fast
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The Math Thread
- Thread starter beero1000
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Bomb#20
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beero1000
Veteran Member
Well, that escalated quickly...
Anyway, I've moved on to fiddling with the problem of Apollonius - re-deriving the constructions that generate the circles tangent to 3 given objects chosen from among {point, line, circle}. There are 10 cases and they all boil down to finding the intersection points of conics.
This is what I do when I don't feel like writing a final exam or doing actual research...
Anyway, I've moved on to fiddling with the problem of Apollonius - re-deriving the constructions that generate the circles tangent to 3 given objects chosen from among {point, line, circle}. There are 10 cases and they all boil down to finding the intersection points of conics.
This is what I do when I don't feel like writing a final exam or doing actual research...
beero1000
Veteran Member
My favorite of the 10 cases: two points and a line (PPL). A bunch of other cases (PLL, LLC, PLC, LCC) can be reduced to solving PPL, which in turn reduces to solving PPP. The construction I've come up with to do the reduction is nice and elegant - I think it's about as simple as possible. If you can find a shorter compass and straightedge construction, I'd like to see it.
Given a line L and two points A and B on the same side of L:
- Draw line AB, say it crosses L at C. (If it does not cross L, we reduce immediately to PPP using the intersection of L and the perpendicular bisector of AB)
- Draw circle CA, say it crosses AB at A and D.
- Draw circle with diameter BD.
- Draw a perpendicular to AB through C, it crosses the circle BD at E.
- Draw the circle CE, it crosses L at F and G.
- The two circles through ABF and ABG are the solutions (again, reduced to PPP).
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Kharakov
Quantum Hot Dog
lol. Bump. And a question. Solve for integer n for fun.
Question: Is there an easy way to generalize the following to non-integer n?
\( \lim_{\epsilon \to 0} \,\, \sum_{k=1}^{n-1} \, \, -cos (\frac{k \pi}{n}+2\epsilon \,\,sin(\frac{k \pi}{n}))/ \epsilon\)
Question: Is there an easy way to generalize the following to non-integer n?
\( \lim_{\epsilon \to 0} \,\, \sum_{k=1}^{n-1} \, \, -cos (\frac{k \pi}{n}+2\epsilon \,\,sin(\frac{k \pi}{n}))/ \epsilon\)
beero1000
Veteran Member
Integrate. It all works out the same anyway...
beero1000
Veteran Member
Well, simplify first. It's too late for Bessel functions.
Use the cosine angle sum identity and small angle approximations:
\(-\frac{1}{\eps}\left(\cos\left(\frac{k\pi}{n} + 2 \eps \sin\left(\frac{k\pi}{n}\right)\right)\right) = -\frac{1}{\eps}\left(\cos\left(\frac{k\pi}{n}\right)\cos\left(2\eps\sin\left(\frac{k\pi}{n}\right)\right) - \sin\left(\frac{k\pi}{n}\right)\sin\left(2\eps\sin\left(\frac{k\pi}{n}\right)\right) \right)\)
\(\approx -\frac{1}{\eps} \left(\cos\left(\frac{k\pi}{n}\right)\left(1 - 2\eps^2\sin^2\left(\frac{k\pi}{n}\right)\right) - \sin\left(\frac{k\pi}{n}\right)\left(2\eps\sin\left(\frac{k\pi}{n}\right)\right) \right) = 2 \sin^2 \left( \frac{k\pi}{n}\right) + 2 \eps \cos \left( \frac{k\pi}{n}\right) \sin^2 \left(\frac{k\pi}{n}\right) - \frac{1}{\eps} \cos \left( \frac{k \pi}{n}\right)\)
Since epsilon is fixed while we sum/integrate and then goes to 0 in the limit, the last two terms disappear and we are left with just the sine squared term. In either case:
\( \sum_{k = 0}^{n} 2\sin^2 \left( \frac{k\pi}{n}\right) = \int_0^n 2 \sin^2 \left(\frac{k\pi}{n}\right)dk = n\)
beero1000
Veteran Member
This?
\(\lim_{\eps \to 0} \int_{0}^n \frac{1}{\eps}\cos\left(\frac{k\pi}{n}\right) dk = \lim_{\eps \to 0} \frac{1}{\eps} \int_{0}^n \cos\left(\frac{k\pi}{n}\right) dk = \lim_{\eps \to 0} \frac{1}{\eps} \cdot 0 = \lim_{\eps \to 0} 0 = 0\)
beero1000
Veteran Member
No problem. For the sum, the result is the same for 1,...,n-1 and 0,...,n. For the integral, there's a strong preference for (0,n), so there's no real reason not to use 0 to n in both.
Kharakov
Quantum Hot Dog
First, I should say that I mangled the math a bit to use cosine instead of sin because I was trying to eliminate a certain part of an algorithm. So... that explains the form I posted here.. a little bit. I have to go do some stuff, this is not a full explanation. It will be edited later...
In this first image there is a gray equilateral triangle with orange stars at its vertices. Relationship between chords and angles is ~ (
\(\frac {chord 1}{\frac {\pi}{6}- angle 1} =\frac {chord 2}{angle 2 - \frac {\pi}{6}} \)
Purple/pink line is the sin of the angles. The chords are infinitesimally close to the equilateral triangle in the image:
Sum of sins = # of points.
Then another point appears at pi, and the angles spread out equally, still keeping the chords infinitesimally close to lines between points. Once points are all equidistant, pi point splits. Once points are equidistant again, another point appears at pi. This new point splits when points are equidistant.
When all points are equidistant, sum of sins = number of points.
Now, I've got to figure out another part of the math problem when I get back from a run.... but that's part of it. I'll need to explain more of it later- basically it will relate the chords to the lines they are infinitesimally far from.
In this first image there is a gray equilateral triangle with orange stars at its vertices. Relationship between chords and angles is ~ (
\(\frac {chord 1}{\frac {\pi}{6}- angle 1} =\frac {chord 2}{angle 2 - \frac {\pi}{6}} \)
Purple/pink line is the sin of the angles. The chords are infinitesimally close to the equilateral triangle in the image:
Sum of sins = # of points.
Then another point appears at pi, and the angles spread out equally, still keeping the chords infinitesimally close to lines between points. Once points are all equidistant, pi point splits. Once points are equidistant again, another point appears at pi. This new point splits when points are equidistant.
When all points are equidistant, sum of sins = number of points.
Now, I've got to figure out another part of the math problem when I get back from a run.... but that's part of it. I'll need to explain more of it later- basically it will relate the chords to the lines they are infinitesimally far from.
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lpetrich
Contributor
Here is a challenge. It involves matrices, but the only ones you will be explicitly working with here are small ones.
The group of all real orthogonal n*n matrices is called O. The group of "special" ones, those with determinant = 1, is called SO.
An orthogonal matrix M has inverse(M) = transpose(M).
* What determinant values can an orthogonal matrix have?
* What are all the O(1) matrices? The SO(1) ones?
* The O(2) matrices might seem to have 2*2 = 4 parameters, one for each matrix element. Can you show that their matrix elements can be specified with a smaller number of parameters?
* What are all the finite subgroups of O(2)? Of SO(2)?
The group of all real orthogonal n*n matrices is called O
. The group of "special" ones, those with determinant = 1, is called SO.An orthogonal matrix M has inverse(M) = transpose(M).
* What determinant values can an orthogonal matrix have?
* What are all the O(1) matrices? The SO(1) ones?
* The O(2) matrices might seem to have 2*2 = 4 parameters, one for each matrix element. Can you show that their matrix elements can be specified with a smaller number of parameters?
* What are all the finite subgroups of O(2)? Of SO(2)?
beero1000
Veteran Member
I like it.
Kharakov
Quantum Hot Dog
I figured out the chord problem the wrong way, using an inverse symbolic calculator and searching for patterns, instead of figuring out the math.
The chords were all from 0 to 2*theta, with 2*theta being the central angle.
\(\theta_k = \frac {k \pi}{n}\)
if k/n>.5 \(\theta_k = \theta_k + \frac { (n-\lfloor{n}\rfloor) \pi}{n}\)
ck = chord length of kth chord=
\( 2 sin (\theta_k - \eps \, c_k)\)
sum of sines:
\(\sum_{1}^{n-1} \, \frac {sin (\eps \, c_k \, + \, (\frac {\pi}{2}-\theta_k))}{\eps}\)
sum of sines function:
\(1 + 2 \lfloor{\frac{n}{2}}\rfloor + \frac{sin(\pi \,\, \frac{3n - 4 \lfloor\frac{n}{2}\rfloor -2} {2n-4\lfloor\frac{n}{2}\rfloor}-\frac{\pi}{2} )} { sin(\pi \,\, \frac {n-2}{2n-4 \lfloor\frac{n}{2}\rfloor}-\frac{\pi}{2})}\)
sum of sins function for n=x/y:
\(1 + 2 \lfloor{\frac{x}{2y}}\rfloor + \frac{sin(\pi \,\, \frac{3x - 4y \lfloor\frac{x}{2y}\rfloor -2y} {2x-4y\lfloor\frac{x}{2y}\rfloor}-\frac{\pi}{2} )} { sin(\pi \,\, \frac {x-2y}{2x-4y \lfloor\frac{x}{2y}\rfloor}-\frac{\pi}{2})}\)
sum of (sines/chordlength^2):
\(\sum_{1}^{n-1} \, \frac {sin (\eps \, c_k \, + \, (\frac {\pi}{2}-\theta_k))}{\eps {c_k}^2}\)
Can you prove that the # of chords = sum of sines/ (chordlengths^2)?
If the chords are added in the opposite direction (chords growing from 0 instead of popping into existence at pi), would the sines be smooth instead of discontinuous? <--
The chords were all from 0 to 2*theta, with 2*theta being the central angle.
\(\theta_k = \frac {k \pi}{n}\)
if k/n>.5 \(\theta_k = \theta_k + \frac { (n-\lfloor{n}\rfloor) \pi}{n}\)
ck = chord length of kth chord=
\( 2 sin (\theta_k - \eps \, c_k)\)
sum of sines:
\(\sum_{1}^{n-1} \, \frac {sin (\eps \, c_k \, + \, (\frac {\pi}{2}-\theta_k))}{\eps}\)
sum of sines function:
\(1 + 2 \lfloor{\frac{n}{2}}\rfloor + \frac{sin(\pi \,\, \frac{3n - 4 \lfloor\frac{n}{2}\rfloor -2} {2n-4\lfloor\frac{n}{2}\rfloor}-\frac{\pi}{2} )} { sin(\pi \,\, \frac {n-2}{2n-4 \lfloor\frac{n}{2}\rfloor}-\frac{\pi}{2})}\)
sum of sins function for n=x/y:
\(1 + 2 \lfloor{\frac{x}{2y}}\rfloor + \frac{sin(\pi \,\, \frac{3x - 4y \lfloor\frac{x}{2y}\rfloor -2y} {2x-4y\lfloor\frac{x}{2y}\rfloor}-\frac{\pi}{2} )} { sin(\pi \,\, \frac {x-2y}{2x-4y \lfloor\frac{x}{2y}\rfloor}-\frac{\pi}{2})}\)
sum of (sines/chordlength^2):
\(\sum_{1}^{n-1} \, \frac {sin (\eps \, c_k \, + \, (\frac {\pi}{2}-\theta_k))}{\eps {c_k}^2}\)
Can you prove that the # of chords = sum of sines/ (chordlengths^2)?
If the chords are added in the opposite direction (chords growing from 0 instead of popping into existence at pi), would the sines be smooth instead of discontinuous? <--
Last edited:
beero1000
Veteran Member
I'm not sure I follow where you're getting those thetas and chord lengths and the definitions seem to be circular.
Once you have the definitions, the general procedure will be similar - trig identities and small angle simplifications.
Once you have the definitions, the general procedure will be similar - trig identities and small angle simplifications.
Kharakov
Quantum Hot Dog
I need animation software so I can show my work better. I'll attempt to explain it a bit, but my brain is a bit mushy at this point, worked on a bunch (for me) of different things today.
I get theta by dividing a circle into 2pi/n sections. Base chords are 'drawn' from 0 (2pi) to each (2*k*pi)/n point. If n is not an integer, to keep symmetry after passing pi (when k/n>.5), you have to add in 2(n-floor
)/n*pi. I drop the 2 from the equations because I'm using 1/2 the central angles for all the calculations I'm doing (so instead of 2pi/n, I use pi/n).If 2<n<=4, there are 2 base chords. If n=4, there are 3 base chords (one is the diameter of the circle). if 4<n<6, there are 4 chords. If n=6 there are 5 base chords (one is the diameter again), if 6<n<8 there are 6 chords...
In the image below, 2<n<4
The circular chord length calculation.. I run a brute force squeeze algorithm, with epsilon=10^-30:
\( 0 = c_k - 2 sin ( \theta_k + c_k \eps) \)
Epsilon has to be the same throughout the whole calculation...
And actually it's not the sum of sines, it's the sum of sines/epsilon because the sum of sines is <10^-29...
beero1000
Veteran Member
I need animation software so I can show my work better. I'll attempt to explain it a bit, but my brain is a bit mushy at this point, worked on a bunch (for me) of different things today.
I get theta by dividing a circle into 2pi/n sections. Base chords are 'drawn' from 0 (2pi) to each (2*k*pi)/n point. If n is not an integer, to keep symmetry after passing pi (when k/n>.5), you have to add in 2(n-floor
If 2<n<=4, there are 2 base chords. If n=4, there are 3 base chords (one is the diameter of the circle). if 4<n<6, there are 4 chords. If n=6 there are 5 base chords (one is the diameter again), if 6<n<8 there are 6 chords...
In the image below, 2<n<4
The circular chord length calculation.. I run a brute force squeeze algorithm, with epsilon=10^-30:
\( 0 = c_k - 2 sin ( \theta_k + c_k \eps) \)
Epsilon has to be the same throughout the whole calculation...
And actually it's not the sum of sines, it's the sum of sines/epsilon because the sum of sines is <10^-29...
I'm not sure that's the best approach... And if you try to substitute in with your previous equations then it doesn't work.
Not sure what epsilon is representing, but if you define the perturbed points via small central angle differences \(\phi_k\) then the adjusted chord has length \(2\sin(\theta_k - \phi_k)\) (Note that your angle= epsilon*adjusted chord length equation is pretty much a small angle approximation of this, the angle is phi/2). Then the sine you are looking at is \(\sin(\theta_k + \phi_k)\).
Also, the sum of the sines can't be zero - unless you're not measuring the angles as shown or counting negative angles somehow?