1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Some insights about the Riemann hypothesis needed

  1. Nov 16, 2013 #1
    The Riemann hypothesis states, whenever the Riemann zeta function hits 0, the real part of the input must be 0.5. Does any input with real part being 0.5 make the function hit 0? Also, assuming the hypothesis is true, would it suffice to prove that if the input's real part is not 0.5, then the function cannot hit 0? Many thanks. :smile:
  2. jcsd
  3. Nov 16, 2013 #2


    User Avatar
    Science Advisor

    The zeroes on the line x = 0.5 are a discrete set, so setting the real part = 0.5 will not lead to a zero most of the time.

    Your second statement is just a restatement of the Riemann hypothesis (with the condition the real part is between 0 and 1).
  4. Nov 16, 2013 #3


    User Avatar
    2017 Award

    Staff: Mentor

    All even negative integers are zeros of the function. They are called "trivial" zeros, and excluded in the Riemann hypothesis.
    Yes, apart from the special points mentioned above.
  5. Nov 16, 2013 #4
    In other words, most of the points on the line where the real part is 0.5 do not have height 0? Also, could the Reimann zeta function evaluate to negative values?
  6. Nov 16, 2013 #5
    Starting at 22:50 of the given link, there's a pretty cool computer depiction of Riemann's zeta function. :cool: Can anyone make a comment as to how realistic this particular depiction might be?

    Last edited by a moderator: Sep 25, 2014
  7. Nov 16, 2013 #6


    User Avatar
    2017 Award

    Staff: Mentor

    "Height"? At most points, the function value is not 0, right.
    Yes it does, it reaches every complex value (and there are even stronger statements how the function can be used as approximation for a wide class of other functions and so on).
  8. Nov 16, 2013 #7
    Some sources refer to values of functions as height. I suppose it's easier for some people to comprehend. :tongue:

    Any possible way to depict graphically what the Riemann zeta function looks like? Or would this be too complex to do, no pun intended :tongue:?
    Last edited: Nov 16, 2013
  9. Nov 17, 2013 #8
    The only depiction I have ever seen was from a semi-pop math book by du Sautoy. He referred to it as the "landscape" of the zeta function (and I'm not sure how common that phrase is elsewhere), and searching for that on Google yields a number of depictions.
  10. Nov 17, 2013 #9


    User Avatar
    2017 Award

    Staff: Mentor

    That is easy to do with real functions, but complex function values cannot be represented by a single real number ("height").

    The google image search shows many ways to do that.
  11. Nov 19, 2013 #10
    Wow! Totally didn't know that. Many thanks! :smile:
  12. Nov 20, 2013 #11
    The complex number is in the form a+bi.
    a being the real part.
    The hypothesis is that all non trivial numbers will lie on the line 1/2+bi. The modern version is all non trivial zeros will be with a real part 1/2+bi=0.
    The zeta function is like any other function such that each different input will put out a output.

    Were they use the term height is such that the zeta function can be drawn in a graph form that it 3d.
    This means using z, y, z axis to do so.
    the trivial and non trivial zeros are of course on the plane formed by the x,y lines.

    Now look at the input form used to get this.
    The reals for the line with a+0i. The imaginary part intersects this at 0 in a perpendicular form, or 0+bi.
    This is a right angle of course.

    Now realize all trivial zeros are at every negative even integer (real).
    So far all known non trivial zeros are on the line of 1/2=bi, or a perpendicular line to the reals.

    See a pattern??

    Input is perpendicular and output may be the same in the form of 1/2+bi to the real line.

    Well as stated the trivial zeros are at every negative even integer starting at negative 2. Lets see if the pattern hold to the original input point. -2+2=0. Yet of course the value of the zeta function at zero is not 0. It is -1/2. Well by subtracting this from zero you get of course 1/2.


    As for the know values of the non trivial zeros and the patterns of the zeta function, it shown a phase relation between the real and imaginary parts of a graph. Both parts can be draws in the wave form. Also the polar form can be used to see this relation.
    To see what I am saying on this point view

    Also the page at
    scroll down the page to
    Riemann Hypothesis Movie

    Of course if some one can prove that instead of the critical strip, it is a critical line, if it can be, then a proof would be found.
    Hope this all helps.
  13. Nov 21, 2013 #12


    User Avatar
    2017 Award

    Staff: Mentor

    It can be written in that way, it does not have to.

    As b is real, this equation cannot be true. "will have a real part of 1/2" is correct.

    What does that mean? The 3D plot is a visualization of the function, it is not the function itself.
  14. Nov 21, 2013 #13
    "As b is real"
    B is a designation for an action upon i. It is a real number.
    Yet it is a statement coupled to i: as bi, which is adding i a real number of times. Example. 0i is no i, which is equivalent to the reals at 0. 1i is 1 times i which is equal to i. 2i is i+i and so on.
    Why people get confused over b is beyond me.

    Part of this I am sure is the short hand of just writing i and leaving of the b part. Same with using the reals in the complex numbers.
    A real is such the form is a+0i. Of course such is equivalent to a real value, so the short hand works if it is remembered it is shorthand.

    B is not a part of the reals, except at a+0i, or when the b part is 0
    Last edited: Nov 21, 2013
  15. Nov 21, 2013 #14
    "The 3D plot is a visualization of the function"

    The lay out of the plot is by input to out put. Each individual input forms the output to the z axis for the 3d Verizon.
    The input of course is the a+bi form. Which is the imaginary part is at a right angle to the reals. Perpendicular.
  16. Nov 21, 2013 #15
    "It can be written in that way, it does not have to."
    Using shorthand versions seems to be the best way to confuse folks.
  17. Nov 21, 2013 #16


    User Avatar
    2017 Award

    Staff: Mentor

    Don't make it complicated, your "actions" are just addition and multiplication in the complex numbers. a+bi is a method to write all complex numbers with the real numbers a and b.
    I don't know why you write wrong equations.

    What do you mean with that?

    You cannot graph the full complex function as a surface in 3 dimensions, you need a 4th dimension or two graphs (showing real and imaginary part, for example, or magnitude and phase) or a color-coded 3D graph or similar things.

    You can plot it with a different angle, that is completely irrelevant for the function.

    How is that related to the quoted statement? In which way is it a "shorthand version" compared to the polar form?
  18. Nov 21, 2013 #17
    getting way off track.
    To answer some confusion of my posts, Some times boards are a problem for me. Dyslexia interferes.

    I should simply ask, how is the zeta function plotted?
    How it the layout ordered in points?
    Does it use the input to output as the plotting positions?
    If so how.
    Can all point of the graph be shown as ordered points of input to the ordered points of output of the zeta function?
    What form does it start with?

    This should be interesting.
  19. Nov 22, 2013 #18
    I should leave this alone. Yet..

    For the complex number set I will use this form for the numbers of the set of complex numbers..
    x+yi. This form also can be written as x+iy.

    To define the real part of this number or the set of reals: I use the form with y set to zero, or x+0i. This form can identify a real number in the subset of reals, Can also be short handed to simply x.

    To define the purely imaginary part: The x is set to 0, or 0+yi. This defines the set of purely imaginary. Shorthand is yi.

    These sets of reals and purely imaginary can be drawn as lines, that exist in the graphical representation of the complex number set. It uses a z and y axis or lines that intersect at 0.
    This intersection of lines is at a 90 degree intersection or right angle intersection. It forms an intersection like + at 0=0i. The reals and the purely imaginarys form lines from the point of origin or the 0+0i point.

    Now when I say the input is in the form of two pure lines distinct in the pure form of reals and imaginaries, at right angle interceptions, does anyone not understand this???
  20. Nov 22, 2013 #19


    User Avatar
    2017 Award

    Staff: Mentor

    Usually with real and imaginary part of the argument on two axes and a "part" of the function value as third axis. This "part" is often the real part, the imaginary part, the magnitude or the complex phase, and drawn as height of a surface.

    Each point in such a graph says "for this specific complex number as function argument, the [plotted quantity] is [value]".

    What do you mean with "ordered"?

    "Can" is important here.

    That's exactly like saying 10 is in the form of 5 and 5, just because 5+5=10.
  21. Nov 22, 2013 #20
    "What do you mean with "ordered"?"
    Poor choice of words on my part, I should say coordinates of input to output. This creates an ordered set which of course has to be done to create a usable graph.

    Quote of mine: "Now when I say the input is in the form of two pure lines distinct in the pure form of reals and imaginaries, at right angle interceptions, does anyone not understand this???"

    Quote of the reply statement:
    "That's exactly like saying 10 is in the form of 5 and 5, just because 5+5=10."

    Seems some folks here have not had basic complex analysis. The pure imaginarys are generally one of the first things taught. Maybe some leave out the concept.

    I take it you can understand that the reals are a line. The pure imaginarys are a line. Both are subsets of the complex numbers. Only one point of the subsets of reals and imaginary union. That is at 0+0i
    They cross at right angles at 0, or 0 in the full form of the complex number being 0+0i.

    mfb you are nitpicking. Please keep doing such. It may prove to be very practical.
    Last edited: Nov 22, 2013
  22. Nov 22, 2013 #21


    User Avatar
    2017 Award

    Staff: Mentor

    I don't think you mean ordered set.

    Be assured that I had more than just basic complex analysis.
  23. Nov 22, 2013 #22
    "Be assured that I had more than just basic complex analysis."


    your nit picking will be usfull maybe.
    Now were was I...
  24. Nov 22, 2013 #23


    Staff: Mentor

    You can use the Quote button to quote someone. What they said will be in a pair of [ quote=<username> ] and [ /quote ] HTML tags (without extra spaces).
  25. Nov 22, 2013 #24


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This thread has gone sufficiently far off topic from the Riemann hypothesis, and it seems like goldust got his answer already.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook