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2581435 | Is it possible to say that product of positive strictly quasiconcave functions is a quasiconcave function?
I have two functions [imath]f_1(x)[/imath] and [imath]f_2(x)[/imath] which are positive over a certain interval and are strictly quasiconcave. Can I say that their product [imath]f(x)=f_1(x)f_2(x)[/imath] is quasiconcave function? Any help in this regard will be much appreciated. Thanks in advance. | 2581375 | Is the product of strictly quasiconcave functions quasiconcave?
I have two functions [imath]f_1(x)[/imath] and [imath]f_2(x)[/imath] which are strictly quasiconcave (since they have positive derivative before a certain point and negative derivative after that point. The points where derivative become zero can be different for each of the functions). I want to ask whether [imath]f(x)=f_1(x)f_2(x)[/imath] will be a quasiconcave function or not? Any help in this regard will be highly appreciated. Thanks in advance. |
2581754 | Prove that [imath]e^x - \ln(x)=2^{2014}[/imath] has exactly two positive real roots
Prove that [imath]e^x - \ln(x)=2^{2014}[/imath] has exactly two positive real roots. I know how to do it by graph. I need equation based solution. I differentiated it twice and got it's strictly increasing on [imath]0[/imath]. How to proceed? | 628320 | Roots of a given equation
How can I show that the equation [imath]e^x-\ln(x)-2^{2014}=0[/imath] has exactly two positive roots? |
2582793 | prove [imath]f(\cup A) = \cup f(A)[/imath]
I know it is a simple theorem that [imath] f(\bigcup_{\gamma\in \Gamma} A_\gamma) = \bigcup_{\gamma \in \Gamma} f(A_\gamma)[/imath] I know how to prove it by the definition of the above things, but I want to prove it by logical proposition using [imath]\forall, \exists[/imath]. \begin{align} &y \in f(\bigcup_{\gamma\in \Gamma} A_\gamma) \\ &\Leftrightarrow \exists x \in \bigcup_{\gamma\in \Gamma} A_\gamma : y=f(x)\\ &\Leftrightarrow \exists x(x \in \bigcup_{\gamma\in \Gamma} A_\gamma \wedge y=f(x) )\\ &\Leftrightarrow \exists x\{(\exists \gamma \in \Gamma: x \in A_{\gamma})\wedge y=f(x)\} \end{align} and then I don't know how to deal with it. Help me to prove the theorem by using predicate logic. *edit. I want to prove it by predicate logic changing the word 'for some x, for all x, and, or' to '[imath]\exists, \forall, \wedge, \vee[/imath]' and using bracket. I wrote that I know how to prove it by the definition of things. | 2411554 | Image of the union is the union of the images
[imath] f\left(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}\right) = \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)[/imath] Let [imath]b \in f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda})[/imath] [imath]\rightarrow b=f(a)[/imath] for some [imath]a \in (\bigcup\limits_{\lambda \in \wedge} A_{\lambda})[/imath]. Since [imath]a \in (\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) \rightarrow a \in A_{\lambda}[/imath] for some [imath]\lambda \in \wedge[/imath]. Since [imath]b=f(a)[/imath], then [imath]b \in f(A_{\lambda})[/imath] for some [imath]\lambda \in \wedge[/imath]. [imath]\rightarrow b \in \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)[/imath] [imath]\rightarrow f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) \subset \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)[/imath]. I used a similar argument to prove [imath]\bigcup\limits_{\lambda \in \wedge} f(A_\lambda) \subset f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda})[/imath]. Which shows [imath] f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) = \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)[/imath]. Please let me know if this is valid. |
2582046 | False proof that [imath]ρe^{iθ} = ρ[/imath] and so complex numbers do not exist?
My professor showed the following false proof, which showed that complex numbers do not exist. We were told to find the point where an incorrect step was taken, but I could not find it. Here is the proof: (Complex numbers are of the form [imath]\rho e^{i\theta}[/imath], so the proof begins there) [imath]\large\rho e^{i\theta} = \rho e^{\frac{i\theta*2\pi}{2\pi}} = \rho (e^{2\pi i})^{\frac{\theta}{2\pi}} = \rho (1)^{\frac{\theta}{2\pi}} = \rho[/imath] [imath]Note: e^{i\pi} = -1, e^{2\pi i} = (-1)^2 = 1[/imath] Since we started with the general form of a complex number and simplified it to a real number (namely, [imath]\rho[/imath]), the proof can claim that only real numbers exist and complex numbers do not. My suspicion is that the error occurs in step [imath]4[/imath] to [imath]5[/imath] , but I am not sure if that really is the case. | 2601389 | Exponentiation rules
we know these properties of exponentiating [imath]x^bx^a=x^{a+b}[/imath] [imath](x^a)^b=x^{ab}[/imath] Now, I've been introduced to complex numbers and this amazing formula: [imath]e^{ix}=\cos{x}+i\sin{x}[/imath] Take [imath]x=2k\pi,k\in Z[/imath] to get [imath]e^{2k\pi i}=1=e^0[/imath] which would then imply that [imath]\forall k\in Z:2k\pi i=0[/imath] which kinda doesn't make sense to me. Another thing is... Take a fourier transform of the function given by (taken from https://en.wikipedia.org/wiki/Fourier_transform) [imath]\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}dx[/imath] All this stuff involving [imath]e^{2k\pi i}[/imath] are kinda confusing to me. One could say the following: [imath]e^{-2\pi ix\xi}=(e^{-2\pi i})^{x\xi}=1^{x\xi}=1[/imath] which kinda doesn't make sense, does it? Why would one include the term [imath]e^{-2\pi ix\xi}[/imath] in a formula then. Could someone explain me, what is actually going on and when these basic properties of exponentiation are valid? |
2583166 | Does [imath]f(\mathbb{Q}^n) \subseteq \mathbb{Q}[/imath] hold only for rational functions?
Let [imath]f[/imath] be a [imath]C^{\infty}[/imath]-function from an open subset [imath]U[/imath] of [imath]\mathbb{R}^n[/imath] to [imath]\mathbb{R}[/imath]. If [imath]f(\mathbb{Q}^n\cap U) \subseteq \mathbb{Q}[/imath] can we conclude that [imath]f[/imath] is a rational function? | 167620 | Functions that take rationals to rationals
What is known about [imath]\mathcal C^\infty[/imath] functions [imath]\mathbb R\to\mathbb R[/imath] that always take rationals to rationals? Are they all quotients of polynomials? If not, are there any that are bounded yet don't tend to a limit for [imath]x\to +\infty[/imath]? If there are, then can we also require them to be analytic? (This is basically just a random musing after I found myself using trigonometric functions twice in unrelated throwaway counterexamples. It struck me that it was kind of conceptual overkill to whip out a transcendental function for the purpose I needed: basically that it had to keep wiggling forever. I couldn't think of a nice non-trancendental function to do this job, however, and now wonder whether that is because they can't exist). |
1328391 | Reading a basis for [imath]\operatorname{Null}(A)[/imath] and [imath]\operatorname{Null}(A^T)[/imath] from the reduced row-echelon form of [imath]A[/imath]
I know that it is possible to read the basis for [imath]\operatorname{Null}(A)[/imath] and [imath]\operatorname{Null}(A^T)[/imath] by simply looking at the reduced row-echelon form (RREF) of the matrix [imath]A[/imath]. I have only an approximate idea of how to do it. For example, if we [imath]A = \begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 &1 &2 \\ 0 &0 &0 &0 \end{bmatrix},[/imath] then one of the basis vectors for [imath]\operatorname{Null}(A)[/imath] will be [imath]\begin{bmatrix}-2 & 1 & 0 &0\end{bmatrix}^T[/imath], and the other one will be [imath]\begin{bmatrix}1 & 0 & -2 &1 \end{bmatrix}^T[/imath]. So, basically we're "going around" over eight digits and alternating signs. But I noticed that this method isn't perfect. Does someone have a better idea of how to read bases for these two spaces from the RREF of [imath]A[/imath]? | 1521264 | Kernels and reduced row echelon form - explanation
The following text is written in my textbook and I don't really understand it: If [imath]A = (a_{ij}) \in[/imath] Mat[imath](m x N, F)[/imath] is a matrix in reduced row echelon form with [imath]r[/imath] nonzero rows and pivots in the columns numbered [imath]j_1 < ... < j_r[/imath], then the kernel ker[imath](A)[/imath] is generated by the [imath]n-r[/imath] elements [imath] w_k = e_k - \sum\limits_{1 \le i \le r, j_i \le k} a_{ik}e_{j_i}[/imath] for [imath]k \in {1, \cdots , n} \ {j_1, \cdots, j_r}[/imath], where [imath]e_1, \cdots, e_n[/imath] are the standard generators of [imath]F^n[/imath]. There are no computational examples and the notation is a bit overwhelming so I don't know how I can use it. Can somebody give an example of this in practice? |
2583348 | Deducing Galois group of [imath]t^6 + 3[/imath] over the rationals
I am asked to deduce, in the following order, 3 facts about [imath]G = Gal(f(t) = t^6+3)[/imath] over the rationals: [imath]G[/imath] is order 6 The elements have orders [imath]1,2,3[/imath] [imath]G \cong S_3[/imath] I have figured these out but in the wrong order so I wanted to write down a different train of thought that I am less sure of to see if there is anything wrong with it. By Eisenstein [imath]f(t)[/imath] is irreducible, and is thus the minimal polynomial of a root [imath]\beta[/imath] of [imath]f[/imath] over [imath]\mathbb Q[/imath], so if [imath]L[/imath] is the splitting field of [imath]f[/imath] over [imath]\mathbb Q[/imath] then as the degree of [imath]f[/imath] is [imath]6[/imath], then the degree of the extension [imath]\mathbb Q\leq L[/imath] is also [imath]6[/imath]. Letting [imath]\xi[/imath] be a primitive [imath]6[/imath]-th root of unity, then we see that over [imath]\mathbb Q(\xi) = \mathbb Q(i\sqrt 3), \; f[/imath] factorises as [imath]f(t) = (t^3 + i\sqrt 3)(t^3 - i\sqrt 3)[/imath] . This means that all permutations [imath]\sigma \in Gal(L/\mathbb Q(\xi))[/imath] permute the roots of each cubic factor amongst themselves. i.e. [imath]\sigma[/imath] is a [imath]3[/imath]-cycle, and there are three of them. Also we note that complex conjugation preserves [imath]\mathbb Q[/imath] in [imath]L[/imath], so [imath]G[/imath] contains an order [imath]2[/imath] permutation. The identity is order [imath]1[/imath]. I am not quite sure how to conclude that there is no element of order [imath]6[/imath], but with that we have that the elements are of orders [imath]1,2,3[/imath] Finally, we notice that composing complex conjugation with one of the [imath]3[/imath]-cycles gives us the inverse of the original [imath]3[/imath]-cycle. This gives a dihedral relation and hence [imath]G[/imath] is the dihedral group of order [imath]6[/imath], i.e. [imath]G \cong S_3[/imath] I would like to clarify why it is exactly that we can conclude that there is no element of order [imath]6[/imath] in [imath]G[/imath]. I can feel that it is a very simple thing that I have overlooked, but I can't figure out what it is and would appreciate a little help or a hint. Additionally I would really appreciate it if anyone could let me know if there's anything flawed with my logic here at all. Thank you. | 156003 | Galois group of [imath]x^6 + 3[/imath] isomorphic to a copy of [imath]S_3[/imath] inside [imath]S_6[/imath]
I have seen the the thread here related to the computation of the Galois group of the same polynomial. However, my question is not about the computation itself but about the group presentation of the Galois group. I will explain. I have determined that the polynomial [imath]x^6 +3 \in \Bbb{Q}[x][/imath] has Galois group of order 6. The splitting field is [imath]\Bbb{Q}(a)[/imath], where [imath]a[/imath] is a root of [imath]x^6 + 3[/imath]. One can take [imath]a = \sqrt[6]{3}\zeta[/imath] where [imath]\zeta = e^{2\pi i/6} = e^{\pi i/6}[/imath]. Now I have determined the rest of the roots to be: [imath]\begin{array}{ccccc} \alpha_1 &=& a &&& \alpha_4 = -\alpha_1 \\ \alpha_2 &=& \frac{a^4 + a}{2}= \zeta a &&& \alpha_5 = -\alpha_2 \\ \alpha_3 &=& \frac{a^4 - a}{2} = \zeta^2 a &&& \alpha_6 = - \alpha_3 \end{array}. [/imath] I have also computed some automorphisms of the Galois group, for example the automorphism [imath]\tau : \alpha_1 \mapsto \alpha_4[/imath] that has order 2, [imath]\sigma : \alpha_1 \mapsto \alpha_2[/imath] that has order 2 and [imath]\rho : \alpha_1 \mapsto \alpha_3[/imath] that has order 3. The presence of two automorphisms of order 2 tells me that the Galois group is isomorphic to [imath]S_3[/imath] However the problem now is if I want to identify my [imath]\tau,\sigma[/imath] and [imath]\rho[/imath] as cycles in [imath]S_6[/imath], I get the cycles [imath](14)[/imath], [imath](12)[/imath] and [imath](132)[/imath]. I don't think these cycles lie in the copy of [imath]S_3[/imath] inside of [imath]S_6[/imath]; what am I misunderstanding here? Thanks. Edit: I made a mistake in the calculations. We actually need [imath]\alpha_1 = a = \sqrt[3]{3}\zeta[/imath] where [imath]\zeta = e^{\pi i/6}[/imath]. I did not take a primitive 6-th root of unity earlier. Now if I write [imath]a = \sqrt[6]{3}e^{\pi i/6}[/imath], then [imath]a^3 = \sqrt{3}i[/imath] and so [imath]\frac{1 + a^3}{2} = \frac{1 + \sqrt{3}i}{2} = \zeta^2[/imath]. So indeed with the redefined [imath]\zeta[/imath] and [imath]a[/imath], the equations now are [imath]\begin{array}{ccccc} \alpha_1 &=& a &&& \alpha_4 = -\alpha_1 \\ \alpha_2 &=& \frac{a^4 + a}{2}= \zeta^2 a &&& \alpha_5 = -\alpha_2 \\ \alpha_3 &=& \frac{a^4 - a}{2} = \zeta^4 a &&& \alpha_6 = - \alpha_3. \end{array} [/imath] After all this mess, I have got the automorphisms [imath]\sigma = (12)(45)(36)[/imath] and [imath]\gamma = (135)(246)[/imath]. We check that [imath]\sigma\gamma = \gamma^2\sigma[/imath]. [imath]\sigma\gamma = (12)(36)(45)(135)(246) = (16)(25)(34)[/imath]. [imath]\gamma^2\sigma = (153)(264)(12)(36)(45) = (16)(25)(34)[/imath] so indeed [imath]\sigma\gamma = \gamma^2\sigma[/imath]. Hence the Galois group has elements [imath]\{1,\gamma,\gamma^2,\sigma,\sigma\gamma, \sigma\gamma^2\} = \{1, (135)(246),(153)(264),(12)(45)(36),(14)(23)(56),(16)(34)(25)\}.[/imath] |
2584021 | Suppose a group G is generated by two elements a and b. If ab=ba, prove that G is abelian.
This question is from Charles Pinter's A book of Abstract Algebra. Pinter defines [imath]G[/imath] to be the set which contains all the possible products of [imath]a,b[/imath] and their inverses, in any order, with repetition of factors permited. He calls this [imath]G[/imath] as the group generated by [imath]a,b[/imath]. Since [imath]ab=ba[/imath], then following things are true: [imath]a^{-1}b=ba^{-1}[/imath], [imath]a^{-1}b^{-1}=b^{-1}a^{-1}[/imath] and [imath]b^{-1}a=ab^{-1}[/imath]. [imath](*)[/imath] Let [imath]x_1,x_2 \in G[/imath]. Then [imath]x_1,x_2[/imath] are any two possible products of [imath]a,b,a^{-1}[/imath] and [imath]b^{-1}[/imath] (as defined above). The product [imath]x_1 x_2[/imath] is also any possible product. We can rearrange the [imath]a,b,a^{-1},b^{-1}[/imath] in [imath]x_1 x_2[/imath] using the equations [imath](*)[/imath] in way such that it equals [imath]x_2x_1[/imath]. Hence G is abelian. Does the above even qualify as proof? What are some alternative ways to prove this? | 1184981 | if G is generated by {a,b} and ab=ba, then prove G is Abelian
if G is generated by {a,b} and ab=ba, then prove G is Abelian All elements of G will be of the form [imath]a^kb^j,\,\, k,j \in \mathbb{Z}^+[/imath] so I need to get [imath]a^kb^j = b^ja^k[/imath] |
2584075 | 6=-6? Is it possible?
Recently my friend showed me this proof. I know this is wrong but I don't know accurate reasons. Can anyone help? His proof: [imath]6 = \sqrt{\frac{-9}{-4}} = \sqrt{-9} \times \sqrt{-4} = 3i × 2i = 6×(i^2) = -6[/imath] | 2047349 | When does [imath]\sqrt{a b} = \sqrt{a} \sqrt{b}[/imath]?
So, my friend show me prove that [imath]1=-1[/imath] by using this way: [imath]1=\sqrt{1}=\sqrt{(-1)\times(-1)}=\sqrt{-1}\times\sqrt{-1}=i\times i=i^2=-1[/imath] At first sight, I stated "No, [imath]\sqrt{ab}=\sqrt{a}\times\sqrt{b}[/imath] is valid only for [imath]a,b\in\mathbb{R}[/imath] and [imath]a,b\geq0[/imath]" But, I remember that [imath]\sqrt{-4}=\sqrt{4}\times\sqrt{-1}=2i[/imath] which is true (I guess). Was my statement true? But, [imath]\sqrt{ab}=\sqrt{a}\times\sqrt{b}[/imath] is also valid if one of a or b is negative real number. Why is it not valid for a dan b both negative? If my statement was wrong, what is wrong with that prove? |
2584370 | Prove [imath]\lim_{x \to \infty} (1+1/x)^x = e[/imath] using sequences
I'm aware of the proof of [imath]\lim_{n \to \infty }(1+\frac 1n)^n = e[/imath] using monotone convergence theorem but I want to prove [imath]\lim_{x \to \infty} (1+\frac 1x)^x = e[/imath] using sequential criterion for limits . How I can write it rigorously ? More precisely : For every [imath]a_n \to \infty[/imath] s.t [imath]a_n \in D_f[/imath] prove [imath](1+\frac 1{a_n})^{a_n} \to e[/imath] [imath]\ \ \ \ \ \ \ \ \ \ (\star)[/imath] | 2554111 | Prove that [imath]\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{a_{n}} \right)^{a_{n}}=e[/imath] if [imath]\lim\limits_{n\rightarrow\infty} a_{n}=\infty[/imath]
What would be the nicest proof of the following theorem: If [imath]\lim\limits_{n \rightarrow \infty} a_{n} = \infty[/imath], then [imath]\lim\limits_{n \rightarrow \infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = e[/imath]. If [imath]\lim\limits_{n \rightarrow \infty}b_{n} = 0[/imath], then [imath]\lim\limits_{n \rightarrow \infty} \left(1 + b_{n} \right) ^ {\frac {1} {b_{n}} } = e[/imath]. I somehow failed to find a proof here on the website and in the literature. |
2584500 | The origin of [imath]\int[/imath]
I am recalling the first day of integration class. The teacher taught us some history about how the particular symbols came into play. Let [imath]y = f(x)[/imath]. Here is how he broke up [imath]\int_a^b y[/imath] [imath]dx[/imath]: Draw two lines [imath]x = a[/imath] and [imath]x = b[/imath] and the polynomial [imath]y = f(x)[/imath]. The area between the three equations and [imath]x[/imath]-axis is the area that can be found using [imath]\int_a^b y[/imath] [imath]dx[/imath]. Now split up the lines into rectangular strips of width [imath]dx[/imath]. The integral is actually now the sum of the products of the lengths and [imath]dx[/imath]. So it can be shown as [imath]\Sigma_{x=a} ^b y[/imath] [imath]dx[/imath]. And he said that the integral sign is found by stretching (?) [imath]\Sigma[/imath] to [imath]\int[/imath]. It did help me understand integration but the previous sentence went over my head. Is it true or was it a harmless prank by my teacher? Thanks in advance! | 299127 | What is this symbol [imath]\int[/imath] called?
Title says it all (I think). I'm betting it's something to do with Standard Integrals though. What is this symbol "[imath]\int[/imath]" called? |
2584550 | show that [imath]f''(a)=\lim _{h\to 0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}[/imath]
Let [imath]I\subset \mathbb{R}[/imath] be open interval ,let [imath]f:I\to \mathbb{R}[/imath] be differentiable on [imath]I,[/imath] suppose [imath]f''(a)[/imath] exist at [imath]a \in I[/imath]. show that [imath]f''(a)=\lim _{h\to 0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}[/imath] how can we approach this .i was thinking of MVT. applying MVT on [imath][a,a+h][/imath], we have [imath]f(a+h)-f(a)=hf'(\alpha_1)[/imath], where [imath]\alpha_1\in(a,a+h)[/imath] similarly [imath]f(a)-f(a-h)=hf'(\alpha_2)[/imath] where [imath]\alpha_2\in(a-h,a)[/imath] subtracting these we get [imath]\frac{f(a+h)+f(a-h)-2f(a)}{h^2}=\frac{(f'(\alpha_1)-f'(\alpha_2))}{h}[/imath] now taking limit [imath]h \to0[/imath] both side, can we somehow connect right hand side to [imath]f''(a)[/imath] any hint?? or some other method .i have seen this question already on stackexchange ,but most of them are using taylor expension up to 2nd order, but we are given function is twice differentiable at [imath]x=a[/imath],we donot know about twice differentiability at other points,so how can we use taylor expansion in this case??? | 235885 | Suppose that [imath]f''(0)[/imath] exists and is finite, show that [imath]f''(0)=\lim\limits_{h \to 0}\frac{f(h)-2f(0)+f(-h)}{h^{2}}[/imath]
I read the following from Chung's Probability: ([imath]f[/imath] is the characteristic function of some distribution function [imath]F[/imath]) Suppose that [imath]f''(0)[/imath] exists and is finite, then we have [imath]f''(0)=\lim\limits_{h \to 0}\frac{f(h)-2f(0)+f(-h)}{h^{2}}[/imath] Could anyone help on how to show this? The problem is that we only assume [imath]f[/imath] is twice differentiable at the point 0. So we can't prove the above simply by Taylor's expansion. Thank you very much! |
2584596 | T0 find the value of [imath]\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^{2}+kn}}[/imath]
What is [imath]\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^{2}+kn}}[/imath].I try to check whether series can be written as telescoping series but it doesn't work. I am getting no idea how to start? | 266322 | How to find limit of the sequence [imath]\sum\limits_{k=1}^n\frac{1}{\sqrt {n^2 +kn}}[/imath]?
How can I find the limit [imath]\lim_{n\to\infty}\displaystyle\sum_{k=1}^n\frac{1}{\sqrt {n^2 +kn}} \quad?[/imath] I have tried to solve it using squeeze theorem: [imath]\sum_{k=1}^n\frac{1}{\sqrt {n^2 +kn}} > \displaystyle\sum_{k=1}^n\frac{1}{\sqrt {n^2 +n^2}} = \displaystyle\sum_{k=1}^n\frac{1}{\sqrt {2n^2}}=\frac{1}{\sqrt {2}} [/imath] and [imath]\sum_{k=1}^n\frac{1}{\sqrt {n^2 +kn}} <\displaystyle\sum_{k=1}^n\frac{1}{\sqrt {n^2}} = 1.[/imath] But I could not find the sequences with the same limits. Please help - how to solve this? |
2584992 | Find [imath]\int^{2}_{0} (x^2+1) d{\lfloor x \rfloor}[/imath]
Find [imath]\int^{2}_{0} (x^2+1) d{\lfloor x \rfloor}[/imath] where [imath]\lfloor x \rfloor[/imath] denotes the greatest integer smaller than or equal to x. I found this question in a book. After reading this question, I am now confused about what an integral is. Please try to clarify what this particular integral means and then solve it | 2580938 | Evalute the definite integral: [imath]\int_{0}^{3} (x^2+1) d[x][/imath]
Evalute the definite integral: [imath]\int_{0}^{3} (x^2+1) d[x][/imath] [imath][x] -[/imath] is integer part of the [imath]x[/imath]. Is the solution correct? |
2585615 | Let [imath]f:[0,\infty) \to[0, \infty)[/imath] be a continuous function such that [imath]\int_0^{\infty} f(t) dt <\infty[/imath] then find the correct option
Let [imath]f:[0,\infty) \to[0, \infty)[/imath] be a continuous function such that [imath]\int_0^{\infty} f(t) dt <\infty[/imath] then which of following are true (1) the sequence [imath]\{f(n)\}[/imath] is bounded (2) [imath]f(n) \to0[/imath] as [imath]n \to \infty[/imath] (3) the series [imath]\sum f(n)[/imath] is convergent i think option 1 and 2 is true, and option 3 is false.but not able to prove 1 and 2 and disprove 3. any hint please | 1735489 | If [imath]f[/imath] is continuous with [imath] \int_0^{\infty}f(t)\,dt<\infty[/imath] then which are correct?
Let [imath]f:[0,\infty)\to [0,\infty)[/imath] be a continuous function such that [imath]\displaystyle \int_0^{\infty}f(t)\,dt<\infty[/imath]. Which of the following statements are true ? (A) The sequence [imath]\{f(n)\}[/imath] is bounded. (B) [imath]f(n)\to 0[/imath] as [imath]n \to \infty[/imath]. (C) The series [imath]\displaystyle \sum f(n)[/imath] is convergent. I am unable to prove directly but I am thinking about the function [imath]f(x)=\frac{1}{1+x^2}[/imath]. For this function all options are correct. Is it correct ? I think not , as I have no proof in general. Please help by giving a proof or disprove the statements. |
2585780 | Show that the sequence [imath]\sqrt{2}, \ \sqrt{2+\sqrt{2}}, \ \sqrt{2+\sqrt{2+\sqrt{2+}}}...[/imath] converges and find its limit.
Setting [imath]a_n=\sqrt{2+\sqrt{2+\sqrt{2+}}}\ [/imath] I get that [imath]a_{n+1}=\sqrt{2+a_n} \quad \quad a_1=\sqrt{2}.[/imath] Clearly all numbers in the sequence are positive and we see that [imath]a_n<a_{n+1} \ \forall \ n[/imath] which implies that the sequence is strictly increasing. To show that it's convergent, I need to show that it's bounded above. We can use the help-function [imath]f(x)=\sqrt{2+x}[/imath] such that [imath]a_{n+1}=f(a_n).[/imath] But since [imath]f'(x)=\frac{1}{2\sqrt{2+x}}=0\Leftrightarrow\text{no real solutions,}[/imath] [imath]f(x)[/imath] never flattens out or decreases, so it can't be convergent? | 3013689 | Question on sequences and induction
I'm trying to show that [imath]a_n[/imath] is increasing where [imath]a_1=1[/imath] and [imath]a_{n+1}[/imath] = [imath] \sqrt{a_n+2}[/imath] I proceeded by induction so showed that [imath]a_{n+1}>a_n[/imath] for all n greater than 1 So I showed it was true for n=1 Then assumed [imath]a_{k+1}>a_k[/imath] for some k greater than 1 Then I add two to both sides and square root to show it's true for k+1. So [imath]a_{k+1}+2>a_k+2[/imath] [imath] \sqrt{a_{k+1}+2}> \sqrt{a_k+2}[/imath] Which is [imath]a_{k+2}>a_{k+1}[/imath] So is this ok to show this statement? Also how would you show that a_n has a limit and compute [imath]lim_na_n[/imath] I think to compute the limit you need to prove that if [imath]b_n[/imath] goes to b then [imath] \sqrt{b_n+2}[/imath] goes to [imath] \sqrt{b+2}[/imath] thank you for your help. |
2585857 | Problem on commutator subgroup
Let [imath]G'[/imath] be the commutator subgroup of [imath]G[/imath]. Prove that if [imath]H[/imath] is a subgroup of [imath]G[/imath] and [imath]G'\subset H[/imath], then [imath]H[/imath] is normal in [imath]G[/imath]. I do not know how to prove it. I have tried different ways but all of them were unsuccessful. However, I know that [imath]G'[/imath] is normal in [imath]G[/imath] and quotient group [imath]G/G'[/imath] is abelian. Can anyone demonstrate how to solve that problem. | 1862869 | H is a subgroup of G and G' is a subgroup of H. Prove H is normal in G.
Question: Let G be a group and let G' be the subgroup of G generated by the set [imath]S=\left \{ x^{-1}y^{-1}xy \mid x,y \in G \right \}[/imath] [imath]\space[/imath] Prove that G' is normal in G. Solved [imath]\space[/imath] Prove that G/G' is abelian. Solved [imath]\space[/imath] Prove that if H is a subgroup of G and G' is a subgroup of H, then H is a normal subgroup of G. I cannot even begin to start on this question. Any hint is appreciated. |
2583415 | Cech Cohomology Hartshorne III 4.4
I´m stuck with the section [imath]c)[/imath] of the following Hartshorne´s exercise I have just proved [imath]a)[/imath] and [imath]b)[/imath]. I wonder if you can help me. Thank you for your time | 2584683 | Cech cohomology (Harthshorne)
I´m studying Cech cohomology through Hartshorne´s algebraic geometry and EGA, and I´m stuck in two exercises, I wonder if you can help me. The first one is the exercise III.4.4, we want to prove that [imath]\varinjlim_{\mathfrak{U}}\check{H}^{1}(\mathfrak{U},\mathcal{F})=H^{1}(X,\mathcal{F})[/imath] where [imath]\mathfrak{U}[/imath] is a covering of [imath]X[/imath]. (I attach a photo of the exercise) Exercise III. 4.4 I have just proved [imath]a)[/imath] and [imath]b)[/imath]. I´m also stuck with the last exercise of this section: Let [imath]X[/imath] be a topological spaces, [imath]\mathcal{F}[/imath] a sheaf of abelian groups, and [imath]\mathfrak{U}=(U_{i})[/imath] an open cover. Assume for any finite intersection [imath]V=U_{i_{0}}\cap\cdots\cap U_{i_{p}}[/imath] of open sets of the covering, and for any [imath]k>0[/imath], that [imath]H^{k}(V,\mathcal{F}_{\mid V})=0[/imath]. Then prove that for all [imath]p\geq 0[/imath], the natural maps [imath]\check{H}^{p}(\mathfrak{U},\mathcal{F})\rightarrow H^{p}(X,\mathcal{F})[/imath] of the last exercise (III. 4.4) are isomorphisms. Thank you for your time. |
2585704 | How to answer this limit question
Evaluate the limit as n tends to infinity of: [imath](\sqrt[n]{n}-1)^n[/imath]. Heres what I tried: [imath](n^\frac{1}{n}(1-{n^{-\frac{1}{n}}})^n = n(1-{n^{-\frac{1}{n}}})^n[/imath] Where do I go from here? | 817999 | Determining [imath]\lim_{n\to\infty}\left(n^{\tfrac{1}{n}}-1\right)^n[/imath] with only elementary math
I am trying to find this limit: [imath]\lim_{n\to\infty}\left(n^{\tfrac{1}{n}}-1\right)^n,[/imath] I tried using exponential function, but I see no way at the moment. I am not allowed to use any kind of differentiation or other topics of advanced math, only induction and school math are possible. Thank you |
2582885 | Finding the number of arrangements of [imath]n[/imath] binary elements ([imath]a[/imath], [imath]b[/imath]) with a maximum of [imath]k[/imath] repeating consecutive elements of one type
Here's an example For [imath]n = 4[/imath] and [imath]k = 1[/imath], there are 8 possible arrangements: [imath]a\ a\ a\ a[/imath] [imath]a\ a\ a\ b[/imath] [imath]a\ a\ b\ a[/imath] [imath]a\ b\ a\ a[/imath] [imath]b\ a\ a\ a[/imath] [imath]b\ a\ b\ a[/imath] [imath]b\ a\ a\ b[/imath] [imath]b\ a\ b\ a[/imath] with [imath]b[/imath] being the element that can't have more than [imath]k[/imath] consecutive appearances. I'm not really well-versed in combinatorics, so [imath]2^n[/imath] being the total number of arrangements is the only piece of info I came up with. Anyway, this is actually a programming challenge from one of those competitive programming sites, but I wanna get a grasp of the math behind the idea before thinking about implementation methods, so I figured I'd ask things here. Also, the input range for both [imath]n[/imath] and [imath]k[/imath] is [imath][1, 10^6][/imath] so a dynamic programming approach is the way to go. | 2045496 | Number of occurrences of k consecutive 1's in a binary string of length n (containing only 1's and 0's)
Say a sequence [imath]\{X_1, X_2,\ldots ,X_n\}[/imath] is given, where [imath]X_p[/imath] is either one or zero ([imath]0 < p < n[/imath]). How can I determine the number of strings, which do contain at least one occurrence of consequent [imath]1[/imath]'s of length [imath]k[/imath] ([imath]0 < k < n[/imath]). For example, a string [imath]\{1, 0, 1, 1, 1, 0\}[/imath] is such a string for [imath]n = 6[/imath] and [imath]k = 3[/imath]. Here I have found an answer for arbitrary [imath]n[/imath] and [imath]k = 2[/imath], ([imath]k = 1[/imath] is trivial), but I need a more general answer for any natural number [imath]k[/imath] smaller than [imath]n[/imath]. |
2586345 | Explain [imath]\sum_{n=0}^\infty 2^n= 1+2+4+8+16+\cdots =-1[/imath]
So I was playing around with, [imath]\sum_{n=0}^\infty 2^n= 1+2+4+8+16+... [/imath] Let [imath]S[/imath] be the sum [imath]S= 1+2+4+8+16+...[/imath] Dividing both sides by [imath]2[/imath] [imath]\frac{S}{2}=\frac{1+2+4+8+16+...}{2}[/imath] Which means [imath]\frac{S}{2}=\frac{1}{2}+1+2+4+8+...[/imath] Or [imath]\frac{S}{2}=\frac{1}{2} + S[/imath] Therefore after solving for [imath]S[/imath] we get, [imath]S=-1[/imath] I believe I'm doing something terribly wrong here but I am convinced that the math checks out. Please help. I am in [imath]10[/imath]-th so please bear with me if this is a stupid idea and tell me why it's wrong, cause I know it's wrong. | 174768 | The sum of powers of two and two's complement – is there a deeper meaning behind this?
Probably everyone has once come across the following "theorem" with corresponding "proof": [imath]\sum_{n=0}^\infty 2^n = -1[/imath] Proof: [imath]\sum_{n=0}^\infty q^n = 1/(1-q)[/imath]. Insert [imath]q=2[/imath] to get the result. Of course the "proof" neglects the condition on [imath]q[/imath] for this formula, and the sum really diverges. However I now noticed an interesting fact: If you use two's complement to represent negative numbers on computers, [imath]-1[/imath] is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left. Now imagine that formally you sign-extend the number [imath]-1[/imath] to infinitely many bits. What you get is an infinite-to-the-left string of [imath]1[/imath]s. Which, using the normal base-2 formula [imath]n = \sum_k b_k 2^k[/imath] (where [imath]b_k[/imath] is the bit k positions from the right, i.e. [imath]b_0[/imath] is the rightmost bit), that infinite string of [imath]1[/imath]s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation. Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential. |
2586371 | If every Left Cosets of H are Right Cosets of H, show that H is normal in G
If H is a subgroup of a group G and if every Left Cosets of H are Right Cosets of H, show that H is normal in G As per the question, consider a left coset [imath]aH[/imath] and a right coset [imath]Hb[/imath] such that [imath]aH=Hb[/imath], [imath]a,b\in G [/imath] and [imath]a\neq b[/imath]. If [imath]a=b[/imath] then H is normal in G, by the definition of normal subgroup. My question is if [imath]aH=Hb[/imath] then does it imply [imath]a=b[/imath] always? Please answer | 283014 | A subgroup such that every left coset is contained in a right coset.
Let [imath]G[/imath] be any group, and [imath]H \leq G[/imath] a subgroup. Suppose that for each [imath]x \in G[/imath], there exists a [imath]y \in G[/imath] such that [imath]xH \subseteq Hy[/imath]. In other words, every left coset of [imath]H[/imath] is contained inside some right coset of [imath]H[/imath]. Question: what can we say about [imath]H[/imath]? In particular, does this imply that [imath]H[/imath] is normal? I know that if [imath]G[/imath] is finite, then [imath]H[/imath] must be normal, because then [imath]xH \subseteq Hy[/imath] implies [imath]xH = Hy[/imath], since [imath]|xH| = |Hy|[/imath]. |
2585584 | Why is [imath]z\mapsto |z|[/imath] not differentiable at [imath](0,0)[/imath]?
Can someone explain to me why the absolute value function [imath]f(z)=|z|[/imath] is not differentiable at [imath](0,0)[/imath] even though Cauchy Riemman conditions are true at [imath](0,0)[/imath] ( using [imath]|z|=\sqrt{x^2+y^2}[/imath] )? | 1122871 | Complex differentiability of [imath]f(z)=|z|[/imath]
Why is the absolute value function [imath]f : \mathbb{C} \rightarrow \mathbb{C}[/imath] given by [imath]f(z) = |z|[/imath] not complex differentiable at any point [imath]z_0[/imath] in the plane? |
2586535 | If [imath]a\mid b[/imath] then [imath]\phi(a)\mid \phi(b)[/imath] for [imath]a,b\in\mathbb{N}[/imath]
Hey I would like to show that [imath]a\mid b\Rightarrow \varphi(a)\mid\varphi(b)\qquad a,b\in\mathbb{N}[/imath] where [imath]\varphi(n)[/imath] is the the totient function. My try: Let [imath]a,b\in\mathbb{N}[/imath] and [imath]a\mid b[/imath]. By Eulers formula we have that [imath] \varphi(a)=a\cdot \prod_{\substack{p \mid a\\ p\ \text{prime}}} \Big(1-\frac{1}{p}\Big). [/imath] Now since [imath]a\mid b[/imath] by the Fundamental Theorem of Arithmetic we can factorise [imath]a[/imath] and [imath]b[/imath] into [imath] a=\prod_{i=1}^{n} p_i^{\alpha_i} \qquad b=\prod_{i=1}^{m} p_i^{\beta_i} [/imath] with [imath]p_i[/imath] being primes, [imath]n \le m[/imath], and [imath]\alpha_i \le \beta_i[/imath] for [imath]1 \le i \le n[/imath]. Using this we can say [imath] \left.\prod_{i=1}^{n} \Big(1-\frac{1}{p_i}\Big) \middle| \prod_{i=1}^{n}\Big(1-\frac{1}{p_i}\Big)\right. [/imath] and this concludes the result. Now my question is that is this correct? If it is is there another way to arrive to this conclusion? My way of describing this felt a bit clucky. Happy new years eve to all! | 634742 | Prove that if [imath]d[/imath] divides [imath]n[/imath] then [imath]φ(d)[/imath] divide [imath]φ(n)[/imath] for [imath]φ[/imath] denotes Euler’s φ-function.
Prove that if [imath]d[/imath] divides [imath]n[/imath] then [imath]φ(d)[/imath] divide [imath]φ(n)[/imath] for [imath]φ[/imath] denotes Euler’s [imath]φ[/imath]-function. I know that [imath]d|n[/imath] mean there exists some integer [imath]k[/imath] such that [imath]n=kd[/imath], but how can I use this to prove [imath]φ(d)[/imath] divide [imath]φ(n)[/imath] |
771466 | Mertens Constant and Euler–Mascheroni constant
I found this titillating equation: [imath]M = \gamma + \sum_{p} \left[ \ln\! \left( 1 - \frac{1}{p} \right) + \frac{1}{p} \right][/imath] where [imath]\displaystyle M=\lim_{n \rightarrow \infty } \left( \sum_{p\,\leq \,n} \frac{1}{p} - \ln(\ln n) \right)[/imath] and [imath]\displaystyle\gamma=\lim_{n \rightarrow \infty } \left( \sum_{k\,=\,1}^n \frac{1}{k} - \ln n \right)[/imath] and [imath]p[/imath] ranges over the prime numbers. Sadly I am unable to prove it. Can anyone help? | 1742395 | Relation between Meissel–Mertens constant and Euler–Mascheroni constant
From the Wikipedia page, the Meissel–Mertens constant [imath]M[/imath] is defined as the limit: [imath]M:=\lim_{n\to\infty}\left(\sum_{p\leq n}\frac{1}{p}-\log\log n\right).[/imath] Why is it equal to [imath]\gamma+\sum_{p}\left(\log(1-\frac{1}{p})+\frac{1}{p}\right)[/imath]? where [imath]\gamma[/imath] is the Euler–Mascheroni constant defined by [imath]\gamma:=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log n\right).[/imath] I tried to prove this, but cannot work it out. |
2586645 | Number of integer triangles.
Number of integer isosceles or equilateral triangle none of whose sides exceed [imath]2c[/imath] is? I substituted [imath]c =3[/imath] checked, and got [imath]27= 3 \times (3^2)[/imath] i.e. [imath]3c^2[/imath] triplets. It gives the right answer([imath]3c^2[/imath]) but how do I write a proper formal proof of this? I tried it this way: We have three vacant places for three integers. First place can be filled in [imath]2c[/imath] ways, 2nd one in [imath]2c[/imath] ways, 3rd one in [imath]2c[/imath] way but it involves over-counting, is it possible to eliminate the extras? | 1501824 | Number of isosceles triangle.
Prove that the number of isosceles triangles with integer sides, no side exceeding n, is [imath] \frac{1}{4}(1+3n^2)[/imath] or [imath]\frac{3}{4}(n^2)[/imath] according to whether n is odd or even. I am able to count the number of isosceles triangles which have the length of equal sides greater than or equal to length of base. I cannot count the remaining triangles. |
2587512 | Is the series [imath]\sum_{k=1}^{\infty}(\sqrt[k]{k}-1)[/imath] divergent or convergent?
I find this series to be a bit troublesum. I can't find a proper method that works. Root test doen't seem to be useful. Ratiotest got ugly. Integral test seems to force me to integrate a function [imath]x^{1/x},[/imath] which doesn't have any elementary primitive. I can't find an apropriate limit to compare it to and do a limit test comparison. | 253335 | Convergence of [imath]\sum_{n=1}^\infty{\left(\sqrt[n]{n}-1\right)}[/imath]
A student was recently asked this question by his instructor: [imath]\sum_{n=1}^\infty{\left(\sqrt[n]{n}-1\right)}[/imath] Converge or diverge? I feel a little dumb for not being able to answer it. The following tests fail to prove convergence or divergence: nth term test for divergence (limit is 0), ratio test (limit is 1), root test (see ratio test), limit comparison with [imath]\sqrt[n]{n}[/imath] (not sure why I thought that'd work) Something I did try was using the fact that [imath]x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\;,[/imath] to rewrite [imath]\sqrt[n]{n}-1[/imath] as [imath]\frac{n-1}{n^{1-1/n}+n^{1-2/n}+\cdots+n^{2/n}+n^{1/n}+1}\;.[/imath] However, I'm not sure what to compare this to. According to wolfram alpha this series "diverges by the comparison test", but comparison to what? There is a similar problem in Baby Rudin, but for [imath](\sqrt[n]{n}-1)^n[/imath], and a simple nth root test resolves that series [convergence] in a hurry. Any ideas? Have any of you encountered a similar looking series before? Thanks. |
2587657 | How to show that [imath]\langle a \rangle[/imath] is a normal Subgroup of a group [imath]G[/imath]
How to show that [imath]\langle a \rangle[/imath] is a normal subgroup of a group [imath]G[/imath], where [imath]a\in Z(G)= \{x \in G ∣ xg = gx ~ \forall g \in G\}, \text{the center of the group [/imath]G[imath]}.[/imath] With the help of How to show that [imath]\langle a \rangle[/imath] is a subgroup of a group [imath]G[/imath], where [imath]a[/imath] is in center of the group [imath]G[/imath], it is possible to show that [imath]\langle a \rangle[/imath] is a subgroup of a group [imath]G[/imath]. How to prove for normality? | 2587565 | How to show that [imath]\langle a \rangle[/imath] is a subgroup of a group [imath]G[/imath], where [imath]a[/imath] is in center of the group [imath]G[/imath]
How to show that [imath]\langle a \rangle[/imath] is a subgroup of a group [imath]G[/imath], where [imath]a\in Z(G)= \{x \in G ∣ xg = gx ~ \forall g \in G\}[/imath], the center of the group [imath]G[/imath]. I am unable to show this result. Although I know that [imath]Z(G)[/imath] is a subgroup of [imath]G[/imath]. |
2587720 | Existence of [imath]c \in [a,b][/imath] Such That [imath]\int_{c-a}^{b-c}f(x)dx=0[/imath]
If [imath]f:[a,b]\to R[/imath] is a continuous function, prove that there exists [imath]c \in [a,b][/imath] such that [imath]\int_{c-a}^{b-c}f(x)dx=0[/imath]. | 2587551 | Given continuous [imath]f[/imath] and [imath]g[/imath], there exists [imath]c \in (a,b)[/imath] such that [imath]\int_{a}^{c} f(x) \ \mathrm dx = \int_{c}^{b} g(x) \ \mathrm dx[/imath]
If [imath]f,g : [a,b] \to (0,\infty)[/imath] are continuous functions, prove that there exists [imath]c \in (a,b)[/imath] such that: [imath]\int_{a}^{c} f(x) \ \mathrm dx = \int_{c}^{b} g(x) \ \mathrm dx[/imath] |
2588013 | Prove inequality [imath](1+x+y)(1+y+z)(1+z+x) \ge 9(xy+xz+yz)[/imath]
I need to prove inequality [imath](1+x+y)(1+y+z)(1+z+x) \ge 9(xy+xz+yz)[/imath] for nonnegative [imath]x,y,z \ge 0[/imath] I can do it with partitional derivates [imath]f(x,y,z)=((1+x+y)(1+y+z)(1+z+x)-9(xy+xz+yz))[/imath] and find [imath] \frac{\partial f}{\partial x}=2xy + 2xz + 2yz + z^2 + y^2 + 2x - 6y - 6z + 2[/imath] [imath] \frac{\partial f}{\partial y}=2xy + 2xz + 2yz + x^2 + z^2 + 2y - 6c - 6x + 2[/imath] [imath] \frac{\partial f}{\partial c}=2xy + 2xz + 2yz + x^2 + y^2 + 2z - 6x - 6y + 2[/imath] minimum value of function [imath]f[/imath] is [imath]f(1,1,1)=0[/imath] that's why [imath](1+x+y)(1+y+z)(1+z+x) \ge 9(xy+xz+yz)[/imath]. But I need the prove without partitional derivates. Any standart inequality (Inequality of arithmetic and geometric means, Cauchy–Schwarz inequality) don't help me. | 1996771 | Prove [imath](1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)[/imath]
How one can prove the following. Let [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] be non-negative real numbers. Then the inequality holds: [imath](1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).[/imath] WLOG one can assume that [imath]0\le a\le b\le c[/imath]. It is not difficult to prove the statement when [imath]0\le a\le b\le c\le 1[/imath] or [imath]1\le a\le b\le c[/imath]. |
2587896 | Determine the convergence radius for [imath]\sum_{n=1}^{\infty}\frac{(n!)^kx^n}{(kn)!}, \quad k\in\mathbb{N}.[/imath]
The ratio test gives [imath]|x|\cdot\lim_{n\rightarrow\infty}\left|\frac{((n+1)!)^k\cdot(kn)!}{(k(n+1))!\cdot(n!)^k}\right|=|x|\cdot\lim_{n\rightarrow\infty}\left|(n+1)^k\cdot\frac{(kn)!}{(k(n+1))!}\right|.[/imath] I'm having trouble simplifying [imath]\frac{(kn)!}{(k(n+1))!}[/imath], so far I have: [imath]\frac{(kn)!}{(k(n+1))!}=\frac{kn(kn-1)...}{(kn+k)(kn+k-1)...(kn)(kn-1)...}=\frac{1}{(kn+k)(kn+k-1)...(kn+1)}.[/imath] This means that I have [imath]|x|\cdot\lim_{n\rightarrow\infty}\frac{(n+1)^k}{(kn+k)(kn+k-1)...(kn+1)},[/imath] since the highest power after expansion of the denominator and numerator are [imath]1[/imath] and [imath]k^k[/imath] respectively, the above is asymptotically equal to [imath]|x|\cdot\lim_{n\rightarrow\infty}\frac{(n+1)^k}{(kn+k)(kn+k-1)...(kn+1)}\sim|x|\cdot\lim_{n\rightarrow\infty}\frac{n^k+\text{lesser powers}}{k^kn^k+\text{lesser powers}}=|x|\frac{1}{k^k},[/imath] thus the radius is [imath]R=k^k[/imath]. I feel really insecure here using these vauge arguments. Is it mathematically correct to reason like this? | 812505 | Finding the radius of convergence for [imath]\sum_{n=0}^{\infty} \frac{(n!)^kx^n}{(kn)!}[/imath]
I'm reviewing some calculus concepts right now, and this series stumped me. I'm trying to figure out what its radius of convergence is, but it's not falling for any of my old tricks. Using the root-test, I want to compute: [imath] L_1=\lim_{n\rightarrow\infty} \left|\sqrt[n]{\frac{(n!)^k}{(kn)!}}\right| [/imath] where [imath]k[/imath] is some positive integer. Passing the limit through the absolute-value bars and [imath]n[/imath]th root symbol, I now want to compute the simpler (?) [imath] L_2 = \lim_{n\rightarrow\infty} \sqrt[n]{\frac{(n!)^k}{(kn)!}}[/imath] Which I can't seem to pin down. Numerical tests indicate it goes to 0, and that jives with an intuitive notion I have that there are always [imath]k[/imath] terms on the numerator, but [imath]kn[/imath] terms in the denominator. But since all of these terms themselves depend on [imath]n[/imath], proving that [imath]L_2 = 0[/imath] escapes me. Once we get this result (if it's correct), then clearly [imath]L_1[/imath] is also 0, and the radius of convergence of the series is [imath]\mathbb{R}[/imath]. Thanks for any ideas you may have. |
2587670 | Is the series [imath]\sum_{k=1}^{\infty}(-1)^k(\sqrt{k+1}-\sqrt{k})[/imath] convergent or divergent?
This is quite infuriating. Just when I feel like I'm grasping this whole series konverging/diverging tests-notion, this one pops up. Here is what I did: [imath]\sum_{k=1}^{\infty}|(-1)^k(\sqrt{k+1}-\sqrt{k})|=\sum_{k=1}^{\infty}(\sqrt{k+1}-\sqrt{k})=\sum_{k=1}^{\infty}\frac{1}{\sqrt{k+1}+\sqrt{k}}\leq\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}=\text{Div.}[/imath] So the series is not absolutely convergent. So this test didin't give me anything. For Leibniz test, that states that if [imath]\{a_k\}^{\infty}_{k=1}[/imath] is a decreasing sequence of positive numbers with limit equal to zero, then the alternating series [imath]\sum(-1)^{k-1}a_k[/imath] is convergent. However, [imath]a_k=\sqrt{k+1}-\sqrt{k}[/imath] is unfortunately increasing. | 327474 | Convergence of [imath]\sum_{n=1}^\infty (-1)^n(\sqrt{n+1}-\sqrt n)[/imath]
Please suggest some hint to test the convergence of the following series [imath]\sum_{n=1}^\infty (-1)^n(\sqrt{n+1}-\sqrt n)[/imath] |
2588145 | Parenthesis in Domain
I just had a quick question about a notation I saw. I know that a [imath]C^2[/imath] function is a function whose second partial derivative exists. But, what does it mean when one writes [imath]C^2(U)[/imath] where [imath]U[/imath] is a subset of [imath]R^n[/imath]? Thanks in advance. | 38769 | Extension of partial derivatives and the definition of [imath]C^k(\overline{\Omega})[/imath]
The following is an excerpt from Folland's Introduction to Partial Differential Equations: Let the open set [imath]\Omega\subset{\mathbb R}^n[/imath], and [imath]k[/imath] be a positive integer. [imath]C^k(\Omega)[/imath] will denote the space of functions possessing continuous derivatives up to order [imath]k[/imath] on [imath]\Omega[/imath], and [imath]C^k(\overline{\Omega})[/imath] will denote the space of all [imath]u\in C^k(\Omega)[/imath] such that [imath]\partial^{\alpha}u[/imath] extends continuously to the closure [imath]\overline{\Omega}[/imath] for [imath]0\leq|\alpha|\leq k[/imath]. As I understand, an extension means that there exists [imath]\widetilde{\partial^{\alpha}u}[/imath] which is defined on [imath]\overline{\Omega}[/imath] and [imath]\widetilde{\partial^{\alpha}u}|_{\Omega}=\partial^{\alpha}u[/imath]. And "extends continuously" means [imath]\widetilde{\partial^{\alpha}u}[/imath] is continuous with respect to the relative topology on [imath]\overline{\Omega}[/imath]. Here are my questions: How do people usually do such extension? When does such extension exist and when does not? Let [imath]\Omega[/imath] be an open subset of [imath]{\mathbb R}[/imath]. Is there any non-trivial counterexample such that this kind of extension does not exist? |
2588161 | Solutions of quadratic equation with complex coefficients, using the quadradic formula.
I have a hard time solving the following equation. [imath]z^{2} + (1 - i) \cdot z - i = 0[/imath] I tried factorization and got [imath]z = -1[/imath] or [imath]z = i[/imath], which I suppose are correct. However, when I try the quadratic formula for solving second degree equations, the results I get don't match with the solutions I get above. | 1908074 | Quadratic Formula in Complex Variables
Let [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] be complex numbers with [imath]a\neq0[/imath]. Show that the solutions of [imath]az^2+bz+c=0[/imath] are [imath]z_1,z_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/imath], just as they are in the case when [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] are real numbers. |
2588742 | how many non-isomorphic group of order 15 are there?
Let [imath]G[/imath] be group of order [imath]15 = 3 \cdot 5 = p \cdot q[/imath] , where [imath]p < q[/imath] are primes. I know that there exists a unique non abelian group of order [imath]pq[/imath] and one abelian non isomorphic group of order [imath]pq[/imath]. I think here are two non isomorphic of order [imath]15[/imath] Am i right or wrong ? | 1211689 | How many different groups of order [imath]15[/imath] there are?
I wanted to share with you my resolution of this exercise. How many different groups of order [imath]15[/imath] there are? My resolution: We're looking for groups such that [imath]|G|=15=3\cdot 5[/imath]. Then: [imath]G[/imath] has an unique Sylow [imath]3[/imath]-subgroup of order [imath]3[/imath] ([imath]P_3[/imath]) and an unique Sylow [imath]5[/imath]-subgroup of order [imath]5[/imath] ([imath]P_5[/imath]). Furthermore, [imath]|P_3 \cap P_5|\ \Large\mid\ \normalsize|P_3|,\ |P_5|[/imath] So [imath]|P_3 \cap P_5|=1\implies P_3 \cap P_5=\{e\}.[/imath] Then:[imath]|P_3\cdot P_5|=\displaystyle\frac{|P_3|\cdot |P_5|}{|P_3 \cap P_5|}=|P_3|\cdot |P_5|=3\cdot 5=|G|,[/imath] and [imath]G=P_3\cdot P_5 \simeq P_3 \times P_5 \simeq \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}.[/imath] It's everything correct and well justified? There's some things that I didn't say, for example that a subgroup a prime [imath]p[/imath] as order is cyclic, and thus isomoprhic to [imath]\mathbb{Z}/p\mathbb{Z}[/imath]. Maybe I did something wrong. I would appreciate someone to look and criticize my exercise. Thank you. |
2588881 | Find a surjective homomorphism of a group
I have difficulties with this task Find a homomorphism [imath]\Phi : (G, \circ) \to (G, \circ)[/imath], that is surjective but not injective. I tried to find such a homomorphism, by finding a surjective function, yet functions I tried were not homomorphisms. And then I end up with [imath]Ker(\Phi)=\{e_G\}[/imath]. So I found an isomorphism. | 1203035 | Surjective, but not injective endomorphism
What is a (simple) example of a group [imath]G[/imath] with an endomorphism [imath]G\rightarrow G[/imath] That is surjective, but not injective? |
2185499 | Let [imath]G[/imath] be a group, and let [imath]n\in\mathbb Z[/imath]. Show that [imath](ab)^n=a^nb^n\iff(ab)^{1-n}=a^{1-n}b^{1-n}[/imath].
Let [imath]G[/imath] be a group, and let [imath]n\in\mathbb Z[/imath]. Show that [imath](ab)^n=a^nb^n\iff(ab)^{1-n}=a^{1-n}b^{1-n}.[/imath] For the [imath]"\implies"[/imath] direction, let's assume [imath](ab)^n=a^nb^n[/imath], and let's consider the inverse of [imath]ab[/imath]. We can write [imath](ab)^{-n}=b^{-n}a^{-n}[/imath]. If we multiply by [imath]ab[/imath], we get [imath] (ab)^{1-n}=(ab)(ab)^{-n}=(ab)b^{-n}a^{-n}=ab^{1-n}a^{-n}. [/imath] But now I need commutativity to conclude the proof. Could someone help me with the last step? | 2588943 | Prove that [imath](ab)^\mathrm{n-1}[/imath] = [imath]b^\mathrm{n-1}*a^\mathrm{n-1}[/imath] where n > 1 is a fixed integer
For all [imath]a,b[/imath] in a group G, show [imath](ab)^\mathrm{n-1}[/imath] = [imath]b^\mathrm{n-1}*a^\mathrm{n-1}[/imath] where n > 1 is a fixed integer. Given [imath](ab)^n = a^nb^n[/imath] I took inverse on both sides and got [imath]a^n b^n = b^n a^n[/imath] but unable to proceed further. My book solution has straightaway started with first step as [imath][b^\mathrm{-1}(ba)b]^\mathrm{n} = b^\mathrm{-1}(ba)^nb[/imath] I don't get how this step is reached. P.S. : My previous similar question has been marked as duplicate with Let [imath]G[/imath] be a group, and let [imath]n\in\mathbb Z[/imath]. Show that [imath](ab)^n=a^nb^n\iff(ab)^{1-n}=a^{1-n}b^{1-n}[/imath]. I am re-asking to clarify that [imath]n[/imath] in this question belongs to any integer whereas mine has a fixed value. Also, I am unable to derieve my own solution even after doing the steps mentioned in same. |
2589228 | All primes for which [imath]p! + p[/imath] is perfect square
Find all prime number [imath]p[/imath] for which [imath]p! + p[/imath] in a perfect square. Of course [imath]p=2[/imath] and [imath]p=3[/imath] are solutions. I think there are not other solutions, but I can’t prove it! | 1208750 | Number Theory: [imath]p!+p=n^2[/imath], [imath]p[/imath] = prime
Find all prime numbers [imath]p[/imath] such that [imath]p!+p[/imath] is a perfect square. I think the only ones are when [imath]p!+p=p^2[/imath], i.e. [imath]p\in \{2,3\}[/imath]. Any ideas at all? |
2589177 | For given [imath]k,N\in\mathbb{N}[/imath], how to compute [imath]\sum_{i=0}^Ni^k[/imath]?
For given [imath]k[/imath] and [imath]N[/imath], [imath]k,N\in\mathbb{N}[/imath], how to compute [imath]\sum_{i=0}^Ni^k[/imath]? We have: [imath]\sum^N_{i=0}i=\frac{N(N+1)}2[/imath] Also according to what I found in the Internet we have [imath]\sum^N_{i=0}i^2=\frac{n(n+1)(2n+1)}{6}[/imath] (I can prove this formula by induction, but I couldn't obtain it myself) But how to generalize this for any [imath]k[/imath]? | 2397918 | Formula for the general Cavalieri Sum: [imath]S_n(p)=\sum\limits_{k=1}^{n} k^p\,\,\,n, p\in\mathbb N[/imath]
What kind of formula is there that can be used for calculating the sum of power of [imath]x[/imath] of numbers from [imath]1[/imath] to [imath]a[/imath]? I know that the sum of numbers from [imath]1[/imath] to [imath]a[/imath] is [imath]\ (n^2 + n)/ 2 \ [/imath] and that the sum of the squared numbers from [imath]1[/imath] to [imath]a[/imath] is [imath]n^3/3+n^2/2+n/6 [/imath]. If possible, please give a solution that excludes Bernoulli numbers and works for negative numbers too. |
1647007 | Smallest [imath]n[/imath] such that [imath]U(n)[/imath] contains a subgroup isomorphic to [imath]\mathbb Z_5 \oplus \mathbb Z_5[/imath]
I solved the following exercise: Find an integer [imath]n[/imath] such that [imath]U(n)[/imath] contains a subgroup isomorphic to [imath]\mathbb Z_5 \oplus \mathbb Z_5[/imath]. Here [imath]U(n)[/imath] is the group of units modulo [imath]n[/imath]. To solve it I used that [imath]U(st) = U(s) \oplus U(t)[/imath] when [imath]s,t[/imath] are coprime and [imath]U(p^n) = \mathbb Z_{p^n - p^{n-1}}[/imath]. I tried the first few primes until I found two with the property that [imath]\mathbb Z_{p^n - p^{n-1}}[/imath] is divisible by [imath]5[/imath]. My result: [imath] U(3025) = U(5^2 \cdot 11^2) = \mathbb Z_{20}\oplus \mathbb Z_{110}[/imath] It seemed clear to me that I had found the smallest number with this property since I chose the first primes I found. Then I looked at the solution which is [imath] n =275[/imath] What other methods are there to solve this question so that I can find the smallest such [imath]n[/imath]? Obviously, while possibly correct, my method does not find the smallest [imath]n[/imath]. | 517517 | Subgroups of [imath]U(n)[/imath] isomorphic to a direct sum of cyclic groups
Let [imath]U(n)[/imath] to be the set of all positive integers less then [imath]n[/imath] and relatively prime to [imath]n[/imath]. Then [imath]U(n)[/imath] is a group under multiplication modulo [imath]n[/imath]. A. Find integer [imath]n[/imath] such that [imath]U(n)[/imath] contains subgroup isomorphic to [imath]Z_5\oplus Z_5[/imath]. B. Show that there is an [imath]U(n)[/imath] containing a subgroup isomorphic to [imath]Z_3 \oplus Z_3[/imath], but not isomorphic to [imath]Z_4 \oplus Z_4[/imath]. |
1880698 | Prove equation using Bessel functions: [imath]\frac{d}{dx}[xJ_n(x)J_{n+1}(x)]=x[J_n^{2}(x)-J_{n+1}^{2}(x)][/imath]
The question is to prove that [imath]\frac{d}{dx}[xJ_n(x)J_{n+1}(x)]=x[J_n^{2}(x)-J_{n+1}^{2}(x)][/imath] I have attempted using the product rule and the definition of Bessel but I just end up with a mess and no end in sight. Any help would be greatly appreciated. | 1608259 | Bessel function equation with derivation: [imath]\frac{d}{dx} (xJ_{\alpha}(x)J_{\alpha+1}(x)) = x(J_{\alpha}^2(x)-J_{\alpha +1}^2(x))[/imath]
Let [imath]J_{\alpha}(x)[/imath]be the Bessel function. Show the equality: [imath]\frac{d}{dx} \left( xJ_{\alpha}(x)J_{\alpha +1}(x)\right) = x \left( J_{\alpha}^2(x) -J_{\alpha +1}^2(x)\right)[/imath] How to start with it? |
2590281 | Give a one-to-one and onto function from [imath]\mathbb{N} \times \mathbb{N}[/imath] to [imath]\mathbb{N}[/imath]
Give a function [imath]g : \mathbb{N} \times \mathbb{N} \to \mathbb{N}[/imath] such that [imath]g[/imath] is one-to-one and onto function. the onto part I could easily solve but every function I think of, is not one-to-one. | 1051456 | Bijection between [imath]\mathbb N \times \mathbb N[/imath] and [imath]\mathbb N[/imath]
Show that [imath]\mathbb{N} \times \mathbb{N} \sim\mathbb{N}[/imath]. I found a bijection such that [imath]g(k,l): \mathbb{N} \times \mathbb{N} \to \mathbb{N}[/imath] by [imath]g(k,l) = {(k+l)(k+l-1) \over 2} - (l-1)[/imath] But I am having trouble showing that it is 1-1 and onto. First I said let [imath]{(k_1+l_1)(k_1+l_1-1) \over 2} = {(k_2+l_2)(k_2+l_2-1) \over 2}[/imath] and I was trying to use that to show [imath]k_1 + l_1 = k_2 + l_2[/imath] but I don't know how to do that, and once I get there I'm not sure how to show that [imath]l_1=l_2[/imath] or [imath]k_1=k_2[/imath] for onto I said let [imath]y \in k+l[/imath]. Then [imath]{y(y-1) \over 2} = y[/imath] so therefore [imath]g(k,l)[/imath] is onto. therefore g is a bijection. |
264037 | Relationship between complex number and vectors
What is the relation between complex numbers and vectors? I have read in some places "a complex number a 2-dimensional vector". One can easily observe that [imath]i\cdot i=-1[/imath] in complex multiplication like the cross product of vectors. Is vector only the generalisation of complex number to n dimensions? Is there any fundamental difference between those two? | 2732176 | Are complex numbers just vectors defined by cosine (x axis) and sine (y-axis)?
There is no clear and standard definition of a complex number, I recall a prof being at a loss when a freshman aske him what is a complex number. I read somewhere that a great mathematicians was not ashamed to say, re the formula [imath]e^xpi = -1[/imath] I think, that nobody knows what it means , but they are proud of it, is that true?* I suppose nobody has a clear answer, I am just asking to specify what is more or different from the regular operations on a vector field taken as a 360° circle, where points are identified by cosine of the relative angle (real part) and sine (imaginary part) Edit The linked question/answer resembles only superficially my question. I stressed the crucial feature that the vector is defined by sin and cos and consequently obeys the rules of duplication etc of angles. I am not familiar with all details of that theory and I am asking , if any, about the differences. If there are none or minor ones, then complex numbers ar not numbers and not complex at all. *I found the quote: Benjamin Peirce, a noted American 19th-century philosopher, mathematician, and professor at Harvard University, after proving Euler's identity during a lecture, stated that the identity "is absolutely paradoxical; we cannot understand it, and we don't know what it means, but we have proved it, and therefore we know it must be the truth * |
2590009 | Proving the divergence of [imath]\sum a_n[/imath] with [imath]a_{n}=\frac{1}{n}\left( 1+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}} \right)[/imath]
How can we prove formally that the series [imath]\sum a_{n}[/imath] diverges whose [imath]n^{th}[/imath] term has been provided below: [imath] a_{n}=\frac{1}{n}\left( 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}} \right),\qquad n\ge1.[/imath] The sequence converges to zero but I am not sure how book has proved that the series diverges? | 1123115 | Convergence of [imath]\sum_{n=1}^\infty\frac{1}{n} [1+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}}][/imath]
Let [imath]t_n= \frac{1}{n} [1+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}}][/imath], n=1,2... Then I am asked whether series[imath] \sum t_n [/imath] converge or diverge. Also whether sequence [imath] t_n [/imath] converge to zero or not. I tried About sequence [imath]t_n[/imath], by Cauchy's 1st theorem, that since 1/[imath]\sqrt{n} [/imath] converges to zero so does t_n. But the seriex part is doubtful to me.? |
2590407 | Is there this notation in mathematics [imath]\mathbb{R^{\frac{1}{2}}}[/imath] and is it meant the space dimension is [imath]\frac{1}{2}[/imath]?
I'm not familiar with algebra theory, but I'm interested to know if there is an ensemble in mathematics for which the power of ensemble could be a real number, we take for example [imath]\mathbb{R^{\frac{1}{2}}}[/imath], Does this meant the space dimension is [imath]\frac{1}{2} [/imath]? Note:I want [imath]\mathbb{R^{\frac{1}{2}}}[/imath] have a metric space structure Thanks for any comments | 1466820 | Vector spaces with fractional dimension
Can the notion of vector space or algebra over a field be meaningfully extended to fractional dimensions, so that for example [imath]\mathbb{R}^{-2/3}[/imath] makes sense? Has this been explored somewhere? I know that super vector spaces can be thought of as one way of generalizing vector spaces to negative integer dimensions. Is there a similar concept for dimensions that are rational numbers? I'm not talking about Hausdorff dimension, because it doesn't allow for negative rationals, and I'm rather looking for extensions from a more algebraic point of view (dimension as the trace of the identity map), without recurring to a given metric. |
2589535 | Continuous [imath]f[/imath] such that [imath]f(x)=f(x^2)[/imath] is constant?
Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be a continuous function such that [imath]f(x)=f(x^2)[/imath] for all [imath]x\in\mathbb{R}[/imath]. I've proven that also [imath]f\left(y^{(2^{-n})}\right)=f(y)[/imath] for all [imath]n\in\mathbb{N}, y\geq0[/imath]. How can I deduce that [imath]f[/imath] is constant? | 2086315 | Proof that a continuous function with [imath]f(x) = f(x^2)[/imath] is constant.
Given a continuous function [imath] f: \mathbb{R} \rightarrow \mathbb{R} \quad \text{with} \quad f(x) = f(x^2) \quad\quad \forall x \in \mathbb{R} [/imath] How can I show that [imath]f[/imath] must be constant? |
2590939 | How to prove [imath]E(Y|X)=E(Y),a.s.[/imath] formally if [imath]X,Y[/imath] are independent?
The question is from A First Look at Rigorous Probability Theory 13.4.9 Let [imath]X,Y[/imath] be two random variable in a probability space [imath](\Omega,\mathcal{F},\mathbf{P})[/imath]. Suppose that [imath]X[/imath] and [imath]Y[/imath] are independent. Show that [imath]E(Y|X)=E(Y),w.p. 1[/imath] I read several books which stated this as a property other than proving it. I understand that [imath]E(Y|X)[/imath] is [imath]\sigma(X)[/imath]-measurable. However, I have no idea how to start this. What exactly [imath]E(Y|X)=?[/imath] How to prove this? | 2572175 | Proving seemingly obvious equations with measure theory
I want to prove very simple equations, but am having some trouble because I have no clue. Definition of Conditional Probability is given as follows: Definition [imath]E(Y|X)[/imath] is a conditonal expectation of [imath]Y[/imath] given [imath]X[/imath] if it is a [imath]\sigma (X)[/imath]-measurable random variable and for any Borel set [imath]S \subseteq R[/imath], we have [imath]E(E(Y|X)1_{X \in S})=E(Y1_{X \in S})[/imath] I want to solve those two problems based on above definition. Suppose [imath]X[/imath] and [imath]Y[/imath] are independent. (a) Prove that [imath]E(Y|X)=E(Y)[/imath] with probability 1. (b) Prove that [imath]Var(Y|X)=Var(Y)[/imath] with probability 1. (c) Explicitly verify the following theorem.(with [imath]\mathcal{G}=\sigma (X)[/imath] in this case) Theorem: Let [imath]Y[/imath] be a random variable, and [imath]\mathcal{G}[/imath] a sub-[imath]\sigma[/imath]-algebra. If [imath]Var(Y)<\infty[/imath], then [imath]Var(Y)=E(Var(Y|\mathcal{G}))+Var(E(Y|\mathcal{G}))[/imath] Let X and Y be jointly defined random variables. (a) Suppose [imath]E(Y|X)=E(Y)[/imath] with probability 1. Prove that [imath]E(XY)=E(X)E(Y)[/imath] (b) Give an example where [imath]E(XY)=E(X)E(Y)[/imath], but it is NOT the case that [imath]E(Y|X)=E(Y)[/imath] with probability 1. 1. For this part, I think I could deal with this by myself if I understand how to prove (a). So please help me with just (a). In order to prove that [imath]E(Y|X)=E(Y)[/imath] with probability 1, I need to show that \begin{equation} E(E(Y|X)1_{X})=E(E(Y)1_{X}). \end{equation} if [imath]X[/imath] and [imath]Y[/imath] are independent. From the above definition, conditional expectation [imath]E(Y|X)[/imath] is defined such that [imath]E(E(Y|X)1_{X \in S})=E(Y1_{X \in S})[/imath]. So maybe it suffices to show that [imath]E(Y1_{X \in S})=E(E(Y)1_{X \in S})[/imath]. But it looks too obvious. My guess is that there is more educated way of showing why [imath]E(Y1_{X \in S})=E(E(Y)1_{X \in S})[/imath] holds if [imath]X[/imath] and [imath]Y[/imath] are independent. 2. I can do (b) on my own. So for (a), "[imath]E(Y|X)=E(Y)[/imath] with probability 1" means that [imath]E(E(Y|X)1_{X})=E(E(Y)1_{X})[/imath]. But, how can I prove [imath]E(XY)=E(X)E(Y)[/imath] with that information? I know how to prove this equation with integral sign or summation sign. But how does information about a.s. convergence give any implication about that equation? |
2589876 | Convergence of recursive sequence given Uo
Please how could I possibly show that the sequence defined by [imath]U_{n+1} = \frac{1}{2} (U_n + \frac{a}{U_n})[/imath] converges Given that [imath]U_0 > 0[/imath] and [imath]a>0[/imath] I have calculated [imath]\frac{U_{n+1}}{U_n}[/imath] which is positive but [imath]U_{n+1}-U_n[/imath] doesn't yield any result and i can't show it is bounded. Thanks in advance | 690332 | Proving a sequence converges by defining the sequence recursively.
Let [imath]a > 0[/imath]. Choose [imath]x_1 >[/imath] [imath]\sqrt a[/imath]. Define a sequence {[imath]x_n[/imath]} recursively as [imath]x_{n+1} = 1/2(x_n + a/x_n)[/imath] for [imath]n > 1[/imath]. Prove that lim[imath]x_n = \sqrt a[/imath]. I think I first want to prove that {[imath]x_n[/imath]} is bounded below by [imath]\sqrt a[/imath]. But, I am having trouble with this. Could someone help me out a little? |
2591098 | In [imath]\Delta ABC[/imath] if [imath]a^2+b^2+c^2-ac-\sqrt{3}ab=0[/imath] Then Prove that Triangle is Right angled
In [imath]\Delta ABC[/imath] if [imath]a^2+b^2+c^2-ac-\sqrt{3}ab=0[/imath] Then Prove that Triangle is Right angled I tried in this way: Multiplying with [imath]2[/imath] both sides we get [imath]2a^2+2b^2+2c^2-2ac-2\sqrt{3}ab=0[/imath] [imath](a-c)^2+(a-\sqrt{3}b)^2+c^2=b^2[/imath] Also [imath](a-c)^2+(b-\sqrt{3}a)^2+c^2=2a^2-b^2[/imath] How to proceed? | 1746491 | In a [imath]\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3[/imath],then the triangle is
In a [imath]\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3[/imath],then the triangle is [imath](A)[/imath]equilateral [imath](B)[/imath]isosceles [imath](C)[/imath]right angled [imath](D)[/imath]none of these The given condition is [imath]a^2+b^2+c^2=ac+ab\sqrt3[/imath]. Using sine rule, [imath]a=2R\sin A,b=2R\sin B,c=2R\sin C[/imath],we get [imath]\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3[/imath] I am stuck here. |
2591438 | Interesting equation over the naturals with powers
I have an interesting equation over the naturals. [imath]x^5=y^2+4[/imath] How do I solve it? | 1748015 | Prove that there does not exist integer solutions for the diophantine equation [imath]x^5 - y^2 = 4[/imath].
Prove that there does not exist an integer solution for the diophantine equation [imath]x^5 - y^2 = 4[/imath]. It's obvious that [imath]x[/imath] and [imath]y[/imath] are of the same parity. We can also claim that if [imath]x[/imath] is odd, then it is [imath]1 \pmod 4[/imath]. Also, if [imath]x[/imath] and [imath]y[/imath] are even, then [imath]y \equiv 2 \pmod 4\text{ since } x^5 \text{ is a multiple of } 32[/imath] and if [imath]y[/imath] were a multiple of [imath]4[/imath] then it would be at a distance of at least [imath]16[/imath] from [imath]x^5[/imath] or be equal to it. These are my observations. How should I proceed with the proof? Please give hints. |
2591543 | Trigonometric limit problem
[imath]\lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^{n-1}_{k=1}\cot^2\left(\frac{k\pi}{n}\right)[/imath] [imath]n>1,n\in N[/imath] Try: [imath]\displaystyle \lim_{n\rightarrow \infty}\cot^2 \frac{k\pi}{n} = \frac{n^2}{k^2 \pi^2}[/imath] (where [imath]\lim_{x\rightarrow 0}\sin x=x, \lim_{x\rightarrow 0}\cos x = 1[/imath]) so [imath]\lim_{n\rightarrow \infty}\frac{n^2}{\pi^2 n^2}\bigg(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \cdots +\frac{1}{(n-1)^2}\bigg) = \frac{1}{\pi^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right) = \frac{1}{6}[/imath] can someone please explain me whats wrong in my try and also explain me right answer | 1562037 | Question regarding [imath]f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)[/imath]
[imath]f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)[/imath] then how to find limit of [imath]\dfrac{3f(n)}{(n+1)(n+2)}[/imath] as [imath]$n\to\infty$[/imath]? I don't know any series like that. Riemann sum is not working. What to do? Update:FYI this appeared in my FIITJEE AITS (JEE MAIN) PAPER 2 FOR 2016. |
2591810 | What is the volume of the solid defined by [imath]x^4+y^4+z^4=1[/imath] enclosed in a box of unit length?
I came across a post on the geometric interpretation of [imath]x^3+y^3+z^3 = t^3[/imath] asking if it was a higher analogue of the Pythagorean theorem. It made me wonder: if we plot [imath]x^2+y^2+z^2=1[/imath], of course we get a sphere. But if we plot [imath]x^4+y^4+z^4=1[/imath], we get this solid instead: [imath]\hskip1.8in[/imath] I know by a result of Elkies that this has infinitely many rational points [imath]x,y,z[/imath]. What else is known about this solid? For one, what is its name? Also, if we snugly enclose it in a box of unit length, what is its volume? P.S. For the case [imath]x^5+y^5+z^5 = 1[/imath] [imath]\hskip1.8in[/imath] | 301506 | Hypervolume of a [imath]N[/imath]-dimensional ball in [imath]p[/imath]-norm
Suppose I have a N-dimensional ball with radius R in p-norm: [imath] \sum_{n=1}^N |x_n|^p = R^p [/imath] Is there a closed formula for its (hyper)volume? I can't find anything. If there isn't, can we at least know how it varies with [imath]R[/imath]? |
2591315 | Sum of all Fibonacci numbers [imath]1+1+2+3+5+8+\cdots = -1[/imath]?
I just found the sum of all Fibonacci numbers and I don't know if its right or not. The Fibonacci sequence goes like this : [imath]1,1,2,3,5,8,13,\dots[/imath] and so on So the Fibonacci series is this [imath]1+1+2+3+5+8+13+\dots[/imath] Let [imath]1+1+2+3+5+8+\dots=x[/imath] [imath]\begin{align} 1 + 1 + 2 + 3 + 5 + \dots &= x\\ 1 + 1 + 2 + 3 + \dots &= x\\ 1 + 2 + 3 + 5 + 8 + \dots &= 2x \text{ (shifting and adding)} \end{align}[/imath] We in fact get the same sequence. But the new sequence is one less than the original sequence. So the new sequence is [imath]x-1[/imath]. But [imath]x-1=2x[/imath] which implies that [imath]x=-1[/imath]. So [imath]1+1+2+3+5+8+\dots=x[/imath] which means... [imath]1+1+2+3+5+8+13+21+\dots=-1[/imath] Is this right or wrong? Can someone please tell? Thanks... | 1769145 | What is the sum of all the Fibonacci numbers from 1 to infinity.
Today I believe I had found the sum of all the Fibonacci numbers are from [imath]1[/imath] to infinity, meaning I had found [imath]F[/imath] for the equation [imath]F = 1+1+2+3+5+8+13+21+\cdots[/imath] I believe the answer is [imath]-3[/imath], however, I don't know if I am correct or not. So I'm asking if someone can show how [imath]F = -3[/imath]. |
2591008 | Functional equation: what function is its inverse's reciprocal?
The fact that so many students confuse functional inverse notation [imath]f^{-1}(x)[/imath] with multiplicative inverse notation [imath][f(x)]^{-1}[/imath] got me to thinking... does there exist a function whose inverse is its inverse? That is, is there a function [imath]f:\mathbb R_+\mapsto \mathbb R_+[/imath] whose functional inverse is also its multiplicative inverse, so that [imath]f^{-1}(x)=[f(x)]^{-1}, \space\space\space \forall x\in\mathbb R_+[/imath] Any ideas? I'll impose the restriction of continuity to deter nasty solutions. | 1585394 | Is there a function whose inverse is exactly the reciprocal of the function, that is [imath]f^{-1} = \frac{1}{f}[/imath]?
Is there a function whose inverse is exactly the reciprocal of the function? That is [imath]f^{-1} = \frac{1}{f}[/imath]. We know that the inverse of a function is not necessarily equal to its reciprocal in general. |
2592169 | How many cardinals exist between [imath]\aleph_0[/imath] and [imath]\mathfrak c[/imath]?
Assuming the negation of the continuum hypothesis, there is a set [imath]G_1 \subset \mathbb{R}[/imath] such that if [imath]\aleph^\prime_1[/imath] is the cardinality of [imath]G_1[/imath], then [imath]\aleph_0 < \aleph^\prime_1 < \mathfrak c[/imath]. Do there exist models of set theory in which there exist another set, say [imath]G_2[/imath] such that [imath]\aleph_0 < \aleph^\prime_1 < \aleph^\prime _2 < \mathfrak c[/imath] with [imath]\aleph^\prime _2[/imath] the cardinality of [imath]G_2[/imath] and so on (that is, the cardinality of [imath]G_2[/imath] is bigger than that of [imath]G_1[/imath] and less than the cardinal of the set if real numbers)? Is it possible to repeat this process and find sets with larger and larger cardinalities which still are less than [imath]\mathfrak c[/imath]? If the answer is yes, what are the consequences of this assumption in analysis? | 444273 | how many infinite cardinals are smaller than [imath]\aleph[/imath]?
What can be said in general (if we do not assume the continuum hypothesis) about the cardinality of the set of all infinite cardinals smaller than [imath]\aleph[/imath]? |
2592401 | If N is a quadratic residue modulo p for all primes p<N, is N a perfect square?
It is known that [imath]N[/imath] is a perfect square if and only if [imath]N[/imath] is a quadratic residue for every prime [imath]p[/imath]. This gives a good probabilistic algorithm for testing if a randomly chosen positive integer is a perfect square - simply compute it's Legendre Symbol for a sufficiently large set of randomly chosen primes. A single result of [imath]-1[/imath] definitively tells you that [imath]N[/imath] is not a perfect square, while repeated results of only [imath]+1[/imath] or [imath]0[/imath] increase the probability that [imath]N[/imath] is a perfect square. This probabilistic test could be much quicker with lower probability of a false positive if we could check only primes less than the integer [imath]N[/imath]. The proofs I've seen that [imath]N[/imath] is a perfect square IFF [imath]N[/imath] is a QR mod p for all primes [imath]p[/imath] cannot be used to prove the result using only the finitely many primes less than [imath]N[/imath]. Has anyone seen or know of a proof of this more specific result, or alternatively, know of a counter-example? I've burned more than a few CPU cycles on Mathematica looking for a counterexample and have not found one yet. | 2509448 | Squares in [imath]\mathbb{Z}_p[/imath] that are not squares in [imath]\mathbb{Z}[/imath]
The following question popped up during a conversation with a friend of mine: is there any integer [imath]n[/imath] such that [imath]n[/imath] is not a square in [imath]\mathbb{Z}[/imath] but it is a square in every field [imath]\mathbb{Z}_p[/imath] with [imath]p>n[/imath]? It sounds like an elementary question, and I am almost sure that the answer is no (mostly because of some distant memories of mine about a theorem of Hasse and Minkowsky), but I was not able to find a definitive answer. I tried to apply some Model Theory to it, but I don't think it can be a fruitful path to follow. Another thing I noticed is that if I can prove that there is an [imath]a[/imath] such that [imath]n=a^2[/imath] in two different fields [imath]\mathbb{Z}_p[/imath] and [imath]\mathbb{Z}_q[/imath], then the the statement must be false: otherwise, writing [imath]a^2=n+bp[/imath] and [imath]a^2=n+cq[/imath] for the smallest possible [imath]b[/imath] and [imath]c[/imath], we would have [imath]bp=cq[/imath], a contradiction since neither [imath]b[/imath] nor [imath]c[/imath] can be [imath]0[/imath]. But again, I got stuck after this consideration. Do you have any ideas or suggestions? |
2591859 | Splitting the exponent of a negative number?
Recently I have been chewing on a bit of a paradox in my mind, and I'm trying to figure out what I'm doing wrong. I am not in high school anymore, I have just graduated from college, and I haven't actively done any non-CompSci related math in a long time aside from youtube videos. So, I'm sorry if this question is really basic, I am versed in this kind of stuff, I am just having problems googling the answer to this one. Alright, so back in middle school Algebra we learned [imath](b^n)^m = b^{nm}[/imath]. What I'm wondering is, say b is a negative number raised to an odd power. Couldn't we write it in such a way that the answer is always positive? [imath]b^n[/imath] where [imath]b[/imath] is negative and [imath]n[/imath] is odd. Can't we always write it in the form [imath](b^2)^{n/2}[/imath]. Following the order of operations we can evaluate the parenthesis first, get a positive number, and from there it's a positive number raised to a positive, rational exponent. Which is always positive. Yet, if you did the expansion of the exponents, the answer is clearly negative. My googling of this property of exponents hasn't led to anywhere that stipulates that this property is only valid when [imath]b > 0[/imath], and from my point of view it's a very trivial question, so what am I doing wrong here? | 1628759 | What are the Laws of Rational Exponents?
On Math SE, I've seen several questions which relate to the following. By abusing the laws of exponents for rational exponents, one can come up with any number of apparent paradoxes, in which a number seems to be shown as equal to its opposite (negative). Possibly the most concise example: [imath]-1 = (-1)^1 = (-1)^\frac{2}{2} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = \sqrt{1} = 1[/imath] Of the seven equalities in this statement, I'm embarrassed to say that I'm not totally sure which one is incorrect. Restricting the discussion to real numbers and rational exponents, we can look at some college algebra/precalculus books and find definitions like the following (here, Ratti & McWaters, Precalculus: a right triangle approach, section P.6): The thing that looks the most suspect in my example above is the 4th equality, [imath](-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}[/imath], which seems to violate the spirit of Ratti's definition of rational exponents ("no common factors")... but technically, that translation from rational exponent to radical expression was not used at this point. Rather, we're still only manipulating rational exponents, which seems fully compliant with Ratti's 2nd property: [imath](a^r)^s = a^{rs}[/imath], where indeed "all of the expressions used are defined". The rational-exponent-to-radical-expression switch (via the rational exponent definition) doesn't actually happen until the 6th equality, [imath](1)^\frac{1}{2} = \sqrt{1}[/imath], and that seems to undeniably be a true statement. So I'm a bit stumped at exactly where the falsehood lies. We can find effectively identical definitions in other books. For example, in Sullivan's College Algebra, his definition is (sec. R.8): "If [imath]a[/imath] is a real number and [imath]m[/imath] and [imath]n[/imath] are integers containing no common factors, with [imath]n \ge 2[/imath], then: [imath]a^\frac{m}{n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m[/imath], provided that [imath]\sqrt[n]{a}[/imath] exists"; and he briefly states that "the Laws of Exponents hold for rational exponents", but all examples are restricted to positive variables only. OpenStax College Algebra does the same (sec. 1.3): "In these cases, the exponent must be a fraction in lowest terms... All of the properties of exponents that we learned for integer exponents also hold for rational exponents." So what exactly are the restrictions on the Laws of Exponents in the real-number context, with rational exponents? As one example, is there a reason missing from the texts above why [imath](-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}[/imath] is a false statement, or is it one of the other equalities that fails? Edit: Some literature that discusses this issue: Goel, Sudhir K., and Michael S. Robillard. "The Equation: [imath]-2 = (-8)^\frac{1}{3} = (-8)^\frac{2}{6} = [(-8)^2]^\frac{1}{6} = 2[/imath]." Educational Studies in Mathematics 33.3 (1997): 319-320. Tirosh, Dina, and Ruhama Even. "To define or not to define: The case of [imath](-8)^\frac{1}{3}[/imath]." Educational Studies in Mathematics 33.3 (1997): 321-330. Choi, Younggi, and Jonghoon Do. "Equality Involved in 0.999... and [imath](-8)^\frac{1}{3}[/imath]" For the Learning of Mathematics 25.3 (2005): 13-36. Woo, Jeongho, and Jaehoon Yim. "Revisiting 0.999... and [imath](-8)^\frac{1}{3}[/imath] in School Mathematics from the Perspective of the Algebraic Permanence Principle." For the Learning of Mathematics 28.2 (2008): 11-16. Gómez, Bernardo, and Carmen Buhlea. "The ambiguity of the sign √." Proceedings of the Sixth Congress of the European Society for Research in Mathematics Education. 2009. Gómez, Bernardo. "Historical conflicts and subtleties with the √ sign in textbooks." 6th European Summer University on the History and Epistemology in Mathematics Education. HPM: Vienna University of Technology, Vienna, Austria (2010). |
2591834 | Let Wolfram Alpha evaluate function for natural numbers only
How can I tell Wolfram Alpha that I only want it to evaluate my function for natural numbers? For example [imath] 2n [/imath], assuming n integer used to work, but not anymore. Is there a new syntax I can use? Edit: Since this was marked as duplicate: The question popped up a couple years ago already, but the syntax seems to be a little different now. The answers given back then don't work anymore. | 547109 | Tell Wolfram Alpha that a variable is a natural number
How can I tell Wolfram Alpha that some variables are natural numbers, when I want to solve a equation? An example of what I want to do: [imath]\binom{n}{k}\cdot p^k \cdot (1-p)^{n-k} = \frac{1}{\sqrt{n\cdot p \cdot (1-p)}}\cdot \frac{1}{\sqrt{2 \cdot \pi}} \cdot e^{-\frac{1}{2} \cdot \left(\frac{x-np}{\sqrt{n \cdot p \cdot (1-p)}}\right)^2}[/imath] solve for [imath]x[/imath] with [imath]n,k \in \mathbb{N}[/imath], [imath]0 \lt p \lt 1[/imath]. |
2592034 | How can I find the number of integral Pythagorean triples with a given hypotenuse?
How can I find the number of integral pythagorean triples if the hypotenuse is [imath]2^{4035}[/imath]? I want to do this without finding them all and counting them. | 1928753 | Find the cardinality of the following set [imath]S_n=\{(x,y)|x^2+y^2=n, \text{where }x,y,n \in \mathbb Z\text{ and }n\geq0 \}[/imath]
We are given with [imath]S_n=\{(x,y)|x^2+y^2=n, \text{where }x,y,n \in \mathbb Z\text{ and }n\geq0 \}[/imath] Now observe that [imath]-\sqrt n\leq x\leq\sqrt n [/imath] and similarily, [imath]-\sqrt n\leq y\leq\sqrt n [/imath]. This means that [imath]-\left \lfloor{\sqrt n}\right \rfloor\leq x\leq \left \lfloor{\sqrt n}\right \rfloor[/imath] and [imath]-\left \lfloor{\sqrt n}\right \rfloor\leq y\leq \left \lfloor{\sqrt n}\right \rfloor[/imath]. Through this result one can easily write code and figure out the cardinality of set [imath]S_i[/imath] but I was wondering whether an expression can be found that would allow one to directly compute [imath]|S_i|[/imath]. |
2592919 | how to evaluate the integral [imath]\int_0^1 \frac{\ln(1+x)}{x} dx[/imath]？
How do I evaluate the integral: [imath] \int_0^1 \frac{\ln(1+x)}{x} dx [/imath] in a simple way, without using complex methods? | 2046219 | Find: [imath] \int^1_0 \frac{\ln(1+x)}{x}dx[/imath]
Find: [imath] \int^1_0 \frac{\ln(1+x)}{x}dx[/imath] There is suppose to be a clean solution (maybe some symmetry involved?) I have tried integration by parts as followed: [imath]\ln(x+1)=u[/imath] ,[imath]\frac{1}{x+1} dx = du[/imath] and also [imath]\frac{1}{x}dx=dv[/imath], [imath]\ln(x)=v[/imath] which means our integral becomes [imath]\int^1_0 \frac{\ln(1+x)}{x}dx=\ln(x+1)\ln(x)|^1_0 - \int^1_0 \frac{\ln(x)}{x+1}[/imath] which does not make this easier. I have also tried using the identity [imath]\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx[/imath] So let [imath]I = \int^1_0 \frac{\ln(1+x)}{x}dx[/imath] and also [imath]I = \int^1_0 \frac{\ln(2-x)}{1-x} [/imath] so [imath]2I= \int^1_0 \ln(1+x)+\ln(2-x)[/imath] which also doesn't make it easier? Any ideas! :) |
2585832 | gcd[imath](a,b)=1[/imath] and [imath]ax=by\Rightarrow a|y[/imath]
Define [imath]gcd(a,d)=1[/imath] as [imath]m|a\wedge m|b\Rightarrow m=\pm 1[/imath]. Of course if we use the unique prime decomposition it is trivial. Is there any other easier proofs avoiding using that theorem? | 904406 | If [imath]a[/imath] divides [imath]bc[/imath] and [imath]\gcd(a,b) = d[/imath] then [imath]\frac a d[/imath] divides c
I'm trying to prove that if [imath]a[/imath] divides [imath]bc[/imath] and [imath]\gcd(a,b) = d[/imath] then [imath]\frac a d[/imath] divides c. I tried using Bezout identity but couldn't get anywhere. |
180150 | Alternating sum of squares of binomial coefficients
I know that the sum of squares of binomial coefficients is just [imath]{2n}\choose{n}[/imath] but what is the closed expression for the sum [imath]{n\choose 0}^2 - {n\choose 1}^2 + {n\choose 2}^2 + \cdots + (-1)^n {n\choose n}^2[/imath]. | 2594102 | Using Binomial Theorem to find the value of [imath] ^nC_0^2 - ^nC_1^2 + ^nC_2^2 - ^nC_3^2... + (-1)^n . ^nC_n ^2 [/imath]
So I was solving problems in my textbook and I got stuck on this question. It says: Prove that: [imath] ^nC_0^2 - ^nC_1^2 + ^nC_2^2 - ^nC_3^2 + ... + (-1)^{n} .^nC_n^2 [/imath] = [imath] 0 [/imath] or [imath] \frac {(-1)^{n/2} n!} {\frac{n}{2}! \frac{n}{2}!} [/imath] , when [imath] n [/imath] is odd/even. So I tried to bring the required terms and my idea was to multiply [imath] (1+x)^n [/imath] with [imath](x - 1)^n[/imath] . [imath] (1+x)^n = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + ... ^nC_n x^n [/imath] [imath] (x-1)^n = ^nC_0x^n - ^nC_1 x^{n-1} + ^nC_2 x^{n-2} + ... (-1)^n .^nC_n [/imath] I can see that if I multiply the two equations, then I will get the terms I need with their required sign. LHS will contain [imath] (x^2 - 1)^n [/imath] . But along with that, there will be many other terms on the RHS which will probably be impossible for me to get rid of. Am I going in the right way or is there an entirely different procedure? |
2110348 | series: [imath]\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots[/imath]
We have the infinite series:[imath]\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots[/imath] This is not my series: [imath]\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{4\cdot 5\cdot 6}+\cdots[/imath] so I cannot use [imath]\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}[/imath] My attempt: I know that this type of series solved by making it telescoping series but here, I am unable find general term of the series. Thank you. | 1663080 | How to solve [imath]\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...[/imath]
As my question says, [imath]\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...[/imath] As far as [imath]\frac 1{1.2}+\frac 1{2.3}+\frac 1{3.4}+...[/imath] is concerned, we can write it as [imath]\frac 1{n-1} - \frac 1{n}[/imath] What can we do here? |
2070975 | Prove that [imath] L = { \{\ a^{i^{2}} : i \in N \ \cup \{0\} \ \}} [/imath] is not regular using pumping lemma
I have this language: [imath] L = { \{\ a^{i^{2}} : i \in N \ \cup \{0\} \ \}} [/imath] I need to prove that the language is not regular by using pumping lemma. I assume that the language is regular so I choose suitable word with which I will be trying to prove that it's not regular language. The word that is suitable should be [imath]w = a^{p^{2}}[/imath] then I can decompose that word to [imath]xyz[/imath] parts [imath]|xy| \leq p \ \land |y| \ge 1[/imath] The decomposition is: [imath]x = a^{l}\land l\ge0\\y= a^{k}\land k\ge1\\z=a^ma^{p^{2}-p}\land m\ge0\\l+k+m=p\ \land l+k\leq p [/imath] Now I will pump the [imath]y[/imath] part in [imath]xy^{i}z[/imath] so the word won't be in the language anymore and that means that the language was not regular. The problem is I have absolutely no idea which [imath]i[/imath] should I choose to break the pumping property and show that the language was not regular. Could anyone explain me for what I should be searching in order "break" pumping property and what [imath]i[/imath] to choose? | 1559724 | Proving that [imath]\mathscr L=\{0^n \big|\text{n is the square of a natural number }\}[/imath] is non regular using the pumping lemma
I need to prove that the language [imath]\mathscr L=\{0^n \big|\text{n is the square of a natural number}\}[/imath] is non regular using the pumping lemma My try: [imath]\mathscr L=\{\overbrace{\epsilon}^{0^2},\overbrace{0}^{1^2},\overbrace{0000}^{2^2},\overbrace{000000000}^{3^2},\overbrace{0000000000000000}^{4^2},\dots\}[/imath] Suppose that [imath]\mathscr L[/imath] is regular so [imath]\exists[/imath] a word '[imath]x[/imath]' with length of at least [imath]n[/imath] [imath]|x|\geq n [/imath] such that [imath](1)\,\,\,|uv|\leq n[/imath] [imath](2)\,\,\,|v|\geq 1[/imath] [imath](3)\,\,\,uv^iw \in \mathscr L[/imath] Now, let as choose the word [imath]\color{blue}{x=0^{n^2}}[/imath] [imath]|x|\geq n[/imath] so we can use [imath](1)-(3)[/imath] [imath]x=\underbrace{000000\dots}_{n^2\text{ times}}[/imath] [imath]x=uv0^kw=\underbrace{uv0^k}_{\text{n zeros }}\overbrace{w}^{n^2-n\text{ zeros }}[/imath] Let as choose [imath]\color{blue}{i=??}[/imath] I'm difficult choosing the word '[imath]\color{blue}x[/imath]' and choosing [imath]\color{blue}i[/imath] to come to contradiction with the lemma |
2593564 | Is [imath]e^{i\theta}[/imath] multivalued or singlevalued?
We know from Euler's formula that for [imath]\theta \in \mathbb{R}[/imath] we have [imath]e^{i\theta} = \cos(\theta) + i \sin (\theta)[/imath] which is a single complex number. However, If I apply the definition of complex power [imath]z^\alpha = e^{\alpha \log z}[/imath] I get the following for [imath]z=e[/imath] and [imath]\alpha = i \theta[/imath]: [imath]e^{i \theta} = e^{i \theta \log e} = e^{i \theta (Log|e| + i\arg e)} = e^{i\theta (1+2k\pi i)} = e^{-2k\pi \theta} \cdot e^{i\theta}[/imath] with [imath]k \in \mathbb{Z}[/imath] This means that [imath](1- e^{-2k\pi \theta}) \cdot e^{i \theta} = 0[/imath]. We know that [imath]e^{i \theta}[/imath] cannot be zero. So it must be that [imath]1- e^{-2k\pi \theta} =0[/imath] for all real [imath]\theta[/imath]. This means that [imath]k=0[/imath]. So [imath]e^{i\theta} = e^{-2k\pi \theta} \cdot e^{i\theta}[/imath] does not hold for all [imath]k \in \mathbb{Z}[/imath]. What is going on here? Also [imath]e^{i \theta}[/imath] is supposed to take infinitely many values since the exponent is neither a real integer, nor a rational fraction. Which way is it? EDIT: If I input [imath]e^{\pi i}[/imath] to Wolfram Alpha: https://www.wolframalpha.com/input/?i=e%5E(pi*i) I get both a single valued result -1 and a multivalued result. | 1319952 | Definition of exponential function, single-valued or multi-valued?
If we define [imath]e^z=1+z+\frac{z^2}{2!}+\cdots[/imath] then it is single-valued. However, if we write [imath]e^z=e^{z\ln e}[/imath] then it is multi-valued. Besides, [imath]a^z[/imath] is multi-valued in general. It is kind of strange if only when the base is [imath]e[/imath] that it is single-valued. My thought: Is it true that there are two exponential functions, let's call them [imath]\exp(z)[/imath] and [imath]e^z[/imath]? Where [imath]\exp(z)[/imath] is defined by [imath]\exp(z)=1+z+\frac{z^2}{2!}+\cdots[/imath] and is single-valued, while [imath]e^z[/imath] is defined by [imath]e^z = \text{exp}(z\ln e)[/imath] and is multi-valued? Here [imath]\ln z[/imath] is defined by [imath]\exp(\ln z)=z[/imath] and is multi-valued. |
2358715 | Is [imath]R=K[\![x^3,x^2y,xy^2,y^3]\!][/imath] a Gorenstein ring?
Let [imath]K[/imath] be a field and [imath]R=K[\![x^3,x^2y,xy^2,y^3]\!][/imath] the ring of formal power series. Is [imath]R[/imath] a Gorenstein ring? [imath]R[/imath] is Cohen-Macaulay of dimension 2. So, I have to check if [imath]Ext^2_{K}(K,R)=K.[/imath] | 2359819 | Is [imath]R=K[|x^3,x^2y,xy^2,y^3|][/imath] a Gorenstein or a regular ring?
Let [imath]K[/imath] be a field and [imath]R=K[|x^3,x^2y,xy^2,y^3|][/imath] the ring of formal power series. Is [imath]R[/imath] a Gorenstein ring? [imath]R[/imath] is Cohen-Macaulay of dimension 2. So, I have to check if [imath]\operatorname{Ext}^2_{K}(K,R)=K.[/imath] I need some hints about that. |
2593538 | Let [imath]U,V \sim R(0,1)[/imath] be independent. Calculate [imath]P(U \leq V)[/imath]
Let [imath]U,V \sim R(0,1)[/imath] be independent. Calculate [imath]P(U \leq V)[/imath]. About notation: [imath]U,V \sim R(0,1)[/imath] mean random variables [imath]U,V[/imath] is equally distributed from interval [imath]0[/imath] to [imath]1[/imath]. I think there is not much to do because we have given [imath]P(U \leq V)[/imath], there is symmetric rule which say [imath]P(U \leq V) = P(V \leq U) = 0.5[/imath] Is this really easy like that? Or maybe that rule is not really useable here? | 1092192 | Let [imath]U, V \sim U(0,1)[/imath] be independent. What is [imath]P(U \leq V)[/imath]?
Let [imath]U, V \sim U(0,1)[/imath] be independent. What is [imath]P(U \leq V)[/imath]? My attempt: We are looking for [imath]P(U-V \leq 0)[/imath]. For a given [imath]t \in [0,1][/imath] this is equivalent to [imath]P(U \leq t, V \gt t)= P(U \leq t)P(V \gt t)=P(U \leq t)(1-P(V\leq t))=t(1-t)=t-t^2[/imath] Then summing over [imath]t[/imath] on [imath][0,1][/imath], I get: [imath]\int_0^1 t-t^2 \, dt = \frac{1}{6}[/imath] Which seems like a reasonable value, but I'm not exactly sure whether my approach with doing this in terms of t was correct, can anybody please validate my solution? |
2594011 | Find pairs of [imath](x, y)[/imath] satisfying the given equation.
Find all pairs [imath](x, y)[/imath] of integers such that [imath]xy+\frac{x^3+y^3}{3}=2007[/imath] [imath]\mathbf{Try:}[/imath] The given expression can be simplified as [imath]x^3+y^3+3xy=6021[/imath] [imath]\Rightarrow (x+y-1)[(x-y)^2+(y+1)^2+(x+1)^2]=12040[/imath] I tried considering the divisors of [imath]12040[/imath] but seems to be a much tedious task. Now how to proceed further. | 188737 | All pairs (x,y) that satisfy the equation [imath]xy+(x^3+y^3)/3=2007[/imath]
How we can find the all pairs [imath](x,y)[/imath] from the integers numbers ,that satisfy the equation : [imath]xy+\frac{x^3+y^3}{3} =2007[/imath] |
2594126 | Open sets having open images in [imath]R^2[/imath]
Let [imath]h:\Bbb R^2\to\Bbb R^2[/imath] be a surjective function such that [imath]\|h (x)-h (y)\|\ge 3 \|x-y\|[/imath]. Prove that the image of every open set in [imath]\Bbb R^2[/imath] under the map [imath]h[/imath] is open in [imath]\Bbb R^2[/imath]. My question is that as the limit of the given function remains greater equals 3 so can it be proved continuous and Then as not every continuous functions show that image is open then?? | 2202025 | [imath]\|h(x)-h(y)\|\geq 3\|x-y\|[/imath] prove that image of open is open
Let [imath]h:\mathbb{R^2}\rightarrow \mathbb{R^2}[/imath] be a surjective map such that [imath]\|h(x)-h(y)\|\geq 3\|x-y\|[/imath] for all [imath] x,y\in \mathbb{R^2}[/imath]. Here [imath]\|\cdot\|[/imath] denotes euclidean norm on [imath]\mathbb{R^2}[/imath]. Prove that the image of every open set (in [imath]\mathbb{R^2}[/imath]) under [imath]h[/imath] is an open set. My try : Every open set in [imath]\mathbb{R^2}[/imath] can be written as union of balls [imath]B_r(a)[/imath] of radius [imath]r[/imath] entered at [imath]a[/imath] and every ball [imath]B_r(a)[/imath] can be obtained by scaling up and translating the center of the unit ball [imath]B_1(0)[/imath] in [imath]\mathbb{R^2}[/imath] .So it is enough to prove that [imath]h(B_1(0))[/imath] is open set . So i have to show that for any [imath](a,b)\in \mathbb{R^2}[/imath] there exists an open ball [imath]B_r(h(a,b)) [/imath] such that [imath]B_r(h(a,b)) \subset h(B_1(0))[/imath] . My problem is that i can't find an appropriate [imath]r[/imath] . Please help from here. If the above approach is completely wrong then provide a solution. Thank you |
2594182 | Find the coefficient of [imath]x^n[/imath] in [imath]\frac{x^4}{(1-x^2)(1-x)^2}[/imath]
I'm solving a combinatorics problem using generating function and reached the following expression: [imath]\frac{x^4}{(1-x^2)(1-x)^2}[/imath] How can I expand it to find a general expression for the coefficient of [imath]x^n[/imath]? Any help would be appreciated. | 2593478 | Find coefficient of [imath]x^{n}[/imath] in [imath]\frac{1}{(1-x)^3(1+x)^2}[/imath]
I want to find a general expression for [imath]x^{n}[/imath] in [imath]\frac{1}{(1-x)^3(1+x)^2}[/imath]. I have tried to use partial fraction but that got complicated and I got stuck. Any help would be appreciated. Note: I would need a closed form expression |
2595035 | Find an abelian group [imath]V[/imath] and a field [imath]F[/imath] such that [imath](V,F)[/imath] can define two distinct vector spaces
This is an exercise of Advanced linear algebra of Roman Find an abelian group [imath]V[/imath] and a field [imath]F[/imath] such that [imath](V,F)[/imath] can define two distinct vector spaces. That is: I must find two distinct definitions of scalar multiplication. What I did was choose [imath]V:=(\Bbb R^2,+)[/imath] as the abelian group, and [imath]F:=\Bbb C[/imath] as the field, and defined scalar multiplication by [imath](a+bi)\cdot (x,y)=(ax-by,by+ax)[/imath], this defines a vector space because its just the complexification of [imath]\Bbb R[/imath] as a vector space. [imath](a+bi)\cdot (x,y)=(by+ax,ax-by)[/imath] this is symmetric from the above definition. But I dont found other example. Can someone show me a non so trivial example, or provide me a general rule to define such alternative scalar multiplications? UPDATE: it seems that, from a known scalar multiplication defined by [imath]d[/imath], I can define an alternative scalar multiplication by [imath]d\circ f[/imath], where [imath]f:V\to V[/imath] is a group automorphism. | 102800 | number of differents vector space structures over the same field [imath]\mathbb{F}[/imath] on an abelian group
My question here raised another one. How many differents vector space structures over a field [imath]\mathbb{F}[/imath] we may have on an abelian group? I know that there are abelian groups that we can not endow it with a structure of vector space over any field, for example [imath]\mathbb{Z}_{6}[/imath]. But if an abelian group has a structure of a vector space over a field [imath]\mathbb{F}[/imath], is there an upper bound for ways we can define a different structure? Such as the number of automorphisms of the field [imath]\mathbb{F}[/imath]. I am conjecturing according to the answer given before. |
2594814 | existence of a complex sequence
Let [imath]z_n[/imath] be a sequence of nonzero complex numbers such that [imath]z_{n+1}=z_n^2+z_n[/imath] for all [imath]n\in\mathbb{Z}[/imath]. Can such a sequence satisfy [imath]\lim_{n\to +\infty}z_n=\lim_{n\to -\infty}z_n=0[/imath] ? A friend from the university showed me this question saying it's a student competition problem from the 90's. None of us and our colleagues was able to solve it, nobody had even an idea how to start the solution. ======================= It is NOT a duplicate. There is a big difference between real and complex numbers and sequences, especially when one wants to use the fact that being bounded and monotonic implies convergence, which was the main observation in the linked question! | 661631 | Question about convergence of a simple recursive sequence
Define the recursive sequence [imath](a_n)[/imath] as [imath]a_{n+1} = a_n^2 + a_n[/imath] where [imath]n \in \mathbb{N}[/imath]. We want to find [imath]a_1 \in \mathbb{R}[/imath] such that [imath](a_n)[/imath] converges to a limit that depends on [imath]a_1[/imath]. Attempt: Let [imath]a_1 = k[/imath], then [imath]a_2 = k^2 + k[/imath], [imath]a_3 = (k^2 + k)^2 + (k^2 + k)[/imath], ... clearly if we have |k| > 1 then the sequence [imath](a_n)[/imath] will diverge. From this we deduce the interval of convergence must be when [imath]|k| \le 1[/imath]. Suppose [imath](a_n) \to L[/imath] then [imath]a_{n+1} = a_n^2 + a_n[/imath] [imath]\implies L = L^2 + L \implies L = 0[/imath]. Surely if [imath]k = 0[/imath] this will occur as [imath]a_n = 0[/imath] for all n. But it is not clear to me if there are other [imath]k[/imath] such that the recursive sequence converges. |
2595002 | Prove algebraically [imath]\vec{x} = \vec{a} + s\vec{b}, s\in \mathbb R[/imath] and [imath]\vec{x} = \vec{c} + t\vec{d}, t\in \mathbb R[/imath] intersect
Given two lines [imath]\vec{x} = \vec{a} + s\vec{b}, s\in \mathbb R[/imath] and [imath]\vec{x} = \vec{c} + t\vec{d}, t\in \mathbb R[/imath] in [imath]\mathbb R^2[/imath], where [imath]\vec{d} \not= k\vec{b}, k\in \mathbb R[/imath], prove algebraically these two lines intersect. I know logically that this is true since if two lines are not parallel then they must intersect eventually, but how do I show this algebraically? | 2594592 | Proving two lines in vector form intersect algebraically
let [imath]x,y,a,b,c,d[/imath] be vectors and [imath]r,k \in \Bbb{R}[/imath] with vector [imath]d[/imath] not being a scalar multiple of vector [imath]b[/imath] How would you prove that the line [imath]\vec{x}=\vec{a}+r\vec{b}[/imath] intersects the line [imath]\vec{y}=\vec{c}+k\vec{d}[/imath] algebraically in [imath]\Bbb{R^2}[/imath]? Geometrically In [imath]\Bbb{R^2}[/imath] since [imath]\vec{b}[/imath] and [imath]\vec{d}[/imath] are not scalar multiples then they are not parallel and since they are in the x-y plane then obviously the two lines will not intersect, however i am having trouble expressing this algebraically |
2595558 | What does [imath]2^π[/imath] : the multiplication of [imath]2 \pi[/imath] many times?
This is rather an intuitive question, in the sense that indeed a real number raised to the power of an irrational doesn't make sense but I wanted to know what does it mean intuitively. If [imath]2^7=2*2*2*2*2*2*2, 2^{\frac{1}{2}}=\frac{1}{\sqrt{2}}[/imath] and [imath]a^b=a*a*a*a*a*...*a\, \, b[/imath] many times, I'm curious as to what does [imath]2^π[/imath] mean. Obviously it's equal to [imath]8.8249778271...[/imath] but what's the meaning in the aforementioned scenario? | 2577354 | Approximation of real number to the power of irrational
Say I have two numbers, [imath]c\in\Bbb R^+\,,q\in\Bbb{R\setminus Q}[/imath]. Is there a way to approximate the value of [imath]c^q[/imath]? This question bugging me because of how simple it sound in compare to how complicated it is. First I want to understand what a power of irrational number is, when talking about [imath]\Bbb R^+[/imath](I use this domain for sake of simplicity) we have very well defined actions for power a rational number, we can even use some theorems to get [imath]c^{a/b}=\sqrt[b]{c^a}[/imath], but when I am dealing with irrational number I can not do this. I know that we have some definition for this because the solution of the equation [imath]2^x=3[/imath] is irrational. So what exactly does it mean to take something to a irrational power and is there a way to approximate a numeric value for this? |
2595354 | Is ture that [imath]L/K[/imath] is normal extension?
Considering a field tower [imath]K \subset E \subset L [/imath], [imath]E/K[/imath] is separable extension,[imath]L/E[/imath] is normal extension, is ture that [imath]L/K[/imath] is normal extension? | 2592898 | Is it Galois extension?
If [imath]E/F[/imath] is separable extension,[imath]N/E[/imath] is normal extension,is [imath]N/F[/imath] a galois extenstion? I guess the answer is no.I do not know how to prove it.If [imath]E_s[/imath] is the separable closure of [imath]F[/imath],[imath]N[/imath] is normal closure of [imath]E_s[/imath],is the [imath]N/F[/imath] is galois extension? |
2595426 | [imath]I = [0,1][/imath] with euclidean topology. [imath]f : I \rightarrow I^2 = I \times I[/imath]. If f is surjective then f is not injective
I have this problem which seems trivial, but I do not know how to get to the answer. I have the set [imath][0,1][/imath] with the Euclidean topology, and an arbitrary continuous function [imath]f : I \rightarrow I^2 = I \times I[/imath]. I need to prove that if it is surjective then it cannot be injective. I tried to think to the problem in this way: if [imath]f[/imath] is surjective and also injective, than we would have a homeomorphism between [imath]I[/imath] and [imath]I \times I[/imath]. I wanted to say that one is compact and the other is not to reach a contradiction. However in my opinion they are both compact because product of compacts is compact as well, so I do not see why it would be a problem to have the injective property. Any tips? | 1266211 | On the existence of a continuous bijection [imath]f\colon [0,1]\to [0,1]\times [0,1][/imath]
Let [imath]f[/imath] be a continuous function on [imath][0,1][/imath] such that [imath]f([0,1])=[0,1]\times[0,1].[/imath] Then show that [imath]f[/imath] is not one-one. Hints will be appreciated. |
2595424 | Proof in complex
Prove that for each [imath]z\in \mathbb{C}[/imath] we have [imath]\overline{\cos z} = \cos \overline{z}[/imath] using [imath] \cos z = {e^{iz}+e^{-iz}\over 2}[/imath]. Can you please give me a link or show this proff using the equation i wrote. My english sucks so i can really search for it. | 2485640 | Conjugate to [imath]\cos z[/imath] is not equal to [imath]\cos z[/imath] conjugate [imath]\overline{\cos(z)}\neq\cos(\overline{z})[/imath]
I have a dilemma. I have a task where I'm supposed to show that [imath]|\cos z|^2+|\sin z|^2=1[/imath] if and only if [imath]$z\in \mathbb{R} $[/imath]. In my argument I have that [imath]\cos z\cdot\overline{\cos z} + \sin z\cdot\overline{\sin z}=\cos z\cdot \cos\overline{z} + \sin z\cdot \sin\overline{z}[/imath] but my teacher said that [imath]\overline {\cos z} \neq \cos \overline{z}[/imath]. Is this true? But [imath]\overline{\cos(z)} = \overline{\frac{e^{iz}+e^{-iz}}{2}}=\frac{e^{-i\overline{z}}+e^{i\overline{z}}}{2}=\cos(\overline{z})[/imath] |
2595818 | If [imath]a,b,c \in \mathbb{Z}[/imath] and [imath]a^2 - b^2 = c[/imath] then [imath]a = \frac{m+n}{2}, b = \frac{m-n}{2}[/imath]
Let [imath]a,b,c \in \mathbb{Z}[/imath]. Proove that if [imath]a^2 - b^2 = c[/imath] then exists [imath]m,n \in \mathbb{Z}[/imath] which are both even/odd such that [imath]a = \frac{m+n}{2}, b = \frac{m-n}{2}, c = mn[/imath] I think I should use Fermat's theorem, but I'm not sure how to do it | 588521 | Integer solutions to [imath]c=a^2-b^2[/imath]
I have been working on the following problem: For a given [imath]c\in \Bbb Z^+[/imath], find [imath]a,b \in \Bbb Z^+: c=a^2-b^2[/imath]. I have already figured out and proved a number of things: A way to directly determine if a particular instance is solvable: [imath]\exists a,b \in \Bbb Z^+: c=a^2-b^2 \iff c \not\equiv 2 \mod 4[/imath] An upper bound on the values for a (and b): [imath]b \lt a\le\lfloor\frac{c-1}{2}\rfloor[/imath] And a simple algorithm to find a solution: [imath] \begin{align*} (a_0, b_0) &= (0, 0) \\ (a_{n+1}, b_{n+1}) &= \begin{cases} (a_n+1, &b_n&)&\text{if } a_n^2-b_n^2 < c\\ (a_n, &b_n+1&)&\text{if } a_n^2-b_n^2 > c\\ \end{cases} \end{align*} [/imath] This can be iterated until a solution is found or the upper bound is reached. I am currently trying to work out a more direct way to find an answer, but I have been unable to figure out any other ways to approach the problem. How can I more directly find values for [imath]a[/imath] and [imath]b[/imath] for a given value of [imath]c[/imath]? |
2595771 | [imath]r[/imath]-separated sequence in Hilbert spaces
I'm trying (with no good results) to prove the follwing: Prop. Let [imath]B^1\subset H[/imath] be the unit ball in the Hilbert space [imath]H[/imath]. If [imath](x_n)_{n\in\mathbb{N}}\subset B^1[/imath] is a [imath]r[/imath]-separated sequence, i.e. [imath]\vert\vert x_j-x_k\vert\vert\geq r[/imath], for any [imath]j\neq k[/imath], then [imath]r\leq \sqrt{2}[/imath]. Can you help me to prove the reult above? | 2594699 | Sequences in Hilbert spaces
We have a sequence [imath](x_{n})_{n=1}^\infty[/imath] in a Hilbert space [imath]H[/imath] and we know that: for every [imath]n \in \mathbb{N}[/imath] holds [imath]\|x_{n}\|\le1[/imath] for every [imath]m\ne n[/imath] holds [imath]\|x_{n}-x_{m}\|\ge r > 0[/imath] How I can show that [imath]r \le \sqrt2[/imath] ? |
2579460 | Is there any theorem that states an indefinite integral is left and right differentiable everywhere?
Suppose we take some function [imath]f(x)[/imath] and have the integral [imath]\int_0^x f(x) = F(x)[/imath]. Note, [imath]f[/imath] is not continuous so the first fundamental theorem need not apply to this instance. Suppose that [imath]f[/imath] was integrable everywhere and has no vertical asymptotes. Is there any theorem that states [imath]F[/imath] is left and right differentiable everywhere? I'm trying to find such a theorem. | 2576488 | Verifying a Superior Integration Method for Step Functions
I should note that this is an integration algorithm and therefore intermediate steps DO appear to be unjustified. The purpose of this question is to justify or reject this algorithm as always giving correct solutions. Suppose we have some piecewise continuous function [imath]f[/imath] and that it can be written in the form [imath]f(x) = G(x,\lfloor g_1(x) \rfloor,\lfloor g_2(x) \rfloor,\cdots, \lfloor g_n(x) \rfloor)[/imath], for some functions [imath]g_1(x), g_2(x), \cdots g_n(x)[/imath] and [imath]G(x,y_1,y_2,\cdots, y_n)[/imath] such that [imath]\lfloor g_1(x) \rfloor,\lfloor g_2(x) \rfloor,\cdots, \lfloor g_n(x) \rfloor[/imath] are piecewise constant functions and [imath]G[/imath] is piecewise continuous with respect to [imath]x[/imath]. Now we have the algorithm to compute the general integral for [imath]f[/imath]: INTEGRATE-WITH-FLOOR(f) Let [imath]H(x,y_1,y_2,\cdots, y_n) = \int G(x,y_1,y_2,\cdots, y_n) dx[/imath] be a new function. Let [imath]F(x) = H(x,\lfloor g_1(x) \rfloor,\lfloor g_2(x) \rfloor,\cdots, \lfloor g_n(x) \rfloor)[/imath] be a new function. Let [imath]C(x)[/imath] be a new piecewise constant function such that [imath]F(x) + C(x)[/imath] is continuous. Return [imath]F(x) + C(x)[/imath]. Now I know this is more like an algorithm from a Computer Science class than a Math forum but I think it works and I seek verification. Furthermore: Does this algorithm compare to any existing methods. I was taught this method back when I was in calculus a few years ago but I have never seen it since then nor understood if it was justified or not. Is there any way to extend this to differential equations at all or would it be too difficult? Note: not looking for algorithms. Just general discussion of how it would be extended is just fine as well. |
983043 | Is [imath]\frac{\vert x+y\vert}{1+\vert x+y\vert}\leq\frac{\vert x\vert}{1+\vert x\vert}+\frac{\vert y\vert}{1+\vert y\vert}[/imath] true?
Is [imath]\frac{\vert x+y\vert}{1+\vert x+y\vert}\leq\frac{\vert x\vert}{1+\vert x\vert}+\frac{\vert y\vert}{1+\vert y\vert}[/imath] true for all [imath]x,y\in\mathbb{R}[/imath]? If not, how can I prove that [imath]\int\frac{\vert f-h\vert}{1+\vert f-h\vert}\leq\int\frac{\vert f-g\vert}{1+\vert f-g\vert}+\int\frac{\vert g-h\vert}{1+\vert g-h\vert}[/imath]? I tried C-S: [imath]$$\sum_{cyc}\frac{|x|}{1+|x|}=\sum_{cyc}\frac{x^2}{|x|+x^2}\geq\frac{(x+y)^2}{x^2+y^2+|x|+|y|}.$$[/imath] Thus, it's enough to prove that [imath]$$\frac{(x+y)^2}{x^2+y^2+|x|+|y|}\geq\frac{(x+y)^2}{|x+y|+(x+y)^2}$$[/imath] or [imath]$$|x+y|+(x+y)^2\geq x^2+y^2+|x|+|y|,$$[/imath] which is wrong for [imath]$(x,y)=(1,-1).$[/imath] | 2648468 | Use Triangle Inequalities to Prove an Inequality
Use the triangle inequalities to prove that: [imath]\frac{|a+b|}{1+|a+b|}\le\frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}[/imath] for all [imath]a,b\in\mathbb{R}[/imath]. I've made it this far, but it doesn't seem to be helping much; what I end up with at the end isn't very useful: Clearly I'm taking a wrong turn somewhere, but I'm not sure what else to try. Any advice? |
2595869 | Does Stokes theorem depend on the definition of a closed curve?
The two-dimensional Stokes theorem, written in vector notation, reads [imath]\oint_\mathcal{C} \vec{F} \cdot d \vec{r} = \int_\mathcal{S} (\vec{\nabla} \times\vec{F})\cdot d \vec{S},[/imath] where the closed curve [imath]\mathcal{C}[/imath] is the boundary of the surface [imath]\mathcal{S}[/imath], [imath]\mathcal{C} = \partial \mathcal{S}[/imath]. Now, the usual definition of a closed plane curve [imath]\gamma[/imath] is [imath]\gamma:[a,b] \to \mathbb{R}^2,[/imath] with [imath]\gamma(a) = \gamma(b)[/imath]. However, wouldn't it make more sense to define a closed curve as [imath]\gamma:[a,b\rangle \to \mathbb{R}^2,[/imath] with the condition [imath]\lim_{t \to b} \gamma(t) = \gamma(a)?[/imath] In this way, no point on the curve is parametrized twice (assume [imath]\gamma[/imath] is injective), and this is consistent with, e.g., the polar coordinate [imath]\varphi[/imath] which has the domain [imath]\varphi \in [0,2\pi\rangle[/imath]. My question: would the change in the definition of a closed curve alter Stokes theorem in any way? | 2595048 | Proper parametrization of a closed curve
Let [imath]\gamma:I\to\mathbb{R}^2[/imath] be a closed plane curve, for simplicity, a unit circle. Therefore, we have [imath]\gamma(\varphi) = (\cos \varphi, \sin \varphi).[/imath] What is the proper domain of [imath]\varphi[/imath]? Wikipedia says it's [imath]\varphi \in [0,2\pi][/imath] with [imath]\gamma(0) = \gamma(2\pi)[/imath]. What is the advantage of this domain rather than a domain [imath]\varphi\in[0,2\pi\rangle[/imath] with the additional condition that [imath]\lim_{\varphi \to 2\pi}\gamma(\varphi)=\gamma(0)?[/imath] It seems to me that the second definition is much more natural since no point on the curve is repeating, and the usual angle variable in the polar cordinates uses this domain. |
2596014 | Determine if the sequence [imath]a_{n+1}=\arctan{a_n}[/imath] is convergent and compute its limit.
Determine if the sequence [imath]\{a\}_{n=1}^{\infty}[/imath] defined by [imath]a_1>0\\a_{n+1} =\arctan{a_n}[/imath] is convergent and compute its limit. Will Banachs Fixed Point theorem be of any help here? My attempt. We note that the recurrence relation is of the form [imath]x_{n+1}=f(x_n),[/imath] let's introduce the help function [imath]f(x)=\arctan{x}.[/imath] I know that the equation [imath]f(x)=0[/imath] has a root [imath]x=0[/imath], so I'll choose an interval [imath]I=[0,\frac{1}{2}].[/imath] In order to use the fixed point theorem, the following criterias have to be met: [imath]f(x)[/imath] has to be defined and continuous on [imath]I[/imath] (OK!) [imath]f(x)\in I[/imath] for each [imath]x\in I.[/imath] (OK!) There should exist a [imath]0\leq k<1[/imath] such that [imath]|f'(x)|\leq k<1.[/imath] If all of these are met, it follows that [imath]a_n[/imath] converges to the unambiguous root of the equation [imath]x=f(x)[/imath]. The third criteria is where I run into problems. I get that [imath]|f'(x)|=\frac{1}{1+x^2}\leq1,[/imath] So my [imath]k[/imath] is [imath]1[/imath], but I need it to be less than one! How can I go around this problem? | 1421820 | Convergence of a Sequence Involving arctan - Is my solution correct?
Here's my question: Let [imath](a_{n})[/imath] be a sequence where [imath](a_{1}) > 0[/imath], defined as: [imath](a_{n+1})=\arctan*(a_{n})[/imath] for all [imath]n[/imath]. Prove that [imath]a_{n}[/imath] has a limit [imath]L[/imath] and calculate it. Solution: Prove that x>[imath]arctan (x)[/imath] for all x. Prove that above we can conclude that [imath]a_{n+1}<a_{n}[/imath], since they are equal we multiply [imath]a_{n}[/imath] by [imath]arctan(x)[/imath]. [imath]arctan(x)=0 \space iff \space x=0[/imath], and that doesn't happen here since [imath]a_{1}[/imath] is positive. Therefore, we can use the follwoing theorem: Let r be a number such that [imath]r \in \Bbb R[/imath] and [imath]0<r<1[/imath], so: [imath]IF \space\space \left|\frac{a_{n+1}}{a_{n}}\right| \le r \Rightarrow \lim \limits_{n\to\infty}(a_{n})=0[/imath] Sorry for the last theorem - I don't know how it's named in English (I'm not a native speaker). Is my solution correct? Thanks, Alan |
769677 | Why [imath]\mathbb{Q}[/imath] is not a projective [imath]\mathbb{Z}[/imath]-module?
From the fact that [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0[/imath], how do we conclude that [imath]\mathbb{Q}[/imath] is not a projective [imath]\mathbb{Z}[/imath]-module? | 506256 | Prove that [imath]\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0[/imath] and show that [imath]\Bbb{Q}[/imath] is not a projective [imath]\Bbb{Z}[/imath]-module.
1) Prove that [imath]\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0[/imath]. 2) Show that [imath]\Bbb{Q}[/imath] is not a projective [imath]\Bbb{Z}[/imath]-module. 1) We know that [imath]\Bbb{Q}[/imath] is an injective [imath]\Bbb{Z}[/imath]-module. This implies that every short exact sequence [imath]0 \rightarrow \Bbb{Q} \xrightarrow{f} A \xrightarrow{g} B \rightarrow 0[/imath] is split exact. In particular, [imath]0 \rightarrow \Bbb{Q} \xrightarrow{f} \Bbb{Z} \xrightarrow{g} A \rightarrow 0[/imath] is split exact. But this implies that we have a homomorphism [imath]k: \Bbb{Z} \rightarrow \Bbb{Q}[/imath] such that [imath]kf = 1_{\Bbb{Q}}[/imath]. So we need to look at what possible homomorphisms we could have from [imath]\Bbb{Z}[/imath] to [imath]\Bbb{Q}[/imath]. Since [imath]\Bbb{Z}[/imath] is cyclic, the homomorphism [imath]k[/imath] is determined by where it sends [imath]1[/imath]. Suppose that we send [imath]1[/imath] to [imath]q \in \Bbb{Q}[/imath] such that [imath]q \not= 1[/imath] and [imath]q \not= 0[/imath]. Let [imath]m \in \Bbb{Z}[/imath]. Then [imath]k(m) = mk(1) = mq[/imath] But then [imath]qmq = k(1)k(m) \not= k(m) = mq[/imath] So either [imath]1 \in \Bbb{Z}[/imath] must be sent to [imath]1 \in \Bbb{Q}[/imath] or [imath]k[/imath] can be the trivial homomorphism. But if [imath]1_{\Bbb{Z}}[/imath] is sent to [imath]1_{\Bbb{Q}}[/imath], then [imath]k[/imath] must be the inclusion map. However in order to for [imath]kf = 1_{\Bbb{Q}}[/imath] to hold, [imath]f[/imath] must also sent [imath]z[/imath] to [imath]z[/imath] for all [imath]z \in \Bbb{Q}[/imath]. But if that's the case, then for any [imath]a/b \in \Bbb{Q}[/imath] where [imath]b \not= 0, 1[/imath], we have (for [imath]n \in \Bbb{Z}[/imath]) [imath]f(a/b) = n = f(n)[/imath] and this contradicts the fact that [imath]f[/imath] must be injective (since the sequence is exact). So the only possible [imath]\Bbb{Z}[/imath]-module map from [imath]\Bbb{Z}[/imath] to [imath]\Bbb{Q}[/imath] is [imath]0[/imath]. Do you think my answer is correct? 2) I was wondering if anybody could give a hint on this one, because I couldn't really get started. Thanks in advance. |
2596611 | Characteristic of an integral domain without unit element must be prime?
I want to show that if an integral domain [imath]D[/imath] has finite characteristic, its characteristic must be prime. This is easy enough if we assume that [imath]D[/imath] has a unit element, but the textbook I'm using (Herstein, Topics in Algebra) does not require there to be a unit element in an integral domain. | 111425 | Characteristic of a Non-unital Integral Ring
If [imath]R[/imath] is a unital integral ring, then its characteristic is either [imath]0[/imath] or prime. If [imath]R[/imath] is a ring without unit, then the char of [imath]R[/imath] is defined to be the smallest positive integer [imath]p[/imath] s.t. [imath] pa = 0 [/imath] for some nonzero element [imath]a \in R[/imath]. I am not sure how to prove that the characteristic of an integral domain without a unit is still either [imath]0[/imath] or a prime [imath]p[/imath]. I know that if [imath]p[/imath] is the char of [imath]R[/imath], then [imath]px = 0 [/imath] for all [imath]x \in R[/imath]. If we assume [imath] p \neq 0 [/imath] and [imath]R[/imath] has nonzero char, and [imath]p[/imath] factors into [imath]nm[/imath], then [imath] (nm) a = 0 [/imath] , which means [imath] n (ma) = 0 [/imath]. Well [imath]ma \neq 0[/imath], because this would contradict the minimality of [imath]p[/imath] on [imath]a[/imath]. But I don't know where to go from this point w/o invoking a unit. Edit: I had left out the assumption that [imath]R[/imath] is assumed to be a integral domain. This has been corrected. |
2596339 | Differentiability at a point x=0
I road in a book that the function [imath]f(x)=\sqrt{x} \sin(x)[/imath] is differentiable at 0. Personally, I don't think so since f is undefined on the left of [imath]x=0[/imath]. Can you confirm my thought or the book? | 1295181 | Can a function be differentiable at the end points of its (closed interval) domain?
Assume [imath]f[/imath] has a domain of [imath][a,b][/imath]. Is it possible that [imath]f[/imath] is differentiable on the closed interval [imath][a,b][/imath], or must the maximal domain for [imath]f'[/imath] be [imath](a,b)[/imath]? |
1836928 | Can the coefficients of a Dirichlet series be recovered?
Specifically if I have a known function [imath]F(s)[/imath] is there some way I can find a function [imath]f(n)[/imath] that satisfies this equation? [imath]F(s) = \sum_{n=1}^\infty \frac{f(n)}{n^s}[/imath] I'm imagining something similar to finding the coefficients of a Fourier series. | 1417284 | For all Dirichlet series, is [imath]a_n[/imath] unique to [imath]f(s)[/imath]?
For any Dirichlet series, [imath]f(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}[/imath] is the sequence, [imath]a_n[/imath], always unique to [imath]f(s)[/imath]? In other words, is it possible to show that [imath]a_n[/imath] is the only sequence that will ever satisfy the series being equal to [imath]f(s)[/imath]? If this is not true, could someone try to provide a counter example if possible? |
382259 | Prove an identity for the continuous integral solution of the conservation law
This is an exercise in Evans, Partial Differential Equations (1st edition), page 164, problem 13: Assume [imath]F(0) = 0, u[/imath] is a continuous integral solution of the conservation law [imath] \left\{ \begin{array}{rl} u_t + F(u)_x = 0 &\mbox{ in $\mathbb{R} \times (0,\infty)$} \\ u=g &\mbox{ on $\mathbb{R} \times \left\{t=0\right\} $} \, , \end{array} \right. [/imath] and [imath]u[/imath] has compact support in [imath]\mathbb{R} \times [0,\infty][/imath]. Prove [imath] \int_{-\infty}^{\infty} u(\cdot,t)\,dx = \int_{-\infty}^{\infty}g \,dx [/imath] for all [imath]t>0[/imath]. How to solve it? | 2810833 | Evans PDE Conservation Law Integral
Let [imath]u[/imath] be a weak solution to the conservation law: [imath] \begin{cases} u_t + F(u)_x = 0 & (x,t) \in \mathbb{R}\times(0, \infty) \\ u = g & \mathbb{R} \times \{0\} \end{cases} [/imath] Let us assume that [imath]u[/imath] is compactly supported in [imath]\mathbb{R}\times [0,T][/imath] for all [imath]T[/imath], and that [imath]F(0) = 0[/imath]. Prove then that for any fixed [imath]t >0[/imath] we have that: [imath] \int_\mathbb{R}u(x,t) \mathrm{d}x = \int_\mathbb{R} g(x) \mathrm{d}x [/imath] My attempt: I know that for [imath]u[/imath] to be a continuous integral solution/weak solution, it must satisfy : [imath] \int_0^\infty\int_{-\infty}^\infty u \phi_t + F(u) \phi_x \mathrm{d}x\mathrm{d}t + \int_{-\infty}^\infty g\phi \mathrm{d}x = 0 [/imath] For any [imath]C^1[/imath], compactly supported test function. How can I prove this identity? |
2597388 | Lim of [imath]\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}[/imath]
I'm working in this exercise: Find [imath]\lim_{n \to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}.[/imath] I'm thinking about partitions and seeing this like an integral but I don't know which limits of integration should I use, maybe [imath]0[/imath] to [imath]1[/imath] since it's kinda close. Any help is appreciated. | 105631 | Help needed with definite integral for [imath]\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}[/imath]
My foreign school book is not clearest at this point, what is really needed with this? What does it mean that "assign definite integral for [imath]\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}[/imath]"? I am not sure, whether I should assign it like this [imath]\lim_{n\rightarrow \infty} \int_{1}^{n} \frac{1}{n+x} dx[/imath] and noticing that it is continuous then trying to find borders so that [imath]F(a) - F(b)[/imath]? (the book messes up all kind of Leibniz stuff at this point, a bit messy -- and just stating [imath]\int f(x)\,dx= F(a)-F(b)[/imath], but not even paying attention to different borders. As far as I know, it is important to specify whether borders depend on the integration factor) |
2597571 | Question regarding Logic sign circle +
I am doing a multiple choice test, and they say that [imath](p\oplus q)\land q[/imath] is logically equivalent with [imath]\neg (p \land q)[/imath] I am confused about this, because normally I can verify with a truth trable, But I do not understand [imath]\oplus[/imath] could someone explain this for me? | 935916 | Truth table and the meaning of [imath]\oplus[/imath] in propositional logic
Could someone show me the truth table for this proposition? I think I have the last two down, but I'm not sure what the symbol in the following one is: [imath]p\oplus (p\wedge q)[/imath] |
2597329 | Show that [imath]\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}[/imath] is closed
Let [imath]a_n[/imath] be a sequence in set of real numbers. Assume [imath]a_n[/imath] converges to [imath]a[/imath] in [imath]\mathbb{R}[/imath]. Show that [imath]\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}[/imath] is closed Recall. [imath]X\in\mathbb{R}[/imath] is closed if and only if the complement of [imath]X[/imath] is open. Let [imath]X=\left\{a_{n}:n\in\mathbb{N}\right\} \cup \left\{ a\right\}[/imath] be a set. I need to show that [imath]X^c[/imath] is open. Let [imath]t\in X^c[/imath]. Then, [imath]t\not\in X[/imath]. So what should I do? | 125299 | Prove that a set consisting of a sequence and its limit point is closed
Can someone please check whether the following simple proof is "mathematical"? Is it correct, complete, rigid? Can it be simplified? I'm a complete autodidact so I'm looking for someone to give me feedback to gain experience in writing proofs... This is also my first question on MSE. The proposition: Let [imath](X, d)[/imath] be a metric space and [imath]x_n \to x[/imath] where each [imath]x_n \in X[/imath] and [imath]x \in X[/imath]. Let [imath]A[/imath] be the subset of [imath]X[/imath] which consists of [imath]x[/imath] and all of the points [imath]x_n[/imath]. Prove that [imath]A[/imath] is closed in [imath](X, d)[/imath]. My tentative to prove this: We first show that all infinite sequences in [imath]A[/imath] converge to [imath]x[/imath]: Let [imath]y \in X[/imath], [imath]y \ne x[/imath]. Then there is some open ball [imath]B_\epsilon(x)[/imath] with [imath]\epsilon < d(x,y)[/imath] containing all but finitely many elements of [imath]A[/imath]. As [imath]y \notin B_\epsilon(x)[/imath] there can be no infinite sequence in [imath]A[/imath] converging to [imath]y[/imath]. Consequently all infinite series in [imath]A[/imath] converge to a point in [imath]A[/imath] which therefore must be a closed set. Edited: As rightly pointed out in the comments, I should have written in the first sentence "...sequences with infinitely many distinct terms and which converge to some point of [imath]X[/imath]" and the last sentence should be "Consequently all infinite sequences...". |
2596620 | [imath]A \in \mathbb{Z} ^{n \times n}[/imath]. Show that [imath]\det(A) = \pm 1[/imath].
Im stuck with this question because I'm probably misunderstanding it. There is given that [imath]\mathbb{Z}^{n \times n} = \left\{A \in \mathbb{R}^{n \times n} \mid (A)_{ij} \in \mathbb{Z} \text{ for }i,j \in \left\{1,\cdots,n\right\}\right\}[/imath] I have to prove that for a matrice [imath]A \in \mathbb{Z}^{n \times n}[/imath] always: [imath]\det(A) = \pm1 \Longleftrightarrow A^{-1} \in \mathbb{Z}^{n \times n}[/imath] [imath]A[/imath] is invertible here. I were thinking that if you take [imath]A \in \mathbb{Z}^{n \times n}[/imath] then [imath]A^{-1} = \frac{\operatorname{adj}(A)}{\det(A)}[/imath] Now if [imath]\det(A)[/imath] is different from [imath]\{-1,1\}[/imath] then you would get elements in [imath]A^{-1}[/imath] that are not in [imath]\mathbb{Z}[/imath]. Or not because elements in [imath]\operatorname{adj}(A)[/imath] could be multiples of [imath]\det(A)[/imath]. The question is: how do I know that [imath]\operatorname{adj}(A)_{ij} \neq c \cdot \det(A)[/imath] for all elements in [imath]\operatorname{adj}(A)[/imath] where [imath]c[/imath] is an integer? | 19528 | Integer matrices with integer inverses
If all entries of an invertible matrix [imath]A[/imath] are rational, then all the entries of [imath]A^{-1}[/imath] are also rational. Now suppose that all entries of an invertible matrix [imath]A[/imath] are integers. Then it's not necessary that all the entries of [imath]A^{-1}[/imath] are integers. My question is: What are all the invertible integer matrices such that their inverses are also integer? |
2596386 | [imath]\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \ge \frac{1}{2}(x+y+z)[/imath]
let [imath]x,y,z[/imath] be positive real numbers [imath]\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} \ge \frac{1}{2}(x+y+z)[/imath] then how to prove above by Cauchy Schwarz -Inquality | 1841084 | If [imath]x,y,z\gt 0[/imath] and [imath]xyz=1[/imath] Then minimum value of [imath]\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}[/imath]
If [imath]x,y,z\gt 0[/imath] and [imath]xyz=1[/imath] Then find the minimum value of [imath]\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}[/imath] [imath]\bf{My\; Try::}[/imath]Using Titu's Lemma [imath]\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}[/imath] and equality holds when [imath]x=y=z=1[/imath] My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality. Please explain me. Thanks |
2598330 | Evaluating a definite integral involving [imath]\log(1+x)[/imath]
Find [imath]\int_0^1\frac{x\ln(1+x)}{1+x^2}dx[/imath] I think it has not an anti-derivative because i tried it with different ways such as using the sub [imath]x=\tan{y}[/imath] to be [imath]\int\tan{y} \ln\left(1+\tan{y}\right)dx[/imath] so what is the procedure to solve it? | 2182816 | Prove that [imath]\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}[/imath]
We know that [imath]\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,[/imath] but how about [imath]\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?[/imath]Prove that [imath]\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.[/imath] |
2598342 | How to solve higher power of matrix?
Given a matrix [imath]A=\begin{pmatrix} 1 & 0 \\ {-1} & 1\end{pmatrix}[/imath] We are asked to show [imath]A^2 -2A +I = 0[/imath] and hence find [imath]A^{50}[/imath]. I've proved that [imath]A^2 -2A +I = 0[/imath] but couldn't figure out how to proceed for the next part. | 962294 | If [imath]A=\pmatrix{1 &0\\-1&1}[/imath], show that [imath]A^2-2A+I_2=0[/imath]. Hence find [imath]A^{50}[/imath]
If [imath]A=\pmatrix{1 &0\\-1&1},[/imath] show that [imath]A^2-2A+I_2=0,[/imath] where [imath]I_{2}[/imath] is the [imath]2x2[/imath] Identity matrix. Hence find [imath]A^{50}[/imath]. We have [imath]A^2-2A+I_2=A(A-2I_3)+I_=\pmatrix{1 &0\\-1&1}\pmatrix{-1 &0\\-1&-1}+I_2 =-I_2+I_2=0.[/imath] How can I show the second part? |
2598686 | Evaluate [imath]\int_{-\infty}^{\infty}\cos(x)\operatorname{sech}(x)dx[/imath]
I'm having trouble finding the definite integral: [imath]\int_{-\infty}^{\infty}\cos(x)\operatorname{sech}(x)dx[/imath] I know the answer is [imath]\pi\operatorname{sech}(\frac{\pi}{2})[/imath]. There isn't an indefinite integral in terms of elementary functions. The indefinite integral given by Wolfram Alpha is in terms of hypergeometric functions, but I wanted to know if you could get the answer without that. Maybe some differentiating under the integral jazz? Or maybe series. Not too sure where to start, but I would appreciate an explanation. Also, I personally don't know Complex Analysis, but if that's the only way, that wouldn't be a bad answer for others on this site. | 1275759 | Solve [imath]\int \limits_{0}^{\infty} \frac{\cos(x)}{\cosh(x)} dx[/imath] without complex integration.
Solve [imath]\int \limits_{0}^{\infty} \frac{\cos(x)}{\cosh(x)} dx[/imath] without complex integration. This integral can be very easily solved with contour integration, but how can you solve it without taking a tour in the complex plane? |
2598392 | Polynomial at least one root between 0 and 1
[imath]p(x)=a_0+a_1x+...a_nx^n[/imath] with [imath]a_0+\frac{a_1}2+ \cdots +\frac{a_n}{n+1}=0[/imath] Heuristically, I feel like this should have at least one root between 0 and 1, because at least one of the coefficients will have a different sign than the others making the graph of the function have a kink. but I am very unsure where to start. | 1906956 | If [imath]a_0+\frac{a_1}{2}+\dots + \frac{a_n}{n+1}=0[/imath] then [imath]a_nx^n+\dots + a_0[/imath] has a root in [imath][0,1][/imath]
Let [imath]P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_0[/imath] be a real polynomial such that [imath]a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0[/imath]. Show that [imath]P[/imath] has a root in [imath][0,1][/imath]. I would like proof verification and alternative proofs. My solution: notice that the antiderivative of [imath]P(x)[/imath] is [imath]\frac{a_nx^{n+1}}{n+1}+\frac{a_{n-1}x^n}{n}+\dots + \frac{a_1x^2}{2}+a_0x+C[/imath]. From here [imath]\int_{0}^1P(x)dx=a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0[/imath]. If [imath]P[/imath] had no roots then [imath]P[/imath] would be a strictly positive(or negative) continuous function, so its integral would be strictly positive ( or negative), a contradiction. |
2598891 | [imath]L \otimes_K L[/imath] is isomorphic to [imath]\prod_{\sigma \in G} L[/imath]
Let [imath] K \subset L[/imath] be a finite Galois extension. Let [imath]G[/imath] be the Galois group. Let [imath]L^G = \prod_{\sigma \in G} L[/imath] denote copies of [imath]L[/imath] indexed by the group [imath]G[/imath]. I want to show that the map \begin{align*} L \otimes_K L &\to L^G\\ a \otimes b &\mapsto (a\sigma(b))_{\sigma \in G} \end{align*} is an isomorphism. How can I show that? | 263192 | Why is this isomorphism [imath]M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}[/imath] an isomorphism of [imath]M[/imath] - algebras?
Suppose that [imath]L/K[/imath] is a finite separable extension of fields and let [imath]M[/imath] denote the Galois closure of [imath]L[/imath]. Let [imath]\textrm{Hom}_K(L,M)[/imath] denote the set of all [imath]K[/imath] - algebra homomorphisms from [imath]L[/imath] to [imath]M[/imath]. Since [imath]L/K[/imath] is separable we know that the number of elements in [imath]\textrm{Hom}_K(L,M)[/imath] is equal to [imath][L:K][/imath]. Now I want to prove that we have an isomorphism of [imath]M[/imath] - algebras [imath]\varphi : M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}[/imath] Proof that they are isomorphic as [imath]K[/imath] - modules : Write [imath]L = K(a)[/imath] for some [imath] a\in L[/imath] (we can do this via the primitive element theorem). Then [imath]\begin{eqnarray*} M \otimes_K L &=& M \otimes_K K[a] \\ &\cong& M\otimes_K K[x]/(f) \hspace{3mm} \text{where $f$ is the minimal polynomial of $a$ over $K$} \\ &\cong& M[x]/(f)\\ &\cong& M[x]/(f_1\ldots f_{[L:K]}) \hspace{3mm} \text{where the $f_i$ are the distinct}\\ && \hspace{1.5in} \text{irreducible factors of $f$ since $M/L$ is Galois} \\ &\cong& M^{[L:K]}\end{eqnarray*}[/imath] where the last step was using the Chinese remainder theorem. For the third last step, we consider the ses [imath] 0 \to (f) \to K[x] \to K[x]/(f) \to 0[/imath] and tensor with the exact functor [imath]-\otimes_K M[/imath] to get [imath]0 \to (f) \otimes_K M \to K[x] \otimes_K M \to K[x]/(f) \otimes_K M \to 0[/imath] and so [imath]\begin{eqnarray*} M \otimes_K K[x]/(f) &\cong& K[x]/(f) \otimes_{K} M \\ &\cong& \frac{K[x] \otimes_{K} M}{f \otimes_K M}\\ & \cong& M[x]/(f) \end{eqnarray*}[/imath] where [imath](f)[/imath] is now viewed as an ideal of [imath]M[x][/imath]. My question is: The tensor product [imath]M \otimes_K L[/imath] is a left [imath]M[/imath] - module, but why is it also an [imath]M[/imath] - algebra? Also why are the isomorphisms above isomorphisms of [imath]M[/imath] - algebras and not just [imath]K[/imath] - modules? |
2598685 | Detail missing in exercise on factor groups
The goal is to compute the factor group of [imath]\mathbb{Z}_4\times\mathbb{Z}_6/\langle(2,3)\rangle[/imath]. Now, [imath]H=\langle(2,3)\rangle[/imath] has order [imath]2[/imath] and thus the factor group must have [imath]12[/imath] elements, and since [imath]\mathbb{Z}_4\times\mathbb{Z}_6[/imath] is abelian the factor group must be abelian. Furthermore, the only possible abelian groups of order [imath]12[/imath] are [imath]\mathbb{Z}_4\times\mathbb{Z}_3[/imath] and [imath]\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3[/imath]. Up to here everything is clear. It is also clear that [imath]\mathbb{Z}_4\times\mathbb{Z}_3[/imath] has an element of order [imath]4[/imath] and [imath]\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3[/imath] does not. Why based the order of these couple of groups I can decide that [imath]\mathbb{Z}_4\times\mathbb{Z}_6/\langle(2,3)\rangle[/imath] is isomorphic to [imath]\mathbb{Z}_4\times\mathbb{Z}_3[/imath], and not the other one? EDIT: While this questions has been marked as duplicate, the original link does answer all of points asked from the same exercise, and therefore THIS IS NOT a duplicate. Consider reading the question and the answer, you may have the same as this, or both of them. | 684998 | Compute factor group [imath]\dfrac{\mathbb{Z}_4 \times \mathbb{Z}_6}{\langle(2,3)\rangle}[/imath] - Fraleigh p. 147 Example 15.11
(1.) Why's there a 'great temptation' to set [imath]2 \bmod 4[/imath] and [imath]3 \bmod 6[/imath] to 0? (2.) Why are you authorized to set [imath]2 \bmod 4[/imath] and [imath]3 \bmod 6[/imath] to 0? [imath]2 \bmod 4 \neq 0[/imath] and [imath]3 \bmod 6 \neq 0[/imath], hence does this even make sense? (3.) How do you envisage and envision to work with [imath](1,0) + H = \{(1, 0), (3, 3)\}[/imath]? I know the second paragraph starts with the Fundamental Theorem of Abelian Groups, but my course doesn't include this hence I can't use it. I also know: [imath]| \dfrac{\mathbb{Z}_4 \times \mathbb{Z}_6}{\langle(2,3)\rangle} | = 6 \times 4 = 24[/imath], order [imath](\dfrac{\mathbb{Z}_4 \times \mathbb{Z}_6}{\langle(2,3)\rangle}) = lcm(4, 6) = 12.[/imath] |
2591974 | Eigenvectors of 3D image data (eigenfaces)
This is in regards to eigenface decomposition with [imath]M[/imath] images of size [imath]N \times N[/imath]. In short: My question lies in the following statement (bottom of page 4 in above link): "If the number of data points in the image space is less than the dimension of the space [imath](M<N^2)[/imath], there will be only [imath]M-1[/imath], rather than [imath]N^2[/imath], meaningful eigenvectors". Essentially, if there are 30 images, we can simply use the [imath]M \times M[/imath] COV matrix, rather than the [imath]N^2 \times N^2[/imath] COV matrix to find the meaningful eigevectors. I am assuming this is a simple linear algebra result, which I have long forgotten. Let's suppose we have a matrix A of size 256 x 256 x 30 (i.e. 30 images). As a Casorati matrix the size is now 65536 x 30. The covariance matrix can be calculated as [imath]C = A A^T[/imath], which results in a matrix of 65536 x 65536, yielding 65536 eigenvectors. As outlined in the paper linked above (bottom of page 4), we can instead use the following form to find [imath]M = 30 - 1[/imath] meaningful eigenvectors: [imath]A \, A^{T} \, A \,\, \vec{v} = \lambda \, A \, \vec{v}[/imath] such that [imath]A \vec{v}[/imath] are the eigenvectors of [imath]A \, A^T[/imath]. My question lies in the following statement: "If the number of data points in the image space is less than the dimension of the space [imath](M<N^2)[/imath], there will be only [imath]M-1[/imath], rather than [imath]N^2[/imath], meaningful eigenvectors. I am having difficulty understanding why the above statement is true. It makes sense, when I think about having [imath]M[/imath] images, but how does this relate to the 65536 eigenvectors that would be calculated using the traditional COV formalism? I hope my question is clearly stated. | 2087297 | Why is finding [imath]M[/imath] eigenvectors on smaller matrix valid?
I am following this article on face recognition. In "calculating eigenfaces" section, the authors present a solution for the problem of calculating a very big matrix: Let [imath]A_{N^2\times M}[/imath] be an [imath]M[/imath] sized dataset, where each column in an [imath]N\times N[/imath] image. Instead of calculating the [imath]M[/imath] large eigenvectors of the [imath]N^2\times N^2[/imath] co-variance matrix they calculate the [imath]M[/imath] eigenvectors of [imath]L=A^TA[/imath] matrix which is of size [imath]M\times M[/imath]. Why is this a valid\good enough solution? What are the criteria for a largest vector? larger in which seance? |
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