Author Topic: Forum currently undergoing development.  (Read 299 times)

huntersair

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Forum currently undergoing development.
« on: June 22, 2017, 02:26:06 PM »
Hi All,

This is a temporary thread to showcase what messages look like when posted.
At the moment the forum itself is going through some changes and is in its very alpha stages.

Best,
Nick

huntersair

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Re: Forum currently undergoing development.
« Reply #1 on: June 22, 2017, 03:47:49 PM »
[tex]\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k^2}= \frac{\pi^2}{6}[/tex]

huntersair

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Re: Forum currently undergoing development.
« Reply #2 on: June 22, 2017, 04:54:30 PM »
\[  E=mc^2  \]

Tenchberry

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Re: Forum currently undergoing development.
« Reply #3 on: June 22, 2017, 04:56:06 PM »
Hey there,

Looks perfect. Good job!

Tenchberry

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Re: Forum currently undergoing development.
« Reply #4 on: June 22, 2017, 04:59:16 PM »
\(E = mc^2\)

Tenchberry

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Re: Forum currently undergoing development.
« Reply #5 on: June 22, 2017, 04:59:44 PM »
\(H\Psi = E\Psi \)

Tenchberry

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Re: Forum currently undergoing development.
« Reply #6 on: June 22, 2017, 05:02:31 PM »
\begin{align}
H\Psi = E\Psi
\end{align}

Tenchberry

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Re: Forum currently undergoing development.
« Reply #7 on: June 22, 2017, 05:04:57 PM »
\begin{align}
H_{elec} = \sum_{n}{} \frac{Zq}{4\pi \epsilon_0 r} + ...
\end{align}

huntersair

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Re: Forum currently undergoing development.
« Reply #8 on: June 22, 2017, 05:07:35 PM »
[tex]\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{k^2}= \frac{\pi^2}{6}[/tex]

^test.

huntersair

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Re: Forum currently undergoing development.
« Reply #9 on: June 22, 2017, 05:08:15 PM »
have to delete the previous screwup.
Worried about package clashing and possible clobbered variables.

Tenchberry

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Re: Forum currently undergoing development.
« Reply #10 on: June 22, 2017, 05:10:05 PM »
\begin{align}
E_{opt} = \frac{\left< \Psi | H | \Psi \right>}{\left< \Psi | \Psi \right>}
\end{align}
« Last Edit: June 22, 2017, 05:11:40 PM by Tenchberry »

huntersair

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Re: Forum currently undergoing development.
« Reply #11 on: June 22, 2017, 05:25:37 PM »
Done.
The source code no longer recognises [tex] [/tex] as a tag.

Tenchberry

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Re: Forum currently undergoing development.
« Reply #12 on: June 22, 2017, 05:30:53 PM »
\begin{align}
\lim_{n \to \infty}\sum_{k=1}^{n} \frac{1}{k^2}= \frac{\pi^2}{6}
\end{align}

huntersair

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Re: Forum currently undergoing development.
« Reply #13 on: June 22, 2017, 11:07:23 PM »
Very impressive.
But can you solve the same sum with n^2 replaced with n^3? ;)

Keynarx

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Re: Forum currently undergoing development.
« Reply #14 on: November 22, 2017, 12:41:13 PM »
  \( \sum_1^N (N-i)^{D}  = \sum_1^{N-1} (N-i)^{D} = \sum_1^{N-1} i^{D}  \)           \( \forall (N,D) \in \mathbb N^{2} \)

\( D=2   \Rightarrow \sum_1^{N-1} (N-i)^{2} = \sum_1^{N-1} i^{2} \Leftrightarrow \sum_1^{N-1} (N^{2}-2Ni+i^{2}) = \sum_1^{N-1} i^{2} \)

\( \Leftrightarrow \sum_1^{N-1} (N^{2}-2Ni) = 0  \Leftrightarrow(N-1)N^{2} -2N \sum_1^{N-1} i = 0 \)

\(\Leftrightarrow \sum_1^{N-1} i = \frac{(N-1)N}{2} \Leftrightarrow  \sum_1^{N} i = \frac{(N+1)N}{2} \)