Trisecting an angle

[mfgoode]

Originally Posted by alpertron

By the way, I found that the polynomial



has the roots sin 3°, sin 21°, sin 33°, sin 39°, sin 51°, sin 57°, sin 69°, sin 87° and their negatives.

This is remarkable, and excellent work Alpertron
This is a characteristic polynomial of degree 16 with eigenvalues sin3*,sin21*etc.
If you could give us the 16 x 16 matrix and trace of the matrix it will be very valuable. Thank you.
Mally

[mfgoode]

Originally Posted by nibble4bits

If I took ANY binary number and used the technique I showed you, then yes adding a series of halves+4ths+8ths+16ths+...(a_n)/(2^n) would generate approximations.
k/f(x) -> k=integer, f(x)=sum of (a_0)+(a_1)/2+...(a_n)/(2^n)
This formula should work. :)

I was thinking of stations such as 92.9 and 94.3 which are close to each other - the closer ones are the ones I'd worry about. 98.1, 98.5, and 98.9 are also deliberately arranged that way also.

I showed angles A and B in my diagram as coinciding every 8th (or 9th depending on point of view) cycle. In reallity they would take a lot longer period of time to even come close to alignment. I was lazy - it's was easy to draw. :P

For most people working with radio waves, c can be considered constant. If I'm making an antenna for 20 MHz then I know I need a length of twice the wavelength. "All else being equal" applies here - most people need to just simplify the process to make it effecient. Yes, I know c as a constant is misleading when you take account of nonuniform frames of reference.

<EDIT> Actually an even simpler formula is k*f(x)*2^-n where k and f(x) are integers for a constant and a sum, and n is the number of times to halve that constant.

Quote/ "Yes, I know c as a constant is misleading when you take account of nonuniform frames of reference." /unquote
'c' is a constant regardless of the frame of reference uniform or non-uniform.
Mally

[alpertron]

Originally Posted by mfgoode

This is remarkable, and excellent work Alpertron
This is a characteristic polynomial of degree 16 with eigenvalues sin3*,sin21*etc.
If you could give us the 16 x 16 matrix and trace of the matrix it will be very valuable. Thank you.
Mally

Thanks Mally, but there is no need for using matrices here.

If we want to find a polynomial that has a root cos 3°, just use the relation:

cos (20 * 3°) = cos 60° = 1/2

If x = cos 3°, we find that T₂₀(x) = 1/2 where T_n is the Chebyshev polynomial.

Thus p(x) = 2*T₂₀(x) - 1 is a degree 20 polynomial that has the requested root.

We can notice that cos(20 * 15°) = cos(20 * 75°) = 1/2.

The four numbers cos(15°), cos (-75°) and their negatives are roots of the polynomial q(x) = 16x⁴ - 16x² + 1. The polynomial I wrote in my last post is just p(x) / q(x).

[mfgoode]

[QUOTE=alpertron]Thanks Mally, but there is no need for using matrices here.

If we want to find a polynomial that has a root cos 3°, just use the relation:

cos (20 * 3°) = cos 60° = 1/2

If x = cos 3°, we find that T₂₀(x) = 1/2 where T_n is the Chebyshev polynomial./Unquote

Thank you very much Alpertron, for introducing me to the wonders of the Chebyshev polynomials.
Kindly bear with me, and tolerate my questions, so that I can get a better idea of its properties and please clarify.
Now given the first few Chebyshev polynomials of the first kind, I find that the ‘first’ coefficients are powers of 2 such that say T (6) =2^(6-1) -…. Giving 32.
I also find a relationship to successive terms as an integration of the previous terms thus
T3 (x) = 4* x^3 -3*x +
T4 (x) = 8* x^4 -20x^2 +1.
Kindly explain to me how exactly the coefficients are obtained? Such as the ‘2nds’ 3 , 20, etc.
Is there an elementary rule for obtaining these coefficients or does the process require a direct integration of the Contour Integral?
In other words given T (n) can we find T (n+1) or T (n-1) ? by elementary integration or have we to integrate the Contour Integral every time? And follow a simple rule for the coefficients? Sloane's A008310 did not have your coefficients and I couldnt open it further.
Thank you,
Mally

[alpertron]

Using identity (21): T_n+1 (x) = 2 x T_n (x) - T_n-1 (x) I coded the following program in UBASIC (this interpreted language can operate with polynomials):

10   input "Degree";N
20   PrevPoly=1
30   ActualPoly=_X
40   for I=2 to N
50   NewPoly=2*_X*ActualPoly-PrevPoly
60   PrevPoly=ActualPoly
70   ActualPoly=NewPoly
80   next I
90   print ActualPoly

In our example, if I enter 20 I get:

524288*X^20 - 2621440*X^18 + 5570560*X^16 - 6553600*X^14 + 4659200*X^12 - 2050048*X^10 + 549120*X^8 - 84480*X^6 + 6600*X^4 - 200*X^2 + 1

As an exercise for you, you could try to find a 48-degree polynomial that has the root sin 1° among others.

[nibble4bits]

Originally Posted by mfgoode

'c' is a constant regardless of the frame of reference uniform or non-uniform.

Actually what I was refering to is what if I told most people that several galaxies are moving away from us at 60% of light speed. They'de think that the any two in opposite directions must be moving away from each other at 1.2c. Talking about this sort of stuff is disgusting, LOL.

[mfgoode]

Originally Posted by alpertron

Using identity (21): T_n+1 (x) = 2 x T_n (x) - T_n-1 (x) I coded the following program in UBASIC (this interpreted language can operate with polynomials):

10   input "Degree";N
20   PrevPoly=1
30   ActualPoly=_X
40   for I=2 to N
50   NewPoly=2*_X*ActualPoly-PrevPoly
60   PrevPoly=ActualPoly
70   ActualPoly=NewPoly
80   next I
90   print ActualPoly

In our example, if I enter 20 I get:

524288*X^20 - 2621440*X^18 + 5570560*X^16 - 6553600*X^14 + 4659200*X^12 - 2050048*X^10 + 549120*X^8 - 84480*X^6 + 6600*X^4 - 200*X^2 + 1

As an exercise for you, you could try to find a 48-degree polynomial that has the root sin 1° among others.

:surprised Thanks Alpertron for identity (21).
I have to confess that I'm nil in programming and its a great drawback for me.
However I must get some one knowledgeable here who can work it out for me.
Mally

[mfgoode]

Originally Posted by nibble4bits

Actually what I was refering to is what if I told most people that several galaxies are moving away from us at 60% of light speed. They'de think that the any two in opposite directions must be moving away from each other at 1.2c. Talking about this sort of stuff is disgusting, LOL.

Come to think of it nibble4bits how would you explain it to the people, never mind what they think.! Id like to hear from you. Can anyone explain this to me?
Mally

[wblipp]

Originally Posted by mfgoode

Come to think of it nibble4bits how would you explain it to the people, never mind what they think.! Id like to hear from you. Can anyone explain this to me?
Mally

Here's what I say when this comes up:

The Michelson-Morley experiments show that the universe is a stanger place than was expected. When you try to figure out a way for the universe to behave consistently with this experimental observation, this is what you end up with.

[cheesehead]

Originally Posted by nibble4bits

Actually what I was refering to is what if I told most people that several galaxies are moving away from us at 60% of light speed. They'de think that the any two in opposite directions must be moving away from each other at 1.2c.

Well, they are moving away from each other at 1.2c, in our frame of reference.

But they're not moving away from each other at 1.2c in either of their own frames of reference.

[garo]

In a manner of speaking yes I think they are as no light from one ever reaches the other. I'm no expert so anyone can explain why I am wrong I'd appreciate it.