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August 10th, 2016, 12:19 PM   #1
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Question I have some simple questions about the Riemann Hypothesis

I have watched and read about Riemann's Zeta function, but in the weeks I've been trying to learn it there are some things I can't quite get my head around. Two things to be exact.

One, The Trivial and non trivial Zeta functions that equal to 0. Do they converge to 0?

Two, what does this gap on the vertical line mean on Graphs?
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After first seeing Riemann's translated original paper I was shocked to see that he didn't have a visual graph of his function.
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August 11th, 2016, 06:03 AM   #2
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Thinking about the trivial zeroes with complex numbers (-2+i0)

I look at 2/(2^-2) = 0.25 and I think oh ok so it's going down to zero. But then I look at

1
/
2^-2 and this equals 4.

Is the 1 at the top just supposed to represent that it's a fraction of one quarter (1/4)?

EDIT: I guess what I'm asking is, how do the trivial and non trivial zeros become zero?

Last edited by HawkI; August 11th, 2016 at 06:10 AM.
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September 1st, 2016, 11:57 AM   #3
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In order to see the trivial zeros of the zeta function. Look at the functional equation

$$\zeta(s)=2(2\pi)^{s-1}\Gamma(1-s)\sin(\tfrac12\pi s)\zeta(1-s).$$

You can see that $\zeta (-2n) =0 $.
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September 10th, 2016, 11:48 PM   #4
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If I'm not mistaken, (since the series 1 + 1/2^s + 1/3^s... converges for s > 1), and this can be proven as follows

$\displaystyle \int $ a to inf dx/x^p, a, p > 0 and not 1

$\displaystyle \int $ a to B dx/x^p = ( 1/(1-p) 1/(x^(p-1))) taken at B minus taken at a (Leibniz-Newton formula) = 1/(1-p) (B^(1-p) - a^(1-p))

If 1-p < 0, the integral converges to 1/(p-1) a^(1-p) because B^(1-p) -> 0 when x -> inf.

Now all that remains is to prove Cauchy's integral test for series convergence

Let f:I->[0, inf) be a continuous function on I = [1, inf) and let it decrease on I.

$\displaystyle \sum f (n) $ n € N

converges iff $\displaystyle \int $ 1 to inf f(x) dx converges.

Proof.

Let

s_n = f(1)+...+f(n)

be nth partial sum of the series and let

F(x) = $\displaystyle \int $ 1 to x f(t) dt, x € [1, inf)

Since f decreases

f(k+1) <= f(t) <= f(k)

for all k € N and all t € [k, k+1], so

f(k+1) <= $\displaystyle \int $ k to k+1 f(t) dt <= f(k), k € N

So

F(n+1) = $\displaystyle \int $ 1 to n+1 f = $\displaystyle \int $ 1 to 2 f +...+ $\displaystyle \int $ n to n+1 f <= f(1) + ...+ f(n) = s_n

<= f(1) + $\displaystyle \int $ 1 to 2 f +...+ $\displaystyle \int $ n-1 to n f =

= f(1) + F(n)

So

F(n+1) <= s_n <= f(1) + F(n).

If the integral converges, since F grows, it follows from the last inequalities

s_n <= f(1) + F(n) <= f(1) + lim n tends to inf F(n) =

= f(1) + $\displaystyle \int $ 1 to inf f(x) dx,

which shows that the sequence s_n in bounded from above. Since it grows, it's convergent, so the inequalities for n->inf imply

$\displaystyle \int $ 1 to inf f(x) dx <= $\displaystyle \sum f (n) $

<= f(1) + $\displaystyle \int $ 1 to inf f(x) dx. So, the series converges.

So we get a set that has a point (I don't know the English term, I'll call it limiting point) such that for its any neighbourhood there's an element of it. (Any point s => 1). By a theorem from complex analysis, which states:

Let f and g be analytic functions on a domain O (in this case it means an unempty open subset of C such that for any 2 points a, b in it there are finitely many points a = z0, z1, ..., zn = b such that [z0, z1], ...[zn-1, zn] lie in O) If f, g are equal on an infinite set which has a limiting point in O, then theyre equal on O, i.e. f =g.

So this means since the series is an analytic function of s > 1) if we extend it to C\{1} that whichever function equals it for s > 1, they're the same. Or, in other words, we can uniquely perform analytic continuation, if we can do it at all. (Which we can and is exectly what Riemann did first I think in the legendary paper you mentioned). And since every analytic function is continuous, by a theorem which states that any value of a continuous function equals its limit at that point, the zeros equal the limit of the function at them.
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September 11th, 2016, 01:32 AM   #5
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Quote:
Originally Posted by raul21 View Post
If I'm not mistaken, (since the series 1 + 1/2^s + 1/3^s... converges for s > 1), and this can be proven as follows

$\displaystyle \int $ a to inf dx/x^p, a, p > 0 and not 1

$\displaystyle \int $ a to B dx/x^p = ( 1/(1-p) 1/(x^(p-1))) taken at B minus taken at a (Leibniz-Newton formula) = 1/(1-p) (B^(1-p) - a^(1-p))

If 1-p < 0, the integral converges to 1/(p-1) a^(1-p) because B^(1-p) -> 0 when x -> inf.

Now all that remains is to prove Cauchy's integral test for series convergence

Let f:I->[0, inf) be a continuous function on I = [1, inf) and let it decrease on I.

$\displaystyle \sum f (n) $ n € N

converges iff $\displaystyle \int $ 1 to inf f(x) dx converges.

Proof.

Let

s_n = f(1)+...+f(n)

be nth partial sum of the series and let

F(x) = $\displaystyle \int $ 1 to x f(t) dt, x € [1, inf)

Since f decreases

f(k+1) <= f(t) <= f(k)

for all k € N and all t € [k, k+1], so

f(k+1) <= $\displaystyle \int $ k to k+1 f(t) dt <= f(k), k € N

So

F(n+1) = $\displaystyle \int $ 1 to n+1 f = $\displaystyle \int $ 1 to 2 f +...+ $\displaystyle \int $ n to n+1 f <= f(1) + ...+ f(n) = s_n

<= f(1) + $\displaystyle \int $ 1 to 2 f +...+ $\displaystyle \int $ n-1 to n f =

= f(1) + F(n)

So

F(n+1) <= s_n <= f(1) + F(n).

If the integral converges, since F grows, it follows from the last inequalities

s_n <= f(1) + F(n) <= f(1) + lim n tends to inf F(n) =

= f(1) + $\displaystyle \int $ 1 to inf f(x) dx,

which shows that the sequence s_n in bounded from above. Since it grows, it's convergent, so the inequalities for n->inf imply

$\displaystyle \int $ 1 to inf f(x) dx <= $\displaystyle \sum f (n) $

<= f(1) + $\displaystyle \int $ 1 to inf f(x) dx. So, the series converges.

So we get a set that has a point (I don't know the English term, I'll call it limiting point) such that for its any neighbourhood there's an element of it. (Any point s => 1). By a theorem from complex analysis, which states:

Let f and g be analytic functions on a domain O (in this case it means an unempty open subset of C such that for any 2 points a, b in it there are finitely many points a = z0, z1, ..., zn = b such that [z0, z1], ...[zn-1, zn] lie in O) If f, g are equal on an infinite set which has a limiting point in O, then theyre equal on O, i.e. f =g.

So this means) if we extend the series to C\{1} that whichever function equals it for s > 1, they're the same. Or, in other words, we can uniquely perform analytic continuation, if we can do it at all. (Which we can and is exectly what Riemann did first I think in the legendary paper you mentioned). And since every analytic function is continuous, by a theorem which states that any value of a continuous function equals its limit at that point, the zeros equal the limit of the function at them.
Just corrected an error.
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