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• Today, 07:48
Not just for the next term, but for the whole sequence! In fact, your prediction and mine both say that the $(n+1)$th term minus the $n$th term is...
8 replies | 105 view(s)
• Today, 06:37
Hi hm2018April, and welcome to MHB! Mathematicians hate problems like this, because there usually isn't a single, "correct" solution. You need to...
8 replies | 105 view(s)
• Yesterday, 08:57
Let $A = \begin{bmatrix}0&1\\1&1\end{bmatrix} = \begin{bmatrix}\varphi_0&\varphi_1\\ \varphi_1&\varphi_2\end{bmatrix}$. By induction, \$A^n =...
2 replies | 58 view(s)
• April 22nd, 2018, 10:17
MarkFL replied to a thread Olinguito in Introductions
Hello, and welcome to MHB! (Wave)
8 replies | 102 view(s)
• April 22nd, 2018, 09:57
Opalg replied to a thread Banach space in Analysis
$$|f_n(x)-f_m(x)|=\left| \frac{\sin^2\left (\frac{2\pi}{T}x\right )}{\frac{1}{n}+\left |\sin\left (\frac{2\pi}{T}x\right )\right |}-\frac{\sin^2\left... 6 replies | 78 view(s) • April 22nd, 2018, 04:36 Hi Olinguito, and welcome to MHB! We look forward seeing your problems. 2x^2y^2-2xy+x^2+y^2-2x-2y+3 = 2(xy-1)^2 + (x+y-1)^2. If that is zero then... 4 replies | 69 view(s) • April 22nd, 2018, 04:19 Opalg replied to a thread Unsolved Challenge imaginary integral in Challenge Questions and Puzzles Rationalise the fraction: \frac{\sin t}{ \sin t+ i\sqrt{n+\cos^2 t}} = \frac{\sin t\bigl( \sin t - i\sqrt{n+\cos^2 t}\bigr)}{1+n}. 1 replies | 57 view(s) • April 21st, 2018, 16:02 Reminds me of a recent xkcd cartoon. 19 replies | 514 view(s) • April 21st, 2018, 14:15 There was no inconvenience, except to you trying to post and to those trying to read the thread. It was bad coding on my part, done before I knew the... 19 replies | 514 view(s) • April 20th, 2018, 22:16 Update - Version 1.3.1: Bug fix for usernames with special characters. 2 replies | 224 view(s) • April 20th, 2018, 19:16 I have added code to that external script so a single quote in a username won't throw an error, and I edited the post so that only the username is... 19 replies | 514 view(s) • April 20th, 2018, 12:00 Here is this week's POTW: A small bug is placed between two blocks of masses m_1 and m_2 (m_1>m_2) on a frictionless table. A horizontal... 0 replies | 38 view(s) • April 20th, 2018, 11:59 No one answered this week's problem, and my solution is as follows: A formula for power P we can apply here is: P=\frac{1}{2}\mu\omega^2A^2v... 1 replies | 75 view(s) • April 17th, 2018, 10:04 Opalg replied to a thread Most ever users in Chat Room It looks as though Spring Break has brought out the users again! It now says Most users ever online was 439, April 15th, 2018 at 19:20. 2 replies | 321 view(s) • April 17th, 2018, 03:51 Hi Alexthexela, and welcome to MHB! In the number 983,389^{389}, the base is 983,389 and the exponent is 389. The problem tells you to... 1 replies | 94 view(s) • April 16th, 2018, 12:25 In the UK, it is the leading supplier of accounting software packages for small businesses, https://uk.sageone.com/accounts/. I would not... 7 replies | 119 view(s) • April 15th, 2018, 20:01 Awesome work, as always Chris! (Yes) Using your post, where the base case and induction statement have been given, then the induction step would... 6 replies | 133 view(s) • April 15th, 2018, 13:59 Opalg replied to a thread A Difficult Partial Sum in Pre-Calculus Writing those as$$S_0= \frac 14 = \frac12 - \frac14 = \frac12 - \frac28 = \frac12 - \frac{2^1}{3^{2^1} - 1}, \\ S_1= \frac 9{20} = \frac12...
6 replies | 133 view(s)
• April 15th, 2018, 12:02
Here are the choices: A.) \frac{1}{2} B.) \frac{1}{2}-\frac{2^{2006}}{3^{2^{2006}}-1} C.) \frac{1}{2}-\frac{2^{2005}}{3^{2^{2005}}-1} D.)...
6 replies | 133 view(s)
• April 14th, 2018, 10:56
The actual sum he's given is: S=\sum_{k=0}^{2006}\left(\frac{2^k}{3^{2^k}+1}\right) And he's given several choices for the result. I was hoping...
6 replies | 133 view(s)
• April 14th, 2018, 10:10
Hello MHB! (Wave) A young man in high school I know has been essentially tasked with finding the following partial sum: ...
6 replies | 133 view(s)
• April 12th, 2018, 11:47
MarkFL replied to a thread Hello from Pan in Introductions
Hello, and welcome to MHB panpan! (Wave)
6 replies | 94 view(s)
• April 12th, 2018, 11:15
Hello, MHB Community! (Wave) anemone has asked me to stand in for her for a few weeks, so please be gentle. (Bigsmile) Here is this week's...
1 replies | 75 view(s)
• April 12th, 2018, 11:11
Hello MHB Community! (Wave) I am going to stand in for anemone for a few weeks. Congratulations to the following for their correct submissions:...
1 replies | 133 view(s)
• April 9th, 2018, 21:16
Yes, that was a bad post. The OP stated an unfamiliarity with limits, so I "winged it." What I would actually do is: ...
7 replies | 108 view(s)
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