MX2 Marathon (1 Viewer)

HeroWise

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I have solved the Riemann zeta question. I had inspiration from the heavens. God spoke to me directly and helped me solve it. He remarked "Trivial, you should be able to solve it after you do it with me,,,"
Oh well, I can post the solution but this comment box is too small to contain it.
 

stupid_girl

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I have solved the Riemann zeta question. I had inspiration from the heavens. God spoke to me directly and helped me solve it. He remarked "Trivial, you should be able to solve it after you do it with me,,,"
Oh well, I can post the solution but this comment box is too small to contain it.
Did you prove or disprove the hypothesis?:p
 

Checkmate

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Can anyone help me on this one?

Consider the Riemann zeta function , defined as

where is complex with a real part greater than 1.
Prove that the real part of every non-trivial zero of the Riemann zeta function is .

If you work it out, please don't post it here but DM me.

I'll pay you $10, and I'm already taking a risk here. Also, after you send it to me, you must destroy your copy (this part is VERY IMPORTANT and non-negotiable).
I'm willing to take more of a risk and will pay you $10000 if you DM only me ;)
 

HeroWise

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Damn I got screenshots Ill send it to DMCA right this moment
 

stupid_girl

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(a) Consider the graph of .
(i) Is (0,0) a stationary point?
(ii) Is (0,0) a minimum point?
(iii) Is (0,0) a maximum point?
(iv) Is (0,0) a point of inflection?

(b) Consider the graph of .
Is (0,0) a minimum point?

(c) Consider the graph of .
Is (0,0) a point of inflection?
 
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stupid_girl

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(a) Consider the graph of .
(i) Is (0,0) a stationary point?
(ii) Is (0,0) a minimum point?
(iii) Is (0,0) a maximum point?
(iv) Is (0,0) a point of inflection?

(b) Consider the graph of .
Is (0,0) a minimum point?

(c) Consider the graph of .
Is (0,0) a point of inflection?
It seems no one has attempted this interesting question yet.

(a)(i) Yes. When x=0, dy/dx=0.
(ii) No. For any interval -𝛿<x<𝛿, there exists x such that y<0.
(iii) No. For any interval -𝛿<x<𝛿, there exists x such that y>0.
(iv) No. The graph is neither concave nor convex near x=0.

(b) Yes. For any interval -𝛿<x<𝛿, y≥0.
(0,0) is actually one of the global minimum points.

(c) No. The graph is neither concave nor convex near x=0.
 
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HazzRat

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Can someone smarter than me plz tell me how they did these two steps? If it helps it's for this question:
tyjhregwfqd.PNG

Working out
tjyrhegwfqd.PNG
 

scaryshark09

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for the first one, i think its meant to be x^2, instead of 2^x

all they do is split up the 2x into x+x and then they rearrange the algebra for the first step
 

scaryshark09

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for the second one they are just factorising, but i think its meant to be xm instead of just x
 

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