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An equation that takes Pythagoras to a new level
- jimhefferon 7y agoThe caption "The equation 1⁰² + 1¹² + 1²² = 1³² + 1⁴², whose answer is that both sides equal 365, was immortalized in a different form in this 1895 painting: “Mental Arithmetic. In the Public School of S. Rachinsky.” (NIKOLAY BOGDANOV-BELSKY)" has funky exponents. Medium requires that I make an account to say that, so I am saying it here.
- mkl 7y agoThat formatting error is repeated in at least one other place too.
- jkinudsjknds 7y agoOn the subject of that painting, can anyone tell me what the kids are wearing on their feet?
- CliffStoll 7y agoMuckluks. Or the 1890 Russian version
- norrius 7y agoThese are bast shoes [0], known in Russian as lápti, which were the typical footwear in villages until the 20th century. They are strapped to the leg so that they don't fall off. [0]: https://en.wikipedia.org/wiki/Bast_shoe https://en.wikipedia.org/wiki/Bast_shoe
- CliffStoll 7y agoI think the equation on the board is ( 10^2 + 11^2 + 12^2 +13^2 + 14^2 ) / 365 which solves for 2. The students are expected solve this with "Mental Arithmetic". Try that in your head!
- jimhefferon 7y agoThanks. I cannot see that on my screen. The powers of 1 caught my eye, though.
- CliffStoll 7y agoHappy Pi-Day!
- CliffStoll 7y agoLet a=12 then (a-2)^2 + (a-1)^2 + a^2 + (a+1)^2 + (a+2)^2 /365 and a binomial a+b squared yields a^2 + 2ab + b^2 so a^2 + 4 + a^2+1 + a^2 + a^2+1 + a^2 + 4 add up all those a^2 's to get 5a^2 + 10 replace a by 12 and 720 + 10 yields 730 Divide by 365 2 That's what the sharp kid in the foreground is doing at the moment of the painting.
- delhanty 7y agoAssuming one's memorized a table of squares already. 10^2 = 100 11^2 + 13^2 = 121 + 169 = 290 12^2 + 14^2 = 144 + 196 = 340 100 + 290 + 340 = 730 730 / 365 = 2
- lonelappde 7y agoYou can also "walk it up" if you haven't memorized them: 10^2*5 + (10+11)*4 + (11+12)*3 + (12+13)*2 + (13+14) ------------ 500 + 84 + 69 + 50 + 27 ----- 730 It's a bit of detail to juggle in your head but not too bad if you hang on to your current subtotal and remember where you are in the sequence to compute the next addend, and you know it's easier to add multi-digit numbers left-to-right than-right to-left. You only carry from the ones place twice.
- stevespang 7y agoFUCK MEDIUM, assholes
- Avshalom 7y agoPtolemy's theorem is another Pythagoras on steroids https://m.youtube.com/watch?v=bJOuzqu3MUQ https://m.youtube.com/watch?v=bJOuzqu3MUQ
- chx 7y agoThe linked paper https://fermatslibrary.com/s/proof-without-words-pythagorean-runs https://fermatslibrary.com/s/proof-without-words-pythagorean... gives you the formula and I am much more a symbols guy than a geometry guy so I will prove it that way, first rearrange it: (4T_n)^2=(4T_n+1)^2-(4T_n-1)^2+....(4T_n+n)^2-(4T_n-n)^2 Now, (4T_n+k)^2-(4T_n-k)^2=16T_nk (this, of course, shows up in the geometric proof as four rectangles where one side is 4T_n and the other is k) so this is equivalent to 16T_nT_n = 16 T_n (1+...n) which is equvialent to: T_n=1+...n But that is the very definition of T_n. Q.e.d. Also if you want to take Pythagoras to a new level, Edsger W. Dijkstra who is much better known for his work in CS has a very interesting formulation https://www.cs.utexas.edu/users/EWD/transcriptions/EWD09xx/EWD975.html https://www.cs.utexas.edu/users/EWD/transcriptions/EWD09xx/E... proving sgn(alpha+beta-gamma)=sgn(a^2+b^2-c^2) which includes and extends the Pythagoras theorem.
- imglorp 7y agoWow, fermatslibrary has gone further downhill. It's almost unusable now with all the engagement stuff.
- RockofStrength 7y agoAlso 3^3 + 4^3 + 5^3 = 6^3
- rk06 7y agoMedium has paywalled me. Any non-paywall links?
- weibing 7y agoThat is the reason I really don’t like Medium. So many click-bait articles and paywall
- gordaco 7y agoI thought this was going to help me with Project Euler 261 [0], but it's not exactly the same, so no luck so far. Cool article anyway; linking geometry and algebra is always satisfactory. [0] https://projecteuler.net/problem=261 https://projecteuler.net/problem=261
- lonelappde 7y agoHow much CPU time and number theory do problems in that range generally require? Brute force search of 10^10 (with some automatic monitoring of how wide a range of 'm' to scan at each candidate) is a lot but not a lot of a lot, if you code in a CPU efficient language and maybe use parallel computation and are willing to wait a while.
- gordaco 7y agoThe difficulty and the required run time vary a lot starting from problem 150 or so. This one seems particularly difficult (it's rated at 85% difficulty, which is also the highest I've managed to solve). The problem here is that the search space is not 10^10, but higher, since if you iterate over k you don't know which values of m or n need to be tested. For m=1 it's clear that you can just use Pythagorean triples, but for bigger values I don't know if there is a reliable procedure to generate solutions.
- johnr2 7y ago> Medium has paywalled me. Any non-paywall links? The link works here in Firefox with NoScript, also in Dillo (no JS support).
- dr_dshiv 7y agoBeautiful. Pythagoreans believed that the world was made of math -- and that the harmonies of mathematics revealed the harmonies of the universe. Pretty good for 6th century BC.