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We're so glad you're here! Meatsgains^(talk) 02:57, 14 February 2018 (UTC)[reply]

I saw your message on organelle, and while I'm not an admin, the proper place to request this is at Wikipedia:Requests for page protection. It does indeed look like it needs protection, I must say. --AmaryllisGardener ^talk 04:19, 25 February 2018 (UTC)[reply]

This is your only warning; if you add an inappropriate image to Wikipedia again, you may be blocked from editing without further notice. Bedivere (talk) 13:11, 30 January 2023 (UTC)[reply]

You are not a Wikipedia administrator. Furthermore, I have also opened a topic on the talk page to discuss this particular entry, so discuss it there, instead of on my talk page. Do not make empty threats to regular users of Wikipedia. Captain Chicky (talk) 18:07, 30 January 2023 (UTC)[reply]

I am not an administrator but I can always report you. Please remember to be civil. Your edits were reverted by several users. If you continue to engage in edit wars you will be blocked by an administrator. Bedivere (talk) 19:44, 30 January 2023 (UTC)[reply]

Also, toast has been removed from the list so the point of the thread on the talk page is moot. Bedivere (talk) 20:08, 30 January 2023 (UTC)[reply]

Fair enough. Let’s end it here then. Captain Chicky (talk) 05:44, 31 January 2023 (UTC)[reply]

Likewise, I advise you to be civil and not make empty threats. If you may observe, the edit you are referring to is literally the first edit I have made on the page, with provided reason and an extension to discussion. Please be more observant instead of directly accusing and threatening another user. Captain Chicky (talk) 05:44, 31 January 2023 (UTC)[reply]

Arctangent of tangent is a trick, to evaluate this integral, you have to evaluate its piece-wise, to do it, you have to put Dottie number into the integration limits. In other words, it is like ${\displaystyle \int _{0}^{D}1dx}$, an itegral, which is equal to the Dottie number but trivial. In other words, to evaluate it you should already know Dottie number.

Let look into it more precise. Arctangent of tangent is a sawtooth function with discontinuinty at ${\displaystyle \pi /2}$: https://www.wolframalpha.com/input?i=Arctan%5BTan%5Bx%5D%5D. So, the integral ${\displaystyle {\frac {1}{\pi }}\int _{0}^{\pi }\arctan \left(\tan \left({\frac {t-\sin t+{\frac {\pi }{2}}}{2}}\right)\right){\text{ d}}t}$ should be represented as ${\displaystyle {\frac {1}{\pi }}\int _{0}^{N}{\frac {t-\sin t+{\frac {\pi }{2}}}{2}}{\text{ d}}t+{\frac {1}{\pi }}\int _{N}^{\pi }{\frac {t-\sin t+{\frac {\pi }{2}}}{2}}-\pi {\text{ d}}t}$ where the number N is the solution of ${\displaystyle {\frac {N-\sin N+{\frac {\pi }{2}}}{2}}=\pi /2}$ which itself needs Dottie number to be expressed. Reciprocist (talk) 20:32, 8 February 2024 (UTC)[reply]

Icic

I checked the floor function integral now that you’ve mentioned this and it’s pretty much the same thing lol so makes sense

thanks Captain Chicky (talk) 21:22, 8 February 2024 (UTC)[reply]