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.)
April 20, 2008
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.
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