New Findings
I have cataloged all of the Richey Numbers between 1 and 1000, and have come to some new conclusions.
First, I noticed that given a solution, N = a^2 + b^2 = c^2 + d^2, 2N is also a solution, and with some intuition I realized that 2N = (a+b)^2 + (a-b)^2 = (c+d)^2 + (c-d)^2. The solution I worked with first was 65- I noticed that 130 was also a solution.
65 = 8^2 + 1^2 = 4^2 + 7^2
130 = 9^2 + 7^2 = 11^2 + 3^2
As you may notice, 9 is the sum of 8 and 1, 7 is the difference of 8 and 1, 11 is the sum of 7 and 4, and 3 is the difference of 7 and 4.
This result is easily proven.
N = a^2 + b^2
2N = (a+b)^2 + (a-b)^2
= a^2 + 2ab + b^2 + a^2 - 2ab + b^2
= 2a^2 + 2b^2
= 2N
The same also follows for the other part, c^2 + d^2.
The next thing I began experimenting with was the prime factorizations of Richey Numbers. I looked mainly at the numbers I thought to be "pure"; those that are not iterations of other Richey Numbers. For example- 130 is only 65*2, and the prime factorization gave that of 65 with a 2 in it as well. Other examples include any Richey Number multiplied by a square #, because of another property I discovered: Given a solution N = a^2 + b^2 = c^2 + d^2, (k^2)N = (ka)^2+(kb)^2 = (kc)^2 + (kd)^2, an easily verifiable result. Therefore, numbers like 200, 260 and 450 are reiterations: 50*4, 65*4, 50*9 respectively. This finding, coupled with the principle noted above, also led me to include Pythagorean triplets into my definition of Richey Numbers. (This basically means that I would allow a, b, c, or d to be zero, something I hadn't allowed before.) I realized that if a^2+b^2 = c^2 was also considered as a solution, that may explain the prime factorizations of some of the numbers I already came up with, such as 50 and 388. This is because their prime factorizations turned out to be 2*k^2, for some integer k- in these cases, 2*5^2 and 2*169^2 respectively. At first, I was puzzled as to how something like 25, whose components (3^2+4^2 = 5^2) include a zero (in terms of Richey Numbers) could lead to a legitimate Richey Number. I quickly realized that when multiplying 25 by 2 and using the rule I stated above, the sum of 5 and 0 and their difference are both 5- both non-zero numbers, thus yielding 50, a more recognizeable Richey Number.
This analysis left four different kinds of prime factorizations, and I list them here in order of frequency, most frequent to least (N is a given Richey Number):
1. N = 5*Pk, where Pk is some prime in a specific set of primes.
2. N = Pa*Pb, where Pa and Pb are primes of the same set as in 1.
3. N = Pk^2, where Pk is a given prime in the same set as in 1.
4. N = 5^x, given a whole x>1.
The "specific set of primes" I refer to here was found only by studying the prime factorizations. These pure Richey Numbers' prime factorizations used these primes only, and I believe that there is some general relationship all these primes share (though I haven't yet found it). Here's a list of the first few pure Richey Numbers, their prime factorizations, and the categories they fall into:
25 = 5^2 (4)
65 = 5*13 (1)
85 = 5*17 (1)
125 = 5^3 (4)
145 = 5*29 (1)
169 = 13^2 (3)
185 = 5*37 (1)
205 = 5*41 (1)
221 = 13*17 (2)
265 = 5*53 (1)
...
And the list of primes:
13,17,29,37,41,53,61,73,89,97,101,109,113,137,149,157,173,181,193,197...
I hope this will lead somewhere interesting, as this list of oddly spaced primes does seem, well, odd.
January 3, 2009
December 30, 2008
Taking a Break From Number Theory
I came up with a formula that has to do with extending the arcsine function beyond limited input. My interest was sparked when I noticed that there is a simple pattern among the commonly known values for sine:
sin(π/6) = sqrt(1)/2 = 1/2
sin(π/4) = sqrt(2)/2
sin(π/3) = sqrt(3)/2
sin(π/2) = sqrt(4)/2 = 1
As the angles increase, the value for sin(Θ) increases as sqrt(n)/2, for n = 1,2,3,4. I then asked: For what angle will the sine of the angle be sqrt(5)/2?
Clearly, the angle must be imaginary: the range of sine for real input is [-1,1]. Here's how I went about solving it. (The solution I give here will give the angle for any given n, not just 5- though that is how I started out.)
I started with the identity
sin(Θ) = [e^(iΘ) - e^(-iΘ)]/2i
(If you are not familiar with this, it is easily derived with the knowledge that
e^iΘ = cosΘ + isinΘ- I'll leave the proof of sin(Θ) = [e^(iΘ) - e^(-iΘ)]/2i up to you.)
Continuing,
i*Sqrt(n) = e^iΘ - e^-iΘ, since sin(Θ) = Sqrt(n)/2
e^2iΘ - i*Sqrt(n)*e^iΘ -1 = 0
This is quadratic in e^iΘ, so
e^iΘ = {i*Sqrt(n) +- Sqrt(4-n)}/2
Θ = -i*log({iSqrt(n)+/- Sqrt(4-n)}/2)
This says that
arcsin[Sqrt(n)/2] = -i*log({iSqrt(n)+/- Sqrt(4-n)}/2)
Or, substituting n = 4k^2,
arcsin[k] = -i*log(ik+-Sqrt(1-k^2))
This is essentially an analytic continuation of the arcsin function, whose domain is usually only [-1,1]; this continuation defines arcsin(k) for any and all k, including complex values. You may notice that for values of k on [-1,1], the Sqrt(1-k^2) part is real, and within the log we are left with a familiar quantity. For example, take k = 1/2. Then
arcsin[1/2] = -i*log(i/2 + Sqrt(3)/4)
Since i/2 + Sqrt(3)/4 = e^(iπ/6),
arcsin[1/2] = -i*iπ/6
arcsin[1/2] = π/6, which is true.
For my original example of k = Sqrt(5)/2, the angle Θ is -i*log(iφ), where φ is phi, the golden ratio, so
sin(-ilogiφ) = sqrt(5)/2, an elegant result.
Beyond just arcsin, I also created analytic continuations for arccos and arctan, both derived in almost the exactly same manner.
arccos[k] = -i*log(k +- Sqrt(k^2-1)), derived from cos Θ = (e^iΘ + e^-iΘ)/2
arctan[k] = -i/2*log{(1+ik)/(1-ik)}, derived from tan Θ = (e^iΘ-e^-iΘ)/(ie^iΘ+ie^-iΘ)
Happy New Year!
I came up with a formula that has to do with extending the arcsine function beyond limited input. My interest was sparked when I noticed that there is a simple pattern among the commonly known values for sine:
sin(π/6) = sqrt(1)/2 = 1/2
sin(π/4) = sqrt(2)/2
sin(π/3) = sqrt(3)/2
sin(π/2) = sqrt(4)/2 = 1
As the angles increase, the value for sin(Θ) increases as sqrt(n)/2, for n = 1,2,3,4. I then asked: For what angle will the sine of the angle be sqrt(5)/2?
Clearly, the angle must be imaginary: the range of sine for real input is [-1,1]. Here's how I went about solving it. (The solution I give here will give the angle for any given n, not just 5- though that is how I started out.)
I started with the identity
sin(Θ) = [e^(iΘ) - e^(-iΘ)]/2i
(If you are not familiar with this, it is easily derived with the knowledge that
e^iΘ = cosΘ + isinΘ- I'll leave the proof of sin(Θ) = [e^(iΘ) - e^(-iΘ)]/2i up to you.)
Continuing,
i*Sqrt(n) = e^iΘ - e^-iΘ, since sin(Θ) = Sqrt(n)/2
e^2iΘ - i*Sqrt(n)*e^iΘ -1 = 0
This is quadratic in e^iΘ, so
e^iΘ = {i*Sqrt(n) +- Sqrt(4-n)}/2
Θ = -i*log({iSqrt(n)+/- Sqrt(4-n)}/2)
This says that
arcsin[Sqrt(n)/2] = -i*log({iSqrt(n)+/- Sqrt(4-n)}/2)
Or, substituting n = 4k^2,
arcsin[k] = -i*log(ik+-Sqrt(1-k^2))
This is essentially an analytic continuation of the arcsin function, whose domain is usually only [-1,1]; this continuation defines arcsin(k) for any and all k, including complex values. You may notice that for values of k on [-1,1], the Sqrt(1-k^2) part is real, and within the log we are left with a familiar quantity. For example, take k = 1/2. Then
arcsin[1/2] = -i*log(i/2 + Sqrt(3)/4)
Since i/2 + Sqrt(3)/4 = e^(iπ/6),
arcsin[1/2] = -i*iπ/6
arcsin[1/2] = π/6, which is true.
For my original example of k = Sqrt(5)/2, the angle Θ is -i*log(iφ), where φ is phi, the golden ratio, so
sin(-ilogiφ) = sqrt(5)/2, an elegant result.
Beyond just arcsin, I also created analytic continuations for arccos and arctan, both derived in almost the exactly same manner.
arccos[k] = -i*log(k +- Sqrt(k^2-1)), derived from cos Θ = (e^iΘ + e^-iΘ)/2
arctan[k] = -i/2*log{(1+ik)/(1-ik)}, derived from tan Θ = (e^iΘ-e^-iΘ)/(ie^iΘ+ie^-iΘ)
Happy New Year!
December 9, 2008
Pell's Equation Rediscovered
One interesting equation I decided to study (for a reason I can't remember) was 2x^2+1=y^2, for positive integers x and y. I came across some very interesting conclusions in the relationships between x and y and patterns as the solutions grow larger and larger.
First, however, I would like to note that this equation has already been studied, quite extensively, by quite a few mathemetitions. It is known as Pell's Equation, and in a more general form is written as y^2-Nx^2 = 1. Traditionally it has been studied by looking at the smallest solution for x and y given a specific N. I approached it in a different manner, and found some relationships they seem to have missed. (I also eventually generalized the problem to Nx^2+1 = y^2 after getting bored with just N=2. I also tried out Nx^2 + m = y^2, given the flexibility that C++ allows.)
There were a few provable relationships. First, I noticed that as the solutions for x and y got larger and larger, there seemed to be some constant ratio between the two. (I considered y/x.) By approximation on my calculator, It seemed to approach the square root of 2. Fascinated, I realized that one could solve the given equation for y, yielding y = Sqrt[2x^2+1]. Then, all I had to do was take the limit as x goes to infinity of y/x, or
Lim[x-->infinity] (Sqrt[2x^2+1]/x)
Which is easily simplified to the square root of two.
Another strange set of relationships I noticed was between consecutive solutions. For example, consider one solution to the equation in (x,y); call it (a,b). Then, the solution immeadietly following that one is (c,d). As it turns out, a+b = d-c. I found this a striking result, and still I can't think of a way to relate a given solution to the solutions before and after it.
Determined to find something useful, I pressed on and came across some extremely far-fetched patterns I never would have suspected. It starts out simple. For each solution (x,y), x+y can be written as the sum of two square numbers. (I got excited, expecting some connection to Richey numbers; however, while some of these numbers can in fact be written as the sum of two squares in multiple ways, I haven't yet established a notable connection.) Then, those numbers began appearing all over the page- in old solutions! The relationship is quite difficult to explain, (in fact there are two), so I'll write out the solutions I looked at to formulate it.
(2,3) => Sum = 5 = 1^2 + 2^2
(12,17) => Sum = 29 = 2^2 + 5^2
(70,99) => Sum = 169 = 5^2 + 12^2
(408,577) => Sum = 985 = 12^2 + 29^2
(2378,3363) => Sum = 5741 = 29^2 + 70^2
(13860,19601) => Sum = 33461 = 70^2 + 169^2
...
I've bolded the important numbers, and it isn't difficult to see where the squared terms that compose the sum come from. The first, easier relationship is this: one of those terms comes from the sum of the solution before it. (I've written it in a way where you can see this: the 2 comes down from the first solution, the 5 comes down from the second solution, and so on.) However, you may notice where the "new" numbers are coming from. An easy way to put it is that the sum is re-written as u^2+v^2 in the (2n)th solution, where u is the x of the nth solution and v is the sum x+y of the nth solution. Also, the intermediate step (the 2n-1st solution) takes u from the sum of the nth solution and v from the x term of the n+1st solution. Though it may be difficult to sort out, the relationship is there.
There is more! Another, yet odder pattern I noticed is one concerning only the x part of the solutions. I saw that there seemed to be some common ratio of the x of the nth solution and the x of the n-1st solution, that Lim[x-->](Xn/Xn-1) == Some quantifiable value. Using calculators and some intuition, I found that this ratio (for N=2) is 3+2*Sqrt[2]. Furthermore, some ratio appears in the sequence of x solutions for all equations with a given N- for N=3 the ratio was Sqrt[3]+2; for N=5 it was 8*Sqrt[5], and so on. (N cannot equal a square number, because when it does there are no solutions to Nx^2+1 = y^2. I have not yet proven this result, but someone has (according to Mathworld) and I do think it is interesting.) Finally, these ratios do have somthing of a common pattern. Each one, given a particular N, contains the square root of N in it; it appears in the form a + b*Sqrt[N] for integral a and b. (Also, as far as I've noticed, a and b happen to be positive.)
Another part, which I haven't looked at in depth just yet, is something mentioned on wolfram's mathworld site. It says there that the solutions to the equation can in some way be quantified by continued fractions. I don't know much about continued fractions, and I plan on looking into it to see if anything can help me out.
(Check out http://mathworld.wolfram.com/PellEquation.html.)
One interesting equation I decided to study (for a reason I can't remember) was 2x^2+1=y^2, for positive integers x and y. I came across some very interesting conclusions in the relationships between x and y and patterns as the solutions grow larger and larger.
First, however, I would like to note that this equation has already been studied, quite extensively, by quite a few mathemetitions. It is known as Pell's Equation, and in a more general form is written as y^2-Nx^2 = 1. Traditionally it has been studied by looking at the smallest solution for x and y given a specific N. I approached it in a different manner, and found some relationships they seem to have missed. (I also eventually generalized the problem to Nx^2+1 = y^2 after getting bored with just N=2. I also tried out Nx^2 + m = y^2, given the flexibility that C++ allows.)
There were a few provable relationships. First, I noticed that as the solutions for x and y got larger and larger, there seemed to be some constant ratio between the two. (I considered y/x.) By approximation on my calculator, It seemed to approach the square root of 2. Fascinated, I realized that one could solve the given equation for y, yielding y = Sqrt[2x^2+1]. Then, all I had to do was take the limit as x goes to infinity of y/x, or
Lim[x-->infinity] (Sqrt[2x^2+1]/x)
Which is easily simplified to the square root of two.
Another strange set of relationships I noticed was between consecutive solutions. For example, consider one solution to the equation in (x,y); call it (a,b). Then, the solution immeadietly following that one is (c,d). As it turns out, a+b = d-c. I found this a striking result, and still I can't think of a way to relate a given solution to the solutions before and after it.
Determined to find something useful, I pressed on and came across some extremely far-fetched patterns I never would have suspected. It starts out simple. For each solution (x,y), x+y can be written as the sum of two square numbers. (I got excited, expecting some connection to Richey numbers; however, while some of these numbers can in fact be written as the sum of two squares in multiple ways, I haven't yet established a notable connection.) Then, those numbers began appearing all over the page- in old solutions! The relationship is quite difficult to explain, (in fact there are two), so I'll write out the solutions I looked at to formulate it.
(2,3) => Sum = 5 = 1^2 + 2^2
(12,17) => Sum = 29 = 2^2 + 5^2
(70,99) => Sum = 169 = 5^2 + 12^2
(408,577) => Sum = 985 = 12^2 + 29^2
(2378,3363) => Sum = 5741 = 29^2 + 70^2
(13860,19601) => Sum = 33461 = 70^2 + 169^2
...
I've bolded the important numbers, and it isn't difficult to see where the squared terms that compose the sum come from. The first, easier relationship is this: one of those terms comes from the sum of the solution before it. (I've written it in a way where you can see this: the 2 comes down from the first solution, the 5 comes down from the second solution, and so on.) However, you may notice where the "new" numbers are coming from. An easy way to put it is that the sum is re-written as u^2+v^2 in the (2n)th solution, where u is the x of the nth solution and v is the sum x+y of the nth solution. Also, the intermediate step (the 2n-1st solution) takes u from the sum of the nth solution and v from the x term of the n+1st solution. Though it may be difficult to sort out, the relationship is there.
There is more! Another, yet odder pattern I noticed is one concerning only the x part of the solutions. I saw that there seemed to be some common ratio of the x of the nth solution and the x of the n-1st solution, that Lim[x-->](Xn/Xn-1) == Some quantifiable value. Using calculators and some intuition, I found that this ratio (for N=2) is 3+2*Sqrt[2]. Furthermore, some ratio appears in the sequence of x solutions for all equations with a given N- for N=3 the ratio was Sqrt[3]+2; for N=5 it was 8*Sqrt[5], and so on. (N cannot equal a square number, because when it does there are no solutions to Nx^2+1 = y^2. I have not yet proven this result, but someone has (according to Mathworld) and I do think it is interesting.) Finally, these ratios do have somthing of a common pattern. Each one, given a particular N, contains the square root of N in it; it appears in the form a + b*Sqrt[N] for integral a and b. (Also, as far as I've noticed, a and b happen to be positive.)
Another part, which I haven't looked at in depth just yet, is something mentioned on wolfram's mathworld site. It says there that the solutions to the equation can in some way be quantified by continued fractions. I don't know much about continued fractions, and I plan on looking into it to see if anything can help me out.
(Check out http://mathworld.wolfram.com/PellEquation.html.)
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