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beero1000
Veteran Member
For cos(36d):
For P(p,n,-) and P(p,n,+):
I recently attended a talk on the following problem: Take a 'partially eaten pie', i.e. a unit radius disk with a sector of angle 2pi - A removed. We want to slice the pie into sectors and arrange (without distortion or overlap) them on a plate with radius R. For each given A, what is the smallest possible value of R?
The above problem is currently still open, so a (provably) optimal solution would be worth a published paper.
For P(p,n,-) and P(p,n,+):
I recently attended a talk on the following problem: Take a 'partially eaten pie', i.e. a unit radius disk with a sector of angle 2pi - A removed. We want to slice the pie into sectors and arrange (without distortion or overlap) them on a plate with radius R. For each given A, what is the smallest possible value of R?
The above problem is currently still open, so a (provably) optimal solution would be worth a published paper.
Kharakov
Quantum Hot Dog
lpetrich
Contributor
Kharakov, you got it.
M5(x) = cos(5*arccos(x)) = 16*x5 - 20*x3 + 5*x
If x = cos(π/5), then M5(x) is cos(5*(π/5)) = cos(π) = -1. However, it also has that value for x = cos(3π/5), cos(5π/5), cos(7π/5), and cos(9π/5). So M5(x) = -1 has two solution with value cos(π/5), two with value cos(3π/5), and one with value cos(π) = -1:
M5(x) + 1 = 0
(x + 1)*(4*x2 - 2*x + 1)2 = 0
Its solutions: two of cos(π/5) = (1 + sqrt(5))/4, two of cos(3π/5) = (1 - sqrt(5))/4, one of cos(π) = -1
Let's now consider more generally these polynomials:
T(n,x) = cos(n*arccos(x))
U(n,x) = sin((n+1)*arccos(x))/sin(arccos(x))
One can find them with a recurrence derived from trig identities:
Let y = n*x. Then
cos(y+x) + cos(y-x) = 2*cos
*cos(x)
sin(y+x) + sin(y-x) = 2*sin*cos(x)
giving
T(n+1,x) - 2*x*T(n,x) + T(n-1,x) = 0
U(n+1,x) - 2*x*U(n,x) + U(n-1,x) = 0
the same for both sets of polynomials.
Their initial values are different, however.
T(0,x) = 1, T(1,x) = x
U(0,x) = 1, U(1,x) = 2*x
Let's suppose that T(n,x) = C
*xn + O(xn-2) and similarly for U(n,x).
Then for n > 0, C(n+1) = 2*C, giving C = 2n-1 for T(n,x) and 2n for U(n,x).
There are some further relations, all of which can be derived from the trig definition of these polynomials. See if you can do so for at least some of them.
2*(1-x2)*d(T(n,x),x) = n*(T(n-1,x) - T(n+1,x))
2*(1-x2)*d(U(n,x),x) = (n+2)*U(n-1,x) - n*U(n+1,x)
For y = T(n,x), (1-x2)*(d2y/dx2) - x*(dy/dx) + n2*y = 0
For y = U(n,x), (1-x2)*(d2y/dx2) - 3*x*(dy/dx) + n*(n+2)*y = 0
Orthogonality:
\( \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}} T_n(x) T_m(x) dx = \frac{\pi}{2}(1 + \delta_{n,0})\delta_{n,m} ,\ \int_{-1}^{1} \sqrt{1-x^2} U_n(x) U_m(x) dx = \frac{\pi}{2}\delta_{n,m} \)
Special values:
T(n,-x) = (-1)n*T(n,x)
U(n,-x) = (-1)n*U(n,x)
T(n,1) = 1
U(n,1) = n+1
Generating functions:
\( - \frac12 \log (1 - 2xt + t^2) = \sum_n \frac1n t^n T_n(x) ,\ \frac{1 - xt}{1 - 2xt + t ^2} = \sum_n t^n T_n(x) ,\ \frac{1}{1 - 2xt + t^2} = \sum_n t^n U_n(x) \)
These are the Chebyshev polynomials, T being the first kind and U being the second kind.
M5(x) = cos(5*arccos(x)) = 16*x5 - 20*x3 + 5*x
If x = cos(π/5), then M5(x) is cos(5*(π/5)) = cos(π) = -1. However, it also has that value for x = cos(3π/5), cos(5π/5), cos(7π/5), and cos(9π/5). So M5(x) = -1 has two solution with value cos(π/5), two with value cos(3π/5), and one with value cos(π) = -1:
M5(x) + 1 = 0
(x + 1)*(4*x2 - 2*x + 1)2 = 0
Its solutions: two of cos(π/5) = (1 + sqrt(5))/4, two of cos(3π/5) = (1 - sqrt(5))/4, one of cos(π) = -1
Let's now consider more generally these polynomials:
T(n,x) = cos(n*arccos(x))
U(n,x) = sin((n+1)*arccos(x))/sin(arccos(x))
One can find them with a recurrence derived from trig identities:
Let y = n*x. Then
cos(y+x) + cos(y-x) = 2*cos
*cos(x)sin(y+x) + sin(y-x) = 2*sin
*cos(x)giving
T(n+1,x) - 2*x*T(n,x) + T(n-1,x) = 0
U(n+1,x) - 2*x*U(n,x) + U(n-1,x) = 0
the same for both sets of polynomials.
Their initial values are different, however.
T(0,x) = 1, T(1,x) = x
U(0,x) = 1, U(1,x) = 2*x
Let's suppose that T(n,x) = C
*xn + O(xn-2) and similarly for U(n,x).Then for n > 0, C(n+1) = 2*C
, giving C = 2n-1 for T(n,x) and 2n for U(n,x).There are some further relations, all of which can be derived from the trig definition of these polynomials. See if you can do so for at least some of them.
2*(1-x2)*d(T(n,x),x) = n*(T(n-1,x) - T(n+1,x))
2*(1-x2)*d(U(n,x),x) = (n+2)*U(n-1,x) - n*U(n+1,x)
For y = T(n,x), (1-x2)*(d2y/dx2) - x*(dy/dx) + n2*y = 0
For y = U(n,x), (1-x2)*(d2y/dx2) - 3*x*(dy/dx) + n*(n+2)*y = 0
Orthogonality:
\( \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}} T_n(x) T_m(x) dx = \frac{\pi}{2}(1 + \delta_{n,0})\delta_{n,m} ,\ \int_{-1}^{1} \sqrt{1-x^2} U_n(x) U_m(x) dx = \frac{\pi}{2}\delta_{n,m} \)
Special values:
T(n,-x) = (-1)n*T(n,x)
U(n,-x) = (-1)n*U(n,x)
T(n,1) = 1
U(n,1) = n+1
Generating functions:
\( - \frac12 \log (1 - 2xt + t^2) = \sum_n \frac1n t^n T_n(x) ,\ \frac{1 - xt}{1 - 2xt + t ^2} = \sum_n t^n T_n(x) ,\ \frac{1}{1 - 2xt + t^2} = \sum_n t^n U_n(x) \)
These are the Chebyshev polynomials, T being the first kind and U being the second kind.
Kharakov
Quantum Hot Dog
I used the Chebyshev polynomials a few years back to streamline a calculation I was doing. Because I was using both the real and imaginary components, I found that using complex numbers was a bit faster (at least in the program I was using at the time- probably had some assembly language streamlining for its implementation of complex integer powers).
I'd just divide the complex number by its magnitude, then raise it to whatever power I wanted (of course, I only did this for integer powers, otherwise I'd still be using the exponential function to generate the results).
So... now I'm reading this, which reiterates what you've said.
Be back in thread later- gave away my next question below. If you didn't read the hidden part of this post already, don't. I'm asking a question that you'll like to solve, and the hidden part gives it away.
I used to check constants I discovered on google, until a few years back I found:
The Inverse Symbolic Calculator Plus
Although I used Plouffe's inverter for a year or 2 before switching to that one. And now I'm thinking about creating a generating function for integer sequences with the plastic constant... and its x^4-x -1 = 0 relatives.
I'd just divide the complex number by its magnitude, then raise it to whatever power I wanted (of course, I only did this for integer powers, otherwise I'd still be using the exponential function to generate the results).
So... now I'm reading this, which reiterates what you've said.
Be back in thread later- gave away my next question below. If you didn't read the hidden part of this post already, don't. I'm asking a question that you'll like to solve, and the hidden part gives it away.
I used to check constants I discovered on google, until a few years back I found:
The Inverse Symbolic Calculator Plus
Although I used Plouffe's inverter for a year or 2 before switching to that one. And now I'm thinking about creating a generating function for integer sequences with the plastic constant... and its x^4-x -1 = 0 relatives.
Last edited:
Kharakov
Quantum Hot Dog
Given the following Taylor series for log (I refer to natural log, or ln, as log).
\(\log{(\frac{x}{y})} = \log{(x)}-\log{(y)} = \sum_{n=1}^\infty \, {(\frac {x-y}{x})}^n \,\, \times \, n^{-1}\)
What is the relationship of this sum to the Zeta function?
What do you get when you subtract \(\zeta{(1)}-1\) from \(\lim_{x \to \infty} \, \log{(x)}\)?
big big hint... don't look unless you want a spoiler.
Further: If the series is convergent for \( \infty> x > y >0 \) and \(0 < x < y <= x +1 < \infty \), what other function is the sum related to?
\(\log{(\frac{x}{y})} = \log{(x)}-\log{(y)} = \sum_{n=1}^\infty \, {(\frac {x-y}{x})}^n \,\, \times \, n^{-1}\)
What is the relationship of this sum to the Zeta function?
What do you get when you subtract \(\zeta{(1)}-1\) from \(\lim_{x \to \infty} \, \log{(x)}\)?
big big hint... don't look unless you want a spoiler.
Further: If the series is convergent for \( \infty> x > y >0 \) and \(0 < x < y <= x +1 < \infty \), what other function is the sum related to?
lpetrich
Contributor
If you set y = 0, you get ζ(1), which is infinite.
That's infinity - infinity, and that's dependent on how you take the limit to infinity for each operand.
There's a much better way to find what the zeta function is for an argument near 1. Consider an alternating-sign version:
\( \zeta(s) = \sum_n \frac{1}{n^s} ,\ \zeta_{alt}(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s} \)
One can easily show that
\( \zeta(s) = \frac{1}{1 - 2^{1-s}} \zeta_{alt}(s) \)
Near s = 1,
\( \zeta(s) \sim \frac{1}{(s-1) \log 2} \zeta_{alt}(1) \)
But
\( \zeta_{alt}(1) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \)
and from x = 1 and y = 2 in Kharakov's logarithm expression,
\( \log 2 = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \)
This means that
\( \zeta_{alt}(1) = \log 2 \)
and
\( \zeta(s) \sim \frac{1}{s-1} \)
One can show that it has an O(1) term added to it.
No idea. I'd need a hint as to where to look.
lpetrich
Contributor
Now for P(p,n,+-) = pn - 1 proofs. Which constraints on n yield only primes?
If p is odd, then P(p,n,+-) is even, and since the only even prime is 2, there is only one value of p that gives 2. Likewise, n is also constrained to only one value. The answer: P(3,1,-) = 2.
Turning to P(2,n,-), we consider what happens when n is composite: a*b.
2a*b - 1 = (2a - 1) * (2a*(b-1) + 2a*(b-2) + ... + 2a + 1)
That is obviously composite, so we find that P(2,n,-) being prime implies that n is prime. When n is prime, M = P(2,n,-) is a Mersenne number. But only some Mersenne numbers are primes, with the first composite one being M(11). They have been found for n as high as it has been possible to do primality tests on the resulting Mersenne numbers, but they grow more and more scarce as one continues onward. It is unknown whether there is a finite or an infinite number of Mersenne primes, though the latter is a rather obvious conjecture.
Turning to P(2,n,+), we consider what happens when n has an odd factor.
2a*b + 1 = (2a - 1) * (2a*(b-1) - 2a*(b-2) + ... + 2a - 1)
with the final minus sign only being possible if b is odd. Thus n has only even factors, and is thus a power of 2. The resulting number, F(m) = P(2,2m,+) is a Fermat number. Only the first five of them are known to be primes, and they have been checked up to m = 32. As with Mersenne primes, it is unknown whether there are a finite or infinite number of them, though here, it's the former that is a rather obvious conjecture.
The Fermat numbers have various interrelationships, and my favorite is
F(m) = F(0)*F(1)*F(2)* ... * F(m-1) + 2
Can anyone here prove that?
It also has a certain consequence about whether any Fermat numbers share common factors. What is it?
If p is odd, then P(p,n,+-) is even, and since the only even prime is 2, there is only one value of p that gives 2. Likewise, n is also constrained to only one value. The answer: P(3,1,-) = 2.
Turning to P(2,n,-), we consider what happens when n is composite: a*b.
2a*b - 1 = (2a - 1) * (2a*(b-1) + 2a*(b-2) + ... + 2a + 1)
That is obviously composite, so we find that P(2,n,-) being prime implies that n is prime. When n is prime, M
= P(2,n,-) is a Mersenne number. But only some Mersenne numbers are primes, with the first composite one being M(11). They have been found for n as high as it has been possible to do primality tests on the resulting Mersenne numbers, but they grow more and more scarce as one continues onward. It is unknown whether there is a finite or an infinite number of Mersenne primes, though the latter is a rather obvious conjecture.Turning to P(2,n,+), we consider what happens when n has an odd factor.
2a*b + 1 = (2a - 1) * (2a*(b-1) - 2a*(b-2) + ... + 2a - 1)
with the final minus sign only being possible if b is odd. Thus n has only even factors, and is thus a power of 2. The resulting number, F(m) = P(2,2m,+) is a Fermat number. Only the first five of them are known to be primes, and they have been checked up to m = 32. As with Mersenne primes, it is unknown whether there are a finite or infinite number of them, though here, it's the former that is a rather obvious conjecture.
The Fermat numbers have various interrelationships, and my favorite is
F(m) = F(0)*F(1)*F(2)* ... * F(m-1) + 2
Can anyone here prove that?
It also has a certain consequence about whether any Fermat numbers share common factors. What is it?
Kharakov
Quantum Hot Dog
True. Try it with positive integer values for x and y, x>y:
\(log(2) - log(1) = \frac {1}{2} + \frac {1}{2^2 \times 2} + \frac {1}{2^3 \times 3} + \frac {1}{2^4 \times 4}...\)
\(log(3) - log(2) = \frac {1}{3} + \frac {1}{3^2 \times 2} + \frac {1}{3^3 \times 3} + \frac {1}{3^4 \times 4}...\)
\(log(4) - log(3) = \frac {1}{4} + \frac {1}{4^2 \times 2} + \frac {1}{4^3 \times 3} + \frac {1}{4^4 \times 4}...\)
\(log(5) - log(4) = \frac {1}{5} + \frac {1}{5^2 \times 2} + \frac {1}{5^3 \times 3} + \frac {1}{5^4 \times 4}...\)
\(log(6) - log(5) = \frac {1}{6} + \frac {1}{6^2 \times 2} + \frac {1}{6^3 \times 3} + \frac {1}{6^4 \times 4}...\)
Notice a pattern emerging? As log(x) approaches infinity, it is equal to (can figure it out yourself first):
Yeah, but there is a constant that emerges when you do that. Wolfram alpha won't let me do the actual thing, but I can do (sum of (zeta(k)/k - 1/k) k=2 to 1500), and then subtract that from 1 to get the...
Euler Mascheroni constant. Yeah. That's the
Dirichlet eta function. Nice work. I think you already found part of it (it's the Dirichlet Eta function \(\eta{(1)}\)). If you shift all the logs, you end up with the alternating sign versions similar to the log(2) version you posted above. You can do a trick with them too...'
\(log(2) - log(1) = \frac {1}{1} - \frac {1}{2} + \frac {1}{3} - \frac {1}{4}...\)
\(log(3) - log(2) = \frac {1}{2} - \frac {1}{2^2 \times 2} + \frac {1}{2^3 \times 3} - \frac {1}{2^4 \times 4}...\)
\(log(4) - log(3) = \frac {1}{3} - \frac {1}{3^2 \times 2} + \frac {1}{3^3 \times 3} - \frac {1}{3^4 \times 4}...\)
\(log(5) - log(4) = \frac {1}{4} - \frac {1}{4^2 \times 2} + \frac {1}{4^3 \times 3} - \frac {1}{4^4 \times 4}...\)
\(log(6) - log(5) = \frac {1}{5} - \frac {1}{5^2 \times 2} + \frac {1}{5^3 \times 3} - \frac {1}{5^4 \times 4}...\)
So you can add the alternating sign log
-log(2) to the other log(n-1) and find a nice little pattern. You can also subtract the alternating sign log - log(2) from the normal sign log(n-1) and find a neat little pattern as well. Although now that I'm thinking about it, it looks different then what I thought I had worked out before. 
I'm getting (for the second idea above):
log (6)-log(2)-log(5)=
\( \frac {1}{2^2 \times 1} + \frac {1}{2^4 \times 2}+ \frac {1}{2^6 \times 3}...\)
\( \frac {1}{3^2 \times 1} + \frac {1}{3^4 \times 2}+ \frac {1}{3^6 \times 3}...\)
\(\frac {1}{4^2 \times 1} + \frac {1}{4^4 \times 2}+ \frac {1}{4^6 \times 3}...\)
\(\frac {1}{5^2 \times 1} + \frac {1}{5^4 \times 2}+ \frac {1}{5^6 \times 3}...\)
Which should mean that I can calculate the value of every other zeta function added together (zeta (2k-1)/k)... or something like that.
Anyway... time for me to get some sleep. Thanks for the fun thoughts.
lpetrich
Contributor
There is a much better way to state that result. Define a partial zeta function:
\( \zeta(s,n) = \sum_{m=1}^{n} \frac{1}{m^s} \)
Then one can show that
\( \log n = \sum_{k=1}^{\infty} \frac{\zeta(k,n) - 1}{k} \)
BTW, the Euler-Mascheroni constant is defined as
\( \gamma = \lim_{n->\infty} (\zeta(1,n) - \log n) \)
There's also a theorem that states
\( \zeta(s) = \frac{1}{s-1} + \gamma + O(s-1) \)
\( \zeta(s,n) = \sum_{m=1}^{n} \frac{1}{m^s} \)
Then one can show that
\( \log n = \sum_{k=1}^{\infty} \frac{\zeta(k,n) - 1}{k} \)
BTW, the Euler-Mascheroni constant is defined as
\( \gamma = \lim_{n->\infty} (\zeta(1,n) - \log n) \)
There's also a theorem that states
\( \zeta(s) = \frac{1}{s-1} + \gamma + O(s-1) \)
Kharakov
Quantum Hot Dog
Nice. But look at where the cutoff lies for your partial zeta function. I made that mistake the other day. Hahah... Nevermind. I read the same mistake I made into your partial zeta function.. hhahahaha!!
That's why I think it's so important to show your work, so that one sees that log(8) is equal to:
\(\frac {1}{2} + \frac {1}{2^2 \times 2} + \frac {1}{2^3 \times 3} + \frac {1}{2^4 \times 4} + \frac {1}{2^5 \times 5} + \frac {1}{2^6 \times 6}...\)
\(\frac {1}{3} + \frac {1}{3^2 \times 2} + \frac {1}{3^3 \times 3} + \frac {1}{3^4 \times 4} + \frac {1}{3^5 \times 5} + \frac {1}{3^4 \times 6}...\)
\(\frac {1}{4} + \frac {1}{4^2 \times 2} + \frac {1}{4^3 \times 3} + \frac {1}{4^4 \times 4} + \frac {1}{4^5 \times 5} + \frac {1}{4^4 \times 6}...\)
\(\frac {1}{5} + \frac {1}{5^2 \times 2} + \frac {1}{5^3 \times 3} + \frac {1}{5^4 \times 4} + \frac {1}{5^5 \times 5} + \frac {1}{5^4 \times 6}...\)
\(\frac {1}{6} + \frac {1}{6^2 \times 2} + \frac {1}{6^3 \times 3} + \frac {1}{6^4 \times 4} + \frac {1}{6^5 \times 5} + \frac {1}{6^6 \times 6}...\)
\(\frac {1}{7} + \frac {1}{7^2 \times 2} + \frac {1}{7^3 \times 3} + \frac {1}{7^4 \times 4} + \frac {1}{7^5 \times 5} + \frac {1}{7^5 \times 6}...\)
\(\frac {1}{8} + \frac {1}{8^2 \times 2} + \frac {1}{8^3 \times 3} + \frac {1}{8^4 \times 4} + \frac {1}{8^5 \times 5} + \frac {1}{8^6 \times 6}...\)
The other day, when I asked for a mathematical demonstration from you (in the thread about rotations), I wanted something that had EVERY step worked out. Now, it's still good work, but if you work out EVERY step, a majority of people can follow the work, and learn from it. I understand wanting to write out your ideas, get them out there for everyone, but if you don't present them in a way that lots of people can understand (by showing the work... actually crunching numbers through an equation), you're going to lose a lot of people.
In fact, a lot of notation, I'm not completely familiar with, which doesn't mean I don't understand the basics- it means I don't understand the connection between the notation and the math being used. Having something written out, where numbers crunch through, makes the connection. Otherwise, I've got to work things out myself, until I understand a concept, and can see a corollary between what others have done and the work I've done<-- hell, this is exactly how I learned the math I know today. It wasn't that I was taught information in the correct order, it was that I discovered things, played around, searched google, mentioned them in a forum, and that person said "ohh, those are Taylor series for XXXX" or "that is XXX" and I learned that what I was doing was called something else.
Not to mention using shorthand (writing the summation formula), instead of a written demonstration of a summation (like what I wrote above), creates problems... for me.
lpetrich
Contributor
First, I concede:
\( \zeta_{alt}(s) = \eta(s) \)
the Dirichlet eta function.
It should be easy to prove that
η(s) = (1 - 21-s) * ζ(s)
There are some interesting integral expressions:
\( \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx ,\ \eta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x + 1} dx \)
\( \zeta_{alt}(s) = \eta(s) \)
the Dirichlet eta function.
It should be easy to prove that
η(s) = (1 - 21-s) * ζ(s)
There are some interesting integral expressions:
\( \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx ,\ \eta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x + 1} dx \)
lpetrich
Contributor
The Euler gamma function Γ(s) is an analytic continuation of the factorial function: n! = Γ(n+1). It is given by this integral:
\( \Gamma(s) = \int_0^\infty x^{s-1} e^{-x} dx \)
It's easy to prove that Γ(s+1) = s*Γ(s).
Using
\( e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n \)
we get Euler's infinite product
\( \log \Gamma(s+1) = \lim_{n \to \infty} \left[ s \log n - \sum_{k=1}^n \log \left( 1 + \frac{s}{k} \right) \right] \)
and
\( \log \Gamma(s+1) = - s \gamma - \sum_{k=1}^\infty \left[ \log \left( 1 + \frac{s}{k} \right) - \frac{s}{k} \right] \)
One can prove a reflection formula:
\( \Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin s \pi} \)
and Gauss's multiplication formula:
\( \Gamma(ns) = (2\pi)^{(1-n)/2} n^{ns - 1/2} \prod_{k=0}^{n-1} \Gamma\left( s + \frac{k}{n} \right) \)
For proving the reflection formula with Euler's infinite product, one can use
\( \sin s = s \prod_{k=1}^\infty \left( 1 - \left( \frac{s} {k\pi} \right)^2 \right) \)
There's a related one,
\( \cos s = \prod_{k=0}^\infty \left( 1 - \left( \frac{s} {(k+1/2)\pi} \right)^2 \right) \)
From the reflection formula, one can prove that Γ(1/2) = sqrt(π).
Gauss's multiplication formula is more difficult. A partial formula from Euler's infinite product is
\( \Gamma(ns) = n^{ns-1} \prod_{k=0}^{n-1} \Gamma\left( s + \frac{k}{n} \right) / \prod_{k=1}^{n-1} \Gamma\left(\frac{k}{n} \right) \)
From the reflection formula, the denominator is
\( \prod_{k=1}^{n-1} \Gamma\left(\frac{k}{n} \right) = \sqrt{ \prod_{k=1}^{n-1} \frac{\pi}{\sin (k\pi/n)} } \)
I'm stumped there, but an alternate proof of Gauss's multiplication theorem can be found with Stirling's formula:
\( \log \Gamma(s) = s \log s - (1/2) \log s - s + (1/2) \log (2\pi) + O(1/s) \)
This gives us
\( \prod_{k=1}^{n-1} \sin \frac{k\pi}{n} = \frac{n}{2^n} \)
Yet another formula for the Euler gamma function comes from integrating Abramowitz and Stegun 6.3.21:
\( \log \Gamma(s) = \int_0^\infty \left( (s-1)e^{-x} - \frac{e^{-x} - e^{-sx}}{1 - e^{-x}} \right) \frac{dx}{x} \)
One can prove Gauss's multiplication formula from it also, and one finds this rather curious formula:
\( \frac12 \log (2\pi) = \int_0^\infty \left( \frac1x - \frac12 e^{-x} - \frac{e^{-x}}{1-e^{-x}} \right) \frac{dx}{x} \)
\( \Gamma(s) = \int_0^\infty x^{s-1} e^{-x} dx \)
It's easy to prove that Γ(s+1) = s*Γ(s).
Using
\( e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n \)
we get Euler's infinite product
\( \log \Gamma(s+1) = \lim_{n \to \infty} \left[ s \log n - \sum_{k=1}^n \log \left( 1 + \frac{s}{k} \right) \right] \)
and
\( \log \Gamma(s+1) = - s \gamma - \sum_{k=1}^\infty \left[ \log \left( 1 + \frac{s}{k} \right) - \frac{s}{k} \right] \)
One can prove a reflection formula:
\( \Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin s \pi} \)
and Gauss's multiplication formula:
\( \Gamma(ns) = (2\pi)^{(1-n)/2} n^{ns - 1/2} \prod_{k=0}^{n-1} \Gamma\left( s + \frac{k}{n} \right) \)
For proving the reflection formula with Euler's infinite product, one can use
\( \sin s = s \prod_{k=1}^\infty \left( 1 - \left( \frac{s} {k\pi} \right)^2 \right) \)
There's a related one,
\( \cos s = \prod_{k=0}^\infty \left( 1 - \left( \frac{s} {(k+1/2)\pi} \right)^2 \right) \)
From the reflection formula, one can prove that Γ(1/2) = sqrt(π).
Gauss's multiplication formula is more difficult. A partial formula from Euler's infinite product is
\( \Gamma(ns) = n^{ns-1} \prod_{k=0}^{n-1} \Gamma\left( s + \frac{k}{n} \right) / \prod_{k=1}^{n-1} \Gamma\left(\frac{k}{n} \right) \)
From the reflection formula, the denominator is
\( \prod_{k=1}^{n-1} \Gamma\left(\frac{k}{n} \right) = \sqrt{ \prod_{k=1}^{n-1} \frac{\pi}{\sin (k\pi/n)} } \)
I'm stumped there, but an alternate proof of Gauss's multiplication theorem can be found with Stirling's formula:
\( \log \Gamma(s) = s \log s - (1/2) \log s - s + (1/2) \log (2\pi) + O(1/s) \)
This gives us
\( \prod_{k=1}^{n-1} \sin \frac{k\pi}{n} = \frac{n}{2^n} \)
Yet another formula for the Euler gamma function comes from integrating Abramowitz and Stegun 6.3.21:
\( \log \Gamma(s) = \int_0^\infty \left( (s-1)e^{-x} - \frac{e^{-x} - e^{-sx}}{1 - e^{-x}} \right) \frac{dx}{x} \)
One can prove Gauss's multiplication formula from it also, and one finds this rather curious formula:
\( \frac12 \log (2\pi) = \int_0^\infty \left( \frac1x - \frac12 e^{-x} - \frac{e^{-x}}{1-e^{-x}} \right) \frac{dx}{x} \)
lpetrich
Contributor
I have another challenge.
Euclid's Algorithm is a snazzy way of finding the greatest common denominator of two positive integers. Here it is, starting with a and b:
Notice that it does not require factoring the input numbers.
Can anyone prove that it works?
Extra: prove that for positive integers a and b, there exist integers r and s such that r*a + s*b = gcd(a,b)
Can you show how to construct them?
Euclid's Algorithm is a snazzy way of finding the greatest common denominator of two positive integers. Here it is, starting with a and b:
Code:
repeat
r = remainder(a,b)
if r == 0 then
-- The output value
gcd = b
break
end if
-- push the values back
a = b
b = r
end repeatCan anyone prove that it works?
Extra: prove that for positive integers a and b, there exist integers r and s such that r*a + s*b = gcd(a,b)
Can you show how to construct them?
lpetrich
Contributor
Here's a site that some of you people may like: ProofWiki - a big collection of mathematical proofs. Like Square Root of 2 is Irrational - ProofWiki.
Now to constructible polygons. I've found Constructibility of Regular Polygons, something that also discusses a more general version.
An N-gon is constructible with ruler and compass if it is possible to start with the rational numbers and do a finite number of arithmetic operations and square roots and get the Nth roots of unity:
cos(2π/N) + i*sin(2π/N)
One can find either of the two trig functions, and the other one from it using a well-known trig identity. That identity can easily be shown to satisfy that constructibility criterion.
Gauss and Wantzel had proved that this is only possible if N = 2n * product of non-repeated Fermat primes. Let's see how this works out for the lower Fermat primes.
N = 3
\( \cos \frac{2\pi}{3} = - \frac{1}{2} \)
N = 5
\( \begin{array}{rcl} \cos \frac{2\pi}{5} & = & \frac{1}{4} \left( -1 + \sqrt{5} \right) \\ \cos \frac{4\pi}{5} & = & \frac{1}{4} \left( -1 - \sqrt{5} \right) \\ \end{array} \)
N = 17
\( \begin{array}{rcl} \cos \frac{2\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{2(17-\sqrt{17})} + \sqrt{2(2(17+3\sqrt{17}) - (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \\ \cos \frac{4\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} - \sqrt{2(17-\sqrt{17})} + \sqrt{2(2(17+3\sqrt{17}) + (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \\ \cos \frac{6\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} + \sqrt{2(17+\sqrt{17})} + \sqrt{2(2(17-3\sqrt{17}) - (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{8\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{2(17-\sqrt{17})} - \sqrt{2(2(17+3\sqrt{17}) - (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \\ \cos \frac{10\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} + \sqrt{2(17+\sqrt{17})} - \sqrt{2(2(17-3\sqrt{17}) - (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{12\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} - \sqrt{2(17+\sqrt{17})} + \sqrt{2(2(17-3\sqrt{17}) + (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{14\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} - \sqrt{2(17+\sqrt{17})} - \sqrt{2(2(17-3\sqrt{17}) + (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{16\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} - \sqrt{2(17-\sqrt{17})} - \sqrt{2(2(17+3\sqrt{17}) + (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \end{array}\)
(I'm sure that there are no typos)
I won't give the N = 257 and N = 65537 cases, for rather obvious reasons.
Notice how one gets the two solutions for N = 5 by reversing the sign of the square root in it, and the eight solutions for N = 17 by reversing the signs of the three kinds of square root in it. The N = 17 case required solving three nested quadratic equations. How many in general:
[table="class:grid"]
[tr] [td]N[/td] [td]#[/td] [/tr]
[tr] [td]3[/td] [td]0[/td] [/tr]
[tr] [td]5[/td] [td]1[/td] [/tr]
[tr] [td]17[/td] [td]3[/td] [/tr]
[tr] [td]257[/td] [td]7[/td] [/tr]
[tr] [td]65537[/td] [td]15[/td] [/tr]
[/table]
Now to constructible polygons. I've found Constructibility of Regular Polygons, something that also discusses a more general version.
An N-gon is constructible with ruler and compass if it is possible to start with the rational numbers and do a finite number of arithmetic operations and square roots and get the Nth roots of unity:
cos(2π/N) + i*sin(2π/N)
One can find either of the two trig functions, and the other one from it using a well-known trig identity. That identity can easily be shown to satisfy that constructibility criterion.
Gauss and Wantzel had proved that this is only possible if N = 2n * product of non-repeated Fermat primes. Let's see how this works out for the lower Fermat primes.
N = 3
\( \cos \frac{2\pi}{3} = - \frac{1}{2} \)
N = 5
\( \begin{array}{rcl} \cos \frac{2\pi}{5} & = & \frac{1}{4} \left( -1 + \sqrt{5} \right) \\ \cos \frac{4\pi}{5} & = & \frac{1}{4} \left( -1 - \sqrt{5} \right) \\ \end{array} \)
N = 17
\( \begin{array}{rcl} \cos \frac{2\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{2(17-\sqrt{17})} + \sqrt{2(2(17+3\sqrt{17}) - (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \\ \cos \frac{4\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} - \sqrt{2(17-\sqrt{17})} + \sqrt{2(2(17+3\sqrt{17}) + (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \\ \cos \frac{6\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} + \sqrt{2(17+\sqrt{17})} + \sqrt{2(2(17-3\sqrt{17}) - (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{8\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} + \sqrt{2(17-\sqrt{17})} - \sqrt{2(2(17+3\sqrt{17}) - (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \\ \cos \frac{10\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} + \sqrt{2(17+\sqrt{17})} - \sqrt{2(2(17-3\sqrt{17}) - (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{12\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} - \sqrt{2(17+\sqrt{17})} + \sqrt{2(2(17-3\sqrt{17}) + (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{14\pi}{17} & = & \frac{1}{16} \left( -1 - \sqrt{17} - \sqrt{2(17+\sqrt{17})} - \sqrt{2(2(17-3\sqrt{17}) + (3-\sqrt{17})\sqrt{2(17+\sqrt{17})})} \right) \\ \cos \frac{16\pi}{17} & = & \frac{1}{16} \left( -1 + \sqrt{17} - \sqrt{2(17-\sqrt{17})} - \sqrt{2(2(17+3\sqrt{17}) + (3+\sqrt{17})\sqrt{2(17-\sqrt{17})})} \right) \end{array}\)
(I'm sure that there are no typos)
I won't give the N = 257 and N = 65537 cases, for rather obvious reasons.
Notice how one gets the two solutions for N = 5 by reversing the sign of the square root in it, and the eight solutions for N = 17 by reversing the signs of the three kinds of square root in it. The N = 17 case required solving three nested quadratic equations. How many in general:
[table="class:grid"]
[tr] [td]N[/td] [td]#[/td] [/tr]
[tr] [td]3[/td] [td]0[/td] [/tr]
[tr] [td]5[/td] [td]1[/td] [/tr]
[tr] [td]17[/td] [td]3[/td] [/tr]
[tr] [td]257[/td] [td]7[/td] [/tr]
[tr] [td]65537[/td] [td]15[/td] [/tr]
[/table]
lpetrich
Contributor
Constructibility of Regular Polygons discusses a generalization. If you allow a neusis or a marked ruler that pivots around some point and slides past it, you can get the solutions of cubic and quartic equations.
Doing a parallel constructibility analysis, one finds that the possible values of N are
N = 2n1*3n2 * product of non-repeated Pierpont primes.
A Pierpoint prime is a generalization of a Fermat prime: prime numbers with form 2n1*3n2 + 1.
Here are the lowest Pierpoint primes: 2, 3, 5, 7, 13, 17, 19, 37, 73, 97, 109, 163, 193, 257, 433, 487, 577, 769, ...
Here are the lowest non-Pierpoint ones: 11, 23, 29, 31, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 89, ...
So one can construct a regular heptagon (7-gon) with a neusis, or a regular 13-gon or a regular 19-gon, but not a regular 11-gon.
If one can use quintic (degree-5) polynomial equations, the starting prime products become 2n1*3n2*5n3, and one can find regular 11-gons, 31-gons, 41-gons, etc.
If one can use septic (degree-7) polynomial equations, one can find regular 29-gons, 43-gons, 71-gons, etc.
If one can use degree-11 polynomial equations, one can find regular 23-gons, 89-gons, etc.
Etc.
Doing a parallel constructibility analysis, one finds that the possible values of N are
N = 2n1*3n2 * product of non-repeated Pierpont primes.
A Pierpoint prime is a generalization of a Fermat prime: prime numbers with form 2n1*3n2 + 1.
Here are the lowest Pierpoint primes: 2, 3, 5, 7, 13, 17, 19, 37, 73, 97, 109, 163, 193, 257, 433, 487, 577, 769, ...
Here are the lowest non-Pierpoint ones: 11, 23, 29, 31, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 89, ...
So one can construct a regular heptagon (7-gon) with a neusis, or a regular 13-gon or a regular 19-gon, but not a regular 11-gon.
If one can use quintic (degree-5) polynomial equations, the starting prime products become 2n1*3n2*5n3, and one can find regular 11-gons, 31-gons, 41-gons, etc.
If one can use septic (degree-7) polynomial equations, one can find regular 29-gons, 43-gons, 71-gons, etc.
If one can use degree-11 polynomial equations, one can find regular 23-gons, 89-gons, etc.
Etc.
I had looked at the polygon question afew months back. I needed the dimensions of a hex nut. They are readily available in tables. I wondered what the relationship between faces and diameter was.
Given the diameter of the circle withan inscribed hexagon what is the length of each face?
A practical real world problem.
An electronic component is made of a rectangular molded plastic case with three circular pins protruding on one side. The component must be able to fit in three holes in a circuit board without bending any of the pins.
For a representative picture.
http://www.digikey.com/product-detail/en/66WR5KLF/987-1494-ND/3587252
Sketch a rectangle on paper with dimensions 0.4 x 0.2 inches. This is the outline of the component and assume they have no variation for the problem. The rectangle is afixed area on the board the part must fit in.
With [0,0] at the lower left corner of the rectangle draw points at
[0.1,0.2]
[0.2,0.2]
[0.3,0.2]
The points represent the theoreticalperfect center point of each pin relative to [0.0].
The pin center points above can vary due to manufacturing variation in x and y by as much as 0.01 inches.
The minimum and maximum diameters of the three pins are .018 and .022 inches.
What is he minimum diameter of holes ateach point such that the part can be inserted in the board withoutbent pins?
Given the diameter of the circle withan inscribed hexagon what is the length of each face?
A practical real world problem.
An electronic component is made of a rectangular molded plastic case with three circular pins protruding on one side. The component must be able to fit in three holes in a circuit board without bending any of the pins.
For a representative picture.
http://www.digikey.com/product-detail/en/66WR5KLF/987-1494-ND/3587252
Sketch a rectangle on paper with dimensions 0.4 x 0.2 inches. This is the outline of the component and assume they have no variation for the problem. The rectangle is afixed area on the board the part must fit in.
With [0,0] at the lower left corner of the rectangle draw points at
[0.1,0.2]
[0.2,0.2]
[0.3,0.2]
The points represent the theoreticalperfect center point of each pin relative to [0.0].
The pin center points above can vary due to manufacturing variation in x and y by as much as 0.01 inches.
The minimum and maximum diameters of the three pins are .018 and .022 inches.
What is he minimum diameter of holes ateach point such that the part can be inserted in the board withoutbent pins?
beero1000
Veteran Member
He's given the diameter of the inscribed, not the circumscribed circle. For any regular polygon, the side length s, inradius r and circumradius R are related by the Pythagorean theorem, 4R2 = 4r2 + s2.
On the contrary your response is a bit simplistic.
Just observing there are N equilateral triangles does not get you to a solution of the problem as stated, the relationship between diameter and length of a face.
I outlined it.
How would you reduce to a function l=f(d,n) where d is diameter, n is number of faces of the polygon, and l is the length of each face.
In my actual problem quantitatively given the length of a face what is the distance between opposing vertices were, or the diameter to ensure I had adequate clearance or the nut to turn.
Please, no help from the peanut gallery.