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!
December 30, 2008
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.)
Richey Numbers: An Overview
After reviewing this problem further for a while, I came to some new conclusions.
First, (I stole this one from a book by Paul J. Nahin), there is an alternate way to generate numbers of the form a^2+b^2 = c^2+d^2. His method uses the product of the sum of two squares:
(u^2+v^2)(x^2+y^2)
The theorem says that a product of this form (for integral values u, v, x and y) will always be equal to a number that can be expressed in two distinct ways as the sum of two squares. This can be verified by some algebra and the solutions can be found in terms of u, v, x and y.
This, along with a new ability I have maintained to use C++ code to find solutions to some of these number theory problems, led me to write a program that found solutions, and it works quite well. This leads me to another, quicker conclusion. The smallest Richey number that is not a multiple of 5 is 221, not 533 as I had said earlier. (221 = 5^2+14^2 = 10^2 + 11^2)
Finally, there is another element of these numbers that I have not yet studied. There appear some relationships between the numbers which compose my numbers. One example, which I noted last time, was the relationship that a+b-c-d was always an even number. There appear some more of these strange relationships, but I haven't yet determined anything with proof.
After reviewing this problem further for a while, I came to some new conclusions.
First, (I stole this one from a book by Paul J. Nahin), there is an alternate way to generate numbers of the form a^2+b^2 = c^2+d^2. His method uses the product of the sum of two squares:
(u^2+v^2)(x^2+y^2)
The theorem says that a product of this form (for integral values u, v, x and y) will always be equal to a number that can be expressed in two distinct ways as the sum of two squares. This can be verified by some algebra and the solutions can be found in terms of u, v, x and y.
This, along with a new ability I have maintained to use C++ code to find solutions to some of these number theory problems, led me to write a program that found solutions, and it works quite well. This leads me to another, quicker conclusion. The smallest Richey number that is not a multiple of 5 is 221, not 533 as I had said earlier. (221 = 5^2+14^2 = 10^2 + 11^2)
Finally, there is another element of these numbers that I have not yet studied. There appear some relationships between the numbers which compose my numbers. One example, which I noted last time, was the relationship that a+b-c-d was always an even number. There appear some more of these strange relationships, but I haven't yet determined anything with proof.
September 9, 2008
Taxicab Numbers Extended: Extended
I liked my extension of taxicab numbers so much that I've composed an addition to it.
These three formulas are ones I came up with to generate an infinite number of the numbers I talked about last time.
5(2n^2+2n+1)
5(n^2+4n+5)
13(2n^2+10n+13)
The first only works for n greater than or equal two, whereas the other two work for positive integral n.
These formulas were found by looking at the problem as a difference of squares problem, and from there as this: Finding two strings of consecutive odd integers that are equal.
The third formula also allowed me to disprove one of my earlier conjectures; that all of the numbers are multiples of 5. This formula found a whole set of them, the smallest of which is 533. (23^2 + 2^2 and 22^2 + 7^2.)
There is still more to do, since I haven't yet found a formula that will determine every single number of this type, though having an infinite number of them is, well, pretty damn cool.
I liked my extension of taxicab numbers so much that I've composed an addition to it.
These three formulas are ones I came up with to generate an infinite number of the numbers I talked about last time.
5(2n^2+2n+1)
5(n^2+4n+5)
13(2n^2+10n+13)
The first only works for n greater than or equal two, whereas the other two work for positive integral n.
These formulas were found by looking at the problem as a difference of squares problem, and from there as this: Finding two strings of consecutive odd integers that are equal.
The third formula also allowed me to disprove one of my earlier conjectures; that all of the numbers are multiples of 5. This formula found a whole set of them, the smallest of which is 533. (23^2 + 2^2 and 22^2 + 7^2.)
There is still more to do, since I haven't yet found a formula that will determine every single number of this type, though having an infinite number of them is, well, pretty damn cool.
July 26, 2008
Taxicab Numbers Extended
Taxicab numbers are numbers that can be written in two distinct ways as the sum of two cubes, the smallest of which is 1729. (1729 = 1^3 + 12^3 = 9^3 + 10^3.)
I decided it would be a good idea to try and find numbers that can be written as the sum of two squares in two distinct ways, and so far I have found 18 of them. In ascending order,
50, 65, 85, 125, 130, 145, 170, 185, 205, 250, 260, 265, 290, 305, 325, 340, 365, 370.
Unfortunately, I haven't determined a uniform way to find these numbers- so, I've just been checking all the multiples of 5. (I checked every number between 1 and 100, and noticing that numbers that satisfy k = 0 mod 5 are the only ones that work, I only checked multiples of 5 from then on.) So, there could be some missing in between. (There might even be some multiples of 5 I missed in the 300's because I kind of got lazy and started writing down the ones that worked and not checking the ones I was pretty sure didn't work.) I also have no idea why only numbers that are multiples of 5 would work, but I have a vague idea.
If we look at the digit that ends up in the digits place when you square the 10 different 1 digit numbers, we get 0, 1, 4, 9, 6, 5, 6, 9, 4, 1. There are 6 digits you can end up with, namely 0, 1, 4, 5, 6 and 9. Combining these, you can make any digit. However, most can only be made with one or two combinations of numbers- 5 and 0 can be made with 3 different combinations. (5 = 1+4 = 0+5 = 9+6, and 0 = 0+0 = 1+9 = 4+6.) This makes these two digits the most likely to be in the digits place, but it is no help in proving that they will be the only digits in the digits place of my numbers.
Another more obscure pattern that I noticed is in the numbers that you square to create my numbers. For example, 50 = 5^2 + 5^2 = 7^2 + 1^2. Notice that 5+5 = 10 is 2 greater than 7+1 = 8. In other words, for a number k that can be written as both a^2 + b^2 and c^2 + d^2 for positive integers a, b, c and d and for a+b > c+d, a+b-c-d = 2. However, this only holds for the first 5- after that, the difference jumps to either 4 or 6 or back to 2. Although less specific, another rule can form- perhaps a+b-c-d = 0 mod 2.
The numbers seem to show up almost randomly, and the problem is a relatively unfamiliar one so I'm not sure where to go from here.
Taxicab numbers are numbers that can be written in two distinct ways as the sum of two cubes, the smallest of which is 1729. (1729 = 1^3 + 12^3 = 9^3 + 10^3.)
I decided it would be a good idea to try and find numbers that can be written as the sum of two squares in two distinct ways, and so far I have found 18 of them. In ascending order,
50, 65, 85, 125, 130, 145, 170, 185, 205, 250, 260, 265, 290, 305, 325, 340, 365, 370.
Unfortunately, I haven't determined a uniform way to find these numbers- so, I've just been checking all the multiples of 5. (I checked every number between 1 and 100, and noticing that numbers that satisfy k = 0 mod 5 are the only ones that work, I only checked multiples of 5 from then on.) So, there could be some missing in between. (There might even be some multiples of 5 I missed in the 300's because I kind of got lazy and started writing down the ones that worked and not checking the ones I was pretty sure didn't work.) I also have no idea why only numbers that are multiples of 5 would work, but I have a vague idea.
If we look at the digit that ends up in the digits place when you square the 10 different 1 digit numbers, we get 0, 1, 4, 9, 6, 5, 6, 9, 4, 1. There are 6 digits you can end up with, namely 0, 1, 4, 5, 6 and 9. Combining these, you can make any digit. However, most can only be made with one or two combinations of numbers- 5 and 0 can be made with 3 different combinations. (5 = 1+4 = 0+5 = 9+6, and 0 = 0+0 = 1+9 = 4+6.) This makes these two digits the most likely to be in the digits place, but it is no help in proving that they will be the only digits in the digits place of my numbers.
Another more obscure pattern that I noticed is in the numbers that you square to create my numbers. For example, 50 = 5^2 + 5^2 = 7^2 + 1^2. Notice that 5+5 = 10 is 2 greater than 7+1 = 8. In other words, for a number k that can be written as both a^2 + b^2 and c^2 + d^2 for positive integers a, b, c and d and for a+b > c+d, a+b-c-d = 2. However, this only holds for the first 5- after that, the difference jumps to either 4 or 6 or back to 2. Although less specific, another rule can form- perhaps a+b-c-d = 0 mod 2.
The numbers seem to show up almost randomly, and the problem is a relatively unfamiliar one so I'm not sure where to go from here.
June 25, 2008
Confusing Myself!
I thought up a problem that I can't solve. Here it is:
Find solutions in integers n and x to the equation n^(x+2) + (n+1)^(x+1) = (n+2)^x
I found one solution, namely (n, x) = (1, 2).
However, for greater n's, the problem becomes nearly impossible to look at with number sense alone. (At any level, it can't be solved algebraically.) I've used Mathematica to determine that there aren't solutions up through n=19, and beyond there x has to be huge. Using the little number theory I know, I determined that all the solutions have to be for odd n. Beyond that, for a specific n, it is possible to find what form x has to take. (For example, when n is 9, x has to be even. This can be determined by looking at the digits value of the powers of 9 and 10, and seeing whether they sum to 1, the digits value of all powers of 11. You end up seeing that the powers of 9 are equal to 1 when x is even, and since the digits value of the powers of 10 is always 0 the sum of the LHS has the same digits value as the RHS, making solutions possible only for those values of x.)
I've also narrowed the solutions down further, and at this point I'm just trying to determine criteria that solutions will have to meet so they can be more easily determined, if they exist at all. (At this point, the outlook on finding more solutions is pretty bleak- I'm feeling pretty confident that there aren't any others- however, I've got to prove it and at the moment that looks to be a pretty difficult task.)
I thought up a problem that I can't solve. Here it is:
Find solutions in integers n and x to the equation n^(x+2) + (n+1)^(x+1) = (n+2)^x
I found one solution, namely (n, x) = (1, 2).
However, for greater n's, the problem becomes nearly impossible to look at with number sense alone. (At any level, it can't be solved algebraically.) I've used Mathematica to determine that there aren't solutions up through n=19, and beyond there x has to be huge. Using the little number theory I know, I determined that all the solutions have to be for odd n. Beyond that, for a specific n, it is possible to find what form x has to take. (For example, when n is 9, x has to be even. This can be determined by looking at the digits value of the powers of 9 and 10, and seeing whether they sum to 1, the digits value of all powers of 11. You end up seeing that the powers of 9 are equal to 1 when x is even, and since the digits value of the powers of 10 is always 0 the sum of the LHS has the same digits value as the RHS, making solutions possible only for those values of x.)
I've also narrowed the solutions down further, and at this point I'm just trying to determine criteria that solutions will have to meet so they can be more easily determined, if they exist at all. (At this point, the outlook on finding more solutions is pretty bleak- I'm feeling pretty confident that there aren't any others- however, I've got to prove it and at the moment that looks to be a pretty difficult task.)
June 1, 2008
Here's a quick and simple one.
The arithmetic average of 2kn^2 and 2k(n+1)^2 is always k greater than their geometric average for positive integers k and n.
There wasn't a uniform way to find those 2 numbers- I had to do it by finding a bunch that worked and looking for a pattern.
Note: The arithmetic average of 2 numbers a and b is (a+b)/2, and their geometric average is the Sqrt(a*b).
The arithmetic average of 2kn^2 and 2k(n+1)^2 is always k greater than their geometric average for positive integers k and n.
There wasn't a uniform way to find those 2 numbers- I had to do it by finding a bunch that worked and looking for a pattern.
Note: The arithmetic average of 2 numbers a and b is (a+b)/2, and their geometric average is the Sqrt(a*b).
April 20, 2008
I decided I'd share a cute little power-summation formula I came up with.
Sum[k=0, n-1](x^k) = (x^(n)-1)/(x-1)
I came across it by accident. I noticed that the formula works for the x=2 case, and tried it for x=3. However, I excluded the (x-1) term since I hadn't yet discovered it- for x=2, it disappears. (It's just a 1.) So, I took a shot and added the term- and it worked! Magic!
Nothing special, but it's cool the way the terms cancel. (If you write out the sum and multiply by x-1, you can see that all but the first and last terms cancel out.)
Sum[k=0, n-1](x^k) = (x^(n)-1)/(x-1)
I came across it by accident. I noticed that the formula works for the x=2 case, and tried it for x=3. However, I excluded the (x-1) term since I hadn't yet discovered it- for x=2, it disappears. (It's just a 1.) So, I took a shot and added the term- and it worked! Magic!
Nothing special, but it's cool the way the terms cancel. (If you write out the sum and multiply by x-1, you can see that all but the first and last terms cancel out.)
April 16, 2008
Newest Interest: Relatively prime #'s. (Quick definition in case you don't know: Relatively prime numbers are numbers that have no common factors except for 1. Relatively prime numbers don't have to be prime; for example, 6 and 35 are relatively prime, although neither one is prime itself.)
First and foremost, I addressed this question: Is there an infinite number of numbers to which a given whole n is relatively prime?
The answer is yes. Although the proof is easy, I don't want to write it here. The idea is this: For any number, there is always a prime that is greater than that number. (By the infinity of primes theorem.) Since the prime number that is greater then our n has not common factors with n, they are relatively prime.
So, to specify the problem further, I asked this: For any given whole n, is there an infinite number of composite numbers to which it is relatively prime?
This is a little more difficult to solve, and I have proven it for n=odd. Here's how:
Assumption: For a given odd n, rn-1 will always be relatively prime to n for odd r.
Since r can be written as 2m+1 and n as 2k-1, rn-1 = (2k-1)(2m+1)-1 = 4km+2k-2m-1-1 = 2(2km+k-m-1), which is even since it has a factor of 2. Because rn-1 is composite, it fits our requirement. Also, it is not a multiple of n. (rn is a multiple of n, but the multiple of n before rn is (r-1)n or rn-n, which is clearly less than rn-1. So, the multiples of n "skip over" rn-1, and it isn't a multiple of n.) Since rn-1 is not a multiple of n, they share no common factors, and therefore it works! So, any number of the form 2(2km+k-m-1) for whole m and k will always be relatively prime to 2k-1.
The part I have not solved is that for even n's. If I approach it the same way, the issue becomes this- the number rn-1 could be a prime number, and that's not okay. Well, what about rn+1? That too isn't a multiple of n. Unfortunately, it too could be prime. (If it were, then rn-1 and rn+1 would be twin primes.) Because there is no proof regarding the infinity of twin primes, there could be an infinite string of twin primes of the form rn+z, with some odd number z, that would prevent this theorem from working for any and all even n.
Suggestions, as always, are welcome.
First and foremost, I addressed this question: Is there an infinite number of numbers to which a given whole n is relatively prime?
The answer is yes. Although the proof is easy, I don't want to write it here. The idea is this: For any number, there is always a prime that is greater than that number. (By the infinity of primes theorem.) Since the prime number that is greater then our n has not common factors with n, they are relatively prime.
So, to specify the problem further, I asked this: For any given whole n, is there an infinite number of composite numbers to which it is relatively prime?
This is a little more difficult to solve, and I have proven it for n=odd. Here's how:
Assumption: For a given odd n, rn-1 will always be relatively prime to n for odd r.
Since r can be written as 2m+1 and n as 2k-1, rn-1 = (2k-1)(2m+1)-1 = 4km+2k-2m-1-1 = 2(2km+k-m-1), which is even since it has a factor of 2. Because rn-1 is composite, it fits our requirement. Also, it is not a multiple of n. (rn is a multiple of n, but the multiple of n before rn is (r-1)n or rn-n, which is clearly less than rn-1. So, the multiples of n "skip over" rn-1, and it isn't a multiple of n.) Since rn-1 is not a multiple of n, they share no common factors, and therefore it works! So, any number of the form 2(2km+k-m-1) for whole m and k will always be relatively prime to 2k-1.
The part I have not solved is that for even n's. If I approach it the same way, the issue becomes this- the number rn-1 could be a prime number, and that's not okay. Well, what about rn+1? That too isn't a multiple of n. Unfortunately, it too could be prime. (If it were, then rn-1 and rn+1 would be twin primes.) Because there is no proof regarding the infinity of twin primes, there could be an infinite string of twin primes of the form rn+z, with some odd number z, that would prevent this theorem from working for any and all even n.
Suggestions, as always, are welcome.
February 4, 2008
January 27, 2008
Here's an interesting polynomial pattern I picked up.
Sum[n, k=1](1) = n
Sum[n, k=1](2k-1) = n^2
Sum[n, k=1](3k^2-3k+1) = n^3
Sum[n, k=1](4k^3-6k^2+4k-1) = n^4
Sum[n, k=1](5k^4-10k^3+10k^2-5k+1) = n^5
Sum[n, k=1](6k^5-15k^4+20k^3-15k^2+6k-1) = n^6
Can you see the pattern?
.....
Well, although I'm having a big fat load of trouble nesting this polynomial pattern, it can be understood this way: n^x = Sum[n, k=1](Natural(n) - Triangle(n) + Tetrahedral(n) -... +/- 1), where Natural(n) is the nth natural number, Triangle(n) is the nth triangle number, and so on. (I'll explain what triangle numbers and tetrahedral numbers are in a second. {Not to offend those who already know what they are.})
The "..." in this formula represents an extension of the natural, triangle, tetrahedral... sequence. Natural numbers can be written in this way:
The nth natural number = n
Triangle numbers can be written in this way:
The nth triangle number = (n(n+1))/2
Tetrahedral numbers can be written in this way:
The nth tetrahedral number = (n(n+1)(n+2))/6
The relationship between these sequences is such- the nth number in the kth kind of number sequence (where natural numbers correspond to k=0, triangles correspond to k=1, and so on) is equal to Product[k, i=0](n+i)/[(k+1)!]
I'm sure I can use combinations in the nesting process instead of this series, (since combinations are closely related to Pascal's Triangle, where these numbers are directly taken from), but I haven't yet figured out how.
Any form of assistance is appreciated.
Sum[n, k=1](1) = n
Sum[n, k=1](2k-1) = n^2
Sum[n, k=1](3k^2-3k+1) = n^3
Sum[n, k=1](4k^3-6k^2+4k-1) = n^4
Sum[n, k=1](5k^4-10k^3+10k^2-5k+1) = n^5
Sum[n, k=1](6k^5-15k^4+20k^3-15k^2+6k-1) = n^6
Can you see the pattern?
.....
Well, although I'm having a big fat load of trouble nesting this polynomial pattern, it can be understood this way: n^x = Sum[n, k=1](Natural(n) - Triangle(n) + Tetrahedral(n) -... +/- 1), where Natural(n) is the nth natural number, Triangle(n) is the nth triangle number, and so on. (I'll explain what triangle numbers and tetrahedral numbers are in a second. {Not to offend those who already know what they are.})
The "..." in this formula represents an extension of the natural, triangle, tetrahedral... sequence. Natural numbers can be written in this way:
The nth natural number = n
Triangle numbers can be written in this way:
The nth triangle number = (n(n+1))/2
Tetrahedral numbers can be written in this way:
The nth tetrahedral number = (n(n+1)(n+2))/6
The relationship between these sequences is such- the nth number in the kth kind of number sequence (where natural numbers correspond to k=0, triangles correspond to k=1, and so on) is equal to Product[k, i=0](n+i)/[(k+1)!]
I'm sure I can use combinations in the nesting process instead of this series, (since combinations are closely related to Pascal's Triangle, where these numbers are directly taken from), but I haven't yet figured out how.
Any form of assistance is appreciated.
January 12, 2008
January 9, 2008
Happy New Year!
Well, after taking a short break for the holidays I'm back on track with a new strange discovery. I noticed that the infinite sum Sum(infinity, n=0) {1/(e^n)} = (1+rt(5))/2, or the golden ratio. (While trying to evaluate the sum with raw computing power on my calculator I noticed that the terms completely stopped growing at around n=80 or so.) I can't seem to come up with any reason that e and the golden ratio would be related, and the constant I reached could be close to the value of the golden ratio simply by coincidence. (This actually seems pretty likely because the when the terms stopped growing the constant was slightly lower than the golden ratio- 1.581976707. {The golden ratio is equal to 1.618033989.})
Any ideas?
Well, after taking a short break for the holidays I'm back on track with a new strange discovery. I noticed that the infinite sum Sum(infinity, n=0) {1/(e^n)} = (1+rt(5))/2, or the golden ratio. (While trying to evaluate the sum with raw computing power on my calculator I noticed that the terms completely stopped growing at around n=80 or so.) I can't seem to come up with any reason that e and the golden ratio would be related, and the constant I reached could be close to the value of the golden ratio simply by coincidence. (This actually seems pretty likely because the when the terms stopped growing the constant was slightly lower than the golden ratio- 1.581976707. {The golden ratio is equal to 1.618033989.})
Any ideas?
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