• Yesterday, 14:11
Yes. And it also assumes we take $\nu=2$ steps with the damped Jacobi method. (Nod) It's not quite clear from the text to me. With a choice...
16 replies | 396 view(s)
• Yesterday, 12:19
We can freely choose to use $\omega=2/3$ or a different $\omega$, or not use the damped Jacobi method at all. The chapter explains that it's giving...
16 replies | 396 view(s)
• Yesterday, 06:19
Thanks steenis ... No worries at all ... Thanks for all your help ... Peter
10 replies | 163 view(s)
• Yesterday, 06:04
Thanks Steenis ... That proof seems really clear ... Will work through it again shortly... Peter
4 replies | 73 view(s)
• Yesterday, 05:18
Sorry Steenis ... I don't understand you ... Can you give me a hint as to what is wrong ...? Peter
10 replies | 163 view(s)
• Yesterday, 04:51
Thanks steenis ... most helpful ... Can see that the short exact sequence $0\rightarrow \text{ker } f \overset{i}{ \rightarrow}R^{(n)}... 10 replies | 163 view(s) • Yesterday, 01:22 ======================================================================== Since I could not see any specific errors, I have completed the proof... 4 replies | 73 view(s) • June 17th, 2018, 14:46 I have thought of the following algorithm: We put an antenna$k$meters east of the westernmost house. We continue to the east, by placing an... 1 replies | 36 view(s) • June 17th, 2018, 12:39 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations I can't imagine trying to use the internet on a telephone. I'm sorry scrolling on a telephone is such a chore...they should fix that. 8 replies | 106 view(s) • June 17th, 2018, 11:00 Hello!!! (Wave) We consider a long country road with$n$houses placed along of it (we think of the road as a big line segment). We want to put... 1 replies | 36 view(s) • June 17th, 2018, 05:04 Peter started a thread Deveno ... in Chat Room Deveno is much missed ... especially by those who frequent the Linear and Abstract Algebra Forum ... Deveno's pedagogical abilities were as... 0 replies | 42 view(s) • June 17th, 2018, 04:40 I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ... I am currently studying Chapter 10: Introduction to Module Theory ...... 4 replies | 73 view(s) • June 17th, 2018, 00:59 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations Are you using Tapatalk by any chance? I have code in place to let me know when posts have been edited, so I don't miss added content (it's better to... 8 replies | 106 view(s) • June 17th, 2018, 00:52 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations When you multiply by$\mu(x)$, you get: \sec(x)y'+\tan(x)\sex(x)y=\sec(x)\sin(2x) This can be written as: ... 8 replies | 106 view(s) • June 16th, 2018, 23:10 Thanks Steenis ... You have shown that$R^{(n)} / N \cong M$where$N = \text{ Ker } f$... ... ... ... ... (1) ... and we have by... 10 replies | 163 view(s) • June 16th, 2018, 19:51 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations -\ln(\cos(x))=\ln\left((\cos(x))^{-1}\right)=\ln(\sec(x)) 8 replies | 106 view(s) • June 16th, 2018, 19:38 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x) 8 replies | 106 view(s) • June 16th, 2018, 00:29 300+300\cdot\frac{1666}{100}=300(1+16.66)=300\cdot17.66=5298 Here, we have taken 300, and added 1666% of 300 to it. However if we multiply 300 by... 4 replies | 98 view(s) • June 16th, 2018, 00:03 Thanks steenis ... but not sure if I follow .. ... but will try ... as follows ... We have an epimorphism$f:R^{(n)} \longrightarrow M$... 10 replies | 163 view(s) • June 15th, 2018, 16:28 \mu(x)=\exp\left(\int\frac{1}{x}\,dx\right)=e^{\ln(x)}=x 3 replies | 59 view(s) • June 15th, 2018, 15:53 We can see the integrating factor is$\mu(x)=x$and so the ODE will become: \frac{d}{dx}(xy)=x\sin(x) Upon integrating, we get: ... 3 replies | 59 view(s) • June 15th, 2018, 14:14 That's fine giving$x$as a function of$y$...I just chose to give$y$as a function of$x$since the original equation has$x\$ as the independent...
4 replies | 69 view(s)
• June 15th, 2018, 12:44
We could also simply state: \d{x}{y}=e^y-x \d{x}{y}+x=e^y Now, our integrating factor is: \mu(y)=\exp\left(\int\,dy\right)=e^y
4 replies | 69 view(s)
• June 15th, 2018, 03:08
I would begin here with the substitution: u=e^y\implies u'=uy' And so the ODE becomes: \frac{1}{u}u'=\frac{1}{u-x} Or:
4 replies | 69 view(s)
• June 14th, 2018, 11:23
Hello and welcome to MHB! (Wave) I've moved this thread to our elementary algebra forum, since this is a better fit for the problem and I've give...
2 replies | 74 view(s)
• June 14th, 2018, 06:50
I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.2 Free Modules ... ... I need help with some...
1 replies | 53 view(s)
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