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Messages  phenomist
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31
« on: April 18, 2011, 11:07:02 pm »
Another, hopefully easy feature request: in the compiler, can the thing say how many pages the app is? Something like "Generated a 147456 byte (9 page) app." would be great
If the program tells you how many bytes the app is, simply dividing by 16384 should give page count
32
« on: April 12, 2011, 08:42:45 pm »
Ok: suppose the roots that you get by running the quadratic formula are r and s (note that there is a plus/minus sign; hence, there are two roots). The quadratic factorized is a(xr)(xs), where a is the coefficient of the quadratic (leading) term.
33
« on: April 08, 2011, 07:05:56 pm »
Three possible interpretations:
Radius is 8/sin(50) about 10.4432583147 = R
1) There is no upper part. Answer is 2sin(80)R = 20.5692 2) The perpendicular is an extension to the other side of the triangle. Answer is 2sin(65)R = 18.9296 3) The sides are extensions to each other. Answer is 2sin(40)R = 13.4256
Basically, the answer is 2R*sin(A/2) where A is the vertex angle of the bottom triangle.
It's underconstrained.
EDIT: By reverseengineering your answer (@OP), it appears that you assume that the angle at the bottom is 120 degrees.
34
« on: April 08, 2011, 06:57:35 pm »
Some exploratory questions to ask (sorted by difficulty?):
1) Prove that every digit in numbers of this series are either "1", "2", or "3". 2) Prove that for every term, the last digit is always 1. 3) Prove that you cannot have 3 3's in a row. 4) Does the length of the terms ever decrease? Find the first instance if it does. 5) Asymptotically speaking, what is the length of each term of the sequence? 6) As the terms grow large, what will the distributions for each digit be?
The first three are definitely doable.
35
« on: April 04, 2011, 08:36:10 pm »
Hm, I'm planning to write a Turn Based Strategy Quadratic Solver in TIBasic for next year
36
« on: April 02, 2011, 03:25:01 am »
Well, you don't need to formalize stuff just like in "everybody else's way". (Just take Ramanujan, for instance.) You can do some pretty amazing stuff independently that nobody else has considered.
37
« on: March 31, 2011, 07:57:02 pm »
I would imagine that the process is similar, because the derivative of a polynomial is easy to compute, regardless of the coefficients.
38
« on: March 31, 2011, 06:34:53 pm »
Yeah, some indefinite integrals for elementary functions output nonelementary functions. Though for a ndegree solver, perhaps use the Newton method for approximating all the roots, complex and noncomplex? (for n>4)
39
« on: March 31, 2011, 06:28:09 pm »
Also, the set {0,1,0,1,0,1,0,1,...} could probably be described by Sine or Cosine, but I think the function (2)^(x mod 2)1 is much simpler.
How about the function (x mod 2)? @Xeda: might help: http://en.wikipedia.org/wiki/Newton_polynomial. If you independently worked all of that out, that's pretty impressive
40
« on: March 31, 2011, 02:04:29 am »
I believe this is Newton interpolation. Note that the noncoefficients can be reexpressed as x/0!, x(x+1)/1!, x(x+1)(x+2)/2!, etc. Unless you already noticed that
41
« on: March 30, 2011, 12:42:19 am »
Let A=2C, B=2D sin2C + sin2D = 2sin(C+D)cos(CD) = LHS = 2sin(C)cos(C)+2sin(D)cos(D) RHS = 2[sin(C)cos(D)+sin(D)cos(C)][cos(C)cos(D)+sin(C)sin(D)] Crossmultiply RHS. 2[sin(C)cos(D)cos(C)cos(D)+sin(C)cos(D)sin(C)sin(D)+sin(D)cos(C)cos(C)cos(D)+sin(D)cos(C)sin(C)sin(D)] = 2[sin(C)cos(C)cos(D)cos(D)+sin(D)cos(D)sin(C)sin(C)+sin(D)cos(D)cos(C)cos(C)+sin(C)cos(C)sin(D)sin(D)] = LHS by pythagorean identity (sin^2(x)+cos^2(x) = 1) Dirty, but it works. Alternatively, reexpress sin((A+B)/2) = cos(90  (A+B)/2). Now let 90(A+B)/2 = X and (AB)/2 = Y. XY = 90A, X+Y = 90B. Look familiar? (Hint: use cosAcosB = .5(cos(AB)+cos(A+B)) )
42
« on: March 29, 2011, 08:34:36 pm »
Now that we're talking about base converters, here's my converter from around October 2010 or so. Sorry for invading your thread It currently has support for base 2 to 62 (with lowercase), and also has support for decimals and fractions (e.g. say you want to know 1/13 in binary).
43
« on: March 27, 2011, 11:52:42 pm »
Assuming that the angle can be anywhere from 0 to 90 degrees, uniformly distributed, using a bit of calculus we get that the ratio is on average pi/4, or .785398. 11/14 is a sufficient approximation for this. So (abs(V)+abs(W))*11/14 provides a fairly good approximation, erring at most by 10% for angles between 10 and 80 degrees.
If you need a linear approximation that reduces the error necessary, the constant that you are looking for is e^(2C/piln(2)/2) (C being Catalan's constant), or about .789348. 15/19 is a sufficient approximation for this.
44
« on: March 27, 2011, 01:43:40 am »
I reversed the bit order, so 0001 <=> 1000, 0010 <=> 0100, etc.
This was done to get rid of the (3C).
(I do believe that you can perform a similar optimization to drop 4 bytes off the displayandforget version, as well.)
45
« on: March 27, 2011, 12:36:27 am »
Oh really?
Ok then. But my optimization actually required a bit of thinking _
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