User talk:Qria

In response to your feedback

I have several helpful links on my user page for editing. Feel free to contact me with any questions!

―Ross coolguy 16:08, 30 January 2013 (UTC)[reply]

June 2013

Hello, I'm DVdm. Your recent edit to the page Regular polygon appears to have added incorrect information, so I removed it for now. If you believe the information was correct, please cite a reliable source or discuss your change on the article's talk page. If you think I made a mistake, or if you have any questions, you can leave me a message on my talk page. Thanks. DVdm (talk) 08:50, 10 June 2013 (UTC)[reply]

Note - The expression that you provided, $4 (1+ \sqrt 2+ \sqrt 2 (2+ \sqrt 2))$ , aka $4\left(1+{\sqrt {2+{\sqrt {2\left(2+{\sqrt {2}}\right)}}}}\right)$ , produces 12.59127551. Cheers - DVdm (talk) 09:00, 10 June 2013 (UTC)[reply]

Note - O.t.o.h. using the identity

\tan {\tfrac {1}{2}}\theta ={\frac {\tan \theta }{1+{\sqrt {1+\tan ^{2}\theta }}}}

it's easy to prove that

{\frac {4}{\cot {\frac {\pi }{16}}}}=4\left(1+{\sqrt {2}}+{\sqrt {2\left(2+{\sqrt {2}}\right)}}\right)

So the correct expression is $4 (1+ \sqrt 2 + \sqrt 2(2+ \sqrt 2))$ , which does indeed produce 20.10935797. ~~I'll make the change.~~ I notice you already made the correction. Looks like it was a typo. Cheers - DVdm (talk) 11:50, 10 June 2013 (UTC)[reply]