imath] where [imath]x(0)=x_0[/imath] [imath]x(t)=xβR^n[/imath] where [imath]A(t)[/imath] is a [imath]nΓn[/imath] matrix Assume there exists [imath]M[/imath] such that [imath]||A(t)||\le M[/imath] for all [imath]t\ge 0[/imath] and that [imath]A(t)[/imath] is a piecewise continuous function. Show that, for any [imath]T > 0[/imath], a solution exists in the interval [imath][0,T][/imath]. I know that [imath]x(t)=x_0+\int^t_0A(Ο)x(Ο) dΟ[/imath] must be satisfied if the solution does indeed exist. I am supposed to use Picard iterations to prove this, which I am not familiar with yet. But I know that they are of the form: [imath]x^{[k+1]}=x_0+\int^t_0A(Ο)x^{[k]}(Ο) dΟ[/imath]. Where [imath]x^{[0]}(t)=x_0[/imath] Can I use the standard Picard proof to prove this or do I need to start somewhere else? I was also asked to show that this method satisfies [imath]|x^{[k+1]}(t)-x^{[k]}(t)|\le |x_0| [/imath][imath]\frac{Ο^{k+1}(t)}{(k+1)!}[/imath] where [imath]Ο(t)=\int^t_0||A(Ο)||dΟ[/imath]. I have no idea where to start with this, but hopefully it will become clearer once I finish the first part. |
2149340 | How to prove this [imath]\pi[/imath] formula?
I am hoping to find out where the formula [imath]\frac{\pi}{2}=\sum_{k=0}^{\infty}\frac{k!}{\left(2k+1\right)!!}[/imath] comes from. I can't see how one could begin to prove it. | 77607 | How to sum this series for [imath]\pi/2[/imath] directly?
The sum of the series [imath] \frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}\tag{1} [/imath] can be derived by accelerating the Gregory Series [imath] \frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2} [/imath] using Euler's Series Transformation. Mathematica is able to sum [imath](1)[/imath], so I assume there must be some method to sum the series in [imath](1)[/imath] directly; what might that method be? |
2570027 | Whether range of matrix [imath]A^TA[/imath] is the same as that of [imath]A^T[/imath]
I cannot ensure whether column space of [imath]A^TA[/imath] is the same as [imath]A^T[/imath], if it is, how to prove? | 1272572 | Row space and column space of [imath]A^T A[/imath] and [imath]A A^T[/imath]
Suppose we have a [imath]n \times m[/imath] rectangular matrix [imath]A[/imath] with row space [imath]\text{row}(A)[/imath] and column space [imath]\text{col}(A)[/imath]. What are the row spaces and the column spaces of [imath]A^T A[/imath] and [imath]A A^T[/imath]? |
2570114 | Why don't we use the proper terminology for linear and affine?
Why do we call y = mx + b linear (in high school)? Why don't we call it affine? and linear what is actually linear [imath]f(x+y)=f(x)+f(y)[/imath] [imath]f(ax)=af(x)[/imath] | 1342907 | Is [imath]y=mx+b[/imath] linear?
Consider [imath]f(x) = mx+b[/imath]. Let [imath]b\ne 0[/imath] If [imath]f[/imath] is linear, [imath]f(0)[/imath] should yield [imath]0[/imath] [imath]f(0) = m(0)+b = b[/imath] Therefore [imath]f(x)=mx+b[/imath] is nonlinear. Question: Why is [imath]y=mx+b[/imath] called a "linear equation"? Remark: Consider [imath]f(x,y) = y + -mx = b[/imath] \begin{align} \alpha f(x,y) &= \alpha b \\ f(\alpha x,\alpha y) &= \alpha y +-\alpha mx = \alpha (y +-mx) =\alpha b \end{align} \begin{align} f(x+\beta,y+\beta) &= y + \beta+ - m (x +\beta) \\&= (y+-mx)+(\beta +-m\beta)\\ &= f(x,y)+ f(\beta,\beta) \end{align} Therefore, [imath]f(x,y) = y +-mx[/imath] and [imath]f(x,y) =b[/imath] are linear. Question: Why is this not a contradiction? |
2567514 | How to prove this inequality [imath]f(a+b)\leq f(a) +f(b)[/imath] for [imath]\frac{f(x)}x[/imath] monotone decreasing
[imath]f(x)[/imath] is defined on[imath](0,\infty)[/imath], and [imath]\frac{f(x)}{x}[/imath] is monotone decreasing, how to prove [imath]\forall a>0,b>0[/imath] [imath]f(a+b)\leq f(a) +f(b)[/imath]. Thanks pretty much in advance | 1433028 | If [imath]\frac{f(x)}{x}[/imath] is decreasing in [imath](0,\infty)[/imath], prove that [imath]f(a+b) \leq f(a) + f(b)[/imath] for [imath]a,b >0[/imath]
If [imath]\frac{f(x)}{x}[/imath] is decreasing in [imath](0,\infty)[/imath], prove that [imath]f(a+b) \leq f(a) + f(b)[/imath] for [imath]a,b >0[/imath] |
2570826 | Proof that an automorphism of an extension field of [imath]\mathbb Q[/imath] fixes rationals
At this point of the lecture on field automorphisms by Professor Macauley, the proof that an automorphism, [imath]\phi,[/imath] on an extension field [imath]F[/imath] of [imath]\mathbb Q[/imath] fixes the rational numbers, i.e. [imath]\phi(q)=q\quad\quad\quad \forall \; q \in \mathbb Q[/imath] is given for the number [imath]1,[/imath] asserting that if [imath]\phi(1)=q,[/imath] [imath]q[/imath] can't be [imath]0[/imath] since, if it were, [imath]\phi(2)=\phi(1+1)=\phi(1)+\phi(1)=0[/imath] and [imath]\phi(\dot)[/imath] would not be a bijection. Likewise, since [imath]q=\phi(1)=\phi(1\cdot1)=\phi(1)\cdot\phi(1)=q^2[/imath] and [imath]q=\phi(1)=\phi(1\cdot1\cdot 1)=\phi(1)\cdot\phi(1)\cdot\phi(1)=q^3[/imath] we have to conclude that [imath]\phi(1)=1.[/imath] This is clear, but the question is How does this proof generalize to the rest of rational numbers [imath]\phi(n)=n;\;n>1,n=\frac{a}{b}; \, a,b\in \mathbb Q?[/imath] | 1454351 | Automorphism of the group of rational numbers under addition
Let [imath]\mathbb \phi [/imath] be an automorphism of the group of rational numbers [imath]\mathbb \ Q[/imath] under addition. Prove that [imath]\mathbb \phi(x)=x \phi(1), \forall x \in Q. [/imath] |
2570415 | Prove that [imath]\lim_{x \to y} \frac{x^k-y^k}{x-y} = k*y^{k-1}[/imath] for [imath]y\in\mathbb R[/imath] and [imath]k\in\mathbb N[/imath]
how can I prove the following: Prove that [imath] \lim_{x \to y} \frac{x^k-y^k}{x-y} = ky^{k-1} [/imath] for a fixed [imath]y\in\mathbb R[/imath] and [imath]k\in\mathbb N[/imath]. I already tried a lot but somehow I don't come to the correct conclusion | 1943786 | Proving [imath]\lim \frac{x^n-a^n}{x-a}=n\cdot a^{n-1}[/imath]
Please note that this question was asked by one of my students who doesn't know differentiation yet nor Lhopital nor mean value theorems. We teach limits before all these topics like differentiation , MVT , Lhopital , etc [imath]\lim_{ x \to a} \frac{x^n-a^n}{x-a}=n\cdot a^{n-1}[/imath] I can prove this result for [imath]n \in \mathbb Z[/imath] And for [imath]n \in \mathbb Q [/imath] , that is when [imath]n =\frac{p}{q}[/imath] , I can prove the result using the result for [imath]n \in\mathbb Z[/imath]. But my question is this : Since [imath]\mathbb Z \subset \mathbb Q[/imath] , why can't we prove this result only for [imath]n \in \mathbb Q[/imath] ? Is there a method to prove [imath]\lim_{ x \to a} \frac{x^\frac{p}{q}-a^\frac{p}{q}}{x-a}=\frac{p}{q}\cdot a^{\frac{p}{q}-1}[/imath] without the result for [imath]n \in \mathbb Z [/imath] ? |
2569297 | Partition of vertex set of a graph containing less than a third of all edges each
I am asked to prove that for a graph [imath]G[/imath] with vertex set [imath]V[/imath]: [imath]\exists \text{ a partition } V_1 \cup V_2 \text { of } V \text{ such that } e(G[V_1]) + e(G[V_2]) \leq \frac{1}{2}e(G)[/imath] We may also require that [imath]e(G[V_1]), \;e(G[V_2]) \leq \frac{1}{3}e(G)[/imath] Using an inductive argument I have proven the first part, but am unsure of how to continue to prove the second. I want to use an inductive argument on the number of vertices by first removing a vertex, finding a partition of the remaining subgraph, and then appropriately adding the original vertex and its edges back in. However, my problem is that I don't know how I could justify that no matter how many edges the removed vertex has, there is a way to place it in either [imath]V_1[/imath] or [imath]V_2[/imath] such that the conditions are still met. In fact I suspect it might not actually be always possible, however I am unsure how else I may approach this question. | 876091 | Graph partition that span a third of edges
Given a graph G is easy to see that we have a partition [imath]V=V_1 \cup V_2[/imath] so that [imath]e(G[V_1])+e(G[V_2])\leq e(G)/2[/imath]. How can we improve this result showing that we can choose [imath]V_i[/imath] such that [imath]e(G[V_i])\leq e(G)/3[/imath] for [imath]i=1,2.[/imath] |
2570411 | Calculate expected value of random variable [imath]Z=\min\left\{X,Y\right\}[/imath]
Hi math people in my last question I asked the question [imath]X,Y[/imath] are independent, identical distributed with [imath]P(X=k) = P(Y=k)=\frac{1}{2^k} \,\,\,\,\,\,\,\,\,\,\,\, (k=1,2,...,n,...)[/imath] Calculate the probabilities [imath]P(\min\left\{X,Y\right\} \leq x)[/imath] I calculate it correct, here is calculation: [imath]\begin{split}F_M(x) &= P(\min\left\{X,Y\right\} \leq x) \\ &= 1-P(x<\min\left\{X,Y\right\} ) \\ &= 1-P(x<X, x<Y) \\ & = 1-P(X>x)\,P(Y>x)\\ & = 1-(1-P(X \leq x))\,(1-P(Y \leq x))\\ & = 1-(1-F_X(x))\,(1-F_Y(x))\end{split}[/imath] But now how calculate expected value of random variable [imath]Z=\min\left\{X,Y\right\}[/imath] So we have [imath]Z = 1-(1-F_X(x))\,(1-F_Y(x))[/imath] Then [imath]E(Z) = \int_{0}^{\infty}\left(1-(1-F_X(x))\,(1-F_Y(x))\right) \,\,dx[/imath] Is it good like this? | 2570327 | Calculate probability [imath]P(\min\left\{X,Y\right\} \leq x)[/imath] and [imath]P(\max\left\{X,Y\right\} \leq x)[/imath]
[imath]X,Y[/imath] are independent, identical distributed with [imath]P(X=k) = P(Y=k)=\frac{1}{2^k} \,\,\,\,\,\,\,\,\,\,\,\, (k=1,2,...,n,...)[/imath] Calculate the probabilities [imath]P(\min\left\{X,Y\right\} \leq x)[/imath] and [imath]P(\max\left\{X,Y\right\} \leq x)[/imath] For the minimum I do like this: [imath]\begin{split}F_M(x) &= P(\min\left\{X,Y\right\} \leq x) \\ &= 1-P(x<\min\left\{X,Y\right\} ) \\ &= 1-P(x<X, x<Y) \\ & = 1-P(X>x)\,P(Y>x)\\ & = 1-(1-P(X \leq x))\,(1-P(Y \leq x))\\ & = 1-(1-F_X(x))\,(1-F_Y(x))\end{split}[/imath] Is this correct for minimum? I'm not sure how do it for [imath]\max[/imath]? Maybe I do it too complicated because they are equal these [imath]P(X=k)=P(Y=k)[/imath] maybe you can do it more elegant? But I don't know how? |
2567884 | Value of [imath]a[/imath] such that [imath]ax^2-4x+9[/imath] has integer roots
The question is to find out the value of [imath]a[/imath] such that [imath]ax^2-4x+9=0[/imath] has integer roots. If [imath]a[/imath] is 1 then the roots if they are rational are integers.But the discriminant in such case is -ve so [imath]a\neq 1[/imath] Similarly by rational root theorem [imath]a \neq -1[/imath] So [imath]a[/imath] must be a fraction of the form [imath]1/b[/imath] such that [imath]b[/imath] is a integer. Now the condition of the equation [imath]x^2-4bx+9b[/imath] to have rational roots and thereby integers because the leading coefficient is unity is that its discriminant must be a perfect square.so [imath]16b^2-36b[/imath] is a perfect square.I couldn't get how to determine [imath]b[/imath] from this.Any ideas?Thanks This link Elementary number theory and quadratic doesn't have the attempt in which I am unable to proceed. | 1935515 | Elementary number theory and quadratic
Find the value(s) of [imath]a[/imath] for which the equation [imath]ax^2-4x+9=0[/imath] has integral roots(i.e. [imath]x[/imath] is an integer). My attempt: By hit and trial I am getting answer as [imath]a=\frac{1}{3}[/imath]. No information about nature of [imath]a[/imath] is given. |
2570773 | bijective function over [imath]\mathbb{N}[/imath] that isn't [imath]f(X) = (X)[/imath]
I'm struggling to come up with a good example for a bijective function over [imath]\Bbb N[/imath]. I've come up with: [imath]f(n) =\begin{cases} 1,& n=0\\ 0, &n=1 \\ n, &\text{otherwise}\\ \end{cases} [/imath] Firstly, can you confirm this is acceptable, and secondly, does anyone have any other examples? (Perhaps a more beautiful one!) | 1903416 | Is there a bijective function mapping Natural numbers to Natural numbers, other than [imath]f(n) = n[/imath]?
Is there a function that can be bijective, with the set of natural numbers as domain and range, other than [imath]f(n) = n[/imath]? |
2567760 | Orthonormal system close to Orthonormal basis is basis.
[imath]\{e_n\}[/imath] is an orthonormal basis in Hilbert space. [imath]\{f_n\}[/imath] is an orthonormal system in H, such that [imath]\sum ^\infty_1 \| e_n - f_n \| < 1[/imath]. Prove that [imath]\{f_n\}[/imath] is a basis. It looks natural to use projections here, but I don't understand how to do it. | 64676 | orthonormal system in a Hilbert space
Let [imath]\{e_n\}[/imath] be an orthonormal basis for a Hilbert space [imath]H[/imath]. Let [imath]\{f_n\}[/imath] be an orthonormal set in [imath]H[/imath] such that [imath]\sum_{n=1}^{\infty}{\|f_n-e_n\|}<1[/imath]. How do I show that [imath]\{f_n\}[/imath] is also an orthonormal basis for [imath]H[/imath]? |
2571649 | what is the meaning of eigenvalue of Dirichlet laplacian in the real life (physics...)
I am asking a question that I have already asked, but which did not have any success for the answers let [imath]\lambda_{1}(\Omega),\lambda{2}(\Omega),\lambda_{3}(\Omega)...[/imath] the eigenvlues of the laplacian Operator with Dirichlet condition on the boundary on [imath]\Omega [/imath] the classical spectrale optimisation problem is : [imath] min\{\lambda_{k}(\Omega),\; \Omega\subset \mathbb{R}^N open ,\; |\Omega| = c\;\},[/imath] where [imath]|.|[/imath]denoted the Lebesgue measure , [imath]c>0[/imath], [imath]k \in \mathbb{N}^{\ast}[/imath] and [imath]\lambda_{k}[/imath] is the [imath]k-[/imath]eme eigenvalue of Dirichlet-lapalacian on [imath]\Omega[/imath]. I would like to know what is the application of this Problem (For the laplacian or another Operator) in the physics or the natural sciences.. thank you for your answers | 2569272 | whats is the applications of the minimization of eigenvalue in the real life ( physics,the natural sciences...)
let [imath]\lambda_{1}(\Omega),\lambda{2}(\Omega),\lambda_{3}(\Omega)...[/imath] the eigenvlues of the laplacian Operator with Dirichlet condition on the boundary on [imath]\Omega [/imath] the classical spectrale optimisation problem is : [imath] min\{\lambda_{k}(\Omega),\; \Omega\subset \mathbb{R}^N open ,\; |\Omega| = c\;\},[/imath] where [imath]|.|[/imath]denoted the Lebesgue measure , [imath]c>0[/imath], [imath]k \in \mathbb{N}^{\ast}[/imath] and [imath]\lambda_{k}[/imath] is the [imath]k-[/imath]eme eigenvalue of Dirichlet-lapalacian on [imath]\Omega[/imath]. I would like to know what is the application of this Problem (For the laplacian or another Operator) in the physics or the natural sciences.. thank you for your answers |
2569330 | Monotony of a sequence
How do I prove that the function [imath]n\to a_n:=\left(1+\frac{1}{n}\right)^{n+1}[/imath] is decreasing, where [imath]n[/imath] is a positive integer? P.s.: I'm 11th grade,and I didn't get to derivatives, we are currently studying limits of functions operations. Thanks in advance! | 2071492 | The function [imath]f (n) = (1 + 1 / n) ^ {n+1}[/imath] is decreasing
I cannot prove that the function [imath]f (n) = \left(1 + \frac1n\right) ^ {n + 1},[/imath] defined for every positive integer [imath]n[/imath], is strictly decreasing in [imath]n[/imath]. I already tried to prove by induction and also tried to prove by calculating the difference between [imath]f (n + 1)[/imath] and [imath]f (n)[/imath]. I need help. |
2572234 | Prove that if [imath] f \in L^1[0, +\infty][/imath], then [imath]\lim\limits_{n \to \infty}\frac{1}{n}\int_0^n xf(x)=0[/imath]
Prove that if [imath] f \in L^1[0, +\infty][/imath], then [imath]\lim\limits_{n \to \infty}\frac{1}{n}\int_0^n xf(x)=0.[/imath] I tried to use integration by parts, but it does not help. | 527976 | Prove that [imath]\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0[/imath].
Let [imath]f[/imath] be a continuous, nonnegative, real-valued function and [imath]\int_0^{\infty}f(x)dx<\infty.[/imath] Prove that [imath]\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0.[/imath] A start: If [imath]\lim\limits_{n\to\infty}\int_0^{n}xf(x)dx[/imath] is finite, then it's obvious. Otherwise, perform L'Hopital's rule, we get [imath]\lim nf(n)[/imath], which we want to show is [imath]0[/imath]. |
2572831 | The minimum of [imath]F(u)=\int_0^1 |u'(x)-u(x)| dx[/imath]
Let [imath]C=\{u\in C^1[0,1]\mid u(0)=0,u(1)=1\}[/imath]. What is the minimum of the functional [imath]F(u)=\int_0^1 |u'(x)-u(x)| dx[/imath]. What I guess is that the minimum is attained when [imath]u(x)=x[/imath]. What follows is my try. Let [imath]C_0=\{u\in C^1[0,1]\mid u(0)=u(1)=0\}[/imath]. For any function [imath]u\in C[/imath],one has [imath]u-x\in C_0[/imath]. Then we only need to consider the minimization of [imath] \int_0^1|v'(x)-v(x)-x| dx=\int_0^1|x+v(x)-v'(x)| dx [/imath] over [imath]C_0[/imath]. But, I do not know how to process further. | 1041713 | Show [imath]\inf_f\int_0^1|f'(x)-f(x)|dx=1/e[/imath] for continuously differentiable functions with [imath]f(0)=0[/imath], [imath]f(1)=1[/imath].
Let [imath]C[/imath] be the class of all real-valued continuously differentiable functions [imath]f[/imath] on the interval [imath][0,1][/imath] with [imath]f(0)=0[/imath] and [imath]f(1)=1[/imath]. How to show that [imath]\inf_{f\in C}\int_0^1|f'(x)-f(x)|dx=\frac{1}{e}?[/imath] I have been able to show that [imath]1/e[/imath] is a lower bound. Indeed, [imath]\begin{align*} \int_0^1|f'(x)-f(x)|dx &= \int_0^1|f'(x)e^{-x}-f(x)e^{-x}|e^xdx \\ &\geq \int_0^1\left(f'(x)e^{-x}-f(x)e^{-x}\right) dx \\ &= \int_0^1 \frac{d\left(f(x)e^{-x}\right)}{dx}dx \\ &= f(1)e^{-1}-f(0)e^{0}\\ &=\frac{1}{e}. \end{align*}[/imath] But how to show this is the infimum? Is there a function [imath]f\in C[/imath] such that we get [imath]\int_0^1|f'(x)-f(x)|dx=1/e[/imath]? |
2572765 | Counterexample for Monotone convergence theorem
Let [imath]f_n : [0,1] \rightarrow \mathbb{R}[/imath] be a sequence of monotone decreasing measurable functions [imath]f_n \geq f_{n+1}[/imath] that converges pointwise to [imath]f: [0,1] \rightarrow \mathbb{R}[/imath]. What would be the counterexample which shows that Monotone convergence theorem does not apply for such sequence? Besides original version of MCT (non-negative monotone increasing sequence) we also stated slightly different version where MCT applies also if [imath]{f_n}[/imath] are measurable, monotone increasing and [imath]f_1[/imath] is integrable. My intuition is that I should come up with the sequence of functions that are measurable but not integrable (they need to converge though), however I did not manage to come up with something so far. Thanks for your help in advance! | 1647106 | Why is the Monotone Convergence Theorem restricted to a nonnegative function sequence?
Monotone Convergence Theorem for general measure: Let [imath](X,\Sigma,\mu)[/imath] be a measure space. Let [imath]f_1, f_2, ...[/imath] be a pointwise non-decreasing sequence of [imath][0, \infty][/imath]-valued [imath]\Sigma-[/imath]measurable functions, i.e. for every [imath]k\ge 1[/imath] and every [imath]x[/imath] in [imath]X[/imath], [imath]0 \le f_k(x) \le f_{k+1}(x).[/imath] Next, set the pointwise limit of the sequence [imath]{f_n}[/imath] to be [imath]f[/imath]. That is, for every [imath]x[/imath] in [imath]X[/imath], [imath]f(x) = \lim_{k\to \infty}f_k(x).[/imath] Then [imath]f[/imath] is [imath]\Sigma-[/imath]measurable and [imath]\lim_{k\to \infty}\int f_k d\mu = \int f d\mu.[/imath] I've noticed that when it comes to monotone convergence theorem (either Lebesgue or general measure), usually its definition restricts the monotone function sequences to be nonnegative. I'm not sure why the 'nonnegative' is necessary. |
2572279 | How do I prove that the sets x = {y}, y = {z} and z = {x} canβt exist with the axiom of regulation/foundation in ZFC?
How do I prove that the sets [imath]x = \{y\}, y = \{z\} \text{ and } z =\{x\}[/imath] canβt exist with the axiom of regulation/foundation in ZFC? Iβm having the damnest time trying to prove this since the sets are not assumed to be transitive sets. Iβve tried using powersets and union operations but these have brought me no luck thus far. Iβm beginning to wonder whether or not this is possible to prove. | 2431594 | Complete Newbie: Axiom of Foundation Exercise
No other axioms. There are [imath]3[/imath] sets such that: [imath]x\in y[/imath]; [imath]y\in z[/imath]; [imath]z\in x[/imath]; These sets violate foundation. I don't get it... For [imath]x[/imath], [imath]z\in x[/imath], so z must be the [imath]\epsilon[/imath]-minimal member but the only member of [imath]z[/imath] is [imath]y[/imath]. [imath]z \cap x = \{y\}\cap\{z\} = \emptyset[/imath]. None of the members of [imath]z[/imath] (only [imath]y[/imath]) are members of [imath]x[/imath] so doesn't that satisfy definition of Foundation? Please help this noob, thanks |
2571850 | [imath]\{x^k\}^\infty_{k=0}[/imath] is not basis in [imath]C[0;1][/imath]
[imath]\{x^k\}^\infty_{k=0}[/imath] is not basis in [imath]C[0;1][/imath]. I know how to show that a system of functions is basis if it is orthogonal (Parseval's identity), but I do not understand how to show it for arbitrary system. Possibly, we should find a continuous function that has some problems with approximation with an infinite sum of monomials, but I have no idea what contradiction we need to obtain. | 695421 | How to show that the monomials are not a Schauder basis for [imath]C[0,1][/imath]
why the monomials are not a Schauder basis for [imath]C[0,1][/imath]? [imath]p_n(x)=x^n[/imath] such that [imath](p_n)[/imath] does not form a Schauder basis for [imath]C[0,1][/imath] span[imath]\lbrace p_n : n\ge 0\rbrace[/imath] is dense in [imath]C[0,1][/imath] by Weierstrass approxmation theorem, but I cannot figure out why they are not a Schauder basis. Could you please explain |
781033 | Integrating : [imath]\int_0^1 {\frac {x^a-x^b} {\ln x} dx}[/imath]
We are given parameters [imath]a > 0, b > 0[/imath]. Task is to integrate that: [imath]\displaystyle \int_0^1 {\frac {x^a-x^b} {\ln x} dx}[/imath]. I have tried approaching problem from different angles with no luck. I tried integration by parts(tried all combinations of possible [imath]v[/imath] and [imath]u[/imath]), u-substitution with no luck. Also I tried to integrate this two similar terms separately. Tried to get some idea of how to go from answer, got nice answer from MATLAB: [imath]\displaystyle \ln{\frac{a+1}{b+1}}[/imath], but no idea how to reach it. I would appreciate some suggestions. | 154881 | Is the integral [imath]\int_1^\infty\frac{x^{-a} - x^{-b}}{\log(x)}\,dx[/imath] convergent?
Is the integral [imath]\int_1^\infty\frac{x^{-a} - x^{-b}}{\log(x)}\,dx[/imath] convergent, where [imath]b>a>1[/imath]? I think the answer lies in defining a double integral with [imath]yx^{(-y-1)}[/imath] and applying Tonelli's Theorem, but the integral of [imath]\frac{x^{-a}}{\log x}[/imath] is still not integrable. Any ideas? |
2573201 | Subgroup of[imath]\mathbb{Z}[/imath] and finite subgroup of [imath]\mathbb{Z}[/imath]
In class, we described a subgroup as follow : A subset of [imath]\mathbb{Z}[/imath] is called a subgroup if it is nonempty, and the sum and difference of any two of its members is also a member. I would like to find the "unique finite subgroup of [imath]\mathbb{Z}[/imath]. I think it is [imath]S=\{0\}[/imath] but I am not sure [imath]S[/imath] is a subgroup of [imath]\mathbb{Z}[/imath] in this case. | 455072 | Subgroups of (Z,+) of order n
Studying groups and subgroups I find this question: Are there subgroups of order [imath]\mathbf 6 \mathbf 5[/imath] in the additive group [imath](\Bbb Z[/imath],[imath]+)[/imath]? I would answer no, because a subgroups of [imath](\Bbb Z,+)[/imath] is the multiple of a Natural number [imath]n [/imath] and it has the form: [imath]n\Bbb Z[/imath]={[imath]na|n \in \Bbb N, a \in \Bbb Z[/imath]} and they have no finite order. But I'm not sure of this answer. Could someone explain if I'm wrong and why? Thanks. [imath]\mathbf {edit}:[/imath] My question comes because discussing with a course mate he argued that exist Z/65Z in (Z,+), the group module 65 and it has order 65 so I'm a bit confused. |
2573017 | Find all positive integers [imath]n[/imath] such that [imath]\phi(n)=10[/imath]
Find all positive integers n such that [imath]\phi(n)=10[/imath]. I have found 2 positive integers: [imath]11[/imath] and [imath]22[/imath]. Is there any more? Thanks! | 127998 | Find all positive integers [imath]n[/imath] such that [imath]\phi(n)=6[/imath].
I am asked to find all positive integers [imath]n[/imath] such that [imath]\phi(n)=6[/imath], and to prove that I have found all solutions. The way I am tackling this is by constructing all combinations of prime powers such that when passed to Euler's function will yield [imath]6[/imath]. For example, I know that [imath]\phi(7)=6[/imath], [imath]\phi(3^2)=6[/imath], and [imath]\phi(2)=1[/imath]. Therefore, the numbers should be [imath]7[/imath], [imath]7\cdot2=14[/imath], [imath]3^2=9[/imath], and [imath]3^2\cdot2=18[/imath]. I believe that there cannot possibly be any others because of the way the [imath]\phi[/imath] function is defined. What do you guys think? |
2561286 | A counterexample of a differentiable function on an interval whose derivative is discontinuous
Let [imath]f(x)[/imath] be a real-valued differentiable function on an interval [imath]I[/imath] of [imath]\mathbb R[/imath], then is [imath]f'(x)[/imath] necessary to be continuous? I don't think so and I'm trying to construct a counterexample. [imath]\sin\frac{1}{x}[/imath] may be the key element but [imath]x\sin\frac{1}{x}[/imath] failed me. | 602568 | If a function is continuous and differentiable everywhere is the derivative continuous?
Suppose [imath]f[/imath] is continuous on [imath][a,b][/imath] and differentiable on (a,b). Does it follow that [imath]f'[/imath] is continuous on [imath](a,b)[/imath]? |
2570913 | [imath]\mu,\nu[/imath] ergodic implies [imath]\mu\perp\nu[/imath]
Let [imath]T:\Omega\to\Omega[/imath] a measurable function and [imath]\mu,\nu[/imath] [imath]T[/imath]-ergodic measures on [imath]\Omega[/imath]. I am trying to prove that [imath]\mu\perp\nu[/imath] (this is, they concentrate in disjoint sets). My attempt was define [imath]w=\mu+\nu[/imath] and use Radon-NikodΓ½m to obtain [imath]f,g\in L^1(\Omega)[/imath] such that [imath]\mu\sim fdw,\ \nu\sim gdw[/imath]. Then I only have to show that [imath]w(\{f,g>0\})=0[/imath], but I couldn't progress much. Note: for [imath]\mu[/imath] being ergodic I mean "[imath]\mu[/imath] is [imath]T[/imath]-invariant and, for every measurable [imath]A[/imath], [imath]\mu(A\triangle T^{-1}A)=0[/imath] implies [imath]\mu(A)\in\{0,1\}[/imath]. Note2: I am not able to use the ergodic theorem. | 2562055 | Show [imath]A \in \cal F[/imath] exists, such that [imath]\mathbb P(A)=1[/imath] and [imath]\mathbb Q(A)=0[/imath].
Let [imath]\mathbb P[/imath] and [imath]\mathbb Q [/imath] different probability measures on [imath](\Omega, \cal F)[/imath] and [imath]T:\Omega \to \Omega[/imath] a measurable function, which is [imath]\mathbb P[/imath] and [imath]\mathbb Q[/imath] ergodic. Show [imath]A \in \cal F[/imath] exists, such that [imath]\mathbb P(A)=1[/imath] and [imath]\mathbb Q(A)=0[/imath]. Since [imath]T[/imath] is [imath]\mathbb P[/imath] and [imath]\mathbb Q[/imath] ergodic [imath]\forall A \in \cal F[/imath] with [imath]A=T^{-1}(A)[/imath], [imath]\mathbb P(A) \in \{0,1\}[/imath] and [imath]\mathbb Q(A) \in \{0,1\}[/imath]. Furthermore I know since T is ergodic [imath]\lim_{n\to\infty}\frac 1n \sum_{k=0}^{n-1}\mathbb P(A \cap T^{-k}(B))=\mathbb P(A)\mathbb P(B)[/imath]. I was wondering if I can use this lemma here. It exists some [imath] B \in \cal F[/imath] such that [imath]P (B) \ne Q (B)[/imath]. And I considered [imath]A[/imath] to be T-invariant, but still I can not lead this in the right direction.. Some help is highly appreciated! |
2574550 | Proving [imath]p^2q+q^2r+r^2p+pq^2+qr^2+rp^2\geq 6pqr[/imath]
I am solving a problem about probability and I need to show this inequality: [imath]p^2q+q^2r+r^2p+pq^2+qr^2+rp^2\geq 6pqr,[/imath] where [imath]p,q,r\in[0,1].[/imath] I tried writing it as [imath]p^2(q+r)+p((q+r)^2-8qr)+qr(q+r)\geq 0,[/imath] or equivalently, [imath]p^2(q+r)+p(q+r-2\sqrt{2qr})(q+r+2\sqrt{2qr})+qr(q+r)\geq 0.[/imath] Since [imath]p,q,r\in[0,1][/imath], the only thing that remains to check is that [imath]q+r-2\sqrt{2qr}[/imath] is non-negative, but this need not be true. I then tried a second approach: I tried using multivariable calculus to find the minimum of [imath]f(p,q,r)=p^2q+q^2r+r^2p+pq^2+qr^2+rp^2-6pqr,[/imath] but then I get [imath]Jf(p,q,r)=\begin{pmatrix}q^2+r^2+2p(q+r)-6qr\\ p^2+r^2+2q(p+r)-6pr\\ p^2+q^2+2r(p+q)-6pq\end{pmatrix},[/imath] and solving [imath]Jf=\mathbf{0}[/imath] seems quite difficult too. Question: Can anyone help with either of the above approaches, or if not, give a hint on how to solve this problem? | 1599042 | How to prove that this polynomial is larger than another for positive reals?
Let [imath]a, b[/imath] and [imath]c[/imath] be positive real numbers. Prove that [imath]a^2 b + b^2 c + c^2 a + ab^2 + bc^2 + ca^2 \ge 6abc[/imath] I have tried factoring with no success for the better part of an hour or so. How can I go about this? |
2574557 | Without using a calculator how to solve [imath]x^x = 100[/imath]?
Without using a calculator how to solve [imath]x^x = 100[/imath] ? A way of finding an approximation to 2 decimals would be good neough. I know about the Lmabert W function but one cannot compute it mentally. This is why I believe my question not to be a duplicate. | 1087365 | How to solve [imath]x^x=100[/imath]?
[imath]x^x = 100[/imath]. I have no clue on how to solve this. If you guys have, please show me your solution as well. |
2574753 | Complex monic polynomial [imath]f(z)[/imath] with [imath]|f(z)|\leq 1[/imath] for [imath]|z|\leq 1.[/imath]
Let [imath]f(z)=z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}[/imath] be a complex polynomial such that [imath]|f(z)|\leq 1[/imath] for [imath]|z|\leq 1.[/imath] I have to prove that [imath]f(z)=z^{n}.[/imath] I tried it as As [imath]|f(z)|\leq 1[/imath] for [imath]|z|\leq 1[/imath] we must have coefficient [imath]a_{0},a_{1}\cdot\cdot\cdot a_{n}[/imath] to be zero because by triangular inequality [imath]|z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}|\leq 1+|a_{n-1}|+\cdot\cdot\cdot+|a_{0}|[/imath] so all of the coefficient [imath]a_{0},a_{1}\cdot\cdot\cdot a_{n}[/imath] to be zero. But it is not exact proof. Please suggest me further improvement. Thanks | 696500 | If [imath]p[/imath] is a polynomial, then either [imath]\,p(z)=z^n\,[/imath] or [imath]\,\max_{|z|=1}|p(z)|>1[/imath].
Let [imath]p(z)=z^{n}+a_{n-1}z^{n-1}+...+a_{0}[/imath] be a polynomial of degree [imath]n\geq1[/imath]. How can you prove that either [imath]p(z)=z^{n}[/imath] or there exists [imath]z'[/imath] with [imath]|z'|=1[/imath] such that [imath]|p(z')|>1[/imath]. Maybe we can use the maximum modulus principle and consider [imath]q(z)=z^{n}p(1/z)[/imath]. |
2574077 | Find the sum of a series involving n roots of unity
Given that [imath]1,w,w^2,w^3,w^4.....w^{n-1}[/imath] are nth roots of unity. find the sum of [imath]\sum_{i=1}^{n-1}\frac{1}{2-w^i}[/imath] The method I used was to write the equation of [imath]w\;\text{as}\,z^n-1=0[/imath] and further are writing this as [imath]z^n-1=(z-1)(z-w)(z-w^2)(z-w^3)....(z-w^{n-1})[/imath] and then replacing z here by 2. From this I was able to find out the denominator of the series in question but I couldn't figure out a way to simplify the numerator | 2209034 | Finding [imath]\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}[/imath], where [imath]\alpha_k[/imath] are the [imath]n[/imath]-th roots of unity
The question asks to compute: [imath]\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}[/imath] where [imath]\alpha_0, \alpha_1, \ldots, \alpha_{n-1}[/imath] are the [imath]n[/imath]-th roots of unity. I started off by simplifiyng and got it as: [imath]=-n+2\left(\sum_{k=0}^{n-1} \dfrac{1}{2-\alpha_k}\right)[/imath] Now I was stuck. I can rationalise the denominator, but we know [imath]\alpha_k[/imath] has both real and complex components, so it can't be simplified by rationalising. What else can be done? |
2574044 | If [imath](X, \Gamma)[/imath] is a cell complex and [imath]e \in \Gamma[/imath], then the image of a characteristic map for [imath]e[/imath] equals [imath]\bar{e}[/imath]
If [imath](X, \Gamma)[/imath] is a cell complex and [imath]e \in \Gamma[/imath], then the image of a characteristic map for [imath]e[/imath] equals [imath]\bar{e}[/imath] I'm trying to prove the above which is a statement in Introduction to Topological Manifolds by John Lee My Attempted Proof: Let [imath](X, \Gamma)[/imath] be a cell complex and choose an open [imath]n[/imath]-cell [imath]e \in \Gamma[/imath] of dimension [imath]n \geq 1[/imath]. Let [imath]\phi[/imath] denote the characteristic map for [imath]e[/imath]. Since [imath]\phi[/imath] is a characteristic map we have [imath]\phi : D \to X[/imath] for some closed [imath]n[/imath]-cell [imath]D \subseteq Y[/imath] (where [imath]Y[/imath] could be any topological space) with the properties that [imath]\phi' : \operatorname{Int}(D) \to e[/imath] which is the restriction of [imath]\phi[/imath] to [imath]\operatorname{Int}(D)[/imath] is a homeomorphism and the restriction of [imath]\phi[/imath] to [imath]\operatorname{Bd}(D)[/imath] denoted by [imath]\phi|_{\operatorname{Bd}(D)}[/imath] maps [imath]\operatorname{Bd}(D)[/imath] into the union of all cells of [imath]\Gamma[/imath] of dimensions strictly less than [imath]n[/imath]. We need to show that [imath]\phi[D] = \bar{e} = \operatorname{Cl}_X(e)[/imath]. By continuity of [imath]\phi[/imath] we have [imath]\phi[\operatorname{Cl}_Y(D)] \subseteq \operatorname{Cl}_X(\phi[D])[/imath] and since [imath]\phi[/imath] is a map from a compact space to a Haursdoff space we have that [imath]\phi[/imath] is a closed map and hence [imath] \operatorname{Cl}_X(\phi[D]) \subseteq \phi[\operatorname{Cl}_Y(D)][/imath] so we have [imath] \operatorname{Cl}_X(\phi[D]) =\phi[\operatorname{Cl}_Y(D)][/imath]. Now if [imath]D = \operatorname{Cl}_Y(D)[/imath] (****) and hence [imath]D[/imath] is a closed set in [imath]Y[/imath] we can go ahead and say that [imath]\phi[D] = \operatorname{Cl}_X(\phi[D])[/imath]. Now note that we have [imath]\phi[\operatorname{Int}(D)] =\phi'[\operatorname{Int}(D)] = e[/imath]. Observe that by continuity of [imath]\phi[/imath] we have [imath]\phi[\operatorname{Cl}_Y(\operatorname{Int}(D))] \subseteq \operatorname{Cl}_X(\phi[\operatorname{Int}(D)]) = \operatorname{Cl}_X(e) [/imath]. Assuming our above assertion holds since [imath]D[/imath] is homeomorphic to [imath]\overline{\mathbb{B}^n} \subseteq \mathbb{R}^n[/imath] with [imath]\operatorname{Int}(D)[/imath] homeomorphic to [imath]\operatorname{Int}(\overline{\mathbb{B}^n}) = \mathbb{B}^n[/imath] and [imath]\operatorname{Cl}_{\mathbb{R}^n}(\mathbb{B}^n) = \overline{\mathbb{B}^n}[/imath] we have [imath]\operatorname{Cl}_Y(\operatorname{Int}(D)) = D[/imath]. Now since [imath]\phi[/imath] is a closed map (by the argument in the above paragraph) we have [imath] \operatorname{Cl}_X(\phi[\operatorname{Int}(D)]) = \operatorname{Cl}_X(e) \subseteq \phi[\operatorname{Cl}_Y(\operatorname{Int}(D))] [/imath], hence we have [imath]\operatorname{Cl}_X(e) =\bar{e} = \phi[\operatorname{Cl}_Y(\operatorname{Int}(D))] = \phi[D][/imath] as desired. Now there is a huge flaw in this proof I think at the (****) part as we have no idea that [imath]D = \operatorname{Cl}_Y(D)[/imath] so we can't conclude that [imath]D[/imath] is an closed set in [imath]Y[/imath]. Sure [imath]D[/imath] is homoemorphic to a closed subspace of [imath]\mathbb{R}^n[/imath], but homeomorphisms don't preserve closures of improper subspaces. For example define [imath]f : [0, 1] \to (0, 1)[/imath] by [imath] f(x) = \begin{cases} x & \text{if} \ \ x \in (0, 1) \\ 0 & \text{if} \ \ x = 1 \ \ \text{or}\ \ x = 0 \\ \end{cases}[/imath] Then the restriction of the domain of [imath]f[/imath] to [imath](0, 1)[/imath] is a homeomorphism, however [imath]\operatorname{Cl}_{[0, 1]} \left((0, 1)\right) = [0, 1][/imath] but [imath]\operatorname{Cl}_{(0, 1)}((0, 1)) = (0,1)[/imath]. So how can I rectify this into a correct proof? I believe that I may have made this a bit too complicated and the proof is probably a lot simpler. Side note: The book that I am using Introduction to Topological Manifolds by John Lee does not assume the closed [imath]n[/imath]-cell [imath]D[/imath] is embedded in a parent topological space, so there isn't any mention of a topological space [imath]Y[/imath] for which [imath]D \subseteq Y[/imath], however if we don't have such a topological space [imath]Y[/imath], then we'd have [imath]\operatorname{Int}(D) = D = \operatorname{Cl}_D(D)[/imath] because we'd be taking the interior of [imath]D[/imath] in the topological space [imath]D[/imath] as opposed to [imath]Y[/imath], and the largest open set in [imath]D[/imath] is obviously [imath]D[/imath]. And we'd arrive at contrdictory stuff like [imath]D[/imath] is homeomorphic to both [imath]\overline{\mathbb{B}^n}[/imath] and [imath]\mathbb{B}^n[/imath]. | 476582 | Characteristic map of a n-cell in a CW complex
I have a problem in understanding the purpose of the definition of a CW complex. What really would help me is to understand the following: Let [imath]\sigma[/imath] be a n-cell and [imath]\Phi_\sigma:\mathbb D^n \to X[/imath] be the characteristic map. Is there any relation between [imath]\Phi_\sigma(\mathbb S^{n-1})[/imath] and [imath]\partial \sigma[/imath] following directly from the Definition? I know, for example, that [imath]\Phi_\sigma(\mathbb S^{n-1})\subset X^{n-1}[/imath], [imath]X^{n-1}[/imath] being the [imath](n-1)[/imath]-Skeleton, but what does this tell me about the boundary of [imath]\sigma[/imath]? |
2575513 | If [imath]f''(a)[/imath] exists, then [imath]f''(a) = \lim_{h\to 0} \frac{ f(a+ h) - 2f(a) + f(a-h) }{h^2}[/imath] without using Darboux Property
In the book of The elements of Real Analysis by Bartle, at page 220, it is asked that If [imath]f''(a)[/imath] exists, then [imath]f''(a) = \lim_{h\to 0} \frac{ f(a+ h) - 2f(a) + f(a-h) }{h^2}[/imath] I have proved a similar result for the first derivative of [imath]f[/imath], but I'm not able to show this result for the second derivative of [imath]f[/imath]. I have seen this this question, and it gives a proof for this question, but what I'm asking is that how can we prove this result without exploiting the Darboux property, and using the given hint in that question, and without using Taylor's theorem. Edit: Note that I want to see how can this result be obtained only using algebraic manipulations. | 1025538 | If [imath]f''(x_0)[/imath] exists then [imath]\lim_{x \to x_0} \frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)[/imath]
Prove: if [imath]f''(x_0)[/imath] exists then [imath]\lim\limits_{x \rightarrow x_0} \dfrac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)[/imath]. I'm not exactly sure how Taylor's theorem fits into all this, but I found the first derivative, and I haven't been able to progress past that. |
2574221 | Does divergence of [imath]\sum a_k[/imath] imply divergence of [imath]\sum \frac{a_k}{1+a_k}[/imath]?
Does divergence of [imath]\sum a_k[/imath] imply divergence of [imath]\sum \frac{a_k}{1+a_k}[/imath]? Note: [imath]a_k > 0 [/imath] I understand that looking at the contrapositive statement, we can say that the convergence of the latter sum implies [imath]\frac{a_k}{1+a_k}\rightarrow 0[/imath] but from here is it possible to deduce that [imath]a_k\rightarrow 0[/imath] because it is not completely straightforward. If we assume [imath]a_k[/imath] to be convergent, this trivially follows but it could diverge in which case this is nontrivial to me. | 1467552 | Prove that if [imath]a_n\gt 0[/imath] and [imath]\sum a_n[/imath] diverges, then [imath]\sum \frac{a_n}{1+a_n}[/imath] diverges.
Prove that if [imath]a_n\gt 0[/imath] and [imath]\sum a_n[/imath] diverges, then [imath]\sum \frac{a_n}{1+a_n}[/imath] diverges. This is the solution to this problem, but I'm having a hard time understanding it. Why does [imath]a_k/(1+a_k)[/imath] not converge to [imath]0[/imath] if [imath]a_k[/imath] doesn't converge to [imath]0[/imath]? I'd appreciate it if anyone could answer this question for me. |
2573633 | prove that if [imath]\lim\limits_{n\to\infty}(a_{n+1}-a_{n})[/imath] exist then [imath]\lim\limits_{n\to\infty}\frac{a_{n}}{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})[/imath]
prove that if [imath]\lim\limits_{n\to\infty}(a_{n+1}-a_{n})[/imath] exist then [imath]\lim\limits_{n\to\infty}\frac{a_{n}}{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})[/imath] So that's what I did so far Let [imath]B_{n}=\frac{(a_{2}-a_{1})+(a_{3}-a_{2})+...+(a_{n+1}-a_{n})}{n}=\frac{(a_{n+1}-a_{n})}{n}[/imath] and we know that if[imath]\ \lim\limits_{n\to\infty}(a_{n+1}-a_{n})[/imath] exist that's mean that [imath]\lim\limits_{n\to\infty}B_{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})[/imath] Thus we have [imath]\lim\limits_{n\to\infty}\frac{(a_{n+1}-a_{n})}{n}=\lim\limits_{n\to\infty}(a_{n+1}-a_{n})[/imath] But I don't know how to go from here | 2439095 | Limit of [imath]b_n[/imath] when [imath]b_n=\frac{a_n}{n}[/imath] and [imath]\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l[/imath]
I have the following problem. I have to find the limit of [imath]b_n = \frac{a_n}{n}[/imath], where [imath]\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l[/imath] My approach: I express [imath]a_n[/imath] in terms of [imath]b_n[/imath], i.e. [imath]a_n=nb_n[/imath] and [imath]a_{n+1}=(n+1)b_n[/imath] We look at the difference: [imath]a_{n+1}-a_n=(n+1)b_{n+1}-nb_n[/imath] Assuming that [imath]b_n[/imath] converges to a real number m, we see that: [imath]l=(n+1)m-nm[/imath], from where I conclude that [imath]m=l[/imath]. What I'm left with is proving that [imath]b_n[/imath] is convergent which I'm not sure how to do. Thanks in advance! |
2575801 | Prove [imath]\frac{1}{f(x)}[/imath] is Riemann integrable
If [imath]f(x)[/imath] is Riemann integrable on [imath][a, b][/imath] and [[imath]\forall x \in [a, b] \ f(x) \ne 0 \land \frac{1}{f(x)}[/imath] is bounded] then prove [imath]\frac{1}{f(x)}[/imath] is Riemann integrable on [imath][a, b][/imath]. I am thinking: Since [imath]f(x)[/imath] and [imath]\frac{1}{f(x)}[/imath] are bounded, [imath]\exists M \gt 0 s.t. |f(x)| \lt M \land \frac{1}{|f(x)|} \lt M[/imath] Therefore [imath]\frac{1}{M} \leq |f(x)| \leq M[/imath]. [imath]\left| \frac{1}{f(x)}-\frac{1}{f(y)} \right|= \left| \frac{f(y)-f(x)}{f(x)f(y)} \right| \leq M^2 |f(y)-f(x)| \ \ (x, y \in [a, b])[/imath] but I don't know what to do. | 1103545 | Prove the reciprocal of a function is integrable if it is bounded
Let [imath]f(x)[/imath] be Riemann integrable on [imath][a,b][/imath] and satisfy [imath]\inf\{\left|f(x)\right|:a\le x \le b\}=d>0[/imath]. How can we prove that [imath]g(x)=\frac{1}{f(x)}[/imath] is Riemann integrable? I know from the statement that [imath]f(x)\neq 0[/imath] since its [imath]\inf[/imath] is [imath]d>0[/imath], so [imath]g(x)[/imath] is bounded on [imath][a,b][/imath] since [imath]f(x)[/imath] is bounded and there is no division by [imath]0[/imath]. Is there a relationship like [imath]M(f)=m(g)[/imath] and [imath]m(f)=M(g)[/imath], since [imath]g(x)[/imath] is the reciprocal? |
2575392 | Diophantine Equation Possible Values of GCD
Suppose that, for given integers [imath]a[/imath] and [imath]b[/imath], there exist integers [imath]x[/imath] and [imath]y[/imath] satisfying [imath]ax + by = 6[/imath]. What are all the possible values of [imath]\gcd(a, b)[/imath]? Progress, we know that since [imath]ax + by = 6[/imath], then [imath]x = (6 - by)/a[/imath]. In order for [imath]x[/imath] to be an integer, [imath]6 - by[/imath] must be some multiple of a. Similarly, [imath]y = (6 - ax)/b[/imath], so [imath]6 - ay[/imath] must be some multiple of [imath]b[/imath]. Then for some integer [imath]q[/imath], [imath]aq = 6 - by[/imath]. For some integer [imath]r[/imath], [imath]br = 6 - ay[/imath]. Then [imath]b = (6 - ay)/r[/imath] and back substituting gives us [imath]y = (6 - ax)/[(6 - ay)/r] = r(6 - ax)/(6 - ay)[/imath]. Thus [imath]6y - ay^2 = r(6 - ax)[/imath] and so we have the quadratic... I'm in a mess, any help and/or advice would greatly be appreciated. | 880744 | Diophantine equation [imath]ax + by = c[/imath] has an integer solution [imath]x_0, y_0[/imath] if and only if [imath]\gcd(a,b)|c[/imath]
Let [imath]a,b,c[/imath] be positive integers. Verify that Diophantine equation [imath]ax + by = c[/imath] has integer solution [imath]x_0, y_0[/imath] if and only if [imath]GCD(a,b)|c[/imath]. Attempt Diophantine [imath]ax + by = c[/imath] has integer solution [imath]x_0, y_0[/imath] implies [imath]GCD(a,b)|c[/imath]. Factoring out [imath] GCD(a,b)[/imath] from each side gives [imath]\frac{1}{GCD(a,b)}(ax + by) = \frac{1}{GCD(a,b)}c[/imath] which must still have an integer solution as [imath]GCD(a,b)[/imath] obviously divides both [imath]a[/imath] and [imath]b[/imath]. If [imath]GCD(a,b)[/imath] does not divide [imath]c[/imath] there is no integer solution. That [imath]GCD(a,b)|c[/imath] implies that [imath]ax + by = c[/imath] has integer solution [imath]x_0, y_0[/imath]. Let [imath]c = GCD(a,b)d[/imath], where [imath]d[/imath] is integer. [imath]ax + by = GCD(a,b)d[/imath]; divide both sides by [imath]GCD(a,b)[/imath]: Obtain [imath]mx + ny = d[/imath] with [imath]GCD(m,n) = 1[/imath]. If [imath]d[/imath] is an integer and [imath]GCD(m,n) = 1[/imath] then [imath]x,y[/imath] must be integers as [imath]mx + ny[/imath] cannot be further divided evenly. or [imath]ax + by = c[/imath] has no solution implies [imath]GCD(a,b)[/imath] does not divide c. [imath]c = GCD(a,b)(\frac{ax}{GCD(a,b)} + \frac{by}{GCD(a,b)}) + r[/imath] with [imath]GCD(a,b) > 0, 0<r<GCD(a,b)[/imath], r integer [imath]\frac{c}{GCD(a,b)} = \frac{ax + by}{GCD(a,b)} + \frac{r}{GCD(a,b)}[/imath] |
2572958 | Proving the convergence of a sequence [imath]e_{n+1}=e_n e_{n-1}[/imath]
The sequence [imath]e_n[/imath] has limit [imath]0[/imath] and all terms are in [imath](0,1)[/imath]. Also, we have that [imath]e_{n+1}=e_n e_{n-1}[/imath] I need to show that we can pick [imath]C[/imath] s.t. [imath]e_{n+1}\le C \cdot e_n^\phi[/imath] where [imath]\phi[/imath] is the golden ratio. Any ideas? As all terms are in [imath](0,1)[/imath] so we know that [imath]e_{n+1}<\min(e_n,e_{n-1})[/imath] but I'm not sure where to go. | 2552273 | Verification of proof involving logarithms and limits
Question: Consider [imath]E_n\in(0,1)[/imath] and [imath]E_n\to0,\text{as }n\to\infty[/imath] Assuming that [imath]E_{n+1} = E_n E_{n-1}[/imath], show that, for a real [imath]C[/imath] independent of [imath]n[/imath], [imath]E_{n+1} \leq C{E_n}^\phi[/imath] where [imath]\phi=\frac{1+\sqrt{5}}{2}[/imath], the Golden Ratio My attempt so far: We can take logarithms of both side to see that \begin{align*}\log(E_{n+1}) &= \log(E_n E_{n-1})\\ &=\log(E_n)+\log(E_{n-1})\end{align*} We can see that this fulfils the conditions of the Fibonacci Sequence, [imath]F_{n+1} = F_{n}+F_{n-1}[/imath] Therefore, we can assume that [imath]\frac{\log(E_{n+1})}{\log(E_n)}\to \phi,\quad\text{as }n\to\infty[/imath] We can then say that \begin{align*}\lim_{n\to\infty}\left|\frac{\log(E_{n+1})}{\log(E_n)}-\phi\right|&=0\\ \lim_{n\to\infty}\left|\log(E_{n+1})-\phi\log(E_n)\right|&=0\\ \lim_{n\to\infty}\left|\log(E_{n+1}) -\log({E_n}^\phi)\right|&=0\\ \lim_{n\to\infty}\left|E_{n+1} - {E_n}^\phi\right|&=1\end{align*} I'm not entirely sure that the last line of this is correct - I have attempted to get rid of the logarithms as I would have done in an equation with no limits or absolute values in. Could someone let me know if I have done anything wrong please. I am also struggling to continue from this point so any guidance would be much appreciated too. |
2576449 | If [imath]f[/imath] is holomorphic on the unit disk and [imath]f(0)=0[/imath], then [imath]\lvert f(z) + f(-z) \rvert \le 2 \lvert z \rvert^2[/imath]
Let [imath]f[/imath] be a complex valued function holomorphic on the unit disk with [imath]f(0) = 0[/imath]. Then [imath]\lvert f(z) + f(-z) \rvert \le 2 \lvert z \rvert^2[/imath]. I was looking at this answer but I still unsure about certain steps. [imath]\textbf{proof attempt.}[/imath] Define [imath]g(z) = \frac{f(z) + f(-z)}{2}[/imath] (which is also holomorphic). Then [imath]g(0) = 0[/imath] and if I could show [imath]\lvert g(z) \rvert \le 1[/imath], then by Schwarz's lemma [imath]\lvert g(z) \rvert \le \lvert z \rvert[/imath]. Now define [imath]h(z) = \frac{g(z)}{z}[/imath], and by Riemann removable singularity theorem we can define [imath]h(z)[/imath] on all of [imath]D(0,1)[/imath] since [imath]h[/imath] is bounded at [imath]z=0[/imath] since [imath]\lim_{z\to0} \frac{g(z)}{z} = g'(0) = 0[/imath]. Thus [imath]h(0) = 0[/imath] and [imath]\lvert h(z) \rvert = \lvert \frac{g(z)}{z} \rvert \le 1[/imath] finally applying Schwarz's lemma to [imath]h[/imath] we obtain [imath]\lvert \frac{g(z)}{z} \rvert \le \lvert z \rvert[/imath] which is the desired result. // How do I show [imath]\lvert g(z) \rvert \le 1[/imath]? | 1853151 | If [imath]f(0)=0[/imath] then prove that [imath]|f(z)+f(-z)| \leq 2|z|^2[/imath]
Let , [imath]\Delta = B(0,1)[/imath]. If [imath]f:\Delta \rightarrow \Delta[/imath] is holomorphic, and [imath]f(0)=0[/imath], prove that [imath]|f(z)+f(-z)| \leq 2|z|^2[/imath]. My attempt so far: [imath]\displaystyle f(z) = \sum_{n=0}^{+\infty} a_n z^n = \sum_{n=0}^{+\infty} a_{2n} z^{2n} + \sum_{n=0}^{+\infty} a_{2n+1} z^{2n+1}[/imath] [imath]\displaystyle f(-z) = \sum_{n=0}^{+\infty} a_n (-z)^n = \sum_{n=0}^{+\infty} a_{2n} z^{2n} - \sum_{n=0}^{+\infty} a_{2n+1} z^{2n+1}[/imath] Then, [imath]\displaystyle f(z)+f(-z) = 2 \sum_{n=0}^{+\infty} a_{2n} (z^2)^n =: 2 g(z^2)[/imath] If [imath]g(\Delta) \subset \Delta[/imath], as [imath]g(0) = 0[/imath], I can invoke Schwarz Lemma and I win. However, I don't know how to prove this. Could anyone help me? Thanks! |
2576166 | Rearranging the alternating harmonic series and finding differences in its sum
I am to prove that if [imath]1-1/2+1/3-1/4+1/5-1/6+...=s[/imath] that [imath]1-1/2+1/3+1/5-1/4+1/7+1/9-1/6+...=3s/2[/imath] My approach: [imath]s/2=1/2-1/4+1/6-1/8+1/10-1/12...[/imath] so that [imath]s+s/2=3s/2=1-(1/2-1/2)+1/3-(1/4+1/4)+1/5-(1/6-1/6)...=[/imath] [imath]1+1/3-1/2+1/5...[/imath] This sort of leads to what I want to approach, but the terms are not quite in the right order. I assume I may not change the order of the last set of terms for it to match what I'm looking for. Any other suggestions? | 2574386 | Show that [imath]\sum (-1)^{n+1}/n[/imath] = [imath]k[/imath] can be expanded to a sum with [imath] 1 - 1/2 + 1/3 + 1/5 - 1/4 +1/7 + 1/9 - 1/6 + ....[/imath] which is [imath]3k/2[/imath]
I need to show that [imath]\sum (-1)^{n+1} / n[/imath] = [imath]k[/imath] (which is [imath]1 - 1/2 + 1/3 - 1/4+...[/imath]) can be expanded to a sum with [imath] 1 - 1/2 + 1/3 + 1/5 - 1/4 +1/7 + 1/9 - 1/6 + .... [/imath] which is [imath]3k/2[/imath]. How can I start? I know that [imath]k/2[/imath] = [imath]\sum (-1)^{n+1}/2n[/imath] How can I go further so I get [imath]3k/2[/imath]? |
2576464 | How can I use GΓΆdel numbers to make a proof?
As we know, any proposition can translate to Godel numbers, such as [imath]( \exists x)\space (x=sy)[/imath]Its Godel numbers is [imath]2^8 Γ 3 ^4 Γ 5^ {13} Γ 7^ 9 Γ 11^8 Γ 13^{13} Γ 17^ 5 Γ 19^ 7 Γ 23^ {17} Γ 29^ 9=a \space big \space number [/imath] We also could get the character sequence "[imath]( \exists x)\space (x=sy)[/imath]" through prime factorization of 243,000,000. Such as[imath]243,000,000=2^6Γ3^5Γ5^6 [/imath] [imath]6\space\space\space\space\space 5\space\space\space\space\space 6 [/imath] [imath]\downarrow\space\space\downarrow\space\space\downarrow[/imath] [imath]0\space\space\space=\space\space\space 0 [/imath] So the Godel numbers of "[imath]0=0[/imath]" is [imath]243000000[/imath] There are four axioms in Godel's PM system, here are two of them: [imath](p\lor p)\subset p[/imath] [imath]p\subset (p\lor q)[/imath] through these two axioms, we can prove "[imath](p\lor p)\subset(p\lor q)[/imath]" is a true theorem. And these two axioms and a theorem all have a corresponding Godel numbers, can we prove this theorem is true through Godel numbers? If it works, how? Many people think it is a question unrelated with Mathematica, but what I think is that if GΓΆdel numbers could make a proof, then we can build an Association between characters and numbers in Mathematica, and add some rules, then we can prove propositions. But Iβm not sure GΓΆdel numbers can make a mathematical proof. | 2576412 | Can I use GΓΆdel numbers to make a proof?
As we know, any proposition can translate to Godel numbers, such as [imath]( \exists x)\space (x=sy)[/imath]Its Godel numbers is [imath]2^8 Γ 3 ^4 Γ 5^ {13} Γ 7^ 9 Γ 11^8 Γ 13^{13} Γ 17^ 5 Γ 19^ 7 Γ 23^ {17} Γ 29^ 9=a \space big \space number [/imath] We also could get the character sequence "[imath]( \exists x)\space (x=sy)[/imath]" through prime factorization of this big number. Such as[imath]243,000,000=2^6Γ3^5Γ5^6 [/imath] [imath]6\space\space\space\space\space 5\space\space\space\space\space 6 [/imath] [imath]\downarrow\space\space\downarrow\space\space\downarrow[/imath] [imath]0\space\space\space=\space\space\space 0 [/imath] So [imath]243000000[/imath] is the Godel numbers of "[imath]0=0[/imath]" There are four axioms in Godel's PM system, here are two of them: [imath](p\lor p)\subset p[/imath] [imath]p\subset (p\lor q)[/imath] through these two axioms, we can prove "[imath](p\lor p)\subset(p\lor q)[/imath]" is a true theorem. And these two axioms and a theorem all have a corresponding Godel numbers, can we prove this theorem is true through Godel numbers? If it works, how? |
2577004 | How to deduce Axiom Schema of Replacement from Axiom of Separation?
I have two questions regarding this issue. In addition to Axiom Schema of Replacement, Which axioms do we need to deduce Axiom of Separation? How to deduce Axiom of Separation from Axiom Schema of Replacement? Axiom of Separation: [imath]\forall w \forall x \exists y \forall z [z \in y \iff (z \in x \land \varphi(z, w, x ) )][/imath] Axiom Schema of Replacement:[imath]\forall w \forall A [ (\forall x \in A \implies \exists ! y \varphi(x,y,w, A)) \implies (\exists B \forall x (x \in A \implies \exists y \in B \varphi(x,y,w, A)))][/imath] Many thanks! | 680376 | Proving Separation from Replacement
I'm trying to show that separation follows from replacement. There are at least two questions here on SE that deal with the question. See The comprehension axioms follows from the replacement schema. and How do the separation axioms follow from the replacement axioms?. I'm not completely satisfied with the answers there. The basic idea for the proof that most people use seems to be this: Say we have some set [imath]D[/imath] and some formula [imath]\varphi(x)[/imath]. To prove separation, we're trying to show that [imath]\{x \in D| \varphi(x)\}[/imath] is a set. Using separation, we need to write a formula [imath]\psi(x,y)[/imath] such that for each [imath]x \in D[/imath] there is exactly one [imath]y[/imath] such that [imath]\psi(x,y)[/imath]. Here's one attempt: [imath]\psi(x,y) = \text{"}x = y \wedge \varphi(x) \text{''}[/imath]. IF this formula satisfies the separation axiom, THEN this should work fine. The problem: at least in the version of the separation axiom I know, we need for there to be a unique [imath]y[/imath] for all [imath]x \in D[/imath], such that [imath]\psi(x,y)[/imath]. So say [imath]x \in D[/imath] is such that [imath]\neg \varphi(x)[/imath], then [imath]\psi(x,y)[/imath] is always false (and so not true for a unique [imath]y[/imath] as required for the use of replacement). Here's another attempt: [imath]\psi(x,y) = \text{''}(\{x\}=y \wedge \varphi(x))\vee y=\emptyset \text{''}.[/imath] Now we have a function whose domain is the entire set [imath]D[/imath]. And we can use replacement and union to get the set we want. But this requires that both [imath]\emptyset[/imath] and [imath]\{x\}[/imath] are sets. This is the usual argument that they are sets. [imath]D[/imath] is a set, so [imath]\{x \in D| x \not = x\}=\emptyset[/imath] is a set. But that's a straightforward use of separation. We have the same problem for [imath]\{x\}[/imath]. Using pairing we get that there is a set [imath]A[/imath] such that [imath]x \in A[/imath], but we don't necessarily have as a result that there is a set that contains only [imath]x[/imath]. Of course we could construct this set with [imath]\{y \in A | y = x\}[/imath], but again we've used separation. It's for the reasons above that I'm not perfectly satisfied with the answers to other questions here on SE. Perhaps I've made some mistake, or there actually is a better answer. |
2576854 | [imath]\int_{E_n}f d \mu \rightarrow 0[/imath] implies [imath]\mu(E_n) \rightarrow 0[/imath]
Let [imath]f:[0,1] \rightarrow \mathbb{R}[/imath] be a positive measurable function and [imath]E_n[/imath] be a sequence of measurable sets, prove that if [imath]\int_{E_n}f d \mu \rightarrow 0[/imath] implies [imath]\mu(E_n) \rightarrow 0[/imath]. I have try to decompose [imath][0,1][/imath] into [imath]\{f\geq 1/n\}[/imath] but I have no idea after that. | 269295 | Showing [imath]\lim_{n \to \infty} m(E_n) = 0[/imath], assuming [imath]f > 0[/imath] a.e. and [imath]\lim_{n \to \infty} \int_{E_n}f \,dm =0[/imath]
Let [imath]f:[0,1] \to \mathbb{R}[/imath] be Lebesgue measurable with [imath]f > 0[/imath] a.e. Suppose that [imath]\{E_n\}[/imath] is a sequence of measurable sets in [imath][0,1][/imath] with the property that [imath]\displaystyle \lim_{n \to \infty}\int_{E_n} f \,dm = 0[/imath]. Prove that [imath]\displaystyle \lim_{n \to \infty} m(E_n) = 0[/imath]. This question is from an old analysis qual I am studying. So far I have tried a proof by contradiction: if [imath]\displaystyle \lim_{n \to \infty} m(E_n) \neq 0[/imath], then there is an [imath]\epsilon > 0[/imath] and a subsequence [imath]\{n_k\}[/imath] so that [imath]m(E_{n_k}) \ge \epsilon[/imath] for all [imath]k[/imath]. I am trying to somehow use this subsequence and show [imath]\displaystyle \lim_{k \to \infty}\int_{E_{n_k}} f \,dm \neq 0[/imath], which would give me a contradiction. Another fact I know from my measure theory course is that, for meausurable [imath]E \subseteq [0,1][/imath], the map [imath]\displaystyle \nu(E) = \int_E f \,dm[/imath] defines a measure on the Lebesgue measurable subsets of [imath][0,1][/imath]. Will this fact be useful to me? |
2576542 | LeL [imath]f: (0, \infty)\to R[/imath] be continuous and [imath]f(x)\leq f(nx)[/imath] prove [imath]\lim\limits_{x\to\infty} f(x)[/imath] exists
Let [imath]f: (0, \infty)\to R[/imath] be continuous such that [imath]f(x)\leq f(nx)[/imath] for all positive [imath]x[/imath] and natural [imath]n[/imath]. I'd like to prove [imath]\lim\limits_{x\to\infty} f(x)[/imath] exists finite or infinite but I'm not quite sure how to persou this way Any piece of advice would be much appreciated | 2063430 | [imath]f\colon(0,\infty)\to \mathbb R[/imath] be continuous ; [imath]f(x)\le f(nx) , \forall n \in \mathbb N , \forall x >0[/imath] , then [imath]\lim_{x\to \infty} f(x)[/imath] exists?
Let [imath]f\colon(0,\infty)\to \mathbb R[/imath] be a continuous function such that [imath]f(x)\le f(nx) , \forall n \in \mathbb N , \forall x >0[/imath] , then is it true that [imath]\lim_{x\to \infty} f(x)[/imath] exists (may be infinite) ? |
2577225 | Hard Geometry problem, need help
ABC is a right triangle at [imath]A[/imath] , [imath]BH , CD[/imath] are the bisectors of angles [imath]\angle {B},\angle { C}[/imath], respectively , if [imath]BH=9[/imath] and [imath]CD=8\sqrt {2}[/imath] , Find the length of [imath]BC[/imath]? | 516166 | Find hypotenuse given acute angle bisectors
In a right triangle [imath]ABC[/imath] (right-angled at [imath]B[/imath]), [imath]D[/imath] and [imath]E[/imath] are points of [imath]\overline{AB}[/imath] and [imath]\overline{BC}[/imath] respectively such that [imath]\overline{CD}[/imath] and [imath]\overline{AE}[/imath] are the angle bisectors of the acute angles of the triangle. Given that [imath]AE=9[/imath] and [imath]CD=8\sqrt{2}[/imath], find the length of the hypotenuse [imath]\overline{AC}[/imath]. |
2577070 | If [imath]X[/imath] and [imath]Y[/imath] are sets, then [imath]X\times Y[/imath] is also set.
If [imath]X[/imath] and [imath]Y[/imath] are sets, then [imath]X\times Y[/imath] is also set. Recall: [imath]X\times Y=[/imath]{[imath](x,y): x\in X, y\in Y[/imath]}. I only proved: [imath](x,y)=[/imath]{{[imath]x[/imath]},{[imath]x,y[/imath]}} is a set. I couldn't prove above question. Can you help? We use only these: 1. Axiom of extensionality 2. Axiom of regularity (also called the Axiom of foundation) 3. Axiom schema of specification (also called the axiom schema of separation or of restricted comprehension) 4. Axiom of pairing 5. Axiom of union 6. Axiom schema of replacement | 1995384 | If [imath]X[/imath] and [imath]Y[/imath] are sets, their cartesian product [imath]X\times Y[/imath] is a set.
In ZFC set theory, every element object is a set. We know or feel that if [imath]X[/imath] and [imath]Y[/imath] are sets, their cartesian product [imath]X\times Y[/imath] is a set. My question is: Why if [imath]X[/imath] and [imath]Y[/imath] are sets, their cartesian product [imath]X\times Y[/imath] is a set? |
2576821 | [imath]w(T)=0 \implies T=0[/imath]?
Let [imath]E[/imath] be a complex Hilbert space. Let [imath]T\in \mathcal{L}(E)[/imath]. It is true that for an arbitrary operator [imath]T\in \mathcal{L}(E)[/imath], we have [imath]w(T) := \sup\big\{\;\left|\langle Tu\;|\;u\rangle \right|,\;\;u \in E\;, \left\| u \right\| = 1\;\big\}=0\Longrightarrow T=0?[/imath] Or [imath]T[/imath] must be self-adjoint operator? Thank you. | 2576899 | [imath]\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0[/imath]?
Let [imath]E[/imath] be a complex Hilbert space. Let [imath]T\in \mathcal{L}(E)[/imath]. I have two questions: Why it is not true that for an arbitrary operator [imath]T\in \mathcal{L}(E)[/imath], we have [imath]\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0[/imath]? And is this property true for normal operators? I think it is true for self adjoint operators because the norm of a self adjoint operators is given by [imath]\left\|T\right\|= \sup\big\{\;\left|\langle Tu\;|\;u\rangle \right|,\;\;u \in E\;, \left\| u \right\| = 1\;\big\}[/imath] Thank you. |
2577238 | find an injection [imath]h:\{\mathbb{N}\to\{0, 1\}\} \to A[/imath] while A = {f : N β Z|βn, m β N.(n > m) β (f(n) < f (m))}
How can I find an injective function from [imath]h β (\mathbb N β {0, 1}) β A[/imath]. [imath]A = \{f : N β Z|βn, m β N.(n > m) β (f(n) < f (m))\}[/imath]? I was trying to define it as follows: [imath]h=-n-f(n)[/imath], but I know that doesn't work since it only maintains [imath]f(n) \leq f (m)[/imath] but not [imath]f(n) < f (m)[/imath]. Would love to hear your thoughts. | 2576603 | Injection from [imath]\mathbb{N}\to\{0,1\}[/imath] to the set of all monotonic decreasing functions from [imath]\mathbb{N}\to\mathbb{Z}[/imath]
I'm trying to think of such an injection, and unfortunately all of those I've thought about are not strictly decreasing. I would love to get some help or clues. Thank you :) |
2577936 | Showing that the integral with respect to an abstract measure is infinite
Suppose [imath]\mu[/imath] is a measure on [imath](\mathbb{R}, \mathcal{B}(\mathbb{R}))[/imath]. Then show that, for [imath]\mu[/imath]-almost every [imath]x[/imath] [imath] \int_\mathbb{R} \frac{1}{(x-t)^2}\,\mathrm{d}\mu(t) = \infty [/imath] I am not sure where to start. Any hints are welcome. | 30703 | A set with a finite integral of measure zero?
Prove, or give a counter example: Let [imath]\mu[/imath] be a finite positive borel measure on [imath]\mathbb{R}[/imath]. Then [imath]\int (x-y)^{-2} d \mu (y) = \infty [/imath] almost everywhere on [imath]\mu[/imath] (for the selection of x's). This is a question I had in an exam, and the answer is supposed to be presented in less than 30 words, so there must be something quite simple I'm missing. |
2575797 | There are ten girls and four boys in Mr. Fat's Combinatorics class, Find the number of. . .
There are ten girls and four boys in Mr. Fat's combinatorics class. In how many ways can these students sit around a circular table such that no boys are next to each other? (Source : Titu Andreescu Combinatorics Textbook). Here is my solution to this problem, however the number I got as an answer was extremely large and I felt I might have made a mistake so I need MSE to help verify and (probably) point out the error(s) in my solutions. Solution 1. Let us use the Subtraction Principle in this case. There are 10 + 4 persons to be sitted around a circle, there are at most 13! ways to carry out this task. Now. Let us find all such sitting arrangements in which it is compulsory for two boys to sit next to each other, in this case, we can regard the two boys as an entity, say boys [imath]B_1[/imath] [imath]B_2[/imath] as [imath]X[/imath] and boys [imath]B_3[/imath] [imath]B_4[/imath] as [imath]Y[/imath]. Then in this case, the problem is this reduced to finding the number of possible cyclic permutations of the set P = {X, Y, 1, . . 10}, where 1, 2, 3, ... 10 denotes the girls. The number of such cyclic permutations is at most 11!. But since we can permutate 2 boys from 4 boys in at most [imath]_4P_2[/imath]/2 = 6 ways. The number of such cyclic permutations is at most 11![imath]\cdot 6[/imath] ways. Hence, by the Subtraction Principle, there are at most 13! - 6(11!) = 5987520000 ways to arrange 10 girls and 14 boys in a circle such that no two boys are sitted together. Solution 2. In this case I used a direct approach by considering possible cases. Case 1. A boy sits between every two girls. In this case we can select 7 out of 10 girl occupy the first 7 sits around the circle. The number of ways to permute 7 girls from 10 is at most [imath]10P_7[/imath]/7 = 240 ways. Since 6! Was is the number of ways to sit them around a circle, these 7 girls can occupy this sit in at most 240(6!) ways. Now the remaining 4 boys can sit anywhere provided that no two persons occupy the same seat, their sitting arrangements can be done in at most 6! ways. Hence for this case, we have a total of 240(6!)(6!) ways possible sitting arrangements. Case 2. A boy sits between every two girls, in this case, we treat the two girls as a single entity like before, say X and Y. And we can permute two girls from 10 in at most [imath]\10P_2[/imath]/2 =45ways. These 5 entities can sit around the 5 available sits in at most 4! Ways. This leaves us with a total of 45(4!) ways to execute the task. Now. The remaining 4 boys can sit on the remaining four sits around the circle in at most 3! ways. This leaves us with a total of 45(4!)(3!) ways. Obviously these are the only two cases we consider, hence for this solution, the answer is thus 240(6!)(6!) + 45(3!)(4!) = 12448080. But this answer doesn't correspond with my initial answer. What's my mistake? | 2016863 | Problem (1.8) A Path to Combinatorics Solution Verification.
Problem (1.8): There are ten girls and four boys in Mr. Fat's combinatorics class. In how many ways can these students sit around a circular table such that no boys are next to each other. My Attempt: Total number of ways in which both boys are girls can be arranged on a circular table is [imath]13![/imath] and the number of ways in which any two boys can be together is [imath]2!*12![/imath]. Therefore the answer is [imath]13!-2*12!=11*12![/imath]. Is this answer correct? I am asking this because I am confused about the exact meaning of the line "no boys are together." Does this mean that no two/three/four boys are together? Note: Meanwhile I have made another attempt on the problem. We can place the [imath]10[/imath] girls on the circular table in [imath]9![/imath] ways. Then we are left with [imath]10[/imath] spots which need to be filled with [imath]4[/imath] boys. This can be done in [imath]P^{10}_4[/imath] ways. Thus, the number of ways is [imath]9!P^{10}_4[/imath]. |
2577709 | A question about [imath]F[/imath]- algebras
I have a question about [imath]F[/imath]- algebras, please help me: If [imath]A[/imath] is a nonzero finite dimensional unital algebra over a field [imath]F[/imath] with [imath]char(F) = 0[/imath], and let [imath]\varphi[/imath] be an automorphism of [imath]A[/imath], how can we prove that [imath]\varphi(a) \neq a +1[/imath] for every [imath]a \in A[/imath]? | 2547688 | How can we show that [imath]\varphi(a) \neq a +1[/imath] for every [imath]a \in A[/imath]?
Let [imath]A[/imath] be a nonzero finite dimensional unital algebra over a field [imath]F[/imath] with [imath]char(F) = 0[/imath], and let [imath]\varphi[/imath] be an automorphism of [imath]A[/imath]. How can we show that [imath]\varphi(a) \neq a +1[/imath] for every [imath]a \in A[/imath]? How can we prove by [imath]L_{\varphi (a)}[/imath] through [imath]\varphi[/imath],[imath]\varphi^{-1}[/imath], and [imath]L_{a}[/imath]? ( [imath]L_{a}[/imath] is left multiplication map [imath]x \mapsto ax[/imath]) . |
2578747 | Factorize [imath]z^5 - 1[/imath] using [imath]\cos[/imath] function
This is a problem from an exam, with which I am struggling for over a month, and trying over and over. I've given up... The problem states (and my exam is tomorrow where I could expect even same problem): [imath](a)[/imath] Prove that: [imath]z^5 - 1 = (z-1)(z^2 +2z\cos(\frac{\pi}5) + 1)(z^2 - 2z\cos(\frac{2\pi}5) + 1)[/imath] [imath](b)[/imath] Find the values for [imath]\cos\frac{\pi}5[/imath] and [imath]\cos\frac{2\pi}5[/imath]. I've managed to prove [imath](b)[/imath] using [imath](a)[/imath], and to find all roots for the polynomial [imath]z^5 - 1[/imath]. Also, I've managed to prove [imath](b)[/imath] using "double angle formula" and then use it for [imath](a)[/imath]. However, I cannot prove it in the stated order (I hope it is possible). | 1422555 | Show that [imath]z^5-1=(z-1)(z^2+2z\cos{\pi \over 5}+1)(z^2-2z\cos{2\pi \over 5}+1)[/imath]
Show that [imath] \quad \quad z^5-1=(z-1)(z^2+2z\cos{\dfrac{\pi}{5}}+1)(z^2-2z\cos{\dfrac{2\pi}{5}}+1)[/imath] and deduce from this the closed formulas for [imath]\cos{\dfrac{\pi}{5}}[/imath] and [imath]\cos{\dfrac{2\pi}{5}}[/imath]. Here's what I did: I assumed that it was correct, and using the fact that [imath]z^5-1=(z-1)(z^4+z^3+z^2+z+1)[/imath], I said that: [imath] \quad z^4+z^3+z^2+z+1=(z^2+2z\cos{\dfrac{\pi}{5}}+1)(z^2-2z\cos{\dfrac{2\pi}{5}}+1)[/imath] Letting [imath]a=\cos{\dfrac{\pi}{5}}[/imath] and [imath]b=\cos{\dfrac{2\pi}{5}}[/imath] [imath] \quad z^4+z^3+z^2+z+1=z^4+(2a-2b)z^3+(2-4ab)z^2+(2a-2b)z+1[/imath] Equating the coefficients: [imath]\quad a-b= {1 \over 2} \quad[/imath] and [imath] \quad ab={1 \over 4}[/imath] Simultaneously solving, and knowing that [imath]0<a<1[/imath] and [imath]0<b<1[/imath]: [imath]\cos{\dfrac{\pi}{5}}=\dfrac{1}{4} + \dfrac{\sqrt{5}}{4} \quad[/imath] and [imath]\quad \cos{\dfrac{2\pi}{5}}=-\dfrac{1}{4} + \dfrac{\sqrt{5}}{4}[/imath] Then I said that, if I can prove the cosine formulas above in a different way, I have indirectly shown that the relationship shown in the question was true. So, since [imath]\cos{\pi}= \cos{(5 Γ (\frac{\pi}{5}))} =-1[/imath] and with a little trig indentity, [imath]\quad 16{\cos^5{\pi \over 5}}-20{\cos^3{\pi \over 5}}+5{\cos{\pi \over 5}}+1=0[/imath] With substitution [imath]x=\cos{\pi \over 5}[/imath]: [imath]16x^5-20x^3+5x+1=0[/imath], a polynomial that has roots at [imath]x=-1[/imath], [imath]x= \frac{1}{4} - \frac{\sqrt{5}}{4}[/imath], and [imath]x=\frac{1}{4}+\frac{\sqrt{5}}{4}[/imath]. The only possible solution is [imath]\cos{\pi \over 5}=\frac{1}{4}+\frac{\sqrt{5}}{4}[/imath], and the double angle formula shows that [imath] \cos{\frac{2\pi}{5}}=-\frac{1}{4} + \frac{\sqrt{5}}{4}[/imath] My question, is there a simpler or more straightforward way of proving this? |
2578935 | Find [imath]\lim_{x\to 0^+}(\sin x)^\frac{1}{\ln x}[/imath]
[imath]\lim_{x\to 0^+}(\sin x)^\frac{1}{\ln x}[/imath] I was trying to change it to [imath]\lim_{x\to 0+}\frac{(\sin x)^\frac{1}{\ln x}}{x^\frac{1}{\ln x}}{x^\frac{1}{\ln x}}[/imath], but I don't know what to do next. Any help would be appreciated. | 605915 | Find the limit [imath]\displaystyle \lim_{x \to 0^+} (\sin x)^\frac1{\ln x}[/imath]
Find the limit [imath]\displaystyle \lim_{x \to0^+} (\sin x)^{\large {\frac1 {\ln x}}}[/imath] Here's what I did: [imath]\begin{align}\lim_{x \to 0^+} (\sin x)^{\large \frac1{\ln x}}&(1) =\lim e^{\large {\ln( { (\sin x)^\frac1{\ln x}})}} \ \ \\ &(2) = \lim_{x \to0^+} e^{\large {{\frac1{\ln x}}\ln( { (\sin x)})}} \ \ \\ & (3)= \lim_{x \to0^+} e^{\large {\frac{\ln\sin x}{\ln x}}} \ \ \\ &(4) = \lim_{x \to0^+} e^{\large {\ln\frac{ \sin x}{ x}}} \ \\ & (5)= \lim_{x \to0^+} \frac{ \sin x}{ x} = 1 \end{align}[/imath] But I know this is wrong because when I plot it it goes to e. I am now stuck on step 3. Any advice please ? PS: we can't use derivation or integration, nor Taylor's theorem because we haven't covered that. |
2579112 | Limit of [imath]\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)/n[/imath]
For [imath]n\in\mathbb{N}^*[/imath], define [imath]a_n=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)/n[/imath]. Does the sequence [imath](a_n)[/imath] converge and how to prove it? | 1757859 | Test for the convergence of the sequence [imath]S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)[/imath]
[imath]S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)[/imath] Show the convergence of [imath]S_n[/imath] (the method of difference more preferably) I just began treating sequences in school, and our teacher taught that monotone increasing sequence, bounded above and monotone decreasing sequences, bounded below converge. and so using that theorem here.. I found the [imath](n+1)_{th} term[/imath], [imath]S_{n+1} = \frac1{n+1} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n+1}\right)[/imath] and then subtracted the (n)th term from it What I was able to get was... [imath]S_{n+1}-S_n = \frac1{(n+1)^2} - \frac{\left(1+\frac12+\dots+\frac1n\right)}{n(n+1)}.[/imath] ...but then this is where I get stucked, but i'm trying to prove that the sequence > 0(i.e Converges) or < 0 (i.e diverges). |
2579185 | Sum of the squares of a sequences converge, does the product?
If we know that [imath]\sum_{n=1}^{\infty}(a_n)^2[/imath] and [imath]\sum_{n=1}^{\infty}(b_n)^2[/imath] both converge can we say that [imath]\sum_{n=1}^{\infty}(a_nb_n)[/imath] converges? | 473572 | Proving convergence of sums: If [imath]\sum_{n=1}^{\infty}a^2_n[/imath] and [imath]\sum_{n=1}^{\infty}b^2_n[/imath] converge then [imath]\sum_{n=1}^{\infty}a_nb_n[/imath] converges
Let [imath]\sum_{n=1}^{\infty}a^2_n[/imath] and [imath]\sum_{n=1}^{\infty}b^2_n[/imath] prove that: [imath]\sum_{n=1}^{\infty}a_nb_n[/imath] converges. I thought about this answer: In order for a sum to converge i has to be bounded, Means Let [imath]M_1\epsilon R[/imath], So that [imath] \sum_{n=1}^{\infty}a_n<M_1 [/imath]. Same goes for [imath]b_n[/imath]. [And lets say [imath]M_2[/imath]]. So is it just: [imath]\sum_{n=1}^{\infty}a_nb_n\le \max(M_1*M_2, \sqrt{M_1*M_2}[/imath])? And one more question: Let [imath]\sum_{n=1}^{\infty}a^2_n[/imath] prove that [imath]\sum_{n=1}^{\infty}\frac{a_n}{n}[/imath] converges. If i go the same as before, I have hard time cause i get something like: [imath]\sum_{n=1}^{\infty}\frac{M_1 }{n}[/imath] which doesn't converge [I think]. Any guidelines please? |
2578697 | McLaurin expansion for [imath]f(x)=e^{\sin{x}}[/imath] of the 4:th order.
By the 4:th order, they mean using the 4:th derivative. But the differentiation gets a bit ugly quite fast, so instead of computing all the derivatives , I should be able to use the standard expansions: [imath]e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+B(t) \\ \sin{x}=x-\frac{x^3}{3!}+B(x),[/imath] Where [imath]B(t),B(x)[/imath] are functions that are bounded close to [imath]x=0.[/imath] Correct me if I'm wrong, but both of the expansions above are of 4:th order since for the sine function we have two terms vanishing because they become zero? I'm not sure how I should fuse the separate expansions of [imath]e^t[/imath] and [imath]\sin{x}[/imath] to one expansion. Setting [imath]t=\sin{x}[/imath] seems to give me an expression tedious tedious to work with. | 1070067 | Mclaurins with [imath]e^{\sin(x)}[/imath]
To evaluate [imath]e^{\sin(x)}[/imath] I use the standard series [imath]e^t[/imath] and [imath]\sin(t)[/imath], combining them gives me: [imath]e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)[/imath] [imath]\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)[/imath] [imath]\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)[/imath] [imath]\iff e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{x^3}{6}+O(x^5) = 1+x +\dfrac{x^2}{2}-\dfrac{x^4}{6} + O(x^5)[/imath] In the last step, I only evaluate up to [imath]\sin^3x[/imath] term, everything above has a grade equal to or greater than [imath]x^5[/imath] However, I'm wrong. According to Wolfram, the series expansion for [imath]e^{\sin x} = 1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+O(x^5)[/imath]. What did I do wrong? I can't seem to find what or where. |
2579025 | Find all "tame" solutions of [imath]f(x+y)+f(x-y)=2(f(x)+f(y))[/imath]
Question: Find all "tame" solutions of [imath]f(x+y)+f(x-y)=2(f(x)+f(y))[/imath]. This is how I've tried to beginβ Plug in [imath]y=0[/imath] to get [imath]f(0)=0[/imath]. [imath]y\to x[/imath] gives [imath]f(2x)=2^2f(x)[/imath]. Also, I've noted that [imath]f[/imath] is an even function by [imath]y\to -x[/imath] substitution. Now, I somewhat think that [imath]f(kx)=k^2f(x)[/imath] but I'm not sure how to land onto there. And at the end of the day, I believe [imath]f(x)=ax^2[/imath] but I cannot proceed into that stage. | 33788 | Find continuous functions such that [imath]f(x+y)+f(x-y)=2[f(x)+f(y)][/imath]
Here is the problem: Find all continuous [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] which satisfy [imath]f(x+y)+f(x-y)=2[f(x)+f(y)]\;\;\;(1).[/imath] Here is my attempt: Fix [imath]\delta>0[/imath] and let [imath]C=\int_{0}^{\delta}2f(y)dy.[/imath] Then [imath]\begin{align*} 2\delta f(x)+C&=\int_{0}^{\delta}2[f(x)+f(y)]dy\\ &=\int_{0}^{\delta}f(x+y)+f(x-y)dy\\ &=\int_{x}^{x+\delta}f(y)dy+\int_{x-\delta}^{x}f(y)dy\\ &=\int_{x-\delta}^{x+\delta}f(y)dy. \end{align*}[/imath] Now since [imath]f[/imath] is continuous, the last expression is a differentiable function of x and thus the first expression must also be differentiable; hence [imath]f[/imath] is differentiable. By induction, f is infinitely differentiable. Differentiating (1) first with respect to y, we arrive at: [imath]f'(x+y)-f'(x-y)=2f'(y)\;\;\;(2).[/imath] Differentiating once more with respect to x, we have: [imath]f''(x+y)=f''(x-y),[/imath] so [imath]f''[/imath] is constant. It follows that [imath]f(x)= ax^2+bx+c[/imath] are the only potential solutions. Substituting [imath]x=y=0[/imath] in (1) and (2) implies [imath]f(0)=f'(0)=0[/imath]; hence [imath]f(x)=ax^2\;\;\;(a\in\mathbb{R})[/imath]. It is easy to check that all such [imath]f[/imath] are indeed solutions. [imath]\square[/imath] Is my proof correct? |
2579770 | Integration of a function involving algebraic and trigonometric functions
Evaluate [imath]f(x) = \int_0^{\pi/2}\frac{1}{(1+x^2)(1+\tan{x})}dx[/imath] My attempt: I could not apply any standard method known to me to solve this integration. The only way I thought of is expressing [imath]\tan(x)[/imath] as an infinite series and expanding into a polynomial. But this will introduce approximation errors. [imath]f(x) = \int_0^{\pi/2}\frac{1}{(1+x^2)(x + \frac{x^3}{3}+\frac{2x^5}{15}+...)}dx[/imath] [imath]or, f(x) = \int_0^{\pi/2}{(1+x^2)^{-1}(x + \frac{x^3}{3}+\frac{2x^5}{15}+...)^{-1}}dx[/imath] Please let me know how to solve this problem. | 364452 | Evaluate [imath] \int_{0}^{\frac{\pi}2}\frac1{(1+x^2)(1+\tan x)}\:\mathrm dx[/imath]
Evaluate the following integral [imath]$$ \tag1\int_{0}^{\frac{\pi}{2}}\frac1{(1+x^2)(1+\tan x)}\,\mathrm dx $$[/imath] My Attempt: Letting [imath]x=\frac{\pi}{2}-x[/imath] and using the property that [imath]$$ \int_{0}^{a}f(x)\,\mathrm dx = \int_{0}^{a}f(a-x)\,\mathrm dx $$[/imath] we obtain [imath]$$ \tag2\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{\left(1+\left(\frac{\pi}{2}-x\right)^2\right)(1+\tan x)}\,\mathrm dx $$[/imath] Now, add equation [imath](1)[/imath] and [imath](2)[/imath]. After that I do not understand how I can proceed further. |
2579837 | Compute the integral [imath]\int_{0}^{\infty} \exp(in\omega t-\gamma^2 t^2)dt[/imath]
How to compute the following integral? [imath]\int_{0}^{\infty} \exp(in\omega t-\gamma^2 t^2)dt[/imath] Please help me. | 2535819 | Show that [imath]\frac{1}{\sqrt{2\pi}} \cdot \left(\int_{- \infty}^{\infty} e^{itx} \cdot e^{-\frac{1}{2}t^2 } dx \right) = e^{-\frac{1}{2}t^2}[/imath]
I am struggling to show that [imath]\frac{1}{\sqrt{2\pi}} \cdot \left(\int_{- \infty}^{\infty} e^{itx} \cdot e^{-\frac{1}{2}x^2 } dx \right) = e^{-\frac{1}{2}t^2}[/imath] In the same paper we defined [imath]F(t):=\left(\int_0^t e^{-x^2}\right)^2 \quad \text{ and} \quad G(t):=\left( \int _0^1\frac{e^{-t^2(1+x^2)}}{1+x^2} \right)[/imath] I showed that both functions are differentiable with [imath]F'= 2e^{-t^2} \cdot \int_0^t e^{-x^2}[/imath] [imath]G' = -2te^{-t^2} \cdot \int_0^1 e^{-x^2t^2}[/imath] It follows that [imath]F' + G' = 0[/imath] and [imath]F + G = \frac{\pi}{4}[/imath] With that I showed that [imath]\int_{-\infty}^{\infty} e^{-x^2} = \sqrt{\pi}[/imath]. Any help is appreciated. |
2579646 | Example of group of order [imath]p^3[/imath]
Let [imath]p[/imath] be a prime number I am trying to find an example of group of order [imath]p^3[/imath] that is non abelian. I know that a group of order [imath]p[/imath] or [imath]p^2[/imath] is abelian, but how to find an example of non abelian group of order [imath]p^3[/imath]? Any insights would be helpful. | 1817166 | An example of a non-abelian group of order [imath]p^3[/imath]
Let [imath]p[/imath] be prime and consider the set of all matrices [imath]\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}[/imath] where [imath]a,b,c \in \mathbb{Z_p}[/imath]. This set forms a non-abelian group of order [imath]p^3[/imath]. Any ideas how to prove using group actions? |
2576681 | Show equivalent form of [imath]\backsim\left(\forall x {\in} S,P(x) \vee Q(x)\right)[/imath]
Question: [imath]{Letβ βS = \{x_1, x_2, x_3, ..., x_n\}}[/imath] Show that [imath]{\backsim\Bigl(\forall βx β{\in} βS,P(x) β β\vee β βQ(x)\Bigr) \Leftrightarrow \exists βx β{\in} βS, β\Bigl( β\backsim P(x) β β\wedge β β\backsim Q(x)\Bigr)}[/imath] This is what I have done so far ... [imath]{\backsim\Bigl(\forall βx β{\in} βS,P(x) β β\vee β βQ(x)\Bigr) \Leftrightarrow \exists βx β{\in} βS, β\Bigl( β\backsim P(x) β β\wedge β β\backsim Q(x)\Bigr)}[/imath] [imath]{\exists βx β{\in} βS, β\backsim P(x) β β\wedge β β\backsim Q(x)\Leftrightarrow \exists βx β{\in} βS, β\Bigl( β\backsim P(x) β β\wedge β β\backsim Q(x)\Bigr)}[/imath] And then I have no I idea how to continue from here. Any hint will be appreciated. | 2579452 | Help understanding predicate logic question/proof
Can someone help me? I can't even understand what the question is asking for. Let [imath]S = \{x_1,x_2,x_3,\ldots,x_n\}[/imath]. Show that [imath]{\sim}[\forall x\in S,P(x)\lor Q(x)]\iff \exists x\in S,[{\sim}P(x)\land {\sim}Q(x)][/imath] |
1209074 | Continuity of a composition of continuous functions
Suppose [imath]f: \mathbb{R} \to \mathbb{R} [/imath] is continuous at [imath]x = 1 [/imath] and [imath]g: \mathbb{R} \to \mathbb{R} [/imath] continuous at [imath]y = f(1) [/imath]. Then [imath]g \circ f [/imath] is continuous at [imath]x = 1 [/imath] Attempt: Let [imath]\epsilon > 0[/imath] be given. We can find [imath]\delta', \delta'' [/imath]such that if [imath]|x - 1| < \delta' [/imath], then [imath]|f(x) - f(1) | < \epsilon [/imath] and we also have that [imath]|f(z) - f(1) | < \delta'' [/imath] implies [imath]| g( f(z) ) - g( f(1)) | < \epsilon [/imath] Isn't this just the proof? It is obvious that feel confused trying to write the proof nicely. IS this a correct proof ? | 527855 | Real Analysis: Continuity of a Composition Function
Suppose [imath]f[/imath] and [imath]g[/imath] are functions such that [imath]g[/imath] is continuous at [imath]a[/imath], and [imath]f[/imath] is continuous at [imath]g(a)[/imath]. Show the composition [imath]f(g(x))[/imath] is continuous at [imath]a[/imath]. My idea: Can I go straight from definition and take [imath]\delta=\min\{\delta_1,\delta_2\}[/imath], where [imath]\delta_1[/imath] is used for the continuity of [imath]g[/imath] at [imath]a[/imath] and [imath]\delta_2[/imath] is used for f being continuous at [imath]g(a)[/imath]. In my proof I just treat [imath]g(a)[/imath] as a point when referring to the composition. So it goes like this: Proof: Given [imath]\epsilon>0[/imath], take [imath]\delta=\min\{\delta_1,\delta_2\}[/imath]. Then [imath]0<|x-g(a)|<\delta[/imath] which implies [imath]|f(g(x))-f(g(a))|<\epsilon[/imath]. |
2580796 | Why [imath]\|S\|=\sup_{\|x\| \leq 1}\|Sx\|?[/imath]
Let [imath]\mathcal{B}(F)[/imath] the algebra of all bounded linear operators on a complex Hilbert space [imath]F[/imath]. It is well known that if [imath]S\in \mathcal{B}(F)[/imath], then [imath]\|S\|:=\sup_{\substack{x\in F\\ x\not=0}}\frac{\|Sx\|}{\|x\|}[/imath] Why [imath]\|S\|=\sup_{\|x\| \leq 1}\|Sx\|?[/imath] | 2616601 | Why [imath]\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\geq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|?[/imath]
Let [imath]E[/imath] be a complex Hilbert space. For [imath]A\in\mathcal{L}(E)[/imath], I want to show that [imath]\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|=\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|.[/imath] Clearly, since [imath]\{\|Ax\|;\;x\in E,\,\|x\|=1\}\subset \{\|Ax\|;\;x\in E,\,\|x\|\leq1\},[/imath] then [imath]\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\leq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|.[/imath] Why [imath]\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\geq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|?[/imath] |
2581021 | Necessary Condition for linear independence of functions
I am trying to prove the following fact, which seems intuitive to me, yet I cannot find a way to show it. Consider the set of functions [imath]\{f_1,\ldots,f_n\}[/imath], [imath]f_i:\mathbb{R}\rightarrow\mathbb{R}[/imath]. Suppose that this set is linearly independent, i.e. [imath]\sum_i f_ic_i=0\implies c_i=0\;\;\; \forall i[/imath] Then there exists numbers [imath](x_i)_{i=1}^n[/imath] such that the matrix \begin{bmatrix} f_1(x_{1}) & f_2(x_{1}) & \dots & \dots & f_n(x_{1}) \\ f_1(x_{2}) & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots \\ f_1(x_{n}) & \dots & \dots & \dots & f_n(x_{n}) \end{bmatrix} has a nonzero determinat. I have tried to reason by contraposition, assuming that the determinant vanishes for any choice of the [imath](x_i)_{i=1}^n[/imath] but I didn't get anywhere. | 2148972 | Matrix rank and linear independence of functions
The following proposition is easy to verify: Given [imath]n[/imath] functions [imath]f_1,...,f_n[/imath], they are linearly independent in an interval [imath]I[/imath] if there exists a set of [imath]n[/imath] points in [imath]I[/imath], namely [imath]x_1,...,x_n[/imath], such that the matrix \begin{pmatrix} f_1(x_1) & \cdots & f_n(x_1) \\ \vdots & \ddots & \vdots \\ f_1(x_n) & \cdots & f_n(x_n) \end{pmatrix} has full rank. My question is: can I replace the if by if and only if? That is, if [imath]n[/imath] functions are linearly independent, is it true that there exists a set of [imath]n[/imath] points such that the above matrix has full rank? I am pretty convinced so, but I am not being able to prove it. |
2575104 | [imath]\int^b_af(x)\,x^k \, dx=\int^b_ag(x)\,x^k \, dx[/imath] , for all [imath]k \in \mathbb{N}[/imath]
If [imath]f, g\in \mathcal{C}([a,b],\mathbb{R})[/imath] and [imath]\int^b_af(x)\,x^k \, dx=\int^b_ag(x)\,x^k \, dx, \text{ for all } k \in \mathbb{N}[/imath] Prove that [imath]f=g[/imath]. I'm trying to prove this but I don't know how to proceed. Any suggestions? | 1692322 | Show that [imath]\int_{a}^{b}{x^{n}f(x)dx}=0[/imath], then [imath]f=0[/imath]
Let [imath]f:([a,b],\vert\vert)\to (\mathbb{R},\vert\vert)[/imath] a continuous function. Show that, if [imath]\int_{a}^{b}{f(x)x^{n}dx}=0[/imath] for all [imath]n\in\mathbb{N},n\geq 0[/imath], then f is identically zero. My attempt: Let [imath]A=\{p\in C[0,1]\vert p(x)=a_{0}+a_{1}x+...+a_{n}x^{n}\}[/imath], then [imath]A[/imath] is a algebra of constant functions and separates points. By Stone-Weierstrass theorem, given [imath]\epsilon>0[/imath], exist [imath]p\in A[/imath] such that [imath]\lVert f-p\rVert<\epsilon[/imath] and as [imath][0,1][/imath] is compact and [imath]f[/imath] is continuous, exist [imath]M>0[/imath] such that [imath]\vert f(x)\vert\leq M[/imath]. So, [imath]\vert\int_{a}^{b}{f^{2}(x)dx}\vert=\vert\int_{a}^{b}{f(x)(f(x)-p(x))dx}+\int_{a}^{b}{f(x)p(x)dx}\vert[/imath] [imath]\leq \int_{a}^{b}{\vert f(x)\vert \vert f(x)-p(x)\vert dx}\leq M\lVert f-p\rVert<\epsilon M[/imath] [imath]\therefore\int_{a}^{b}{f^{2}(x)dx}=0[/imath], then [imath]f\equiv 0[/imath]. So, for what reason if [imath]\int_{a}^{b}{f^{2}(x)dx}=0[/imath], this implies my problem? Thanks |
2130674 | What is the Hahn Banach extension of this map?
[imath]f(x,y)=3x[/imath] on [imath]X=[/imath]{[imath](x,y): x=y[/imath]} is a linear functional on the subspace [imath]X\subset \Bbb R^2[/imath]. Here the norm on [imath]\Bbb R^2 [/imath] is defined as [imath]\|(x,y)\|= |x|+|y|[/imath]. Let [imath]g(x,y)=ax+by[/imath] on [imath]\Bbb R^2[/imath], where [imath]g[/imath] is a Hahn Banach extension of [imath]f[/imath]. Then what must [imath]a-b[/imath] be? I know that if [imath]g[/imath] is restricted to [imath]X[/imath], then [imath]g[/imath] must be equal to [imath]f[/imath] and this implies [imath]a+b=3[/imath]. But how do I proceed further? | 2164986 | Find the value of [imath]\alpha-\beta[/imath] if [imath]g(x,y)=\alpha x+\beta y[/imath] is a Hahn-Banach extension of [imath]f[/imath]
Consider the normed linear space [imath]\Bbb R^2[/imath] with this norm [imath]\|(x,y)\|=|x|+|y|[/imath]. Define [imath]f[/imath] to be a bounded linear functional on [imath]X=\{(x,y):x=y\}[/imath] by [imath]f(x,y)=3x[/imath]. If [imath]g[/imath] is a Hahn-Banach extension of [imath]f[/imath] on [imath]\Bbb R^2[/imath] given by [imath]g(x,y)=\alpha x+\beta y[/imath] then find [imath]\alpha-\beta[/imath]. My try: Since [imath]g[/imath] is an extension of [imath]f[/imath] so [imath]g|_X=f\implies g(x,x)=(\alpha+\beta)x=3x\implies \alpha+\beta =3[/imath] Now [imath]\|f\|=\sup \{f(x,x):\|(x,x)\|=1,(x,x)\in \Bbb R\}=3\implies \|g\|=3[/imath] So [imath]\sup \{g(x,y):\|(x,y)\|=1,(x,y)\in \Bbb R\}=3\implies \alpha x+\beta y\le 3[/imath] where [imath]|x|+|y|=1[/imath]. Take [imath]x=0,y=1\implies \alpha x+\beta y\le 3\implies \beta\le 3[/imath]. How to find the value of [imath]\alpha,\beta [/imath] from above. Please give some hints. |
2581511 | Problem understanding a step in Josephus Problem in Concrete Mathematics
On Page 11 of Second edition of Concrete Mathematics, it reads as The induction step has two parts, depending on whether [imath]l[/imath] is even or odd. If [imath]m > 0[/imath] and [imath]2^m + l= 2n[/imath], then [imath]l[/imath] is even and [imath]J(2^m +l) = 2J(2^{m-1} +l/2)-1 = 2(2l/2+1)-1 = 2l+1[/imath] by (1.8) Here I didn't get how [imath]2J(2^{m-1} +l/2)-1 [/imath] is reduced to [imath]2(2l/2+1)-1[/imath] Please Help. | 305763 | Concrete Mathematics - The Josephus Problem
I've been going through Concrete Mathematics and have a question on the inductive proof for the Josephus problem. Recurrent relation: [imath]J(1) = 1[/imath] [imath]J(2n) = 2J(n) - 1[/imath] [imath]J(2n+1) = 2J(n) + 1[/imath] Closed form hypothesis: [imath]J(2^m + l) = 2l + 1[/imath], for [imath]m \ge 0[/imath] and [imath]0\le l<2^m[/imath] Inductive Proof (for the even case): [imath]J(1)=1[/imath] [imath]J(2^m+l)= 2J(2^{m-1}+l/2)-1=2(2l/2+1)-1=2l+1[/imath] How do the authors get from [imath]2J(2^{m-1}+l/2)-1[/imath] to [imath]2(2l/2+1)-1[/imath]? |
2581066 | How to algebraically solve [imath]\frac{1}{x} < 5[/imath] inequality to obtain two solutions?
Given the inequality: [imath]\frac{1}{x} < 5[/imath] In order to find a solution, I would normally multiply both sides by [imath]x[/imath]: [imath]1 < 5x[/imath] Then I would divide by [imath]5[/imath] [imath]\frac{1}{5} < x[/imath] To obtain the solution: [imath]x > \frac{1}{5}[/imath]. Now, the thing is, the solutions are actually two: [imath]x > \frac{1}{5}[/imath] and [imath]x < 0[/imath] How am I supposed to reach this conclusion algebraically? It seems I'm not able to obtain the second solution ([imath]x < 0[/imath]). Thanks! | 1905422 | Why does [imath]\frac{1}{x} < 4[/imath] have two answers?
Solving [imath]\frac{1}{x} < 4[/imath] gives me [imath]x > \frac{1}{4}[/imath]. The book however states the answer is: [imath]x < 0[/imath] or [imath]x > \frac{1}{4}[/imath]. My questions are: Why does this inequality has two answers (preferably the intuition behind it)? When using Wolfram Alpha it gives me two answers, but when using [imath]1 < 4x[/imath] it only gives me one answer. Aren't the two forms equivalent? |
2580635 | continuity on compact subsets implies continuity
Consider this problem: Let [imath](X,d)[/imath],[imath](Y,\rho)[/imath] be two metric spaces (no further hypothesis). Prove a function [imath]f:(X,d)\rightarrow (Y,\rho)[/imath] is continuous if and only if for every compact [imath]K\subseteq X[/imath] the restriction of [imath]f[/imath] on [imath]K[/imath] is continuous. Now the => part is fairly simple since the restriction of a continuous function is also continuous. What about the <= part? If we choose a convergent sequence [imath]x_n\rightarrow x[/imath] we need only show [imath]f(x_n)\rightarrow f(x)[/imath], but how does compactness come into play? | 915096 | [imath]f[/imath] is continuous on [imath]X[/imath] iff [imath]f[/imath] is continuous on every compact subset
Let [imath](X,d)[/imath] be a metric space,then prove that [imath]f[/imath] is continuous on [imath]X[/imath] iff [imath]f[/imath] is continuous on every compact subset of [imath]X[/imath]. If [imath]f[/imath] is continuous on [imath]X[/imath] then its restriction on each compact subset will be continuous, and conversely if [imath]f[/imath] is continuous on each compact set then for [imath]x\in X[/imath] the set [imath]\{x\}[/imath] is compact and thus [imath]f[/imath] must be continuous at [imath]x[/imath] , as [imath]x[/imath] was arbitrary [imath]f[/imath] will be continuous on [imath]X[/imath]. |
1722883 | If [imath]dF_p[/imath] is nonsingular, then [imath]F(p)\in[/imath] Int[imath]N[/imath]
Here is the problem 4-2 in John Lee's introduction to smooth manifolds: Suppose [imath]M[/imath] is a smooth manifold (without boundary), [imath]N[/imath] is a smooth manifold with boundary, and [imath]F: M \to N[/imath] is smooth. Show that if [imath]p \in M[/imath] is a point such that [imath]dF_p[/imath] is non-singular, then [imath]F(p)\in[/imath] Int[imath]N[/imath]. To approach this problem, let's assume conversely that [imath]F(p)\in[/imath] [imath]\partial N[/imath], then there are two charts [imath](U,\varphi)[/imath] and [imath](V, \psi)[/imath](a boundary chart) centered at [imath]p[/imath] and [imath]F(p)[/imath] respectively. I can't see how the non-singularity of [imath]dF_p[/imath] contradicts to the fact that [imath]F(p)[/imath] is in the boundary chart. Maybe the proof by contradiction is not the right way? Thanks in advance! | 2276096 | Let [imath]M[/imath] be a smooth manifold and let [imath]N[/imath] be a manifold with boundary. If [imath]dF_{p}[/imath] is an isomorphism, then [imath]F\left(p\right)\in\mbox{int}M [/imath].
Let [imath]M[/imath] be a smooth manifold and let [imath]N[/imath] be a manifold with boundary, both with the same dimension [imath]n[/imath]. If [imath]dF_{p}[/imath] is an isomorphism, then [imath]F\left(p\right)\in\mbox{int}N [/imath]. I am trying to prove this theorem to prove a result about smooth embeddings. Here is how I am thinking the problem could be solved. Assume thet [imath]F(p)[/imath] is a boundary point of [imath]N[/imath]. Then there exists a chart [imath](V,\psi)[/imath] at [imath]F(p)[/imath] such that [imath]\psi(V) [/imath] is an open subset of the upper half space [imath]\mathbb{H}^{n}[/imath]. I guess we have to use some fact about [imath]M[/imath] having no boundary and the the fact that [imath]dF_p[/imath] is an isomorphism to show that there is a contradiction, but I cannot figure that out. |
1943830 | Approximating [imath]f(z)=\frac{1-e^{iz}}{z^2}[/imath]
I am reading Stein and Shakarchi's Complex Analysis, and I encountered a problem at page 45 while reading the proof of [imath]\int_{0}^{\infty}\frac{1-\cos{x}}{x^2}dx=\frac{\pi}{2}[/imath]. The proof says that [imath]f(z)=\frac{1-e^{iz}}{z^2}=\frac{-iz}{z^2}+E(z)[/imath], with [imath]E(z)[/imath] a bounded function as [imath]z[/imath] tend to [imath]0[/imath]. I would like to know how this is done. | 2268114 | [imath]\frac{1 - e^{iz}}{z^2} = \frac{-iz}{z^2} + E(z)[/imath] where [imath]E(z)[/imath] is bounded as [imath]z \rightarrow 0?[/imath]
Was reading some notes and it states that [imath]f(z) = \frac{1 - e^{iz}}{z^2}[/imath] can be written as [imath]f(x) = \frac{-iz}{z^2} + E(z)[/imath] where [imath]E(z)[/imath] is bounded as [imath]z \rightarrow 0.[/imath] I don't exactly see why. Help is appreciated. |
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