User talk:Bukovets

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Hello, Bukovets, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

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I hope you enjoy editing here and being a Wikipedian! Please sign your messages on discussion pages using four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question on this page and then place {{helpme}} before the question. Again, welcome! RJFJR (talk) 19:39, 3 November 2009 (UTC)[reply]

British English vs American English[edit]

Do not edit Wikipedia merely to change one syle of English language usage to another. We have a set policy on the matter which you can read about at Wikipedia:Manual of Style#National varieties of English. Rmhermen (talk) 14:37, 29 October 2010 (UTC)[reply]

It is ridiculous to discuss whether the formula: x is a constant multiple of longitude, y is a constant multiple of the sine of latitude, for projecting the surface of a sphere onto a plane while preserving area, was discovered by Gall or Peters; it was discovered by Archimedes, who died in 212 BC and it is engraved on his tombstone.

Archimedes did not have the modern names for the familiar ratios of trigonometry; he worked in terms of similar triangles. Let O be the centre of the sphere. Let P be any point on the surface of the sphere. Call the circle on which the surface of the sphere intersects the plane containing O and perpendicular to OP the “equator”. (So if the sphere is the earth, P is a pole.) Imagine a narrow path around the sphere joining points equidistant from P (i.e. the path following a circle of latitude). Let N be a point on the edge of this path on the side nearer P. Let S be the point on the opposite side of the path, opposite to N so NS is the width of the path. Let A be the point where the circle through P, N and S (first) meets the equator. Through S and N draw lines parallel to AO meeting OP in R and Q respectively. Through S draw a line parallel to OP to meet QN at T and a line in the plane PNS and perpendicular to OS (and so tangential to the circle PNS) to meet QN at U. Then the angles SOA (in modern terms the latitude of S), ORS and UST are the same. So triangles ORS and UTS are similar and TS/US = RS/OS. (In modern terms this is the cosine of the latitude). Now imagine a cylinder having the same radius as the sphere and OP as its axis. The point A lies on the cylinder. Produce RS and QN to meet the cylinder at X and Y respectively. QR = ST = XY. Project the whole sphere onto the cylinder. The narrow path is projected onto a longer and narrower path on the cylinder. It is longer by a factor OA/RS = OS/RS. As the path is “narrow” it is effectively flat and the points N and U are effectively the same. So the path on the cylinder is narrower by a factor ST/SU. So the path on the cylinder has the same area as the path on the sphere. Unroll the cylinder and take A to be the origin. AX = radius of sphere multiplied by sine of OSA: the so-called Gall-Peters formula. The whole sphere is a rectangle whose height is the diameter of the sphere and whose width is the circumference. This is where the 4 pi r-squared formula for the area of the sphere comes from. Children need to know the formula for practical reasons and since it is wrong to teach a child a formula without demonstrating the proof, the construction has been known to everybody for two millennia. Bukovets (talk) 14:05, 21 September 2013 (UTC)[reply]

When Alice calculates 5^6 mod 23 she does not need 5^6 =15625. She calculates 5^2 = 25 = 2 mod 23. Therefore 5^4 = 2 X 2 = 4. Therefore 5^6= 5^4 X 5^2 = 2 X 4 =8. Thus she arrives at A = 8 in three steps instead of five. Generally if the index is N, Eve, who does not know what it is has to take N steps trying N=1, N=2 etc while Alice can find A and later s in about log(N) steps. That is what makes Alice's secret number secret.

Further, to find t, the inverse of s it is enough to know that st = 1 mod p. Euclid’s algorithm for finding gcd(t,p) is the quickest way of doing this. Bukovets (talk) 22:01, 28 September 2013 (UTC)[reply]

It is ridiculous to discuss whether the formula: x is a constant multiple of longitude, y is a constant multiple of the sine of latitude, for projecting the surface of a sphere onto a plane while preserving area, was discovered by Gall or Peters; it was discovered by Archimedes, who died in 212 BC and it is engraved on his tombstone.

Archimedes did not have the modern names for the familiar ratios of trigonometry; he worked in terms of similar triangles. Let O be the centre of the sphere. Let P be any point on the surface of the sphere. Call the circle on which the surface of the sphere intersects the plane containing O and perpendicular to OP the “equator”. (So if the sphere is the earth, P is a pole.) Imagine a narrow path around the sphere joining points equidistant from P (i.e. the path following a circle of latitude). Let N be a point on the edge of this path on the side nearer P. Let S be the point on the opposite side of the path, opposite to N so NS is the width of the path. Let A be the point where the circle through P, N and S (first) meets the equator. Through S and N draw lines parallel to AO meeting OP in R and Q respectively. Through S draw a line parallel to OP to meet QN at T and a line in the plane PNS and perpendicular to OS (and so tangential to the circle PNS) to meet QN at U. Then the angles SOA (in modern terms the latitude of S), ORS and UST are the same. So triangles ORS and UTS are similar and TS/US = RS/OS. (In modern terms this is the cosine of the latitude). Now imagine a cylinder having the same radius as the sphere and OP as its axis. The point A lies on the cylinder. Produce RS and QN to meet the cylinder at X and Y respectively. QR = ST = XY. Project the whole sphere onto the cylinder. The narrow path is projected onto a longer and narrower path on the cylinder. It is longer by a factor OA/RS = OS/RS. As the path is “narrow” it is effectively flat and the points N and U are effectively the same. So the path on the cylinder is narrower by a factor ST/SU. So the path on the cylinder has the same area as the path on the sphere. Unroll the cylinder and take A to be the origin. AX = radius of sphere multiplied by sine of OSA: the so-called Gall-Peters formula. The whole sphere is a rectangle whose height is the diameter of the sphere and whose width is the circumference. This is where the 4 pi r-squared formula for the area of the sphere comes from. Children need to know the formula for practical reasons and since it is wrong to teach a child a formula without demonstrating the proof, the construction has been known to everybody for two millennia. Bukovets (talk) 14:05, 21 September 2013 (UTC) Bukovets (talk) 22:03, 28 September 2013 (UTC)[reply]

Apostrophe[edit]

Welcome to Wikipedia. We welcome and appreciate your contributions, but we cannot accept original research. Original research also encompasses combining published sources in a way to imply something that none of them explicitly say. Please be prepared to cite a reliable source for all of your contributions. Thank you. Materialscientist (talk) 13:27, 18 January 2014 (UTC)[reply]

Please use Talk: page instead[edit]

Hi Bukovets,

Thanks, you're quite right about the article needing to make it a bit clearer that Mahler wrote the words to Lieder eines fahrenden Gesellen himself, but your (nicely signed btw!) suggestion should go on the discussion page (Talk:Lieder eines fahrenden Gesellen; there is a tab at top of page) instead of the article itself. All the best, Sparafucil (talk) 21:39, 8 July 2014 (UTC)[reply]

Thank you Sparafucil. I will try to put my comments in the right plce in future. (I hope this comment is in the right place!)

Why Sparafucil? The assassin in Rigoletto was Sparafucile. Does he become Sparafucil in German translation?

Thanks again

Bukovets

Bukovets (talk) 22:37, 8 July 2014 (UTC)[reply]

You're welcome. My namesake is from Burgundy and sings his own name as four syllables in the opera. I got a laugh out of the recent Met broadcast set in Las Vegas: Gilda's body ends up in the trunk of a car with the vanity licence-plate "SPARAFUC". Sparafucil (talk) 23:17, 8 July 2014 (UTC)[reply]

you edit to Euler's theorem[edit]

I am not an expert about the content of your edit, but what you did should have been placed in the TALK page of Euler's theorem, not the article. Also you should never sign an edit to the article, and always sign an addition to the talk page. Dhrm77 (talk) 13:08, 22 July 2016 (UTC)[reply]

spelling[edit]

Is it "de Groot" or "De Groot" or is it optional? If the children of Mr. and Mrs. De Groot want to be Mr. and Miss de Groot do they have to change their name?