4 ms·

This does not happen. The zeros on the critical line appear as poles in F(s). There are no other poles,in F(s), because The trivial zeros of the zeta function

by sainik 10y ago

This does not happen. The zeros on the critical line appear as poles in F(s). There are no other poles,in F(s), because The trivial zeros of the zeta function which occur at -2,-4,-6 etc actually cancel out from the numerator and the denominator (i.e. from zeta(2s) and from zeta(s)). The paper then shows that the only poles in F(s) are those on the critical line, thus proving R.H.

mathematician23 10y ago
A pole for F(s) means zeta(s) = 0 but zeta(2s) not=0 for some complex s. So yes, the zeros on the critical line appear as poles in F(s). Looking outside the critical line, at negative even integers, you will get 0/0 and using L-Hospitals rule, they may end up cancelling out, meaning F(s) remains analytic at these locations. But my point is how do you know this cancellation isn't occurring at other locations, outside the critical line? I am not saying there are other poles outside the critical line, I am saying the author hasn't ruled out 0/0 happening in the critical strip. This is really just high school calculus we are talking about, 0/0 is an indeterminate form and more work must be done to see what type of singularity it is. The author saying F(s) is analytic outside the critical line but within the critical strip means there could still be removable singularities, i.e. 0/0, that end up cancelling out. These are like holes in our graphs back in Cal I. Riemann's Hypothesis is 157 years old now, it is not going to be proven by a 20 page paper.
mathematician23 10y ago
There are many equivalent statements to the Riemann Hypothesis (RH) and statements that, if true, would imply the RH holds. But F(s) = zeta(2s)/zeta(s) being analytic in the critical strip, outside the critical line, DOES NOT imply the RH is true. That's my point, it doesn't matter if the author really showed F(s) is analytic in that region or not. Showing something is equivalent to the RH is a major area of research and so is finding statements that directly imply the RH. The function F(s) is very well known and a simple function, so if there were any simple statements involving it that directly implies the RH, we would know by now. I hope this helps.
- vskp2016 10y ago
  The paper clearly demonstrates that if F(s) has poles on the critical line, RH is proved.This is because the poles in F(s) correspond to zeros of zeta(s) and it is well known that there are no other zeros outside the critical strip (except for the trivial zeros). The paper has proved that the only poles of F(s),occur on this critical line. Ok bye!
mathematician23 10y ago
One last point. The RH being true implies F(s) is analytic outside the critical line but within the critical strip. If we take what's called the contrapositive of this statement, then if we could somehow show F(s) is not analytic in this region, then this would disprove the RH. So what the paper claims to be true is a necessary but not sufficient condition for the RH to be true. Since most think the RH does hold, the paper's claim most likely is true and may even have already been proven elsewhere. Clearly the key is understanding what the word analytic means in complex function theory. It's basically like saying a real-valued function in Calculus I is differentiable, but now we are talking about what this means for a complex-valued function.

by sainik 10y ago

mathematician23 10y ago

A pole for F(s) means zeta(s) = 0 but zeta(2s) not=0 for some complex s. So yes, the zeros on the critical line appear as poles in F(s). Looking outside the critical line, at negative even integers, you will get 0/0 and using L-Hospitals rule, they may end up cancelling out, meaning F(s) remains analytic at these locations. But my point is how do you know this cancellation isn't occurring at other locations, outside the critical line? I am not saying there are other poles outside the critical line, I am saying the author hasn't ruled out 0/0 happening in the critical strip. This is really just high school calculus we are talking about, 0/0 is an indeterminate form and more work must be done to see what type of singularity it is. The author saying F(s) is analytic outside the critical line but within the critical strip means there could still be removable singularities, i.e. 0/0, that end up cancelling out. These are like holes in our graphs back in Cal I. Riemann's Hypothesis is 157 years old now, it is not going to be proven by a 20 page paper.

mathematician23 10y ago

There are many equivalent statements to the Riemann Hypothesis (RH) and statements that, if true, would imply the RH holds. But F(s) = zeta(2s)/zeta(s) being analytic in the critical strip, outside the critical line, DOES NOT imply the RH is true. That's my point, it doesn't matter if the author really showed F(s) is analytic in that region or not. Showing something is equivalent to the RH is a major area of research and so is finding statements that directly imply the RH. The function F(s) is very well known and a simple function, so if there were any simple statements involving it that directly implies the RH, we would know by now. I hope this helps.

vskp2016 10y ago
The paper clearly demonstrates that if F(s) has poles on the critical line, RH is proved.This is because the poles in F(s) correspond to zeros of zeta(s) and it is well known that there are no other zeros outside the critical strip (except for the trivial zeros). The paper has proved that the only poles of F(s),occur on this critical line. Ok bye!

mathematician23 10y ago

One last point. The RH being true implies F(s) is analytic outside the critical line but within the critical strip. If we take what's called the contrapositive of this statement, then if we could somehow show F(s) is not analytic in this region, then this would disprove the RH. So what the paper claims to be true is a necessary but not sufficient condition for the RH to be true. Since most think the RH does hold, the paper's claim most likely is true and may even have already been proven elsewhere. Clearly the key is understanding what the word analytic means in complex function theory. It's basically like saying a real-valued function in Calculus I is differentiable, but now we are talking about what this means for a complex-valued function.