Recent content by TheOogy

To my opinion, you are correct.

HallsofIvy, (\sqrt{5}+\sqrt{2})^4-6(\sqrt{5}+\sqrt{2})^2+1 = 98.596 i got a different result, for any \sqrt{a}, \sqrt{b} just use (\sqrt{a}+ \sqrt{b})*(\sqrt{a}- \sqrt{b})*(-\sqrt{a}+ \sqrt{b})*(-\sqrt{a}- \sqrt{b}) and expand i haven't read the whole article, just the...

thanks matt grime!

is \sqrt{2}+\sqrt{5} an algebraic number? i used 2 and 5 arbitrarily, try any integers (as long as they are not the same integer, in which case it is algebraic) I tried finding a polynomial with rational coefficients that zeros at this value, but haven't found any.

every rational number can be expressed as a numerator and a denominator (for decimal to fraction conversion see wikipedia: http://en.wikipedia.org/wiki/Fraction_(mathematics)#Converting_repeating_decimals_to_fractions" you should be able to figure out how to do it to other bases) now just...

solamon please, and I'm not saying it because I'm bad, and i don't want to hurt anyone, but please- SHUT UP! p.s. please lock this thread.

Thanks!

i tried using the euler product but it didn't work, but thanks for the eta-gamma integral, can you show me the zeta-gamma integral two and save me the search?

or, is there a way to calculate values in the critical strip without using an alternating series?

yes.

can we express the eta function as a product of primes?

how do i calculate values of the riemann zeta function in the critical strip? because if you only know zeta as a series: \zeta(s) = \sum 1/n^s and the functional equation \zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)...

conjugates don't have the same imaginary part.

i don't have a proof, which is way I'm asking, i thought maybe someone can prove that there cannot be a pair with the same imaginary part but this turns out more complicated then i thought it would be.:frown:

i'm looking for two complex zeros that both have the same imaginary part but have diffrent real part, non with real part half. it differs from the riemann hypothesis because i don't care for a single zero off the strip, just pairs:smile:. MAYBE some one can prove that, like that dude who proved...