| The Riemann Hypothesis - A 'Complex' Problem |
By LinkZelda
Posted 30 Jul 2012 08:14 Category: Miscellaneous |
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Before we start, we have to go through some set theoretical assumptions. A negative number can't have a square root since minus times minus gives plus. Therefore we allow the square root of minus one to equal 'i' (in engineering it's actually 'j'). 2i is the square root of -4 and so on.
A few mathematicians noticed that the number of primes up to x (any number you want) is approximately Π(x) = x / log x (Yes Π is a function. The number is the lower case one: π) Note that when I say Log x I mean to base e rather than base 10. Research logarithms if that made no sense. Gauss found an even better approximation : the logarithmic integral (denoted by Li[x]) Li(x) = Int of dx / log x This is actually a definite integral which I can't write properly. Basically x goes on the top and 2 on the bottom. This was actually difficult to prove. One approach is complex analysis (mathematicians always look at the hard way). A complex number is a real number (anything on the conventional number line) combined with an imaginary number (a square root of a minus number) An example is 2+i (2 plus the square root of -1). How do prime numbers lead to complex numbers? Every number is a product of primes. For example, 30 = 2 x 3 x 5 and 60 = 2^2 x 3 x 5. Interestingly, (1+2+2^2+3^3...) x (1+3+3^2+3^3...) x (1+5+5^2+5^3...), where each bracket goes on forever and each bracket is the next prime number, is equal to 1+2+3+4... Obviously this is useless as it equals infinity, so by using a power of -s (minus makes them into a fraction) and make s large enough then multiplying each power in the bracket nonsense by -s, we get: 1^-s + 2^-s + 3^-s + 4^-s... which converges to a number less than infinity. This is true because 60^-s = 2^-2s x 3^-s x 5^-s. The formula makes even more sense if we let s equal a complex number (really now?) So s= a + ib if a > 1 Riemann's zeta function was basically ζ(s)= a + ib. Then he wrote a big formula for the exact number of primes preceding 'x'. It's to complicated to write here as it involves integrals and sigma notation. However, one variable is interesting which is ρ; the none trivial zeros in the zeta function. A zero of any sort is what makes ζ(s)=0. This is to just understand the function itself. Obvious zeros are negative even integers like -2, -4, etc... His hypothesis was that all the none trivial zeros lie on the 1/2 line. This is fancy talk for: 1/2 ± ib = 0. One zero is 1/2 ± 14.13i. If this is true, then we have an exact approximation for prime numbers. Also there's one million dollars if you can prove it so re-read it if you skimmed through and didn't understand because you weren't paying attention. |
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