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#71
14th August 2011, 02:35 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

http://www.cybertester.com/data/denest.pdf

I also did a few more two-surd-split tests with the expressions 2ab and just a that of course didn't pass. I am too pressured on time at the moment to right now continue reading that page to learn how to test to see if the nested square root expression assigned to b can be de-nested tonight.
#72
17th August 2011, 04:50 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

No fixated information today besides the immense derivation of that document I've collectively interpreted throughout today. I've gotten back into the eight-page routine of ink-flooding devoted to radical squares, their roots, and their rearrangements. The actual problem in this blog is a lot harder to instantiate with the algorithms presented in that document; in fact one of them I am still working on at the present time. It was difficult to interpret some of the theorem explanations and procedures for square roots, but a few deeper algorithms remain to be tried. Besides making brownies today I would say I have been working on nothing else besides eating and exercise today.
#73
18th August 2011, 11:56 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

I stood up all night working on this shit lol, and I'm still not done. I've already successfully split the nested square root involved in the (2ab) expression, but b is proving to be slightly more extended a task.... In essence, I've extracted a few surds so far, but I have to try out a bunch of combinations to eliminate the number of possibilities for splitting the half that is still equally as nested.

I wasn't planning on ending my night like this lol; I thought I'd have it done by now. From what I do know about the de-nested (2ab), though, I know that this is, in fact, making progress towards finding the sine of one degree. It's a good thing I completely analyzed that PDF document's main algorithm, or I would have had to have distributed a greater amount of time to have gotten this far anyway. Bed first, finish later.
#74
19th August 2011, 03:09 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

Alright, shouldn't have taken that long. I spend a lot of time maintaining proportionate handwriting, though at times I didn't have the sanity to at least be that consistent which sped the process up a bit at climactic points of the algebra. The tediousness of applying the same algorithm over and over again, rinsing and repeating to squeeze the split radicals to smaller addends taught me a little about discipline and debugging from mental math of course.

The nested square roots derived for both the variable names "b" and "ab" have been split into non-nested square roots as much as possible.

Further de-nesting is not possible, but for efficiency and extra information to simplify later on they have been converted from factor-based to addend-based. For example, at this point I should be able to find a way to define b in terms of a now--so that I no longer have to keep using a and b all the time in these images. Working with just one variable should completely transform the nature of the excess left in the expression assigned to sixteen of the sine of one degree.

Later though first ice cream and a break.
#75
19th August 2011, 03:12 AM
 Natch Project Supporter Senior Member Join Date: Jan 2011 Location: Nowhere near you. Posts: 5,078

Woo, good job!

wait did you beat the game yet
__________________
#76
24th August 2011, 05:12 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

lol no.

lol k.

Heh, got to stop that XD

Too tired to look at all these updated threads atm

First up, the original algorithm I guessed in this thread in the original post is BS. That approximates to a complex number in terms of i.

Next is that what remains of this problem appears to be nothing short of a long road of de-nesting cube root expressions. I am still in the process of deriving case-specific examples of that manually; I have some property definitions though they are too long and probably of no use at the time for including here.

There were a couple other very interesting patterns in my recent pages of ink that I don't remember atm since I feel tired. I will be able to use them tomorrow.
#77
19th October 2011, 07:07 PM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

I can't apologize for such delay. I haven't had this much fun working with high-level languages to work on emulator stuff since 2007, when I was just a naive little kid and never did anything properly. So after we left Wisconsin, I was happy to make myself at home where we are now, but my adaption caused me to tremendously put off this project.

However, just hours before moving, I found an old \$100 TI-84 Plus(?) calculator I had for school, which supported operations on complex numbers of the form [a + b * sqrt(-1.)] (or a + bi, more commonly known). I used it to approximate these expressions in this thread project.

Code:
```cbrt(a - b) ~= (3.428669203 - 2.0601523i) /* I will use 'm' for this. */
cbrt(a + b) ~= (3.428669203 + 2.0601523i) /* I will use 'n' for this. */```
I was too time-pressed with getting things out of the house before moving to do much of anything else, but I double-checked both of these to make sure they were not evaluated wrongly. Last night, I put forth some effort to get back into this project by using those decimal approximations to verify my research from before.

Quote:
 Originally Posted by Iconoclast
Here is some of my notes.
Code:
```m = cbrt(a - b) ~= (3.4286692 - 2.0601523i)
n = cbrt(a + b) ~= (3.4286692 + 2.0601523i)

for:  n = cbrt(a + b) ~= (3.4286692 + 2.0601523i)
n^2 ~= 7.511545 + 14.127161i
(n^2)i ~= -14.127161 + 7.511545i

/* the top-most statement in the GIF,
having substituted n for (a + b)^0.33333333 */
sin(+1.) = (n^2 + 16 + (i * sqrt(3) * n^2) - (16 * i * sqrt(3))) / -16n
~= (0.957416 + 0.575274i) / (54.858707 + 32.962437i)
~= 0.0174524 + 0.0000000056708804i```
Approximating the sine of one degree with Windows Calculator returns:
0.017452406437283512819418978516316

So this proves that my interpretation of what to fill in for the variables a and b in that GIF is correct. However, I did this desk-work for the bottom-most statement in that GIF as well...I was not able to return the same result. Somewhere in the middle of that picture then I must have screwed up, but I will get to that shortly.

Last edited by HatCat; 19th October 2011 at 07:12 PM. Reason: lol, oops, fixed the image to the one I meant to refer
#78
19th October 2011, 07:13 PM
 Lightbeest Senior Member Join Date: Oct 2011 Posts: 117

Did you beat the game yet?
#79
19th October 2011, 07:21 PM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

lmao I'll have to get back to you on that

Oh and I found that the exact, non-decimal-approximated cube roots of such expressions in that image can be found using DeMoivre's theorem. Damn lol I remember that guy from my High School Pre-Calculus class. Kinda skipped past all the polar coordinate stuff though...which is useful to find the exact roots of any specified complex number.

Ever since we moved I don't know where that TI-83 calculator is, but there is this online script I found that will do a fair selection of operations between complex numbers.
http://www.solvemymath.com/online_ma...mber/index.php

Just gonna index that for now.
#80
19th October 2011, 07:29 PM
 Lightbeest Senior Member Join Date: Oct 2011 Posts: 117

I am please with your progress, if you hit it big, i am the inspiration behind it.

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