Finding the nontrivial zeros of the Riemann Zeta Function using Desmos

In this video I discuss extending the Riemann Zeta Function using the Dirichlet Eta Function, and use this along with Desmos to find the first few nontrivial zeros.
All nontrivial zeros of the Riemann Zeta Function so far calculated by humans (there are about 10^13 of them) all have real part 1/2, which falls in line with the Riemann Hypothesis which says that all of these zeros have real part 1/2.
The Riemann Hypothesis is a currently unsolved math problem, and is attached to a $1,000,000 prize awarded by the Clay Mathematics Institute.

[@Windprinc3]

You made this abstract thing easy to understand. This is the first time I understood the terms “continuation” and “analytic continuation” and the graphs made things much more interesting 👍🏼

[@philip2205]

You are so eloquent and chose great backing music. I also like how you overlayed yourself over the drawing board. Good job!

[@andrewdsotomayor]

Thank you so much. It took time to figure that overlaying part out, but I’m glad I did; I’m really happy with the final result.

[@madhuragrawal5685]

Fascinating, clear, and very well presented! Thank you so much, I'm looking forward to more exciting videos from this channel

[@andrewdsotomayor]

Thanks for the compliments. More to come!

[@Cully132]

Excellent presentation to a challenging subject. You make it seem all so natural. Followup graphics very illustrative.

[@CastIronNutsack]

amazing video that allows for more nerdy young minds (such as myself) to more easily access and understand the mysteries and intricacies of the math world.

[@user-ns8kb6ii4j]

Thank you for the excellent explanation. I'm glad to discover your channel.

[@2piee]

Feels good to see someone use a tool to its full potential. Loved it!

[@hotmole7621]

Thank you for the wonderful explanation, it had a flow and I learned a ton

[@razvanrusan9319]

beautiful video!

[@fibbooo1123]

Very well done!

[@PhoenixReflex]

Thanks. Keep up the hard work.

[@zeus7914]

thanks. good presentation.

[@DolphyWind]

Cool video! I am looking forward to study complex analysis by myself in the future!

[@andrewdsotomayor]

You’re in for a treat, complex analysis is awesome, but HARD!

[@WildEngineering]

damn bro hopefully you see you have a future on KZbin and upgrade your camera! edit* the proof you finished at 4:18 is beautiful

[@tylosenpai6920]

Thanks, i can't find this "Analytic" calculation anywhere else that i can understand

[@owenpawling3956]

I did this a while ago in desmos. Very good!

[@andrewdsotomayor]

Yes, Desmos is amazing!

[@nicolasbertozzo2997]

Thanks for the video! Awesome explanation +1sub

[@rolanchristofferson9363]

nice graphs!

[@Number_Cruncher]

Cool, Thx for sharing

[@nightowl9512]

The two curves always seem to intersect at right angles when you have a riemann-zero. Is that a coincidence?

[@crimsonvale7337]

The Riemann zeta function is part of a special class of functions called “holomorphic” which, as well as a host of other properties, maintain angles between lines on the input space. This means that because the real and imaginary axies are perpendicular, any points they cross on the graph must also be at right angles when you zoom in enough

[@Sir_Isaac_Newton_]

​@@crimsonvale7337axes*

[@CharlieVegas1st]

@@crimsonvale7337 yeah, the specific word is "conformal" which is as you said, a property of holomorphic functions

[@ianweckhorst3200]

This is how I actually do much of my math

[@user-gu2fh4nr7h]

well... purely from the desmos pic... is there a way to "get rid of" the curves that don't cross the y-axis and the ones that have no bounded x value so as to simplify this question?

[@geekjokes8458]

i imagine you want to simplify the graph? in that case, you cant unless you know the specific input that become those lines, so that you can remove them from the calculation if you want to speed it up, _i think_ if you make it bounded, adding a conditional like {0

[@user-gu2fh4nr7h]

@@geekjokes8458 nice playlists fren

[@pentlandite3651]

The red and blue curves always appear to intersect each other at 90 degrees - is this correct? Very interesting presentation.

[@crimsonvale7337]

The Riemann zeta function is part of a special class of functions called “holomorphic” which, as well as a host of other properties, maintain angles between lines on the input space. This means that because the real and imaginary axies are perpendicular, any points they cross on the graph must also be at right angles when you zoom in enough

[@drdca8263]

@@crimsonvale7337but, aren’t zeros of functions specifically the place where there can be an exception to the “angles are preserved” thing, if the zero has degree greater than 1? So... I guess it is because all these zeros are simple zeros? (Are all zeros of the Riemann zeta function simple zeros?)

[@CharlieVegas1st]

@@drdca8263 Where did you learn about the exception to the rule at the zeros? I'm really interested! I never knew that... And yes it's believed that all the zeros are simple but if I remember correctly that's an unproven conjecture (and also a very important unsolved problem).

[@drdca8263]

@@CharlieVegas1st possibly I’m getting mixed up, and thinking of something that doesn’t apply, but like, Well, consider z^3 for example, and consider two lines passing through 0, and their images under z \mapsto z^3 . The angle made by the images of those two lines, well, stuff that goes around 0 once goes around 3 times in the image, so there’s like, a factor of 3 in what angle things make. like, consider the curves z = t, z = exp(2pi i / 3) t z = exp( 2 * 2 pi i / 3) t . Here, adjacent lines make an angle of 2pi/6 with each-other (or 2pi/3 if you only consider the t>0 parts), but under the cubing map, we just get one line, so an angle of either 0 or 2pi/2 with itself. However, if you had the same shape at some other base point, and used the z \mapsto z^3 map, the angles would still be preserved. (so like, (a + r t)^3 = a^3 + 3 a^2 r t + 3 a r^2 t^2 + r^3 t^3 , for r being exp(2pi i/3) or a power of it, for t close to zero, this can be approximated as a^3 + 3 a^2 r t, and so the angles made between the images of the lines t \mapsto a + r t, are the same as before doing the z \mapsto z^3 I’m probably not explaining this very well... it’s been a while

[@crimsonvale7337]

@@drdca8263 oops I forgor sorry Thanks for the reminder, hopefully I’ll do a proper full on course on complex analysis eventually rather than half-remembering youtube vids

[@sleepysnekk]

ty for this video, it was really nice ^^

[@andrewdsotomayor]

Glad you liked it!

[@Wotsit27]

@andrewdsotomayor Re Transcript lines 19:03 to 19:39. I have always been curious about proving rigorously that the real and imaginary parts of the Zeta Function do in fact converge to zero at the exact same point and also on the line Re(s) = 0.5. Here you are showing that you get “really close”, but how can you be sure that if K was unimaginably large that the intersection doesn’t miss the line Re(s) = 0.5? I don’t think this is sufficiently rigorous unless it’s detailed somewhere else. This is only a check on the first non-trivial zero and it seems that you need to go “infinitely far” along your series to nail this. Similar in a sense to the Riemann Hypothesis itself where you have to go “infinitely far” up the critical zone to check the location of all the zeros. Grateful for clarification please. Nonetheless this is an excellent visualisation of the issues. Thank you for this!

[@andrewdsotomayor]

Thanks for the concern of rigor, this is obviously huge and vital in mathematics, and I know there are parts of this and other videos I have made where I could do a better job at this. As it turns out, there’s a result due to the functional equation combined with the fact that zeta is analytic, that essentially says that if the real part of a zero wasn’t 1/2, there would be another zero reflected over the line Re(z) = 1/2. I plan on showing this result, and more in a future video, so keep your eyes open for that.

[@sebastiandierks7919]

I may just be dumb and I know that convergence proofs are often done like that, but why can we conclude at 4:56 that the real part of s has to be greater than 1? Isn't the inequality sign the wrong way round for that? You showed that for Re(s)

[@andrewdsotomayor]

It’s p-series that I’m referring to: the sum from n=1 to infty of 1/n^p, with p real, only converges when p>1, hence the conclusion. Hope this helps!

[@sebastiandierks7919]

@@andrewdsotomayor No, not really unfortunately. I'm totally aware about the convergence or divergence of the series. The "hence the conclusion" part is the problem. Because of the triangle inequality you used, you showed that |zeta(s)|

[@andrewdsotomayor]

It’s less than something that converges, so it’s absolute value is finite, hence it’s convergent as well. Also there is nothing assumed about Re(s), it is not until the final part of the inequality that implies that Re(s)>1

[@sebastiandierks7919]

@@andrewdsotomayor ok I think I understand the problem. I thought you wanted to prove the divergence of the series when Re(s) 1, which of course you did.

[@andrewdsotomayor]

@@sebastiandierks7919 right, the whole point is that Zeta defined as that sum only makes sense when Re(s)>1. After this is when I go on to discuss how it could be defined beyond that region.

[@dsagman]

very very good. but please no music next time.

[@mokranemokrane1941]

5:13 1+3i doesn't work. You just proved Re(s) must be strictly greater than 1 😅

[@andrewdsotomayor]

Right, there are a few other verbal mistakes, but yes, nice catch! I think what I meant was to say 3+i, since all other values, I mentioned were real, I wanted to throw one out there that was nonreal, but misspoke

[@admirljubovic6759]

And the proof is trivial and left to the reader as an exercise

[@eartphoze]

This must adhere closely to harmonics which in turn to resonating frequency, to orbital resonance, so if K < 0, then the egg is hatched a chicken, if K < 1 then egg is not hatched, so two in the eyes and one in mouth , is like Madonna of the rocks, in art history, or imagine the eleventh pig idea, since 1×1 = 1, 1× 0= ?, yet if the eleventh pig, notion concept it's just a concept in nature or biology, then imagine homless drug addics without their drug resource, this is resonance in symmetry for harmonics, and the woman who studied during Nazi war dies from fever after surgery, this reinman non trivial zero is like the brain of a man who knows there are no stupid questions, but the fact that one must first ask in order to play it like a machine in the eyes of world view, vs TV view, vs ideal view and form tye function which tgen is fed a parameter or argument to a 24hr wait period to hear back an answer, like hertz frequency, for transistor buffer as the zeros in wave function, closure, response, correspondence in the age of science , Jack Kirby inventor of transistor is science trivial the way art may seem trivial when compared to value like Robert noise say, like talking to God. Two in the eyes one in mouth. Transistor radio is a world view, it made a crack from a particle, then Tv view a wave, an ideal view would be the audience for which is not part of the trivial zeros on your graph but rather those who take in content must aquire a trait in order to feed their desire to keep grounded with god.

[@miloszforman6270]

What's the use in making math videos with these closeups of, well, _handsome_ men? I'd prefer ladies in case that they are nice to look at, or otherwise clean mathematical writing.

[@XyndraNerd]

please get a better microphone

[@GromitThaGangsta]

his mic is perfect.

[@muffinconsumer4431]

It’s the perfect math teacher prerecorded class microphone

[@pierre-bobkjellen9803]

Can we please stop saying i = sqrt(-1)....

[@drdca8263]

Eh? i is the first square root of -1 when going counterclockwise, so if one wants to make sqrt single valued and extend the domain to include negative numbers, the usual way to do so would be sqrt(-1)=i?

[@pierre-bobkjellen9803]

@@drdca8263 If you listen to the video you'll hear he says "i is THE square root of -1", and thus implies uniqueness. Since the real (as in real numbers) square root function is only defined for non-negative numbers, we should not use it to define i. If he means the complex square root function, he should firstly specify so, and secondly he should specify which branch of the complex square root function. What's wrong with using the correct terminology of "i is the principal branch of the complex square root of -1"?

[@drdca8263]

@@pierre-bobkjellen9803 it’s longer, and probably sounds needlessly intimidating/confusing to people who aren’t familiar with the topic (people who know the general topic already know what is meant by i), and usually doesn’t communicate much useful additional info to the audience. One could say “where i is a formal variable we adjoint to the real numbers, under the constraint that it squares to negative one”, and that would at least clarify some things to some audiences, but “the square root of -1” should efficiently communicate the gist to most people? I suppose “our favorite square root of -1” might be better because it points to the fact that there are two such square roots, but in a way that is easier for people to ignore if they would find thinking about that too much to be distracting, and has the virtue of only being ~3 syllables longer than “the square root of -1”.

[@pierre-bobkjellen9803]

@@drdca8263 And that is precisely the issue. People who already know of course knows of the distinction, but people who don't will not know of the distinction. You shouldn't lead them into the wrong terminology out of fear of them being scared by a longer phrase, how condescending of you to think that. Instead we should use the correct terminology and instead give some short insight as to why we chose to not go with the easy terms.

[@drdca8263]

@@pierre-bobkjellen9803 The concern isn’t them being scared by the longer phrase, but a phrase with terms they are unfamiliar with, and which won’t make sense until they understand the thing we are currently explaining. The consideration about the length is about convenience. They are separate considerations.