• Today, 03:42
Hi Hugo ... Thanks again for your help ... Hope that despite the heat, you enjoy your holiday ... It gets hot down here sometimes ... even...
7 replies | 109 view(s)
• Today, 03:31
Trying Steenis' idea ... Practice values for first statement in the proof of Theorem 12.9 in Browder ... ... that is practice values for n...
3 replies | 91 view(s)
• Today, 02:23
Thanks for such a clear and honest post, Hugo ... Still reflecting on your new answer to the issue ... Does anyone else have an answer ...?
7 replies | 109 view(s)
• Yesterday, 23:43
Thanks Steenis ... your post really got me thinking ... Yes ... important point that ... ... " ... By definition 12.8, you can only multiply two...
7 replies | 109 view(s)
• Yesterday, 22:22
Here's a link to a Desmos slope field for this ODE: https://www.desmos.com/calculator/my0rwuxl5x
3 replies | 71 view(s)
• Yesterday, 00:37
When you determine your integrating factor, you need only determine up to but not including the constant of integration: ...
3 replies | 71 view(s)
• February 20th, 2019, 00:25
THanks Steenis ... Good idea ... Peter
3 replies | 91 view(s)
• February 19th, 2019, 21:44
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ... I am currently reading Chapter 12: Multilinear Algebra ......
7 replies | 109 view(s)
• February 19th, 2019, 20:32
There are some typos/issues with your work. See if you can spot them...:)
4 replies | 84 view(s)
• February 19th, 2019, 15:41
Okay, if we are to consider a solution of the form: y=t^r Then: y'=rt^{r-1} y''=r(r-1)t^{r-2}
4 replies | 84 view(s)
• February 19th, 2019, 10:27
Congratulations, and welcome to the team! (Yes)
4 replies | 85 view(s)
• February 19th, 2019, 01:56
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ... I am currently reading Chapter 12: Multilinear Algebra ......
3 replies | 91 view(s)
• February 19th, 2019, 00:00
Let's try $$x=5$$: \sqrt{5-1}=5-7 2=-2 Is this true?
4 replies | 77 view(s)
• February 18th, 2019, 23:57
Overview: This product is designed to allow you to specify which usergroups will have access to the user ignore system. Users who are not a member...
0 replies | 49 view(s)
• February 18th, 2019, 22:42
What do you get when you substitute each potential solution into the original equation?
4 replies | 77 view(s)
• February 18th, 2019, 20:22
MarkFL replied to a thread Starred Posts in vBulletin Products
I've installed this product on many vB sites without issue. As I said in another thread, I really recommend that you upgrade your vB. It's very old....
4 replies | 770 view(s)
• February 18th, 2019, 20:18
MarkFL replied to a thread Forum Marquee in vBulletin Products
Please send me the login credentials by PM to an admin account on your site, and I will see if I can figure out what the issue on your site is. You...
2 replies | 569 view(s)
• February 18th, 2019, 08:58
Thanks for the extensive help GJA, I really appreciate it ... Still reflecting on what you have said ... Thanks again ... Peter
2 replies | 90 view(s)
• February 18th, 2019, 03:37
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ... I am currently reading Chapter 12: Multilinear Algebra ......
2 replies | 90 view(s)
• February 18th, 2019, 02:58
Thanks GJA ... appreciate your help ... Peter
2 replies | 57 view(s)
• February 17th, 2019, 23:17
I am reading Loring W.Tu's book: "An Introduction to Manifolds" (Second Edition) ... I need help in order to fully understand Tu's section on the...
2 replies | 57 view(s)
• February 15th, 2019, 17:10
How many PIN codes are possible...can you apply the fundamental counting principle to answer this question?
2 replies | 217 view(s)
• February 14th, 2019, 23:20
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ... I am currently reading Chapter 12: Multilinear Algebra and am...
2 replies | 93 view(s)
• February 13th, 2019, 00:02
MarkFL replied to a thread Career Change in Chat Room
Good luck, Adrian! I've had family members as well as myself receive care at the Mayo Clinic in Jacksonville and St. Augustine here in Florida. They...
6 replies | 163 view(s)
• February 12th, 2019, 16:07
Wilmer replied to a thread A rebus puzzle in Chat Room
Yes sirree!!
4 replies | 120 view(s)
• February 12th, 2019, 09:21
Wilmer replied to a thread A rebus puzzle in Chat Room
Nope! Hint: -- -- -- ----
4 replies | 120 view(s)
• February 11th, 2019, 14:01
Wilmer started a thread A rebus puzzle in Chat Room
L L I I V V E E Answer = 4 words
4 replies | 120 view(s)
• February 11th, 2019, 11:49
Do you "see" that you only need : Proxima Centauri is 4.22 light years away from Earth. A light year is equal to 9,461 billion kilometers. ...
1 replies | 115 view(s)
• February 11th, 2019, 02:49
Hi Peter, Glad to hear you're taking a look at Tu, I think it's a good book on the subject. When $V$ is an arbitrary vector space there is...
2 replies | 134 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 (19)
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
Interests:
Number Theory
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India

#### 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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