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• 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, 17:46
Ackbach replied to a thread Jokes in Chat Room
Excellent!
216 replies | 33267 view(s)
• Yesterday, 15:22
You are right. Since $\arctan(-x)=-\arctan(x)$, it is easy to see that $|x_n|=y_n$ for all $n$. You are also right that $y_n$ is decreasing....
1 replies | 61 view(s)
• Yesterday, 14:58
Evgeny.Makarov replied to a thread Jokes in Chat Room
I would say that the distance between opposite vertices in a rectangle is the diameter. This terminology is used in graph theory. Professor: What...
216 replies | 33267 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, 03:31
Hi Tan Tom, welcome to MHB! ;) If I understand correctly, you have $$(1-2j)e^{j\frac\pi 2 n} +a_3 e ^ {-(j\frac\pi 2n)} + 3 + (-1)^{n+1}=sum$$...
1 replies | 58 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, 20:03
Yep. (Nod)
9 replies | 150 view(s)
• February 19th, 2019, 18:40
Correct. (Nod)
9 replies | 150 view(s)
• February 19th, 2019, 17:48
We have the property that the product of an invertible matrix with another matrix has the same rank as that other matrix. Can we prove that...
9 replies | 150 view(s)
• February 19th, 2019, 15:45
Congratulations! :cool:
4 replies | 85 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, 10:21
steenis has shown wonderful acumen in posting great math help, and so it is with great pleasure that we announce the awarding of the MHB Math Helper...
4 replies | 85 view(s)
• February 19th, 2019, 06:51
Hi, and welcome to the forum. Can you figure out the initial position (height above the ground) of the minute hand tip at 10 a.m.? Also, do you...
1 replies | 54 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)
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### 24 Visitor Messages

1. Thanks Fantini
2. Haha. \widehat's cool. Also check out \widetilde. But these sometimes get ugly.
3. Yep. Corrected now.
4. Thanks for the thanks. For what it's worth, I have already let Jameson know that an option to turn off or disable the social media buttons would be a good idea in my opinion.
5. Hey, good catch...that is in fact what I meant to type, but my darn keyboard seems to have a mind of its own at times. Thanks for letting me know so I could fix it.
6. Yes thanks Fantini, edited
7. Not at all...it's GOOD to think about these things.
8. Fantini, have I seen you on SE before?
9. Fantini - Hey, thanks! Yeah, the colors go well together, I think. Bach was the greatest!
10. Hey Fantini . I reported the thread so the Mods in green can decide where to put it. I don't mess with topics I'm not very familiar with haha.
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