Recent Questions - Mathematics Stack Exchange |
- Need help understanding a proof about the Abel Summation formula for the cases $n\leq y$ and $n>y$
- Is a Wajsberg algebra a Heyting algebra?
- Difference between -x+y=3 and -4x+4y=12 lines equations
- Why does this polynomial splits over $\Bbb{F}_2(\sqrt{X})$
- Sobolev embedding: the injection of $H^1(I)$ into $L^2(I)$ is compact
- Constructing three circles perpendicular to each other
- Finding the set of reachable points given time and speed
- The third homology stability of general linear groups over finite fields
- How do I work out this exponential word question?
- Understanding $m$-th degeneracy locus
- Integral representation of confluent hypergeometric function
- Why is $E( \langle f(x),f(y) \rangle \langle g(x),g(y) \rangle) \leq E(\lVert g(x) \rVert^2)$?
- On a comple integral
- Pigeonhole a + b = c + d
- How can the definition of an Indexed Set be explained in terms of predicate logic?
- What is wrong in my counter example?
- Please help me solve this radical. [closed]
- Example of a Chief series
- GNS Construction Proof
- Challenging vector inequality
- Solving $k\left(e^{a(2k+1)}-1\right)=1$ for k?
- Bound central moments of even order with raw moments of same order
- How fast the water level is rising when the ball is half submerged.
- Recursion relation for the Legendre equation
- Baby rudin theorem 8.8
- Find the centre of gravity of an elliptic arc.
- Binomial Distribution for defects
- Field of order 8, $a^2+ab+b^2=0$ implies $a=0$ and $b=0$.
- Stability condition for explicit scheme in finite differences
| Need help understanding a proof about the Abel Summation formula for the cases $n\leq y$ and $n>y$ Posted: 28 Feb 2022 01:05 AM PST
For the following Theorem 3.1. pp.5-6 I am having trouble understanding Fig.3.1 which accompanies the theorem explaining the integral $\int_{y}^{x}\sum_{n\leq t} a(n)f'(t)dt=\int_{max(y,n)}^{x}f'(t)dt$ for the cases $n\leq t$ and $n\geq y$. The difficulties I am having trouble understanding are as follows:
Theorem 3.1 is reproduced below just in case the url link does not link to the correct pages.
Thank you in advance. |
| Is a Wajsberg algebra a Heyting algebra? Posted: 28 Feb 2022 01:01 AM PST More precisely, I'm interested to know if the implication of a Wajsberg algebra is "distributive" (in the sense that $x \to (y \to z) \leq (x \to y) \to (x \to z)$). |
| Difference between -x+y=3 and -4x+4y=12 lines equations Posted: 28 Feb 2022 01:07 AM PST I would like to know whats the difference between these two lines. My question is what it means to multiplicate the line equation by 4, what does it change? My professor said it has something to do with the ´line speed´ |
| Why does this polynomial splits over $\Bbb{F}_2(\sqrt{X})$ Posted: 28 Feb 2022 01:00 AM PST I have the following problem:
But I somehow don't see why it splits into $(t-\sqrt{X})^2$ and not into $(t-\sqrt{X})(t+\sqrt{X})$. As I know it we define $$\Bbb{F}_2(\sqrt{X})=\{a+b\sqrt{X}:a,b\in \Bbb{F}_2\}$$ Then I would say that $-\sqrt{X}$ is equivalent to $\sqrt{X}$ since $[-1]=[1]$ in $\Bbb{F}_2$ but I'm not sure if this is enought to conclude that $(t-\sqrt{X})(t+\sqrt{X})=(t-\sqrt{X})^2$. Could someone maybe help me. |
| Sobolev embedding: the injection of $H^1(I)$ into $L^2(I)$ is compact Posted: 28 Feb 2022 12:57 AM PST Can you help me to explain in detail why we deduce from Theorem 8.8 (see Image) that: the injection of $H^1(I)$ into $L^2(I)$ is compact. Thank you. |
| Constructing three circles perpendicular to each other Posted: 28 Feb 2022 12:56 AM PST How can we construct 3 three circles which are all perpendicular(Orthogonal to each other) |
| Finding the set of reachable points given time and speed Posted: 28 Feb 2022 01:02 AM PST The following problem is from The Method of Coordinates by I.M. Gelfand et al.: What is a good way to approach this? I provide my very rough sketch for part (a) below (which I am not terribly confident about). Everything between the outer boundaries of my figure and the axes is reachable -- my thinking is that, given enough time to travel a distance of $2b$ in the first quadrant, we can reach points within radius 2b in quadrants I and III, and points within radius $b$ in quadrants II and IV. Additionally, it is possible to travel a distance between $b$ and $2b$ along either direction of the vertical axis and then travel into II or IV for the remaining time (these are the right-triangular sections in my sketch, with vertices at $2b$ and $-2b$ along the vertical axis, having side lengths $b/2$ and $b$). Furthermore, is there any insight to be gained from part (b) of the question? Or is it just a matter of scaling down certain regions accordingly? |
| The third homology stability of general linear groups over finite fields Posted: 28 Feb 2022 12:52 AM PST Given a finite field $\mathbb{F}$ with $|\mathbb{F}|=q=p^m\geq4$ where $p=\text{char}(\mathbb{F})$, I'm wondering is there a characterization of the kernel of the map $f:H_3(\text{GL}_3(\mathbb{F}))\to H_3(\text{GL}_4(\mathbb{F}))$? Is it an isomorphism?(Here $H_n(G)$ means the $n$-th integral homology of the group $G$) Here is some material I have known. Sprehn&Wahl says $f$ must be a surjection. In the paper [1], it is shown that $f$ would be an isomorphism if $\mathbb{F}$ is an infinite field. But in the talk [2], it is asserted that the techniques in [1] doesn't apply for finite fields. And with the result of Galatius-Kupers-Randal-Williams, it is deduced that $f$ induces an isomorphism on $p$-primary part. I'm wondering what happens with the $\text{mod } l$ homology where $l\neq p$. The reason I raise this question is that in the Chapter.VI in Weibel's K-book, remark 5.12.1, he asserts the map $\varphi$ could extend to a map $H_3(\text{GL}(\mathbb{F}))\to B(F)$. There one works on a general field $F$ with $|F|>3$. But all I can get now is $ H_3(\text{GL}(\mathbb{F}))=H_3(\text{GL}_4(\mathbb{F})) $ and if we want to extend $\varphi$ to $H_3(\text{GL}(\mathbb{F}))$ we have to work on $\ker(f)$. reference: [1]Yu. P. Nesterenko and A. A. Suslin, Homology of the general linear group over a local ring, and Milnor's K-theory, Izv. Akad. Nauk SSSR Ser. Mat. 53 (1989), no. 1, 121–146. MR 992981 [2]Alexander Kupers, NEW TALK: $E_\infty$-CELLS AND THE HOMOLOGY OF GENERAL LINEAR GROUPS OVER FINITE FIELDS |
| How do I work out this exponential word question? Posted: 28 Feb 2022 12:50 AM PST Sorry if this is really simple to work out, I am a beginner at maths.. The question says "Given the equation model $n(t)=200*2^{t/24}$ (t is time in hours). (A) How long does it take (in hours) for n(t) to double? (B) At what time t (in hours) does it take for n(t) to reach $25600$" I don't know how to work out (B) but for (A) I did it as follows (don't know if it is correct): $n(t)= 2n(0)$ $t=1/2^{t/24}$ Could someone please help? |
| Understanding $m$-th degeneracy locus Posted: 28 Feb 2022 12:43 AM PST Let $g : E \to F$ be a morphism between two vector bundles (of rank $e,f$ respectively) on an irreducible algebraic variety $X$ (over $\mathbb C$). Then for $0 \leq m \leq \{e,f\}$, one defines the $m'th$ degeneracy locus by $D_m(g) :=\{x \in X| \text{rank}(g_x) \leq m\}$, (where I guess $g_x :E_x \to F_x$ is morphism between stalks of $E,F$ at the point $x$, which is therefore $\mathcal O_X,x \cong \mathbb C$ linear map. Please correct me if this notation is wrong) We can note that, a linear map has rank $\leq m$ iff all $(m+1) \times (m+1)$ minors. vanishes. But how do we use that to give a closed subscheme structure on $D_m(g)$? I have come across the following two descriptions : $(1)$ Locally, $g$ can be represented by an $f × e$ matrix with entries in an affine coordinate ring of $X$. One can then consider the ideal generated by the $(m + 1) × (m + 1)$ minor determinants of this local representation. These local ideals patch together to give an ideal sheaf, which gives $D_m(g)$ the structure of a closed subscheme of $X$. Here why it does not depend on choice of local trivialization and what is precisely meant by " these local ideals patch"? Can someone demonstrate this argument with a step by step rigorous proof or a concrete example? $(2)$ How do we see that this is same as the zero locus of the section $\Lambda^{m+1}g \in H^0(X, (\Lambda^{m+1}E)^* \otimes \Lambda^{m+1}F)$? |
| Integral representation of confluent hypergeometric function Posted: 28 Feb 2022 12:19 AM PST The integral representation for tricomi function $U(a,c,z)$ defines a solution in the right half-plane $Re(z)$>0. Under what condition can we accept $Re(z)$=0? |
| Why is $E( \langle f(x),f(y) \rangle \langle g(x),g(y) \rangle) \leq E(\lVert g(x) \rVert^2)$? Posted: 28 Feb 2022 12:25 AM PST Say, $p$ is a probability distribution on $\mathbb{R}^n$ and $f,g:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be measurable functions such that $E_{x\sim p} \left (\lVert f(x) \rVert_2^2 \right )\leq 1$. Then, why is $E_{x,y\sim p} \Big (\langle f(x),f(y) \rangle \langle g(x),g(y) \rangle \Big) \leq E_{x\sim p} \left (\lVert g(x) \rVert_2^2 \right )$, where the $x,y$ are sampled independently for the first expectation? |
| Posted: 28 Feb 2022 12:12 AM PST I am trying to compute the following integral: $$\oint_\gamma\exp(1/z)\cosh(z)dz$$ where $\gamma$ is the unit circle with the obvious orientation. I have tried to find a decent closed form for this but I seem to be running into difficulties. What I have tried: the first thing that comes to mind is to use the residue theorem, and since there is only one singularity (in $0$), this should theoretically be doable. However, since we have an essential singularity, it seems quite nontrivial to me how to actually compute the residue. What I mean by this is: obviously I can write (on $\gamma$) $$\exp(1/z)\cosh(z)=\left(\sum\frac{1}{n!z^n} \right)\left(\sum \frac{z^{2n}}{(2n)!}\right)= \sum_{n+m=0}^\infty\frac{z^{2m-n}}{(2m)!n!}$$ so that the residue in $0$ is generated by those factors in which $2m-n=1$ and so $\text{Res}(f,0)=\sum_{m=0}^\infty \frac{1}{(2m)!(2m+1)!}$ but this is far from being a decent closed form for the residue. I have lookd around on this site and I found a few similar integral but they where all solved by making use of simmetries that my integral seems to lack |
| Posted: 28 Feb 2022 12:05 AM PST Ive got this assignment for uni wher ei need to prove the following: We have a set of 16 different positive integers $\leq 100$, prove that there are always $a,b,c,d$ in this set so that $a + b = c + d$. I think this is to be solved with the pigeonhole principle, but I cant quite figure it out. Does anyone know how this could be done? Any help would be appreciated!! |
| How can the definition of an Indexed Set be explained in terms of predicate logic? Posted: 28 Feb 2022 12:32 AM PST I'm learning about indexed sets and came across the definition of an indexed set to be, "...any set,such that for $i \in I$, we have a set $A_i$ ...how would this definition be written using quantifiers? Please let me know if this definition is wrong as well...thank you! I think it should look like this? but I'm not sure: $$ \underset{i \in I}{\forall} \exists A_i $$ |
| What is wrong in my counter example? Posted: 28 Feb 2022 12:00 AM PST Given $V,W$ vector spaces from finite dimension and given $T:V\to W$ linear transformation. decide if the following statment is true/false: (according to the book the answer is true) If $T$ transfrom basis to an linear independant group, than $T$ is injective. I say that this statement is fasle since I can take $V=\mathbb{R^2}$ and $W=\mathbb{R}$ than $$T(1,0)=1$$ $$T(0,1)=1$$ so the group $1$ is linearly independant and obviously that $T$ is not injective. The answer to this statement is that is true and I dont understand why. |
| Please help me solve this radical. [closed] Posted: 28 Feb 2022 01:04 AM PST Please show me how does $\sqrt{12\sqrt[3]{2} - 15}$ + $\sqrt{12\sqrt[3]{4} - 12}$ = 3. I can't find the answer anywhere else. square root of (12cube root of 2 - 15) + square root of (12cube root of 4 - 12) = 3 The answer in the book also says [square root of (12cube root of 2 - 15) + square root of (12cube root of 4 - 12)]² = 12cube root of 2 + 12cube root of 4 - 27 + 2 = 9 which I don't see in my calculation. I can understand the 12cube root of 2 + 12cube root of 4 - 27 but where does the + 2 come from? Cuz instead of + 2, I've got + 2(square root of(204 - 144cube root of 2 + 180cube root of 4)). I don't see how it's the same number. |
| Posted: 28 Feb 2022 12:24 AM PST Can you give an example of a non solvable group whose at least one Chief factor is a product of more than one simple group (i.e. it should not be a simple group). Edit: Thanks @HallaSurvivor for suggestion. I am a Research scholar in mathematics. I have enough background in group theory. The reason why I am asking this questions is that I have not study this concept of Chief series before. Almost all the example I have seen has the chief factors which are simple. But by definition it can be product of simple. So I just wanted to see such examples. Thank you! |
| Posted: 28 Feb 2022 12:04 AM PST I am currently reading through the proof of the GNS Construction, and there are two parts I simply do not understand. Below is the entire proof, the parts in red are what I do not understand. It would be great if someone could clear these up for me. Let $A$ be a $C^\ast$-algebra with $f\colon A\to\mathbb{C}$ a positive linear functional. Define $$ N=\{a\in A: f(a^\ast a)=0\}. $$ Note that $N$ is not only a vector subspace of $A$, but a left ideal as well. In particular, $A/N$ is a vector space. We let $\xi\colon A\to A/N$ be the quotient map. We have chosen $N$ so that $$ \langle\xi(a),\xi(b)\rangle:=f(b^\ast a) $$ is a well-define inner product on $A/N$. Therefore we can complete $A/N$ to a Hilbert space $\mathcal{H}$. Since $N$ is a left-ideal in $A$, for each $a\in A$ we can define an operator $\pi_0(a)$ on $A/N$ by $$ \pi_0(a)\xi(b):=\xi(ab). $$ Furthermore, $\pi_0$ is a homomorphism of $A$ into the linear operators on $A/N$. Since $a^\ast a\leq\|a\|^2\mathbf{1}_{\tilde{A}}$, we have $b^\ast a^\ast ab\leq\|a^2\|b^\ast b$ for any $a,b\in A$. Therefore $$ \|\pi_0(a)\xi(b)\|^2=\|\xi(ab)\|^2=f(b^\ast a^\ast ab)\leq\|a\|^2f(b^\ast b)=\|a\|^2\|\xi(b)\|^2. $$ It follows that $\pi_0(a)$ is bounded and extends to an operator $\pi(a)$ on $\mathcal{H}$. Clearly, $\pi\colon A\to B(\mathcal{H})$ is an algebra homeomorphism. Since $$ \langle\pi_0(a)\xi(b),\xi(c)\rangle=f(c^\ast ab)=\langle\xi(b),\pi_0(a^\ast)\xi(c)\rangle, $$ we must have $\pi(a)^\ast=\pi(a^\ast)$. Thus $\pi$ is a representation. Now let $\{e_\lambda\}$ be an approximate identity in $A$. If $\mu\leq\lambda$, then $e_\lambda-e_\mu\geq0$ and $\left(\mathbf{1}_{\tilde{A}}-(e_\lambda-e_\mu)\right)\geq0$. This implies that $\|e_\lambda-e_\mu\|\leq1$. Thus, $(e_\lambda-e_\mu)^2\leq(e_\lambda-e_\mu)$ and $$ \|\xi(e_\lambda)-\xi(e_\mu)\|^2=f\left((e_\lambda-e_\mu)^2\right)\leq f\left((e_\lambda-e_\mu)\right). $$ Since $\lim_\lambda f(e_\lambda)=\|f\|=\lim_\mu f(e_\mu)$, $\{\xi(e_\lambda)\}$ is a Cauchy net in $\mathcal{H}$. Hence, there is a $h\in\mathcal{H}$ such that $\xi(e_\lambda)\to h$. This means that $$ \pi(a)h=\lim_\lambda\pi_0(a)\xi(e_\lambda)=\lim_\lambda\xi(ae_\lambda). $$ But $a\mapsto\xi(a)$ is continous because $f$ is: $\|\xi(a)-\xi(b)\|^2=f\left((b-a)^\ast(b-a)\right)$. So we can conclude that $\pi(a)h=\xi(a)$. $\color{red}{\text{Therefore }h\text{ is a cyclic vector for }\pi}$ and $$ \langle\pi(a^\ast a)h,h\rangle=\langle\pi(a)h,\pi(a)h\rangle=\langle\xi(a),\xi(a)\rangle=f(a^\ast a). $$ $\color{red}{\text{By linearity, } f(a)=\langle\pi(a)h,h\rangle\text{ for all }a\in A}$. Since $\pi$ has a cyclic vector, it is definitely nondegenerate, so $\pi(e_\lambda)h\to h$. Thus $$ \|f\|=\lim_\lambda f(e_\lambda)=\lim_\lambda\langle\pi(e_\lambda)h,h\rangle=\|h\|^2. $$ |
| Posted: 28 Feb 2022 12:05 AM PST I came across a particular expression of some vectors I'm working with and I would really need some help proving this: Let a and b two vectors in $\mathbb{R}^n$, $n>3$. Then I am pretty sure it's true, I considered even some examples and it seems to always work. I also proved it in the case $a\cdot b \geq 0$, but I have problems when $a\cdot b \leq 0$. And I think there might be some unifying solution. Thank you! |
| Solving $k\left(e^{a(2k+1)}-1\right)=1$ for k? Posted: 28 Feb 2022 12:00 AM PST I've been studying the structure of partial sums of the Dirichlet eta function and noticed that certain critical points occur when the summing from $n=1$ to $n\approx N(k)$ for $k\in\mathbb{N}$ where $$N(k)=\frac{1}{e^{ak}-1},$$ and $a$ is some constant determined by the input of the eta function. There is one special critical point which happens when $k=N(2k+1)$, which I'd like to isolate (i.e; solve for $k$). Thus,
Note: Trivially, this is generally unsolvable for $k\in\mathbb{N}$, however solving it for $k\in\mathbb{R}$ will still yield useful results for my purpose. I have been fumbling with this equation for quite some time to convert it into a form suitable to apply the Lambert W function to no avail. At this point I'm not sure if this is solvable with such a method. Even WolframAlpha cannot solve it. I presume this has no nice elementary solution, and probably involves some transcendental function / infinite series; which I am content with. Any advice? |
| Bound central moments of even order with raw moments of same order Posted: 28 Feb 2022 01:02 AM PST Let $(\Omega, \mathcal A, P)$ be a probability space and consider a real-valued random variable $X \colon \Omega \to \mathbb{R}$. It holds $$ \mathrm{Var}(X) = \mathbb{E}[(X - \mathbb{E}[X])^2] = \mathbb{E}[X^2] - \mathbb{E}[X]^2 \leq \mathbb{E}[X^2]. $$ Is this true for higher moments of even order? Precisely, does it hold for $p = 1, 2, \ldots$ $$ \mathbb{E}[(X - \mathbb{E}[X])^{2p}] \leq C(p)\mathbb{E}[X^{2p}], $$ for some positive constant $C(p)$ depending only on $p$ and such that $C(1) = 1$? Edit: In this thread it is proved that the inequality above does not hold for $C(p) = 1$ for all $p = 1,2,\ldots$ Still, could it hold with a $p$-dependent proportionality constant? Edit 2: It is possible to prove that the inequality holds with $C(p) = 2^{2p}$ using Hölder's and Jensen's inequality. In particular, Hölder's inequality yields for any real numbers $a$ and $b$ $$ (a-b)^{2p} \leq 2^{2p-1}(a^{2p}+b^{2p}), $$ so that by linearity of $\mathbb{E}$ $$ \mathbb{E}[(X - \mathbb{E}[X])^{2p}] \leq 2^{2p-1}\left(\mathbb{E}[X^{2p}]+\mathbb{E}[X]^{2p}\right). $$ Noew Jensen's inequality yields $\mathbb{E}[X]^{2p} \leq \mathbb{E}[X^{2p}]$ and we can conclude that $$ \mathbb{E}[(X - \mathbb{E}[X])^{2p}] \leq 2^{2p-1} \cdot 2 \mathbb{E}[X^{2p}] =2^{2p}\mathbb{E}[X^{2p}]. $$ Is there any sharper bound? In this way, for $p = 1$ we obtain $C(1) = 4$ instead of $C(1) = 1$. |
| How fast the water level is rising when the ball is half submerged. Posted: 28 Feb 2022 12:46 AM PST Question:
For the 1st question: Volume of the cap of a sphere of radius $R$ at height $\displaystyle h = {πh^2(3R-h^2)\over 3}$ $\displaystyle V = πh^2R - {πh^3\over 3}$ $\displaystyle {d\over dt} V = (2πhR - πh^2){dh\over dt}$ Given that $ \displaystyle {d\over dt} V = 1\text{ in}^3/\text s$ and that $ \displaystyle R = 3\text{ in } \implies {dh\over dt} = {1\over 6πh -πh^2}$ So at $h=1, \displaystyle {dh\over dt} = {1\over 5π}$ For 2nd question I have drawn the figure below:
I can't write down the volume in terms of variables. So I am stuck there. Can you please help me out of this? That is give that formula for volume in terms of variables. |
| Recursion relation for the Legendre equation Posted: 28 Feb 2022 12:59 AM PST I am studying Quantum mechanics and in particular angular momentum. At one point, one needs to solve a DE (the Legendre equation) in power series to get a recursion relation (Legendre polynomials). Consider the Legendre differential equation: $$ \frac{d}{dx}((1-x^2)\frac{dP_l}{dx}) +l(l+1) P_l = 0$$ We attempt a power series solution with the form $P_l(x) = \sum_0^{\infty} a_k x^k$. The recursion relation is supposed to give $\frac{a_{k+2}}{a_k}=-\frac{l(l+1)-k(k+1)}{(k+1)(k+2)}$. Substituting the series into the equation we get: $$ \frac{d}{dx}((1-x^2)\frac{d}{dx}\sum_{k=0}^{\infty} a_k x^k) +l(l+1)\sum_{k=0}^{\infty} a_k x^k = 0$$ Performing the first derivative and multiplying by $1-x^2$: $$ \frac{d}{dx}(\sum_{k=1}^{\infty} ka_k x^{k-1}-ka_k x^{k+1}) +\sum_{k=0}^{\infty}l(l+1) a_k x^k = 0$$ Take the remaining derivative: $$\sum_{k=2}^{\infty} (k(k-1)a_k x^{k-2}-k(k+1)a_k x^{k})+\sum_{k=0}^{\infty}l(l+1) a_k x^k=0$$ Now we need to shift the lower bounds of the first sum. For the first term do a substitution $k'\rightarrow k-2$ and for the second one, the only off term will be a term $2a_1x$. Thus: $$\sum_{k=0}^{\infty} [(k+1)(k+2)a_{k+2}-k(k+1)a_k+l(l+1) a_k ]x^k +2a_1 x= 0$$ If the off-term $2a_1 x$ weren't there, then demanding each coefficient of $x^k$ to vanish for the equation to hold would give the desired recursion relation. However, since it is there, the recursion would not be correct for odd $k$ values. Which step of the substitution into the equation is wrong? |
| Posted: 28 Feb 2022 12:32 AM PST how he found the two inequalities (56 and the last)? Thanks! |
| Find the centre of gravity of an elliptic arc. Posted: 28 Feb 2022 01:00 AM PST Find, without using Pappus' theorem, the centre of gravity of an arc of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ which situated in the first quadrant.
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| Binomial Distribution for defects Posted: 28 Feb 2022 12:33 AM PST I'm stuck on the following problem:
I've got the following at the moment for part a but I get the feeling that I'm doing this slightly wrong and don't want to do all of it incorrectly. a.) $n = 10$ $k($number of success$) = 2$ $p($population proportion$) = .01$ $P(k \le 2) = \binom{n}{k} \times p^k \times (1-p)^(n-k)$ $P(k \le 2) = .00441$ Also I'm unsure of what is being asked for part b. |
| Field of order 8, $a^2+ab+b^2=0$ implies $a=0$ and $b=0$. Posted: 28 Feb 2022 12:41 AM PST I was able to come up with a proof for this problem however, it seems like my argument can work for any field of even order and not just odd powers of 2 so I'm convinced there is something wrong here. Can someone verify or see where the error in reasoning is? Problem: Let $F$ be a field with $2^n$ elements, with $n$ odd. Show that for $a,b \in F$ that $a^2+ab+b^2=0$ implies that $a=0$ and $b=0$. Proof: Suppose $a,b \in F$ and $a^2+ab+b^2=0$. $\implies a^2+2ab+b^2 = ab$ $\implies \frac{2^n}{2}(a^2+2ab+b^2) = \frac{2^n}{2}ab$ $\implies \frac{2^n}{2}a^2+ 2^nab+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$ $\implies \frac{2^n}{2}a^2+\frac{2^n}{2}b^2 = \frac{2^n}{2}ab$ (since F is a group under addition, then every element to the $|F|$ multiple is the identity thus $2^n(ab) = 0$) $\implies \frac{2^n}{2}(a^2+b^2) = \frac{2^n}{n}ab$ $\implies a^2+b^2 = ab$ $\implies a^2-ab+b^2 = 0 = a^2+ab+b^2$ $\implies -ab = ab \implies 2ab=0 \implies ab=0$. Thus, $a=0$ or $b=0$. However, if just one of them is zero, then so is the other ($a=0 \implies a^2+ab+b^2 = 0 \implies b^2 = 0 \implies b=0$). Thus, $a=0$ and $b=0$. QED Anyways, if there is something wrong with this proof, could someone give me a subtle hint perhaps? I've been stuck on this seemingly simple problem for awhile now. |
| Stability condition for explicit scheme in finite differences Posted: 28 Feb 2022 12:03 AM PST I've the following explicit scheme in finite differences (for a one dimensional non uniform diffusion problem), being $k$ the time step, $h$ the space step, $A$ the thermal conductivity at position $i$ and $u_i^n$ the diffused quantity at position $i$ on time $n$: $$u_i^{n+1} = \frac{k}{h^2}A_{i+1/2}u_{i+1} + \frac{k}{h^2}A_{i-1/2}u_{i+1} + [1 - \frac{k}{h^2}(A_{i+1/2} + A_{i-1/2})]u_i$$ We know that $k/h^2 > 0$ and if we assume that $1 - \frac{k}{h^2}(A_{i+1/2} + A_{i-1/2}) \geq 0$ then we can show that the stability condition is: $$ \frac{k}{h^2} \leq \frac{1}{\max(A_{i+1/2} + A_{i-1/2})}\ \forall i$$ I have a proof for this (I don't want to write it all here unless it is necessary) but I want to check it by asking if somebody has used this (or some similar) condition before. Thanks in advance, Federico |
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