/sci/ - Science & Math

Is there any bigger Dunning Kruger Broscience than nutrition?

>>8587034

>Also, why do people seriously like Trump?

Because they're idiots, if you aren't a millionaire/billionaire, a CEO of a fortune 500 company, the Chinese Government (China actually stands more to gain than Russia) or an immediate member of his family you do NOT have a legitimate reason to like Trump.

I mean get "why" people voted for Trump, but outside the tax breaks he's going to throw at companies, he isn't going to do shit.

The immigration issue is bunked because if he kills Obamacare the government is going to be scrambling to find enough money to contain the fallout. This is on top of the fact that healthcare/ social security is already nearly fucked because the "legal" working native population isn't going to be enough to support the cost of growing elderly population.

With the exception of nuking the Middle East/ North Africa the constant civil unrest and religious extremism will continue to grow. Causing more migrants to push into Europe, Israel seems to be almost forfeit (seriously you fucking living in the muslim's backyard of course they want to kick your fucking ass six ways to Sunday for attempting to expand).

Japan and South Korea are going to eventually ready themselves for a possible territorial/ nuclear showdown with North Korea and China.

He's even created animosity between himself and the Intel community.

Fucking Matrix when?

>>8580689
>230+ IQ
Anon, please don't do this.

>set up product integral function [math] f(a)=\displaystyle \prod_{-\infty}^{\infty}\zeta(a+xi)^{dx} [/math]
>for some fixed a, if the zeta function has a zero along the line a+xi, then f(a) is going to be 0
>use real induction and prove that f(a) is nonzero and divergent on the set (0,1) \ {1/2}
>you have just proven the riemann hypothesis

[math] \displaystyle \prod_{-\infty}^{\infty}\zeta(a+xi)^{dx} = \exp(\int_{\mathbb{R}}\ln(\zeta(a+xi))dx [/math]
[math] \displaystyle \zeta(x) = \prod_{p\text{ prime}} \frac{1}{1-p^{-x}} [/math]
[math] \displaystyle \ln(\prod_{p\text{ prime}} \frac{1}{1-p^{-x}}) = \sum_{p \text{ prime}}\ln(\frac{1}{1-p^{-x}}) [/math]
[math] \displaystyle \sum_{p \text{ prime}}\ln(\frac{1}{1-p^{-x}}) = \sum_{p \text{ prime}}(\ln(1)-\ln(1-p^{-x})) = -\sum_{p \text{ prime}}\ln(1-p^{-x})[/math]
[math] \displaystyle \exp(\int_{\mathbb{R}}\ln(\zeta(a+xi))dx = \exp(-\sum_{p \text{ prime}}\int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx)[/math]
[math] \displaystyle \exp(-\sum_{p \text{ prime}}\int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx) = \prod_{p \text{ prime}}\exp(-\int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx) [/math]

it gets messier

[math] \displaystyle \int_{\mathbb{R}}\ln(1-p^{-a-xi)})dx = \frac{i\text{Li}_2(p^{a+bi})}{\ln(p)}+b\ln(1-p^{-a-bi})-b\ln(1-p^{a+bi})+\frac{1}{2}ib^2\ln(p)+C [/math]

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