Difference between revisions of "Talk:1047: Approximations"
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+ | The US population estimate is now off by 7 million, although 2018 just started. Even so, in December 2017, it would have been 4 million off. [[User:625571b7-aa66-4f98-ac5c-92464cfb4ed8|625571b7-aa66-4f98-ac5c-92464cfb4ed8]] ([[User talk:625571b7-aa66-4f98-ac5c-92464cfb4ed8|talk]]) 00:54, 19 January 2018 (UTC) | ||
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+ | : Off by 7 million out of 7-8 billion means that it's accurate to one part in 1,000. That's consistent with it's location in the chart -- next to other values that are accurate to 1 in 1,000. {{unsigned|Redbelly98}} | ||
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+ | The world population estimate is still accurate to within .1 billion. [[Special:Contributions/162.158.63.28|162.158.63.28]] 13:41, 5 May 2017 (UTC) | ||
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They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. '''[[User:Davidy22|<span title="I want you."><u><font color="purple" size="2px">David</font><font color="green" size="3px">y</font></u><sup><font color="indigo" size="1px">22</font></sup></span>]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 14:03, 8 January 2013 (UTC) | They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. '''[[User:Davidy22|<span title="I want you."><u><font color="purple" size="2px">David</font><font color="green" size="3px">y</font></u><sup><font color="indigo" size="1px">22</font></sup></span>]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 14:03, 8 January 2013 (UTC) | ||
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:: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC) | :: This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 00:37, 30 August 2013 (UTC) | ||
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+ | ::: Additionally, the Lewis Carroll idea is only one of several theories about where Douglas Adams got the number from. [[Special:Contributions/162.158.158.87|162.158.158.87]] 20:47, 28 November 2019 (UTC) | ||
"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct. Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it. The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)). --DrMath 06:47, 7 September 2013 (UTC) | "sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct. Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it. The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)). --DrMath 06:47, 7 September 2013 (UTC) | ||
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::::Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked [[545]] lately? I answered your question there, too) | ::::Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked [[545]] lately? I answered your question there, too) | ||
:::::As per the derivation from january 16 , you can use any a,b,c that satisfies this set of equations: 2 a = b - a, a + b = c- a, c + a = π - a. This is due to the fact that sin(x) = sin(π-x), and what was derived the 16th. [[Special:Contributions/173.245.53.199|173.245.53.199]] 12:38, 21 February 2014 (UTC) | :::::As per the derivation from january 16 , you can use any a,b,c that satisfies this set of equations: 2 a = b - a, a + b = c- a, c + a = π - a. This is due to the fact that sin(x) = sin(π-x), and what was derived the 16th. [[Special:Contributions/173.245.53.199|173.245.53.199]] 12:38, 21 February 2014 (UTC) | ||
− | + | :::: Dgbrt: If not convinced by the proofs linked to in the "explanation" part, you might want to try this: [http://www.wolframalpha.com/input/?i=sum_%28k%3D0%29%5E38+cos%281%2F77+%282+k%2B1%29+pi%29]. I'm sure you'll find inspiration for similar formulas using PI over [any odd integer]. Your assumption that Randall used WolframAlpha for this very identity is probably wrong. This is a very well-known formula that appears in many high school books, and I am pretty sure it is part of Randall's culture. And this has nothing to do with 1=1. As for your original post, | |
+ | ::::*√2 = (√2-1)/((4-2)π/2-π)+1 : Is this what you call "matching an equation" to √2? | ||
+ | ::::*So what you mean is that if an equation is true for an irrational number, then it must be for any real number? Like, (√2)^2 = 2, but because √2 is irrational, then x^2=2 (for all x?) | ||
+ | ::::*This one's a bit tough. You will probably agree that γ-√2 is irrational. And so is √2. What about their sum? | ||
+ | ::::*Well, maybe it doesn't to you. But is Σ n<sup>-2</sup> = π^2/6 any better? Well, this one is true (using Fourier's expansion of the rectangular function). | ||
+ | ::::Finally, | ||
+ | ::::*√2 = 3/5 + π/(7-π) is false because it would imply that π is an algebraic number | ||
+ | ::::*cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is true, and proven by many | ||
+ | ::::*γ = e/3<sup>4</sup> + e/5 seems false. But there doesn't seem to be a quick way to disprove. | ||
+ | ::::*Σ 1/n<sup>n</sup> = ln(3)<sup>e</sup> seems false, but I can't see why. [[Special:Contributions/108.162.210.234|108.162.210.234]] 09:15, 11 May 2014 (UTC) | ||
+ | ::Dgbrt, yes, sin(6π/7)=sin(π/7). Simple proof: sin(6π/7)=sin(π-π/7)=sin(π)cos(-π/7)+cos(π)sin(-π/7)=0*cos(-π/7)+(-1)*(-sin(π/7))=0+sin(π/7)=sin(π/7) [[Special:Contributions/108.162.215.89|108.162.215.89]] 02:34, 20 May 2014 (UTC) | ||
;So, still incomplete? | ;So, still incomplete? | ||
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The 'Seconds in a year' ones remind me of one of my favorite quotes: "How many seconds are there in a year? If I tell you there are 3.155 x 10^7, you won't even try to remember it. On the other hand, who could forget that, to within half a percent, pi seconds is a nanocentury" -- Tom Duff, Bell Labs. [[User:Beolach|Beolach]] ([[User talk:Beolach|talk]]) 19:14, 17 April 2014 (UTC) | The 'Seconds in a year' ones remind me of one of my favorite quotes: "How many seconds are there in a year? If I tell you there are 3.155 x 10^7, you won't even try to remember it. On the other hand, who could forget that, to within half a percent, pi seconds is a nanocentury" -- Tom Duff, Bell Labs. [[User:Beolach|Beolach]] ([[User talk:Beolach|talk]]) 19:14, 17 April 2014 (UTC) | ||
:Please do not change former discussions. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:57, 17 April 2014 (UTC) | :Please do not change former discussions. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:57, 17 April 2014 (UTC) | ||
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+ | ;cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2 ??? | ||
+ | Why the hell the divider seven makes the difference? | ||
+ | *cos(pi) + cos(3*pi) + cos(5*pi) = -3 | ||
+ | *cos(pi/8) + cos(3*pi/8) + cos(5*pi/8) = 0.92387953251128675612818318939678828682241662586364... | ||
+ | So why the "magic" prime number seven produces this exact result? I know radians and π/7 is just a small part of a circle which is 2π. One prove claims that sin(6π/7) equals to sin(π/7); my best calculator can't show a difference. Of course sin(6π) equals to sin(π), in radians, BUT sin(6π/8) is NOT equal to sin(π/8). So if the number 7 plays a magic rule here this would be "one of the", no... the BIGGEST mystery in mathematics forever. --[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 23:03, 16 May 2014 (UTC) | ||
+ | :Dgbrt, please see my answer from 11 May 2014 up there. Any odd integer will do, as long as you sum enough of cos(pi/[thing]). | ||
+ | :*Let's try with 5 : cos(pi/5) + cos (3pi/5) = 1/2. | ||
+ | :*With 9 : cos(pi/9)+ cos(3pi/9) + cos (5pi/9) + cos(7pi/9) = 1/2 | ||
+ | : No big mystery around here. Just a beautiful formula :) I think there are similar formulas with cosines and even integers. I'll post them here if I have time. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 09:56, 17 May 2014 (UTC) | ||
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+ | ::You mixing up to different equations and even not prove them. If there is any prove to a mathematician I would accept and include a proper explain for non math people here. We still have to find a prove. And I do not trust my calculators, we just have to explain why even cos(pi/5) + cos (3pi/5) is also nearly the same. This issue is still not explained. So please give us a explain. And a PROTIP: This does not work with Integers, PI is infinite--[[User:Dgbrt|Dgbrt]] ([[User talk:Dgbrt|talk]]) 17:55, 17 May 2014 (UTC) | ||
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+ | :::Okay. If I understood what you said. | ||
+ | :::* I mix up different topics. -> True. From now on, we'll just focus on the cosine one. | ||
+ | :::* You ask for a proof/explanation. -> My opinion is those are two different requests. Maybe that's why you use the distinction between math people/not math people. For a proof, please read further. What I exposed above are just other "fun experiments" we could do. e.g : [http://www.wolframalpha.com/input/?i=cos%28pi%2F11%29%2Bcos+%283pi%2F11%29+%2B+cos+%285pi%2F11%29+%2B+cos+%287pi%2F11%29%2Bcos+%289pi%2F11%29]. | ||
+ | :::* You do not trust your calculators -> Great. I don't either. (Well more accurately, I trust mine to 10^-8, so I would definitely not use it to prove any of the discussed equations in PROTIP). That's why we'll prove the formulas we assert. | ||
+ | :::* "This does not work with integers" -> Well, I got myself misunderstood. It would probably have been better if I had said: the following formula is true for all integer n. sum_{k=0}^{n-1}{cos((2k+1)*pi/(2n+1)). But It's harder to read, so just say. Choose any odd integer, say N=2n+1. Then start the following sum. cos(pi/N) + cos(3pi/N) + … and stop when the numerator is cos((N-2)pi/N). Then the result is 1/2. And that's what we'll prove, a few lines down from here. | ||
+ | :::*"Pi is infinite" -> That's a common misconception. What you mean is, Pi is irrational. (Fun fact: Pi is a transcendental number. Quite difficult theorem. Lindeman proved it in 1882. Hence, if we identify the real number x with the Q-vector space Q[x], it would make sense to say that "x is infinite" because, the Q-vector space Q[x] is indeed of infinite dimension. But then, that's not what mathematicians do). I think Vi Hart made a video where she addresses this issue (or was it someone else?). Anyway, I might come to that point some other time in the future. | ||
+ | :::Okay, so now let's first prove the protip formula. Well first, here is the link that the explainxkcd wiki points to: [http://math.stackexchange.com/questions/140388/how-can-one-prove-cos-pi-7-cos3-pi-7-cos5-pi-7-1-2]. Most of them are correct. Some are more ugly than others. I'll adapt the last one. | ||
+ | ::: We need a [http://en.wikipedia.org/wiki/Complex_number complex numbers]. (I choosed this because I think explainxkcd readers are fine with this. See comic [http://explainxkcd.com/179/ 179]). I will be using dots to show the steps of my proof. Please allow me an extra level of indent for clarity's sake. | ||
+ | :::'''Proof''' | ||
+ | :::: *Let z be a primitive 14-th [http://en.wikipedia.org/wiki/Root_of_unity root of unity] (the reader doesn't need to understand the 3 last words). Just say z = exp(i*pi/7) = cos(pi/7) + i sin(pi/7). Using [http://en.wikipedia.org/wiki/Euler%27s_formula Euler's formula]. | ||
+ | :::: *We have z^14-1 = (exp(i*pi/7))^14-1 = exp(i*2pi) - 1 = 0. Using exponential law for integer powers, as seen in this article: [http://en.wikipedia.org/wiki/De_Moivre%27s_formula De Moivre's formula]. | ||
+ | :::: *Now let's factor: z^14-1 = (z^7-1)(z^7+1) = (z^7-1)(z+1)(z^6-z^5+z^4-z^3+z^2-z+1) = (z^7-1)(z+1)*Phi_14(z). where Phi_14(X)= X^6-X^5+X^4-X^3+X^2-X+1, (see [http://en.wikipedia.org/wiki/Cyclotomic_polynomial cycltomic polynomial]). Now, because z^7-1 = (exp(i*pi/7))^7-1 = exp(i*pi)-1 = -2. And because z is not -1, the two first factors are not 0 so, Phi_14(z) = 0, which is already a pretty awesome equality. | ||
+ | :::: *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright. | ||
+ | :::: *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See [http://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry here]. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1. | ||
+ | :::: * So dividing both sides by 2, we get what we want. Pfew. | ||
+ | ::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9. | ||
+ | ::::* Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0. | ||
+ | ::::* Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1. | ||
+ | ::: Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there. It's another valid straightforward proof. Oh. And Friendly TIP: Don't say protip when you're not pro. [[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 21:50, 17 May 2014 (UTC) | ||
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+ | The valid identity cos(pi/7)+cos(3pi/7)+cos(5pi/7)=1/2 was correctly proved by the writer at 108.162.216.74 above. For a different proof, consider the complex number z = cos(pi/7)+i sin(pi/7) corresponding to rotation of the complex plane by pi/7 radians, i.e., 1/14th of a full rotation. It satisfies z^{14} -1 = 0 (z to the fourteenth is one). Dividing by z-1 gives z^{13} + z^{12} + ... + z + 1 = 0. The same argument, starting with z^2 corresponding to 1/7th of a full rotation, gives z^{12} + z^{10} + ... z^2 + 1 = 0. Taking the difference, we get z^{13} + z^{11} + ... + z^3 + z = 0. Looking only at the real parts, we get cos(13pi/7) + cos(11pi/7) + cos(9pi/7) + cos(7pi/7) + cos(5pi/7) + cos(3pi/7) + cos(pi/7) = 0. Here cos(13pi/7) = cos(pi/7), cos(11pi/7) = cos(3pi/7) and cos(9pi/7) = cos(5pi/7), since cos is even and 2pi-periodic. Finally cos(7pi/7) = -1, so 2(cos(pi/7) + cos(3pi/7) + cos(5pi/7)) - 1 = 0, which you can rewrite as the desired identity. All of this can be clearly visualized using a regular 14-gon, so a proof with pictures is possible. {{unsigned ip|141.101.81.216}} | ||
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+ | ;99 is sexual reference? | ||
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+ | In first explanation it says: "99^8 and 69^8 are sexual references". 69 I understand, but what would 99 refer too? | ||
+ | --[[Special:Contributions/173.245.53.167|173.245.53.167]] 17:38, 18 May 2014 (UTC) | ||
+ | : see [[487: Numerical Sex Positions]][[Special:Contributions/141.101.70.181|141.101.70.181]] 15:33, 20 July 2014 (UTC) | ||
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+ | I'd add pi = (9^2 + (19^2)/22)^(1/4) [[Special:Contributions/198.41.230.73|198.41.230.73]] 02:41, 13 May 2015 (UTC) | ||
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+ | '''Yet another proof of cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2''' — Use the multi-angle formula cos(7θ) = 64(cos θ)^7 − 112(cos θ)^5 + 56(cos θ)^3 − 7(cos θ), | ||
+ | and assume cos(7θ)=−1; then 7θ=π, 3π, 5π, 7π, etc. | ||
+ | Let x=cos θ, then x = cos(π/7), cos(3π/7), cos(5π/7), cos(7π/7), etc.<br /> | ||
+ | Now one could actually solve 64x^7 − 112x^5 + 56x^3 − 7x + 1 = (x+1)(8x^3 − 4x^2 − 4x + 1)^2 = 0, | ||
+ | but it’s easier to argue that cos(π/7), cos(3π/7), cos(5π/7) are the 3 roots of the cubic equation 8x^3 − 4x^2 − 4x + 1, | ||
+ | and so (using the relationship of the roots and the coefficients) their sum is −(−4)/8 = 1/2. | ||
+ | [[User:Yosei|Yosei]] ([[User talk:Yosei|talk]]) 08:19, 17 February 2019 (UTC) | ||
+ | |||
+ | '''One step closer to the elusive log(x)^e | ||
+ | In searching for an error correction term of the Taylor expansion of -x log(x) at degree n around 1, I found the term (1 - x)^(n * e)/n. It felt so close to having an actual log(x)^e appearing in a useful equation... | ||
+ | Hope I would be able to see one someday. [[User:Mumingpo|Mumingpo]] ([[User talk:Mumingpo|talk]]) 13:36, 6 May 2021 (UTC) |
Latest revision as of 13:36, 6 May 2021
The US population estimate is now off by 7 million, although 2018 just started. Even so, in December 2017, it would have been 4 million off. 625571b7-aa66-4f98-ac5c-92464cfb4ed8 (talk) 00:54, 19 January 2018 (UTC)
- Off by 7 million out of 7-8 billion means that it's accurate to one part in 1,000. That's consistent with it's location in the chart -- next to other values that are accurate to 1 in 1,000. -- Redbelly98 (talk) (please sign your comments with ~~~~)
The world population estimate is still accurate to within .1 billion. 162.158.63.28 13:41, 5 May 2017 (UTC)
They're actually quite accurate. I've used these in calculations, and they seem to give close enough answers. Davidy^{22}[talk] 14:03, 8 January 2013 (UTC)
I only see a use for the liters in a gallon one. The rest are for trolling or simple amusement. The cosine identity bit our math team in the butt at a competition. It was painful. --Quicksilver (talk) 05:27, 17 August 2013 (UTC)
Annoyingly this explanation does not cover 42 properly, it does not say that Douglas Adams got the number 42 from Lewis Carroll, who is more relevant to the page because he was a mathematician named Charles Lutwidge Dodgson. He was obsessed with the number forty-two. The original plate illustrations of Alice in Wonderland drawn by him numbered forty-two. Rule Forty-Two in Alice in Wonderland is "All persons more than a mile high to leave the court", There is also a Code of Honour in the preface of The Hunting of the Snark, an extremely long poem written by him when he was 42 years old, in which rule forty-two is "No one shall speak to the Man at the Helm". The queens in Alice Through the Looking Glass the White Queen announces her age as "one hundred and one, five months and a day", which - if the best possible date is assumed for the action of Through the Looking-Glass - gives a total of 37,044 days. With the further (textually unconfirmed) assumption that both Queens were born on the same day their combined age becomes 74,088 days, which is 42 x 42 x 42. --139.216.242.254 02:43, 29 August 2013 (UTC)
- This explanation covers 42 adequately, and would probably be made slightly worse if such information were added. The very widely known cultural reference is to Adams's interpretation, not Dodgson's original obsession. Adding it would be akin to introducing the MPLM into the explanation for the hijacking of Renaissance artists' names by the TMNT. I definitely concede that it does not cover 42 exhaustively, but I think it can be considered complete and in working order without such an addition. If it really irks you, be bold and add it! --Quicksilver (talk) 00:37, 30 August 2013 (UTC)
- Additionally, the Lewis Carroll idea is only one of several theories about where Douglas Adams got the number from. 162.158.158.87 20:47, 28 November 2019 (UTC)
"sqrt(2) is not even algebraic in the quotient field of Z[pi]" is not correct. Q is part of the quotient field of Z[pi] and sqrt(2) is algebraic of it. The needed facts are that pi is not algebraic, but the formula implies it is in Q(sqrt(2)). --DrMath 06:47, 7 September 2013 (UTC)
13/15 is a better approximation to sqrt(3)/2 than is e/pi. Continued fraction approximations are great! --DrMath 07:23, 7 September 2013 (UTC)
How could he forget 1 gallon ≈ 0.1337 ft³?! 67.188.195.182 00:51, 8 September 2013 (UTC)
Worth mentioning that Wolfram Alpha now officially recognizes the White House switchboard constant and the Jenny constant. 86.164.243.91 18:28, 8 October 2013 (UTC)
Maybe we should add the [Extension:LaTeXSVG LaTeX extension] to make it easier to transcribe these equations. -- 108.162.219.220 23:02, 16 December 2013 (UTC)
- Protip - Does anyone see the correct equation?
Maybe this is just an other Wolfram Alpha error, like we recently have had here: 1292: Pi vs. Tau. All equations still look invalid to me.
- √2 = 3/5 + π/(7-π): is impossible because √2 is an irrational number and no equation can match.
- cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2: could only match if cos(x) + cos(3x) + cos(5x) = 1/2 would be valid, because π/7 is also an irrational number.
- γ = e/3^{4} + e/5 or γ = e/54 + e/5: would mean that a sum of two irrational numbers do fit to the Gamma Constant. Impossible.
- √5 = 13 + 4π / 24 - 4π: √5 and π are irrational numbers, there is no way to match them in any equation like this.
- Σ 1/n^{n} = ln(3)^{e}: doesn't make any sense either.
Maybe Miss Lenhart can help. --Dgbrt (talk) 21:41, 17 December 2013 (UTC)
cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is exactly correct.
Let a=π/7, b=3π/7, and c=5π/7, then (cosa+cosb+cosc)⋅2sina=2cosasina+2cosbsina+2coscsina=sin2a+sin(b+a)−sin(b−a)+sin(c+a)−sin(c−a)=sin(2π/7)+sin(4π/7)−sin(2π/7)+sin(6π/7)−sin(4π/7)=sin(6π/7)=sin(π/7)=sina
Hence, cos(π/7) + cos(3π/7) + cos(5π/7) = sin(π/7) / 2sin(π/7) = 1/2 108.162.216.74 01:57, 16 January 2014 (UTC)
- What is this: sin(6π/7)=sin(π/7) ? A new math is born... --Dgbrt (talk) 20:49, 16 January 2014 (UTC)
- Actually it does. My proof is geometric: the sines of two supplementary angles (angle a + angle b = π (in radians)) are equivalent because they necessarily have the same x height in a Cartesian plane. Look on a unit circle, or even a sine function. Also, Calculus and most other mathematics use radians over degrees because they make the functions simpler and eliminate irrationality when a trig function shows up, but physics uses degrees because it's easier to understand and taught first. Anonymous 01:27, 13 February 2014 (UTC)
- As an aside, just how far along in math are you? Radian measure is taught in high school (at least the good ones). Anonymous 13:24, 13 February 2014 (UTC)
- Sure, I was wrong at my last statement. sin(6π/7)=sin(π/7) is correct by using the radian measure. But just change π/7 to π/77 would give a very different result on that formular here. I still can't figure out why PI divided by the number 7 should be that unique, PI divided by 77 should be the same. My fault is: I still can't find the Nerd Sniping here. And we all do know that Randall did use wrong WolframAlpha results here. According to the last question: I'm very well on Math, that's because I want to understand this. This is like 0.999=1. --Dgbrt (talk) 22:01, 13 February 2014 (UTC)
- Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked 545 lately? I answered your question there, too)
- As per the derivation from january 16 , you can use any a,b,c that satisfies this set of equations: 2 a = b - a, a + b = c- a, c + a = π - a. This is due to the fact that sin(x) = sin(π-x), and what was derived the 16th. 173.245.53.199 12:38, 21 February 2014 (UTC)
- Dgbrt: If not convinced by the proofs linked to in the "explanation" part, you might want to try this: [1]. I'm sure you'll find inspiration for similar formulas using PI over [any odd integer]. Your assumption that Randall used WolframAlpha for this very identity is probably wrong. This is a very well-known formula that appears in many high school books, and I am pretty sure it is part of Randall's culture. And this has nothing to do with 1=1. As for your original post,
- √2 = (√2-1)/((4-2)π/2-π)+1 : Is this what you call "matching an equation" to √2?
- So what you mean is that if an equation is true for an irrational number, then it must be for any real number? Like, (√2)^2 = 2, but because √2 is irrational, then x^2=2 (for all x?)
- This one's a bit tough. You will probably agree that γ-√2 is irrational. And so is √2. What about their sum?
- Well, maybe it doesn't to you. But is Σ n^{-2} = π^2/6 any better? Well, this one is true (using Fourier's expansion of the rectangular function).
- Finally,
- √2 = 3/5 + π/(7-π) is false because it would imply that π is an algebraic number
- cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 is true, and proven by many
- γ = e/3^{4} + e/5 seems false. But there doesn't seem to be a quick way to disprove.
- Σ 1/n^{n} = ln(3)^{e} seems false, but I can't see why. 108.162.210.234 09:15, 11 May 2014 (UTC)
- Ah, I see. I think it has to do with the way e^i*π breaks down, as one of the answers shown in the corresponding link explains, but other answers rely on various angle identities (including the supplementary sines one in the proof above). Anonymous 03:10, 14 February 2014 (UTC) (PS, have you checked 545 lately? I answered your question there, too)
- Sure, I was wrong at my last statement. sin(6π/7)=sin(π/7) is correct by using the radian measure. But just change π/7 to π/77 would give a very different result on that formular here. I still can't figure out why PI divided by the number 7 should be that unique, PI divided by 77 should be the same. My fault is: I still can't find the Nerd Sniping here. And we all do know that Randall did use wrong WolframAlpha results here. According to the last question: I'm very well on Math, that's because I want to understand this. This is like 0.999=1. --Dgbrt (talk) 22:01, 13 February 2014 (UTC)
- Dgbrt, yes, sin(6π/7)=sin(π/7). Simple proof: sin(6π/7)=sin(π-π/7)=sin(π)cos(-π/7)+cos(π)sin(-π/7)=0*cos(-π/7)+(-1)*(-sin(π/7))=0+sin(π/7)=sin(π/7) 108.162.215.89 02:34, 20 May 2014 (UTC)
- So, still incomplete?
Where's our (in)complete judge? 199.27.128.186 19:21, 18 December 2013 (UTC)
- The protip is still a mystery. I'm calling for help a few lines above. --Dgbrt (talk) 21:16, 18 December 2013 (UTC)
- The cosine one, in radians, is correct 141.101.88.225 12:54, 28 April 2014 (UTC)
The 'Seconds in a year' ones remind me of one of my favorite quotes: "How many seconds are there in a year? If I tell you there are 3.155 x 10^7, you won't even try to remember it. On the other hand, who could forget that, to within half a percent, pi seconds is a nanocentury" -- Tom Duff, Bell Labs. Beolach (talk) 19:14, 17 April 2014 (UTC)
- cos(pi/7) + cos(3pi/7) + cos(5pi/7) = 1/2 ???
Why the hell the divider seven makes the difference?
- cos(pi) + cos(3*pi) + cos(5*pi) = -3
- cos(pi/8) + cos(3*pi/8) + cos(5*pi/8) = 0.92387953251128675612818318939678828682241662586364...
So why the "magic" prime number seven produces this exact result? I know radians and π/7 is just a small part of a circle which is 2π. One prove claims that sin(6π/7) equals to sin(π/7); my best calculator can't show a difference. Of course sin(6π) equals to sin(π), in radians, BUT sin(6π/8) is NOT equal to sin(π/8). So if the number 7 plays a magic rule here this would be "one of the", no... the BIGGEST mystery in mathematics forever. --Dgbrt (talk) 23:03, 16 May 2014 (UTC)
- Dgbrt, please see my answer from 11 May 2014 up there. Any odd integer will do, as long as you sum enough of cos(pi/[thing]).
- Let's try with 5 : cos(pi/5) + cos (3pi/5) = 1/2.
- With 9 : cos(pi/9)+ cos(3pi/9) + cos (5pi/9) + cos(7pi/9) = 1/2
- No big mystery around here. Just a beautiful formula :) I think there are similar formulas with cosines and even integers. I'll post them here if I have time. Varal7 (talk) 09:56, 17 May 2014 (UTC)
- You mixing up to different equations and even not prove them. If there is any prove to a mathematician I would accept and include a proper explain for non math people here. We still have to find a prove. And I do not trust my calculators, we just have to explain why even cos(pi/5) + cos (3pi/5) is also nearly the same. This issue is still not explained. So please give us a explain. And a PROTIP: This does not work with Integers, PI is infinite--Dgbrt (talk) 17:55, 17 May 2014 (UTC)
- Okay. If I understood what you said.
- I mix up different topics. -> True. From now on, we'll just focus on the cosine one.
- You ask for a proof/explanation. -> My opinion is those are two different requests. Maybe that's why you use the distinction between math people/not math people. For a proof, please read further. What I exposed above are just other "fun experiments" we could do. e.g : [2].
- You do not trust your calculators -> Great. I don't either. (Well more accurately, I trust mine to 10^-8, so I would definitely not use it to prove any of the discussed equations in PROTIP). That's why we'll prove the formulas we assert.
- "This does not work with integers" -> Well, I got myself misunderstood. It would probably have been better if I had said: the following formula is true for all integer n. sum_{k=0}^{n-1}{cos((2k+1)*pi/(2n+1)). But It's harder to read, so just say. Choose any odd integer, say N=2n+1. Then start the following sum. cos(pi/N) + cos(3pi/N) + … and stop when the numerator is cos((N-2)pi/N). Then the result is 1/2. And that's what we'll prove, a few lines down from here.
- "Pi is infinite" -> That's a common misconception. What you mean is, Pi is irrational. (Fun fact: Pi is a transcendental number. Quite difficult theorem. Lindeman proved it in 1882. Hence, if we identify the real number x with the Q-vector space Q[x], it would make sense to say that "x is infinite" because, the Q-vector space Q[x] is indeed of infinite dimension. But then, that's not what mathematicians do). I think Vi Hart made a video where she addresses this issue (or was it someone else?). Anyway, I might come to that point some other time in the future.
- Okay, so now let's first prove the protip formula. Well first, here is the link that the explainxkcd wiki points to: [3]. Most of them are correct. Some are more ugly than others. I'll adapt the last one.
- We need a complex numbers. (I choosed this because I think explainxkcd readers are fine with this. See comic 179). I will be using dots to show the steps of my proof. Please allow me an extra level of indent for clarity's sake.
- Proof
- *Let z be a primitive 14-th root of unity (the reader doesn't need to understand the 3 last words). Just say z = exp(i*pi/7) = cos(pi/7) + i sin(pi/7). Using Euler's formula.
- *We have z^14-1 = (exp(i*pi/7))^14-1 = exp(i*2pi) - 1 = 0. Using exponential law for integer powers, as seen in this article: De Moivre's formula.
- *Now let's factor: z^14-1 = (z^7-1)(z^7+1) = (z^7-1)(z+1)(z^6-z^5+z^4-z^3+z^2-z+1) = (z^7-1)(z+1)*Phi_14(z). where Phi_14(X)= X^6-X^5+X^4-X^3+X^2-X+1, (see cycltomic polynomial). Now, because z^7-1 = (exp(i*pi/7))^7-1 = exp(i*pi)-1 = -2. And because z is not -1, the two first factors are not 0 so, Phi_14(z) = 0, which is already a pretty awesome equality.
- *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright.
- *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See here. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1.
- * So dividing both sides by 2, we get what we want. Pfew.
- Why is 7 so special? Well it isn't. Let's prove it for 9.
- Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0.
- Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1.
- Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there. It's another valid straightforward proof. Oh. And Friendly TIP: Don't say protip when you're not pro. Varal7 (talk) 21:50, 17 May 2014 (UTC)
- Okay. If I understood what you said.
The valid identity cos(pi/7)+cos(3pi/7)+cos(5pi/7)=1/2 was correctly proved by the writer at 108.162.216.74 above. For a different proof, consider the complex number z = cos(pi/7)+i sin(pi/7) corresponding to rotation of the complex plane by pi/7 radians, i.e., 1/14th of a full rotation. It satisfies z^{14} -1 = 0 (z to the fourteenth is one). Dividing by z-1 gives z^{13} + z^{12} + ... + z + 1 = 0. The same argument, starting with z^2 corresponding to 1/7th of a full rotation, gives z^{12} + z^{10} + ... z^2 + 1 = 0. Taking the difference, we get z^{13} + z^{11} + ... + z^3 + z = 0. Looking only at the real parts, we get cos(13pi/7) + cos(11pi/7) + cos(9pi/7) + cos(7pi/7) + cos(5pi/7) + cos(3pi/7) + cos(pi/7) = 0. Here cos(13pi/7) = cos(pi/7), cos(11pi/7) = cos(3pi/7) and cos(9pi/7) = cos(5pi/7), since cos is even and 2pi-periodic. Finally cos(7pi/7) = -1, so 2(cos(pi/7) + cos(3pi/7) + cos(5pi/7)) - 1 = 0, which you can rewrite as the desired identity. All of this can be clearly visualized using a regular 14-gon, so a proof with pictures is possible. 141.101.81.216 (talk) (please sign your comments with ~~~~)
- 99 is sexual reference?
In first explanation it says: "99^8 and 69^8 are sexual references". 69 I understand, but what would 99 refer too? --173.245.53.167 17:38, 18 May 2014 (UTC)
- see 487: Numerical Sex Positions141.101.70.181 15:33, 20 July 2014 (UTC)
I'd add pi = (9^2 + (19^2)/22)^(1/4) 198.41.230.73 02:41, 13 May 2015 (UTC)
Yet another proof of cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2 — Use the multi-angle formula cos(7θ) = 64(cos θ)^7 − 112(cos θ)^5 + 56(cos θ)^3 − 7(cos θ),
and assume cos(7θ)=−1; then 7θ=π, 3π, 5π, 7π, etc.
Let x=cos θ, then x = cos(π/7), cos(3π/7), cos(5π/7), cos(7π/7), etc.
Now one could actually solve 64x^7 − 112x^5 + 56x^3 − 7x + 1 = (x+1)(8x^3 − 4x^2 − 4x + 1)^2 = 0,
but it’s easier to argue that cos(π/7), cos(3π/7), cos(5π/7) are the 3 roots of the cubic equation 8x^3 − 4x^2 − 4x + 1,
and so (using the relationship of the roots and the coefficients) their sum is −(−4)/8 = 1/2.
Yosei (talk) 08:19, 17 February 2019 (UTC)
One step closer to the elusive log(x)^e In searching for an error correction term of the Taylor expansion of -x log(x) at degree n around 1, I found the term (1 - x)^(n * e)/n. It felt so close to having an actual log(x)^e appearing in a useful equation... Hope I would be able to see one someday. Mumingpo (talk) 13:36, 6 May 2021 (UTC)