• Today, 05:18
Oh yes, that's a typo... I meant $k$... (Nerd)
3 replies | 61 view(s)
• Today, 03:50
You mean the derivative in respect to $t$ ? (Thinking)
25 replies | 423 view(s)
• Today, 00:02
You wound up with $x^2\sin(x)$ on the RHS before integrating, but it should be $x\sin(x)$. :)
1 replies | 26 view(s)
• Yesterday, 19:23
When you multiply through by $\mu(x)$, you should have: \frac{d}{dx}(\sin(x)y)=2 And then integrate: \sin(x)y=2x+c_1 ...
2 replies | 33 view(s)
• June 19th, 2018, 19:21
Towards the end, when you divide through by $x$, you want: y(x)=\frac{e^x}{x}+\frac{c}{x} You mistakenly divided the constant by $e^x$.
1 replies | 39 view(s)
• June 19th, 2018, 15:31
Yes, but you used it on the original ODE, not the one in standard linear form. :)
4 replies | 87 view(s)
• June 18th, 2018, 07:29
Wilmer replied to a thread Present Value Maths in Other Topics
Book is correct. Are you using a calculator? You're probably entering the data wrongly...we can't tell... Try it this way: u = (1.02/1.10)^10 v =...
2 replies | 63 view(s)
• June 18th, 2018, 06:19
Thanks steenis ... No worries at all ... Thanks for all your help ... Peter
10 replies | 214 view(s)
• June 18th, 2018, 06:04
Thanks Steenis ... That proof seems really clear ... Will work through it again shortly... Peter
4 replies | 80 view(s)
• June 18th, 2018, 05:18
Sorry Steenis ... I don't understand you ... Can you give me a hint as to what is wrong ...? Peter
10 replies | 214 view(s)
• June 18th, 2018, 04:51
Thanks steenis ... most helpful ... Can see that the short exact sequence $0\rightarrow \text{ker } f \overset{i}{ \rightarrow}R^{(n)}... 10 replies | 214 view(s) • June 18th, 2018, 01:22 ======================================================================== Since I could not see any specific errors, I have completed the proof... 4 replies | 80 view(s) • June 17th, 2018, 14:46 I have thought of the following algorithm: We put an antenna$k$meters east of the westernmost house. We continue to the east, by placing an... 3 replies | 61 view(s) • June 17th, 2018, 12:39 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations I can't imagine trying to use the internet on a telephone. I'm sorry scrolling on a telephone is such a chore...they should fix that. 8 replies | 113 view(s) • June 17th, 2018, 11:00 Hello!!! (Wave) We consider a long country road with$n$houses placed along of it (we think of the road as a big line segment). We want to put... 3 replies | 61 view(s) • June 17th, 2018, 05:04 Peter started a thread Deveno ... in Chat Room Deveno is much missed ... especially by those who frequent the Linear and Abstract Algebra Forum ... Deveno's pedagogical abilities were as... 0 replies | 53 view(s) • June 17th, 2018, 04:40 I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ... I am currently studying Chapter 10: Introduction to Module Theory ...... 4 replies | 80 view(s) • June 17th, 2018, 00:59 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations Are you using Tapatalk by any chance? I have code in place to let me know when posts have been edited, so I don't miss added content (it's better to... 8 replies | 113 view(s) • June 17th, 2018, 00:52 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations When you multiply by$\mu(x)$, you get: \sec(x)y'+\tan(x)\sex(x)y=\sec(x)\sin(2x) This can be written as: ... 8 replies | 113 view(s) • June 16th, 2018, 23:10 Thanks Steenis ... You have shown that$R^{(n)} / N \cong M$where$N = \text{ Ker } f$... ... ... ... ... (1) ... and we have by... 10 replies | 214 view(s) • June 16th, 2018, 19:51 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations -\ln(\cos(x))=\ln\left((\cos(x))^{-1}\right)=\ln(\sec(x)) 8 replies | 113 view(s) • June 16th, 2018, 19:38 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x) 8 replies | 113 view(s) • June 16th, 2018, 09:40 https://percentagecalculator.net/ "percent" means "per hundred" Simple example: 200 increases to 218: that's "9 per hundred", right?... 4 replies | 111 view(s) • June 16th, 2018, 00:29 300+300\cdot\frac{1666}{100}=300(1+16.66)=300\cdot17.66=5298 Here, we have taken 300, and added 1666% of 300 to it. However if we multiply 300 by... 4 replies | 111 view(s) • June 16th, 2018, 00:03 Thanks steenis ... but not sure if I follow .. ... but will try ... as follows ... We have an epimorphism$f:R^{(n)} \longrightarrow M$... 10 replies | 214 view(s) More Activity ### 73 Visitor Messages 1. Yes, you are correct. The first post in a thread may be edited within 2 hours, and all other posts within 24 hours. However, if you run over the 24 hour limit, I will lift the limit for you temporarily since you are adding such great content. 2. Hey Balarka, according the the settings in the ACP it is 5 minutes. 3. Hey Balarka, Daniel is upgrading the forum software to a completely new platform. It should be back up soon. 4. I agree that fun shouldn't introduce wrong concepts. PS: Yes, I'll add one when I start the next post. 5. Ok, then you are looking at rigor but as I said these tutorials will be for fun. I want a high school student to understand it. It would be great to have your comments when I start posting bout that, though. I know I can learn from you as I am haven't read that much about these concepts. 6. What is your definition of Regularization ? 7. We can extend the zeta function to analytic function in the whole complex plane except at 1. By definition we have $$\zeta(s) =\sum \frac{1}{n^s}$$ So zeta on the left is analytic for all$s \neq 1$but the sum on the right does only converge for$Re(s) >1\$.
So the analytic continuation of zeta has enabled us to give values to the divergent sum on the right.
8. Hey Balarka , the idea of regularization is giving a finite value for divergent series or integrals. Consider the following
$$\zeta(-2)= 0$$
$$1+4+9+\cdots = 0$$
9. Okay, thanks. I think the former one is just what I wanted.
10. Hey Balarka, I do not know of a closed for that sum. But there are some identities related to this problem:
\begin{align*} \sum_{k=1}^p \frac{H_k^{(2)}}{k}+\sum_{k=1}^p \frac{H_k}{k^2} &= H_p^{(3)}+H_p^{(2)}H_p \\ \sum_{k=1}^p \frac{H_k^{(2)}}{k}+\sum_{k=1}^p \frac{(H_k)^2}{k} &= \frac{(H_p)^3+3H_p H_p^{(2)}+2H_p^{(3)}}{3} \end{align*}
To prove these, we can use induction.
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#### Basic Information

Date of Birth
January 12, 2000 (18)
Biography:
I know a thing or two about topology. Find number theory interesting but don't know much about it.
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West Bengal, India
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Number Theory
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#### Signature

Some of my notes on number theoretic topics are on a few primality tests and on Hardy & Littlewood's result (incomplete). The other articles are on quintics : about a brief description of Kiepert algorithm and a short introduction on quintic-solving algorithms, respectively. I have also written up an introductory thread on Riemann Hypothesis

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