
#81




Oh yeah, at least the inspiration behind my remembering to get back to this project XD. But yeah man it's much work...thanks for chipping in XD.
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#82




so what are you doing again ._.
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#83




lol, I don't know what happened back there...I can't *quickly* figure out for the life of me what went wrong in that previous derivation of the sine of one degree where I rationalized the denominator.
I can figure that out easily enough, later, but it's timeconsuming to keep doing these decimal approximation tests to see if my expression still holds true. I had the patience to repeat that test when I rederived the reduction elsewise, though. This time, I rationalized the denominator simply by multiplying it, a cube root, by its square, along with the numerator of the ratio. I wrote the statement in four different ways, but the fourth was the last one I did after playing around with it enough and thus is the only one I need to prove that it still holds true. Code:
sin(+1.) = (1 + i * sqrt(3)) / n  ((n + (i * sqrt(3) * n)) / 16 ~= (0.0087262034 + 0.49992385i)  ((n + (3.5682885 + 5.9386293i) / 16 ~= (0.0087262034 + 0.49992385i)  (0.0087262063 + 0.49992385i) = 0.01745241 I can also try that, multiply by the conjugate /* cbrt(a  b) */ to return the cube root of the difference of the square of a and the square of b, which is exactly +4096., later to rederive what I did before, but right now I don't care. It won't help me find the sine expression...so just going to focus on that for the moment. ;P
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#84




Never mind; I guess I must have somehow misverified twice in a row, with the previous formula. Apparently it's still correct when I verify this using the decimal approximation.
16 * sin(+1.) = (1 + sqrt(3)i) * cbrt(a + b) + (1  sqrt(3)i) * cbrt(a  b) 16 * sin(+1.) ~= 0.27923850386 sin(+1.) ~= +0.01745240649125 So whatever...guess I was wrong about being wrong, twice in a row. Effing redundant bs. As I've proven before, the product of cbrt(a + b) and cbrt(a  b) is cbrt(a^2  b^2) = +16. ... a most useful pattern. But I don't think I'll be using that formula. I've tried tons of shit...solving for coefficients to convert the specified cube roots to just square roots (to make denesting easier), various systems of factorization or alternative radical eliminations. I've found out the hard way that I can't simply to the next step of finding the sine of one degree, without directly resolving the cube root expressions of the sum and difference of a and b, by "denesting" those radicals. Denesting them will take tons of time, (Part of the problem is that denesting/splitting nested square root expressions, is over five times as documented than how to do so for cube roots.) so there is no point, IMO, in having both cbrt(a + b) and cbrt(a  b) in the expression. I will just denest only one of them and keep the formula down to this, using only one of them. Again, for reference, the variables a and b, and combinations thereof, I still have precalculated in this image below. I won't square the cube roots, either, because that returns cbrt(a^2 + b^2 + 2ab), and (2ab) is a complex number in terms of the square root of negative one, which only lengthens the expression already formed by the sum of just the squares of both variables. So the easiest way to execute the only remaining option is to just get to work on splitting the nested cube roots as they already purely are in that reference.
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http://theoatmeal.com/comics/cat_vs_internet Last edited by HatCat; 21st October 2011 at 12:37 AM. 
#85




Fantastic news...the code has been broken.
It would have been extremely frustrating to apply, for example, the theorems of Landau to split (denest) the nested cube root expressions, in terms of a and b as defined in that image above. At the same time, I knew that if it was possible, I had to push on until I'd done it. However, before I seized the initiative on that, I eventually started noticing a pattern about the realnumber coefficients of both a and b. I first realized that b is really, actually, simply 64 times the cosine of 3 degrees (and i = sqrt(1.)). Now, I've copied de Moivre's theorem below, which can be used. However, I have to derive the sine of 3 degrees (or, conversely by setting up a 38790 right triangle, 87 degrees in the process) before I can apply the theorem to get the sine of one degree. Fortunately for me, somebody has already derived the sine of 3 degrees. http://intmstat.com/blog/2011/06/exa...indegrees.pdf as noted in this little article http://www.squarecirclez.com/blog/ho...llangles/6212 They also "found" the sine of 1 degree, but it is not completely simplified and is still in terms of the imaginary unit. I will have to fix that myself shortly.
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#86




Landau,do you mean Lando Calrissian
Anyway, congratulations man, that must have taken a lot of time to decipher. 
#87




so then.. you beat the game?
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#88




Sorry for the delay. Just as I had previously updated this thread, I went to work with the plan I described. It wasn't long until I was bugged by a reallife something to finish and old unfinished something (that would have taken months to finish anyway). Biceps were exercised, quadriceps burned, and many constructionlevel dangers pressing against my face ensued. A couple other things just so happened to coincide based on getting a new OS, RAM upgrade, new setup, long conversations that I benefited from, just so much personal time that suddenly I needed to take that happened to take me away from three days. For example, the night before, my connection kept flickering on and off, so while I could have logged on, I would not have had a tolerable Internet experience in the process. Weird, like I just upgrading from a 4yearold 32bit Vista crap system I had the entire time to a 64bit Windows 7 OS or something. Uhh in fact that is a little bit like what happened !
But whatever. None of that matters now. I still had a little time to use triangle setups to derive the sines of 3 and 87degree angles (conversely, then, the cosines). I don't have Microsoft Works (not Word) here anymore (and elected not to transfer the installation here), which used Microsoft Equation 3.0 here. This time I used OpenOffice math typer to produce the image below. Again, using Abraham de Moivre's theorem, the cube root can easily be converted to a base expression, taken to the identity power (or just positive one) by thirding the 87degree angle measures. In the long run, however, this will only succeed in reducing the amount of work for the time in which a cube root denestation algorithm must be applied. In the meantime, need to find some replacement software for creating math GIFs. Quote:
Noticing patterns can be the most easing, efficientonthefuture approach to any logical reallife problem you could apply to. It was an idea that occured to me a little late, but, at the same time, it does not put any of my direct algebraic solutions work in vain. =D And no I mean Susan Landau. Before the turn of the century here, she derived the very first universally successful algorithm for determining when a radical can or cannot be denested. If Renaissance mathematics wasn't enough to make people in current centuries rip their hair out, that was a contribution that succeeded the confusion of many other mathematicians for years.
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#89




WOW epic, glad I am not in your shoes.

#90




Heh...I guess it's a matter of perspective.
Small news, but evaluating the righthand side of the statement rendered in SL1.GIF returned what was really the sine of sixty degrees...instead of just one. I must have jumped the gun when I factored out cbrt(64) = +4. from both cube roots, which may have even affected the applicability of de Moivre's theorem. I will have to try again tomorrow...busy with a new project.
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